COLLEGE ALGEBRA 13TH EDITION (CHAPTER 1_8) BY R. DAVID GUSTAFSON, JEFF HUGHES SOLUTIONS MANUAL

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER R: A REVIEW OF BASIC ALGEBRA

TABLE OF CONTENTS End of Section Exercise Solutions ....................................................................................... 1 Exercises R.1 ..................................................................................................................................1 Exercises R.2 .............................................................................................................................. 19 Exercises R.3 ..............................................................................................................................43 Exercises R.4 ............................................................................................................................. 66 Exercises R.5 ..............................................................................................................................92 Exercises R.6 ............................................................................................................................. 115 Chapter Review Solutions................................................................................................ 144 Chapter Test Solutions .................................................................................................... 167 Group Activity Solutions ...................................................................................................175

END OF SECTION EXERCISE SOLUTIONS EXERCISES R.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Simplify. 10 – 20 Solution

10  20  10

2. Simplify. 5 – (–10) Solution

5   10   15

3. Division by what number is undefined?

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution Division by 0 is undefined 4. Given

5 0 0 , , and . Which one is equivalent to 0? 0 5 0

Solution 0 0 5 5. Write the inequality symbol for greater than. Solution The inequality symbol for greater than is > 6. Write the math symbol for infinity. Solution The math symbol for infinity is Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A ______ is a collection of objects. Solution set 8. If every member of one set B is also a member of a second set A, then B is called a ________ of A. Solution subset 9. If A and B are two sets, the set that contains all members that are in sets A and B or both is called the ________ of A and B. Solution union 10. If A and B are two sets, the set that contains all members that are in both sets is called the ________of A and B. Solution intersection 11. A real number is any number that can be expressed as a __________ . Solution decimal

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

12. A __________ is a letter that is used to represent a number. Solution variable 13. The smallest prime number is __________. Solution 2 14. All integers that are exactly divisible by 2 are called _________integers. Solution even 15. Natural numbers greater than 1 that are not prime are called __________numbers. Solution composite 16. Fractions such as 23 , 82 , and  97 are called _________ numbers. Solution rational 17. Irrational numbers are ________ that don’t terminate and don’t repeat. Solution decimals 18. The symbol ________is read as “is less than or equal to.” Solution  19. On a number line, the __________ numbers are to the left of 0. Solution negative 20. The only integer that is neither positive nor negative is _________. Solution 0 21. The Associative Property of Addition states that  x  y   z  _________. Solution

x   y  z

22. The Commutative Property of Multiplication states that xy = __________. Solution yx

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

23. Use the Distributive Property to complete the statement: 5(m + 2) = ___________. Solution

5m 5  2

24. The statement (m + n) p = p(m + n) illustrates the __________ Property of ________. Solution Commutative, Multiplication 25. The graph of an __________ is a portion of a number line. Solution interval 26. The graph of an open interval has _________ endpoints. Solution no 27. The graph of a closed interval has __________ endpoints. Solution two 28. The graph of a _________ interval has one endpoint. Solution half-open 29. Except for 0, the absolute value of every number is _________. Solution positive 30. The __________ between two distinct points on a number line is always positive. Solution distance Let N = the set of natural numbers W = the set of whole numbers Z = the set of integers Q = the set of rational numbers R = the set of real numbers Determine whether each statement is true or false. Read the symbol  , as “is a subset of.” 31. N  W Solution Every natural number is a whole number, so N  W. TRUE

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

32. Q  R Solution Every rational number is a real number, so Q  R. TRUE 33. Q  N Solution The rational number 21 is not a natural number, so Q | N. FALSE 34. Z  Q Solution Every integer is a rational number, so Z  Q. TRUE 35. W  Z Solution Every whole number is an integer, so W  Z. TRUE 36. R  Z Solution The real number

2 is not an integer, so R | Z FALSE

Practice

,

C

f , e , c , a =

, B 

d n a

g , f , e , d =

e , d , c , b , a =

t e L

A

 . Find each set.

37. A  B Solution A  B  {a, b, c, d, e, f, g} 38. A  B Solution A  B  {d, e}

39. A  C Solution A  C  {a, c, e}

40. B  C Solution B  C  {a, c, d, e, f, g}

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Determine whether the decimal form of each fraction terminates or repeats. 7 41. 16 Solution 7  0.4375; terminates 16 42.

5 8

Solution 5  0.625; terminates 8 43.

5 11 Solution

5 = 0.454545...; repeats 11 44.

7 12 Solution

7  0.583333...; repeats 12 Consider the following set:

7 , 6 , 5 7 . 2 , 2

, 2 , 1 , 0 , 23

, 4 , 5

  

45. Which numbers are natural numbers? Solution natural: 1, 2, 6, 7 46. Which numbers are whole numbers? Solution whole: 0, 1, 2, 6, 7 47. Which numbers are integers? Solution integers: –5, –4, 0, 1, 2, 6, 7 48. Which numbers are rational numbers? Solution rational: 5,  4,  23 , 0, 1, 2, 2.75, 6, 7

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

49. Which numbers are irrational numbers? Solution irrational: 2 50. Which numbers are prime numbers? Solution prime: 2,7 51. Which numbers are composite numbers? Solution composite: 6 52. Which numbers are even integers? Solution even: –4, 0, 2, 6 53. Which numbers are odd integers? Solution odd: –5, 1, 7 54. Which numbers are negative numbers? Solution negative: 5, 4,  23 Graph each subset of the real numbers on a number line. 55. The natural numbers between 1 and 5 Solution

56. The composite numbers less than 10 Solution

57. The prime numbers between 10 and 20 Solution

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

58. The integers from –2 to 4 Solution

59. The integers between –5 and 0 Solution

60. The even integers between –9 and –1 Solution

61. The odd integers between –6 and 4 Solution

62. 0.7, 1.75, and 3 87 Solution

Write each inequality in interval notation and graph the interval. 63. x  2 Solution

x  2   2,  

64. x  4 Solution

x  4   , 4 

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

65. 0  x  5 Solution

0  x  5   0,5

66. 2  x  3 Solution

2  x  3   2, 3

67. x  4 Solution

x  4   4,  

68. x  3 Solution

x  3    , 3

69. 2  x  2 Solution  2  x  2  [  2, 2]

70. 4  x  1 Solution  4  x  1  (  4, 1]

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

71. x  5 Solution x  5  ( ,5]

72. x  1 Solution x  1  [1, )

73. 5  x  0 Solution 5  x  0  ( 5, 0]

74. 3  x  4 Solution 3  x  4  [ 3, 4)

75. 2  x  3 Solution 2  x  3  [ 2, 3]

76. 4  x  4 Solution 4  x  4  [ 4, 4]

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

77. 6  x  2 Solution 6  x  2  2  x  6  [2, 6]

78. 3  x  2 Solution 3  x  2  2  x  3  [ 2, 3]

Write each pair of inequalities as the intersection of two intervals and graph the result. 79. x  5 and x  4 Solution

x  5 and x  4   5,      , 4 

 5,  

 , 4  ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

 5,     , 4  80. x  3 and x  6 Solution

x  3 and x  6  [3, )    , 6 

  3,  

 , 6 

  3,     , 6 

81. x  8 and x  3 Solution x  8 and x  3  [8, )  (, 3]

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

  8,  

 ,  3 __________________________________  8,     ,  3 

82. x  1 and x  7 Solution

x  1 and x  7   1,    ( , 7]

 1,    , 7  ________________________________

 1,     , 7  Write each inequality as the union of two intervals and graph the result. 83. x  2 or x  2 Solution

x  2 or x  2   , 2   2,  

84. x  5 or x  0 Solution

x  5 or x  0  ( , 5]   0,  

85. x  1 or x  3 Solution x  1 or x  3  ( , 1]  [3, )

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

86. x  3 or x  2 Solution

x  3 or x  2   , 3  [2,)

Write each expression without using absolute value symbols. 87. 13 Solution Since 13  0, 13  13. 88. 17 Solution

Since 17  0, 17   17  17. 89. 0 Solution Since 0  0, 0  0. 90.  63 Solution Since 63  0, 63  63.  63    63  63

91.  8 Solution

Since  8  0, 8    8   8.  8    8   8

92. 25 Solution

Since 25  0, 25   25  25.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

93.  32 Solution Since 32  0, 32  32.  32    32   32

94.  6 Solution

Since  6  0, 6    6   6.  6    6   6

95.   5 Solution Since   5  0,

  5     5     5  5   .

96. 8   Solution

Since 8    0, 8    8   . 97.    Solution

   0  0 98. 2  Solution

Since 2  0, 2  2 . 99. x  1 and x  2 Solution

If x  2, then x  1  0. Then x  1  x  1. 100. x  1 and x  2 Solution

If x  2, then x  1  0. Then x  1    x  1 .

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

101. x  4 and x  0 Solution

If x  0, then x  4  0. Then x  4    x  4  .

102. x  7 and x  10 Solution

x  10, then x  7  0. Then x  7  x  7.

103.

x 7 x 7

and x >7

Solution Since x  7 is positive, x  7 is its own absolute value. x 7 x 7

104.

x 8 x 8

x 7 1 x 7

and x  8

Solution

Since x  8 is negative, x  8    x  8

x 8 x 8

x 8

  x  8

 1

Find the distance between each pair of points on the number line. 105. 3 and 8

Solution

distance  8  3  5  5 106. –5 and 12

Solution

distance  12   5   17  17

107. –8 and –3

Solution

distance  3   8   5  5

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

108. 6 and –20

Solution

distance  20  6  26  26 109. –100 and 50

Solution d  ba d  20  6  26 110. –200 and –50

Solution d  ba

d  50   100  50  100  150

Fix It In Exercises 111 and 112, identify the error made and fix it. 111. Let A = {s, n, i, c, k, e, r, s} and B = {t, w, i, x}, find A ∪ B

Solution A  {s, n, i , c, k, e, r , s} and B  {t, w, i, x} A  B  {s, n, i , c, k, e, r , s, t, w, x} 112. Graph the inequality x < 1 on a number line.

Solution Graph x  1

Applications 113. What subset of the real numbers would you use to describe the populations of Memphis and Miami?

Solution Since population must be positive and never has a fractional part, the set of natural numbers should be used. 114. What subset of the real numbers would you use to describe the subdivisions of an inch on a ruler?

Solution Since the subdivisions on a ruler are measured in fractions of an inch, the set of rational numbers should be used.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

115. What subset of the real numbers would you use to report temperatures in London and Lisbon?

Solution Since temperatures are usually reported without fractional parts and may be either positive or negative (or zero), the set of integers should be used. 116. What subset of the real numbers would you use to describe the prices of hoodies at Old Navy?

Solution Since the financial condition of a business is usually described in terms of dollars and cents (fractional parts of a dollar), the set of rational numbers should be used. 117. Temperature The average low temperature in International Falls, Minnesota, in January is –7°F. The average high temperature is 15°F. Determine the degrees difference between the average high and the average low.

Solution change  7  15  22  22 The change is 22° F. 118. Temperature Harbin, China, is one of the world’s coldest cities and known for its ice and snow festivals. In February, the average nightly low temperature is –20°C and the average daily high temperature is –7°C. What is the temperature drop from day to night?

Solution

change  20   7   13  13 The change is 13° C.

Discovery and Writing 119. Explain why – x could be positive.

Solution –x will represent a positive number if x itself is negative. For instance, if x = – 3, then –x = – (–3) = 3, which is a positive number. 120. Explain why every integer is a rational number.

Solution Every integer is a rational number because every integer is equal to itself over 1. 121. Is the statement ab  a  b always true? Explain.

Solution The statement is always true. 122. Is the statement

a a   b  0 always true? Explain. b b

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution The statement is always true. 123. Is the statement a  b  a  b always true? Explain.

Solution The statement is not always true. (For example, let a  5 and b  2.) 124. Explain why it is incorrect to write a  b  c if a  b and b  c.

Solution The statement a  b  c could be interpreted to mean that a  c, when this is not necessarily true. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 125. There are six integers between –3 and 3.

Solution False. There are 5 integers: –2, –1, 0, 1, and 2. 126.

725 is a rational number because 725 and 0 are integers. 0 Solution 725 False. is not a rational number because the denominator cannot equal 0. 0

127. ∞ is a real number.

Solution False. ∞ is not a number at all. 128. a  b  b  a

Solution True.  13  129.   5,  , 3,  4  Solution True. (You cannot find an element in the 1st set that is not in the 2nd set.) 130.   

Solution True. (You cannot find an element in the 1st set that is not in the 2nd set.)

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

131. There are six subsets of 11, 22, 33 .

Solution False. There are eight subsets. 132. A set is always a subset of itself.

Solution True.

EXERCISES R.2 Getting Ready

Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Match each expression with the proper description given below. 5

x7 3 4 5 2 7  x4  3 8 x y x  2  x  x x2 x 

a. b. c. d. e.

 

Product of exponential expressions with the same base Quotient of exponential expressions with the same base Power of an exponential expression Power of a product Power of a quotient

Solution a. Product of exponential expressions with the same base is x 3  x 8 x7 b. Quotient of exponential expressions with the same base is 2 x

 

c. Power of an exponential expression is x 2

d. Power of a product is x 3 y 4  x4  e. Power of a quotient is  2  x 

7

5

5

2. Simplify the expression. x  x  x  x  x  x

Solution x  x  x  x  x  x  x6 3. Simplify the expression.

yyyyyy yyy

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution yyyyyy y6 63  3  y   y3 yyy y

    x  x  x   x

4. Fill in the box. x 2

3

2

2

2

Solution

 x    x  x  x   x 2

3

2

2

2

6

5. If x = –5 what is x3 + x?

Solution

 5   5  125   5  130 3

  x x

6. These look alike. Match each with its correct simplification x 5  x 5 x 5 a. b. c.

5

5

5

2x 5 x 10 x 25

Solution a. 2x 5  x 5  x 5 b.

x 10  x 5  x 5

c.

x 25  x 5

 

5

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. Each quantity in a product is called a __________ of the product.

Solution factor 8. A _________ number exponent tells how many times a base is used as a factor.

Solution natural 9. In the expression (2x)3, ___________ is the exponent and _________ is the base.

Solution 3, 2x 10. The expression xn is called an __________ expression.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution exponential 11. A number is in _________ notation when it is written in the form N  10n , where 1  N  10 and n is an __________ .

Solution scientific, integer 12. Unless __________ indicate otherwise, _________are performed before additions.

Complete each exponent rule. Assume x 

. 0

Solution Answers may vary.

13. x m x n  ________

Solution x m x n  x mn 14.

 x   ________ m

n

Solution

x   x m

15.

n

mn

 xy   _________ n

Solution

 xy   x y n

16.

n

n

xm  __________ xn

Solution xm  x m n n x 17. x 0  _________

Solution x0  1

18. x  n  __________

Solution 1 x n  n x

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21


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Practice Write each number or expression without using exponents. 19. 132

Solution 132  13  13  169

20. 103

Solution

103  10  10  10  1,000 21.  5 2

Solution  52   1  5  5   25

22.  5 

2

Solution

 5   5 5  25 2

23. 4 x 3

Solution 4x3  4  x  x  x 24.  4 x 

3

Solution

 4 x    4 x  4 x  4 x  3

25.  5x 

4

Solution

 5 x    5 x  5 x  5 x  5 x  4

26. 6x 2

Solution 6 x 2  6  x  x

27. 8x 4

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22


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 8 x 4   8  x  x  x  x

28.  8x 

4

Solution

 8x    8 x  8x  8 x  8 x  4

Write each expression using exponents. 29. 7 xxx

Solution 7 xxx  7 x 3

30. 8 yyyy

Solution

8 yyyy  8 y 4 31.

  x   x  Solution

  x   x    1 1 x  x 2

2

   

32. 2a 2a 2a

Solution

 2a  2a  2a   2  2  2  a  8a 3

  

33. 3t 3t 3t

3

Solution

 3t  3t  3t    3 3 3 t  27t 3

3

    

34.  2b 2b 2b 2b

Solution

  2b  2b  2b  2b   1  2  2  2  2  b4  16b4

35. xxxyy

Solution

xxxyy  x3 y 2

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23


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

36. aaabbbb

Solution aaabbbb  a 3 b 4

Use a calculator to simplify each expression. 37. 2.23

Solution 2.23  10.648

38. 7.14

Solution 7.14  2541.1681

39.  0.54

Solution 0.54  0.0625

40.  0.2

4

Solution

 0.2   0.0016 4

Simplify each expression. Write all answers without using negative exponents. Assume that all variables are restricted to those numbers for which the expression is defined. 41. x 2 x 3

Solution x 2 x 3  x 23  x 5 3

42. y y

4

Solution

y 3 y 4  y 3 4  y 7

 

43. z 2

3

Solution

z   z 44.  t  2

6

3

23

 z6

7

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24


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

t   t 6

45.

7

y y  5

2

67

 t 42

3

Solution

y y   y   y 5

2

3

7

3

21

46. a3a6 a 4

Solution

a a  a  a a  a 3

6

4

9

  z 

47. z 2

3

4

4

13

5

Solution

z  z   z z  z 2

3

5

4

  t 

48. t 3

4

5

6

20

26

2

Solution

t  t   t t  t 3

4

5

2

  a 

49. a2

3

4

12 10

22

2

Solution

a  a   a a  a 2

3

4

2

  a 

50. a2

4

3

6

8

14

3

Solution

a  a   a a  a 2

51.

4

3 x 

3

3

8

9

17

3

Solution

 3 x   3 x  27 x 3

3

3

3

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25


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

52.  2 y 

4

Solution

 2 y    2 y  16 y 4

53. x 2 y

4

4

4

3

Solution

x y  x  y  x y 3

2

54. x 3 z 4

3

2

3

6

3

6

Solution

x z   x  z   x z 3

4

 a2  55.    b

6

3

6

4

6

18

24

3

Solution

  a

3 a2  a2     b3  b 

 x  56.  3  y 

3

6

b3

4

Solution 4

 x  x4 x4   3  4 y 12 y  y3

 

57.   x 

0

Solution

x   1 0

58. 4 x 0

Solution 4x0  4  1  4

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26


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

59.  4 x 

0

Solution

4x   1 0

60. 2x 0

Solution 2 x 0  2  1  2

61. z 4

Solution 1 z 4  4 z 62.

1 t 2

Solution 1  t2 2 t 2 3 63. y y

Solution

1

y 2 y 3  y 5 

y5

64.  m 2 m3

Solution  m 2 m 3   m 1   m

65. x 3 x 4

2

Solution

x x   x   x 3

4

66. y 2 y 3

2

1

2

2

4

Solution

y y   y   y 2

3

4

1

4

4

1 y4

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27


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

67.

x7 x3

Solution x7  x 7 3  x 4 x3 68.

r5 r2

Solution r5  r 52  r 3 r2 69.

a 21 a 17

Solution a 21  a 21 17  a 4 a 17 70.

t 13 t4

Solution t 13  t 13  4  t 9 t4

x  71. 2

2

x2 x

Solution

x   x  x 2

2

4

2

x x

72.

x

3

4 3

 x1  x

s9 s 3

s  2

2

Solution s9 s 3 s 12   s 12 4  s8 2 4 2 s s

 

 m3  73.  2  n 

3

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28


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

   

3

3 m3  m3  m9   2   3 n6 n  n2

 t4  74.  3  t 

3

Solution 3

3 3  t4  4 3  t1  t3  3   t t 

a  75. 3

  

2

aa2

Solution

a   a 3

aa

76.

2

2

6

a3

 a 6  3  a 9 

1 a9

r 9r 3

r  2

3

Solution r 9 r 3 r6 6   6   r  r 12 3 6 2 r r

 

 a 3  77.  1  b 

4

Solution

 a3   1  b   t 4  78.  3  t 

4

a   a  b  b 3

1

4

12

4

4

2

Solution

 t 4   3  t 

2

t   t  t  t  t 4 3

2

2

8

2

6

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29


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 r 4 r 6  79.  3 3  r r 

2

Solution 2

2

2  r 4 r 6   r 2  2  r 4  r 3 r 3    r 0   r     1  4 r

x x  80. x x  3

2

2

5

 

2

3

Solution

x x   x   x  x x x  x  x 3

2

2

5

2

1

3

 x 5 y 2  81.  3 2  x y 

3

2

2

3

11

9

1 x 11

4

Solution 4

4

4

3

3

 x 5 y 2   x5 x3   x8  x 32  3 2    2 2    4   16 y x y  y y  y   x 7 y 5  82.  7 4  x y 

3

Solution 3

 x 7 y 5   y5 y4   y9  y 27  7 4    7 7    14   42 x x y  x x  x   5 x 3 y 2  83.  2 3    3x y 

2

Solution  5 x 3 y 2   2 3    3x y 

2

2

2

2

 3 x 2 y 3   3x 2 x 3 y 2   3x5  9 x 10   3 2         3   5y  25 y 2  5x y   5y   

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30


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 3 x 2 y 5  84.  2 6   2x y 

3

Solution  3 x 2 y 5   2 x 2 y 6     3 x 5 y 3  85.  5 3   6x y 

3

3

3

3

 2 x 2 y 6   2y5   2  8      2 2 6    4   2 5  12 3 3 x y 3 x x y 3 x y 27 x y      

2

Solution  3 x 5 y 3   5 3   6x y 

2

2

2

2

 6 x 5 y 3   2y3 y3   2y6  4 y 12   5 3    5 5    10   20 x  3x y   1x x   x 

 12 x 4 y 3 z 5  86.  4 3 5    4x y z 

3

Solution 3

3

3

 12 x 4 y 3 z 5   3y3 y3   3y6  27 y 18          4 3 5   1x 4 x 4 z 5 z 5   x 8 z 10  x 24 z 30  4x y z     

8 z y  87. 5 y z  5 yz  2

2

2

3

1

3

2

1

Solution

8 z y  8 z y 64z y 64z z 64z     5 y z 5 y z 5 y z 25 y y 25 y  5 y z  5 yz  2

2

2

3

3

1

2

2

1

3

 m n p   mn p  88.  mn p   mn p 2

3

4

2

3

2

4

2

3

6

3

6

1

1

1

2

2

3

1

3

4

7

5

4

5

1

6

4

1

2

Solution

 m n p   mn p   m n p  m n p  m n p  m m p p  m p nn n  mn p   mn p m n p  m n p m n p 2

3

2

4

3

2

4

2

3

2

1

4

4

6

8

4

8

12

8

14

4

4

8

12

1

2

1

5

6

13

8

5

4

6

14

13

13

17

20

Simplify each expression. 5 62   9  5   89.   2 4 2  3

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

31


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

5[62   9  5 ] 4  2  3

2



5[36  4] 4  1

2



5[40] 4  1



200  50 4

6[3   4  7  ] 2

90.

5 2  4 2

Solution 6[3   4  7  ] 2

5 2  4

5  2  16 

and z 

6[3  9]

5  14 

6[ 6] 36 18   70 70 35

3

2

, 0

, 2

y

Let x  

2

6[3   3  ]

and evaluate each expression.

91. x 2

Solution x 2   2   4 2

92.  x 2

Solution  x 2    2   1  4  4 2

93. x 3

Solution x 3   2   8 3

94.  x 3

Solution  x 3    2   1   8   8 3

95.   xz 

3

Solution

  xz   [1   2  3]  6  216 3

96.  xz

3

3

3

Solution

 xz 3  1   2  33  2  27  54

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32


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

97.

 x2 z3 z y 2

2

Solution

 x2z3

  [ 2  3 ]  [4  27]  108  12 2

z2  y 2

98.

3

90

32  02

z2 x2  y 2

9

3

x z

Solution

z2 x2  y 2

  3 [ 2  0 ]  9 4  0  9  4  36   3 2

2

2

 8 3

 2  3

x3z

3

24

24

2

2 3 99. 5x  3 y z

Solution 5 x 2  3 y 3 z  5  2   3  0   3   5  4   3  0  3  20  0  20 2

100. 3  x  z   2  y  z  2

3

3

Solution 3  x  z   2  y  z   3  2  3   2  0  3   3  5   2  3   3  25   2  27  2

3

2

3

2

3

 75   54   21

101.

 3 x 3 z 2 6 x 2 z 3

Solution  3 x 3 z 2 1z 3 z 3 3 3 3       2 3 2 3 2 5 5 6x z 2x x z 2x 2  32  64 64 2  2 

 5x z  102. 2

3

2

5 xz 2

Solution

 5x z   25x z 2

3

5 xz

2

2

4

5 xz

6

2

5x 4 2

xz z

6

5x 3 z

4

5  2  4

3

3

5  8  81

40 40  81 81

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33


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Express each number in scientific notation. 103. 372,000

Solution

372,000  3.72  105 104. 89,500

Solution

89,500  8.95  104 105. –177,000,000

Solution

177,000,000  1.77  108 106. –23,470,000,000

Solution

23,470,000,000  2.347  1010 107. 0.007

Solution 0.007  7  10 3

108. 0.00052

Solution 0.00052  5.2  10 4

109. –0.000000693

Solution  0.000000693  6.93  10 7

110. –0.000000089

Solution 0.000000089  8.9  10 8

111. one trillion

Solution

1,000,000,000,000  1  1012 112. one millionth

Solution 0.000001  1  10 6

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34


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Express each number in standard notation. 113. 9.37  105

Solution

9.37  105  937,000 114. 4.26  109

Solution

4.26  109  4, 260,000,000 115. 2.21 × 10-5

Solution 2.21  10 5  0.0000221

116. 2.774  10 2

Solution 2.774  10 2  0.02774

117. 0.00032  104

Solution 0.00032  104  3.2

118. 9, 300.0  104

Solution

9,300.0  104  0.93 119.  3.2  10 3

Solution 3.2  10 3  0.0032

120. 7.25  103

Solution

7.25  103  7,250 Use the method of Example 9 to do each calculation. Write all answers in scientific notation. 121.

65, 000  45, 000  250, 000

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35


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

65,000 45,000  6.5  10  4.5  10   6.5 4.5  10 4

4

4  4 5

250,000

2.5  105

2.5

 11.7  103  1.17  101  103  1.17  104

122.

0.000000045 0.00000012  45, 000, 000

Solution

0.0000000450.00000012   4.5  10  1.2  10    4.5 1.2  10    8

7

8  7  7

4.5  107

45, 000, 000

123.

4.5  1.2  1022

0.00000035 170, 000  0.00000085

Solution

0.00000035 170,000   3.5  10  1.7  10    3.5 1.7   10  7

0.00000085

5

8.5  10

7

7  5   7 

8.5  0.7  105

 7  101  105  7  104 124.

0.0000000144  12, 000  600, 000

Solution

0.0000000144  12,000   1.44  10  1.2  10    1.44  1.2  10  8

4

8  4  5

600,000

6  10

6  0.288  109

5

 2.88  101  109  2.88  1010 125.

 45,000, 000,000 212, 000 0.00018

Solution

 45, 000,000,000 212, 000    4.5  10  2.12  10    4.5 2.12  10 10

0.00018

126.

1.8  10

5

4

10  5   4 

1.8  5.3  1019

0.00000000275  4750  500, 000, 000, 000

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36


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

0.00000000275 4, 750    2.75  10  4.75  10    2.75 4.75  10  9

3

9  3  11

500, 000, 000, 000

5  1011

5  2.6125  1017

Fix It In Exercise 127, identify the step the first error is made and fix it. 3

 3 x 3 y 2  127. Use the properties of exponents to simplify the expression   . 2  x y 

Solution Step 3 was incorrect  x2 y  Step 1:  3 2   3x y 

x y 2

Step 2:

Step 3:

Step 4:

3

 3x y  3

3

2

3

x6 y 3

27 x 9 y 6 y9 27 x 3

In Exercise 128, identify the error made and fix it. 128. 123,456,789 written in scientific notation is 1.23456789

10–8.

Solution

123, 456, 789  1.23456789  108 Applications Use scientific notation to compute each answer. Write all answers in scientific notation. 129. Speed of sound The speed of sound in air is 3.31 104 centimeters per second. Compute the speed of sound in meters per minute.

Solution

3.31  104 cm/sec 



3.31  104 6  101 3.31  104 cm 1m 60 sec    m/min 1 sec 100 cm 1 min 1  102  3.316  104 12 m/min = 1 = 19.86  103 m/min = 1.986  104 m/min

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37


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

130. Volume of a box Calculate the volume of a box that has dimensions of 6000 by 9700 by 4700 millimeters.

Solution



V  lwh   6, 000 mm 9, 700 mm 4, 700 mm  6  103 9.7  103

4.7  10  mm 3

3

  6  9.7  4.7   103  3  3 mm3  273.54  109 mm3  2.7354  1011 mm3

131. Mass of a proton The mass of one proton is 0.00000000000000000000000167248 gram. Find the mass of one billion protons.

Solution

mass  1, 000, 000,000  0.00000000000000000000000167248 g 

 1  109

 1.67248  10 g   1.67248  10 g 24

15

132. Speed of light The speed of light in a vacuum is approximately 30,000,000,000 centimeters per second. Find the speed of light in miles per hour. (160,934.4 cm = 1 mile.)

Solution 30, 000, 000, 00 cm

1 mile 60 sec 60min   1 sec 160, 934.4 cm 1 min 1 hr 3  1010 6  101 6  101  mile/hr 1.609344  105  366  1010 1 15 mile/hr  1.609344  67.11  107 mile/hr  6.711  108 mile/hr

30, 000,000, 000 cm/sec 





133. License plates License plates come in various forms. The number of different license plates of the form three digits followed by three letters, is 10  10  10  26  26  26 . Write this expression using exponents. Then evaluate it and express the result in scientific notation.

Solution

10  10  10  26  26  26  103  263 ; 103  263  17,576,000  1.7576  107 134. Astronomy

The distance d, in miles, of the nth planet from the sun is given by the

formula d  9, 275, 200 3 2n  2  4  To the nearest million miles, find the distance of   Earth and the distance of Mars from the sun. Give each answer in scientific notation.

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38


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution Earth: n  3

d  9, 275, 200[3 2n 2  4] 3 2

 9, 275, 200[3(2

 

)  4]

 9, 275, 200[3 2  4] 1

 9, 275, 200[10]  92, 752,000  93,000,000 9.3  107 mi Mars: n  4

   9, 275, 200[3  2   4]  9, 275, 200[3  2   4]

d  9, 275, 200[3 2n 2  4] 4 2 2

 9, 275, 200[16]  148, 403, 200  148,000,000 1.48  108 mi 135. New way to the center of the Earth The spectacular “blue marble” image is the most detailed true-color image of the entire Earth to date. A new NASA-developed technique estimates Earth’s center of mass within 1 millimeter (0.04 inch) a year by using a combination of four space-based techniques.

NASA Images

The distance from the Earth’s center to the North Pole (the polar radius) measures approximately 6356.750 km, and the distance from the center to the equator (the

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39


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

equatorial radius) measures approximately 6378.135 km. Express each distance using scientific notation. Solution polar radius  6.356750  103 km equatorial radius  6.378135  103 km

136. Refer to Exercise 127. Given that 1 km is approximately equal to 0.62 miles, use scientific notation to express each distance in miles.

Solution polar radius  3.941185  10 mi 3

equatorial radius  3.9544437  10 mi 3

Discovery and Writing Write each expression with a single base. 137. x n x 2

Solution x n x 2  x n2 138.

xm x3

Solution xm  x m3 x3 139.

xm x2 x3

Solution xm x2 x m 2 m 23 x   x m 1 3 3 x x 140.

x 3m 5 x2

Solution x 3m5  x 3m52  x 3m 3 x2 141. x m  1 x 3

Solution x m  1 x 3  x m  1 3  x m  4

142. an3a3

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40


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution a n  3a 3  a n  3  3  a n

143. Explain why –x4 and (–x)4 represent different numbers.

Solution In the expression  x 4 , the base of the exponent is x, while in the expression   x  , 4

the base of the exponent is  x. 144. Explain why –x55 and (–x)55 represent equal numbers.

Solution

  x    1  x 55

55

  x 55

55

145. Explain how to write a number in scientific notation.

Solution Answers will vary. 102 is not in scientific notation.

146. Explain why 32

Solution 32  102 is not in scientific notation because 32 is not a number between 1 and 10. 147. Explain why x 11  x 11  x 121 .

Solution x 11  x 11  x 11 11  x 22 148. Explain why 112  113  1215 .

Solution 112  113  112  3  115

149. Explain why

y 50 y 10

 y5 .

Solution y 50  y 50  10  y 40 y 10 150. Explain why  6 xyz   6 x 6 y 6 z 6 . 6

Solution

6 xyz   6 x y z 6

6

6

6

6

Critical Thinking In Exercises 151–158, determine if the statement is true or false. If the statement is false, then correct it and make it true. 151. 00  1

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41


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution False. 00 is undefined. 152.  0  1

Solution True 153. x  n  

1 xn

Solution 1 xn

False. x  n  154.  x  y 

n

1 xn

1 yn

Solution False.  x  y 

n

1

x  y

n

155. 2 1  2 2

Solution

True. 21  21 , 22  41 156.  2    2  1

2

Solution True.  2    21 ,  2   1

2

 41  

157. Young adults between the ages of 18 and 24 send an average of 110 text messages per day. If there are approximately 31.5 million young adults in the USA in this age group, how many text messages are sent in one year? Write the answer using scientific notation.

Solution

110  365  31,500,000  1.1  102  3.65  102  3.15  107  12.64725  1011  1.264725  101  1011  1.264725  1012 158. Health authorities recommend that we drink eight 8-ounce glasses of water each day. How many glasses of water would you drink over a lifetime of 80 years? Write the answer using scientific notation.

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42


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

8  365  80  8  3.65  102  8  101  233.6  103  2.336  102  103  2.336  105

EXERCISES R.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Write the expressions in radical form. a. 125 1/5 b.

 64 

2/3

Solution a.

1251/5  5 125

b.

 64

2/3

 64  or  64 3

2

3

2

2. Write the expressions in exponential form. 9 64

a.

b.

 16 

5

4

Solution

  9 64

1/2

a.

b.

 16   16 . 4

5

5/4

3. Simplify

x2

a. b. c. d.

3

x3

4

x4

5

x5

Solution

x2  x

a. b.

3

x3  x

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43


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

c.

4

x4  x

d.

5

x5  x

4. a. Determine

16 9 and

144.

b. Based on your answers for a. what can you conclude?

Solution a.

16 9  4  3   12

b.

16 9 = 144

144  12

and

5. Simplify. a.

5 2  10 2  15 2

b.

5 x  10 x  15 x

Solution a.

5 2  10 2  15 2  (5  10  15) 2  10 2

b.

5 x  10 x  15 x  (5  10  15) x  10 x

6. a. What can you multiply

b. What can you multiply

2x by so that the radicand is a perfect square? 2x by so that the radicand is a perfect cube?

Solution a. When you multiply

2x by itself, the radicand will be a perfect square.

2x  2x  4x2 , where 4 x 2 is a perfect square b. When you multiply

2 x by

4 x 2 , the radicand will be a perfect cube.

2x  4x2  8x3 , where 8x 3 is a perfect cube Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If a = 0 and n is a natural number, then a 1/ n  __________.

Solution 0 8. If a > 0 and n is a natural number, then a 1/ n is a __________ number.

Solution positive

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44


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

9. If a < 0 and n is an even number, then a 1/n is _________ a real number.

Solution not 10. 6 2/3 can be written as _________ or _________.

Solution

6  , 6  1/3

2

11.

n

1/3

2

a  _________.

Solution a 1/n

a2  _________.

12.

Solution a

13.

n

a n b  ________.

Solution n

14.

n

ab a  _________. b

Solution n

a

n

b

x  y ________

15.

x

y

Solution

16.

m n

x or

n m

x can be written as _______.

Solution mn

x

Practice Simplify each expression. 17. 9 1/ 2

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45


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

 

91/2  32

1/2

3

18. 8 1/ 3

Solution

 

81/3  23  1  19.    25 

1/3

2

1/2

Solution

 1     25 

1/2

 16  20.    625 

 1 2       5    

1/2

1 5

1/4

Solution

 16     625 

 2 4       5    

1/4

1/4

2 5

21.  811/4

Solution

 

811/4   34  8  22.     27 

1/4

 3

1/3

Solution

 8     27 

 2  3        3    

1/3

23.  10, 000 

1/3



2 3

1/4

Solution

 10,000

1/4

 

 104

1/4

 10

24. 1024 1/5

Solution

 1,024

1/5

 

 45

1/5

4

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46


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 27  25.     8 

1/3

Solution

 27     8 

1/3

 3 3        2    

1/3



3 2

26.  64 1/3

Solution

 

641/3   43 27.  64 

1/3

 4

1/2

Solution

 64

1/2

28.  125 

 not a real number

1/3

Solution

 125

1/3

   5     3

1/3

 5

Simplify each expression. Use absolute value symbols when necessary.

29. 16a2

1/2

Solution

 16a 

1/2

1/2

2

30. 25a4

  4a     2

1/2

4a

Solution

25a  4

31.

 16a  4

1/2

 

2   5a2   

1/2

 5 a2  5a2

1/4

Solution

 16a  4

1/4

  2a     4

1/4

2a

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47


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

32. 64a3

1/3

Solution

 64a  3

33. 32a5

  4a    

1/3

3

1/3

 4a

1/5

Solution

 32a  5

34. 64a6

1/5

  2a     5

1/5

 2a

1/6

Solution

64a  6

  2a    

1/6

35. 216b6

6

1/6

2a

1/3

Solution

 216b  6

36. 256t 8

1/3

3   6b2   

1/3

 6b2

1/4

Solution

256t  8

 16a4  37.  2    25b 

  

  4t 2 

1/4

4

1/4

 4 t 2  4t 2

1/2

Solution

 16a4   2    25b 

 4a2 2       5b    

1/2

 a5  38.   10    32b 

1/2

4a2 4a2  5b 5b

1/5

Solution

 a5    10    32b 

1/5

 a 5     2    2b    

1/5



a 2b2

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48


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 1000 x 6  39.   3    27 y 

1/3

Solution  1000 x 6    3    27 y 

1/3

 10 x 2 3        3 y     

 49t 2  40.  4    100z 

1/3

10 x 2 3y

1/2

Solution

 49t 2   100z 4   

1/2

 7t 2       10z 2    

1/2

 

7t 10z 2 7t 10z 2

Simplify each expression. Write all answers without using negative exponents. 41. 43/2

Solution

  2 8

43/2  41/2

3

3

42. 82/3

Solution

  2 4

82/3  81/3

2

2

43. 163/2

Solution

163/2   161/2 44.  8

   4  64 3

3

2/3

Solution

 8 

2/3

   8  

1/3

2

  2 2  4   

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49


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

45.  1000 2/3

Solution

10002/3   10001/3

    10 2

2

 100 46. 100 3/2

Solution

1003/2  1001/2

  10  1,000 3

3

47. 64 1/2

Solution 64 1/2 

1 1  1/2 8 64

48. 25 1/2

Solution 251/2 

1 1  251/2 5

49. 64 3/2

Solution 643/2 

1 64

3/2

1

64  1/2

3

1 3

8

1 512

1 343

50. 493/2

Solution 493/2 

1 493/2

1

 49  1/2

3

1 73

51. 93/2

Solution 93/2  

1 3/2

9



1

9  1/2

3



1 3

3



1 27

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50


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

52.  27 

2/3

Solution

 27  4 53.   9

2/3

1

 27 

2/3

1  27 1/3     

2

1

 3 

2

1 9

5/2

Solution 5

5  4 1/2  2 32         243  9   3  

5/2

4   9

 25  54.    81 

3/2

Solution 3

3  25 1/2  5 125         81 9 729       

3/2

 25     81 

 27  55.     64 

2/3

Solution

 27     64   125  56.    8 

2/3

 64      27 

2/3

2

2  64 1/3   4 16          9  27    3  

4/3

Solution

 125     8 

4/3

4/3

 8     125 

4

4  8 1/3  2 16         5 625  125      

Simplify each expression. Assume that all variables represent positive numbers. Write all answers without using negative exponents.

57. 100s4

1/2

Solution

 100s  4

1/2

 

 1001/2 s4

1/2

 10s2

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51


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

58. 64u6v 3

1/3

Solution

64u v 

1/3

6

3

59. 32 y 10 z 5

  v  1/3

 641/3 u6

3

1/3

 4u2v

1/5

Solution

32 y z 

1/5

1/4

10

5

1

32 y z  10

60. 625a4 b8

5

1/5

1

  z 

321/5 y 10

1/5

5

1/5

1 2 y 2z

Solution

625a b  4

61.

x y  10

5

8

1/4

1

625a b  4

8

1/4

1

 625

1/4

a   b  1/4

4

8

1/4

1 5ab2

3/5

Solution

x y  10

5

3/5

62. 64a6 b12

 x 30/5 y 15/5  x 6 y 3

5/6

Solution

64a b  6

63. r 8 s 16

12

5/6

 645/6 a30/6 b60/6  641/6

 a b  2 a b  32a b 5

5

10

5

5

10

5

10

3/4

Solution

r s  8

16

3/4

 r 24/4 s 48/4  r 6 s 12 

2/3

 8 x y 

2/3

64. 8 x 9 y 12

1 r s 12 6

Solution 9

12

  8 

2/3

x 18/3 y 24/3 

1

 8 

2/3

x 6 y 8 

1

 2  x y 2

6

8

1 6

4x y 8

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52


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 8a6  65.   9    125b 

2/3

Solution  8a6    9    125b 

 16 x 4  66.  8    625 y 

2/3

 8  

2/3

 2 a  4a  2

a 12/3

1252/3 b18/3

4

52 b6

4

25b6

3/4

Solution  16 x 4   625 y 8   

3/4

 27r 6  67.  12    1000s 

163/4 x 12/4 23 x 3 8x 3  3 6  3/4 24/4 625 y 5 y 125 y 6

2/3

Solution  27r 6   12    1000s 

2/3

 32m10  68.    15   243n 

 1000s 12     6  27r 

2/3

10002/3 s24/3 102 s8 100s8  2 4  272/3 r 12/3 3 r 9r 4

2/5

Solution  32m10    15    243n 

69.

2/5

 243n15    10    32m 

2/5

 243 

2/5

n30/5

322/5 m20/5

 3 n  9n  2

22 m4

6

6

4m4

a2/5a4/5 a1/5

Solution a 2/5a 4/5 a6/5  1/5  a5/5  a a 1/5 a

70.

x 6/7 x 3/7 x 2/7 x 5/7

Solution x 6/7 x 3/7 x 9/7   x 2/7 x 2/7 x 5/7 x 7/7

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53


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Simplify each radical expression.

49

71.

Solution

49  72  7 81

72.

Solution

81  92  9 73.

3

125

Solution

74.

3

125  3 53  5

3

64

Solution 3

75.

3

64  3  4   4 3

125

Solution

76.

3

125  3  5  5

5

243

3

Solution 5

243  5  3   3 5

77. 5 

32 100, 000

Solution 5

5

78. 4

 2  32 2 1   5      100, 000 10 5  10 

256 625

Solution 4

4

256 4  4  4     625 5 5

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54


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Simplify each expression, using absolute value symbols when necessary. Write answers without using negative exponents.

36x2

79.

Solution 36 x 2 

6 x   6 x  6 x 2

80.  25y 2

Solution

 5 y    5 y  5 y 2

 25 y 2  

9y 4

81.

Solution

3 y   3 y  3 y

9y4 

2

2

2

2

a4b8

82.

Solution a4 b8 

a b   a b  a b 2

4

2

2

4

2

4

83. 3 8 y 3

Solution 3

84.

3

8 y 3  3 2 y   2 y 3

27z9

Solution 3

85. 4

27 z 9  3 3z 3

  3 z 3

3

x4 y 8 z 12

Solution 4

x4 y 8 z 12

4

x y  xy 2  xy 2  4  3   3  z z3  z 

2

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55


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

86. 5

a 10 b5 c 15

Solution 5

5

a10 b5 5  a2 b  a2 b      c3  c 15 c3  

Simplify each expression. Assume that all variables represent positive numbers so that no absolute value symbols are needed. 87.

8 2 Solution

8 2  4 2 2 2 2 2  2 88.

75  2 27 Solution 75  2 27  25 3  2 9 3  5 3  2  3  3  5 3  6 3   3

89.

200x2  98x2 Solution

200x2  98x2  100x2 2  49x2 2  10x 2  7 x 2  17 x 2 90.

128a3  a 162a Solution

128a3  a 162a  64a2 2a  a 81 2a  8a 2a  9a 2a  a 2a 91. 2 48 y 5  3 y 12 y 3

Solution

2 48 y 5  3 y 12 y 3  2 16 y 4 3 y  3 y 4 y 2 3 y  2 4 y 2

 3 y  3 y 2 y  3 y

 8y2 3y  6y2 3y  2y2 3y 92. y 112 y  4 175 y 3

Solution

y 112 y  4 175 y 3  y 16 7 y  4 25 y 2 7 y  y  4  7 y  4  5 y  7 y  4 y 7 y  20 y 7 y  24 y 7 y

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56


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3

93. 2 81  3 24 3

Solution 2 3 81  3 3 24  2 3 27 3 3  3 3 8 3 3  2  3  3 3  3  2  3 3  6 3 3  6 3 3  12 3 3

94. 3 32  2 162 4

4

Solution 3 4 32  2 4 162  3 4 16 4 2  2 4 81 4 2  3  2  4 2  2  3  4 2  6 4 2  6 4 2  0

95.

4

768z5  4 48z5

Solution 4

768z5  4 48z5  4 256z4 4 3z  4 16z4 4 3z  4z 4 3z  2z 4 3z  6z 4 3z

96. 2 5 64 y 2  3 5 486 y 2

Solution

2 5 64 y 2  3 5 486 y 2  2 5 32 5 2 y 2  3 5 243 5 2 y 2  2  2  5 2 y 2  3  3  5 2 y 2  4 5 2 y 2  9 5 2 y 2  5 5 2 y 2

8 x 2 y  x 2 y  50 x 2 y

97.

Solution 8 x 2 y  x 2 y  50 x 2 y  4 x 2 2 y  x 2 y  25 x 2 2 y  2x 2 y  x 2 y  5x 2 y  6x 2 y 3 3 98. 3x 18x  2 2x  72x

Solution 3 x 18 x  2 2 x 3  72 x 3  3 x 9 2 x  2 x 2 2 x  36 x 2 2 x  3 x  3 2 x  2 x 2 x  6 x 2 x  9x 2x  2x 2x  6x 2x  5x 2x

99. 3 16 xy 4  y 3 2 xy  3 54 xy 4

Solution 3

16 xy 4  y 3 2 xy  3 54 xy 4  3 8 y 3 3 2 xy  y 3 2 xy  3 27 y 3 3 2 xy  2 y 3 2 xy  y 3 2 xy  3 y 3 2 xy  0

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57


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

100.

4

512x5  4 32x5  4 1250x5

Solution 4

512 x 5  4 32 x 5  4 1, 250 x 5  4 256 x 4 4 2 x  4 16 x 4 4 2 x  4 625 x 4 4 2 x  4 x 4 2x  2x 4 2x  5x 4 2x  7 x 4 2x

Rationalize each denominator and simplify. Assume that all variables represent positive numbers. 3 101. 3 Solution

3 3 102.

3

3

3

3

3 3  3 3

6 5 5

6 5

Solution

6 5

6 5

5

5

2

103.

x

Solution 2 x

2

x

x

2

x

x x

8

104.

y Solution 8 y

105.

8

y

y

y

8 y y

2 3

2

Solution

2 3

2

2 3

2

3

4

3

4

23 4 3

8

23 4 3  4 2

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58


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

106.

4d 3

9

Solution

4d 3

107.

9

4d 3

9

3

3

3

3

4d 3 3

3

27

4d 3 3 3

5a 3

25a

Solution

5a 3

108.

25a

5a 3

25a

3

5a2

3

5a2

3

6c2

3

6c2

4

27a2

4

27a2

5a 3 5a2 3

125a3

5a 3 5a2  3 5a2 5a

7 3 6c2 6c

7 3

36c

Solution

7 3

109.

36c

7 3

36c

7 3 6c2 3

216c3

2b 4

3a2

Solution

2b 4

3a2

2b 4

3a2

2b 4 27a2 4

81a4

2b 4 27a2 3a

x 2y

110.

Solution x  2y

111.

3

x 2y

x 2y

2y 2y

2 xy 2y

2u4 9v

Solution 3

3 3 3 3 2u4 2u4 u 2u 3 3v 2 u 3 6uv 2 u 3 6uv 2      3 3 3 3 9v 3v 9v 9v 3v 2 27v 3

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59


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

112.

3

3s 5 4r 2

Solution

3

113.

3 3 3s5 3s5 s3 3 s2 3 2r s 3 6rs2 s 3 6rs2       3 3 3 3 2r 4r 2 2r 4r 2 4r 2 8r 3

5 10

Solution

5 5 5 5 1     10 10 5 10 5 2 5 y 3

114.

Solution y y   3 3

y y

y 3 y

3

9 3

115.

Solution

9 3 9 3 3 3 27 3 1      3 3 3 3 33 3 33 3 3 3

3

3

116.

16b2 16

Solution 3

3 16b2 16b2 3 4b 3 64b3 4b b      3 3 3 3 16 16 4b 16 4b 16 4b 4 4b

5

16b3 64a

117.

Solution 5

5 5 16b3 16b3 5 2b2 32b5 2b b      5 5 5 64a 64a 2b2 64a 2b2 64a 2b2 32a 5 2b2

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60


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3x 57

118.

Solution

3x  57

3x 57

3x

57

3x

3x

3x

3x

171x

9 19 x

3x 3 19 x

x 19 x

Rationalize each denominator and simplify. 1 1  3 27

119.

Solution 1 1   3 27

120. 3

1 3

1 27

1 3

3 3

1

27

3

3

3 3 3 3    3 3 9 81 3 3 3 2 3    9 9 9

1 3 1  2 16

Solution 3

3 3 3 3 3 1 3 1 1 1 1 34 1 4 34 4 4 34            2 16 3 2 3 16 3 2 3 4 3 16 3 4 3 8 3 64 2 4

x  8

121.

23 4 3 4 33 4   4 4 4

x x  2 32

Solution x  8

x  2

x  32

x 8

x 2

x 32

 

x 8 2x

2

2

2x

x 2

2 2 2x

x 32

2 2

 16 4 64 2x 2x 2x    4 2 8 2 2x 4 2x 2x 2x     8 8 8 8

122. 3

y 3 y y  3 4 32 500

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61


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 3 3 3 3 3 3 3 y y y y 32 y 32 y y 3 y y 2  3          3 3 3 3 3 3 3 3 3 4 32 500 4 32 500 4 2 32 2 500 2

3

3

2y 3

8

3

2y

3

64

3 3

2y

1, 000

2y 2y 2y   2 4 10 10 3 2 y 5 3 2 y 2 3 2 y 13 3 2 y     20 20 20 20 

3

3

3

Simplify each radical expression. 123.

4

9

Solution

124.

 

4

9  91/4  32

6

27

1/4

 32/4  31/2  3

Solution

125.

 

6

27  27 1/6  33

10

16x6

1/6

 33/6  31/2  3

Solution

126.

10

16 x 6  16 x 6

6

27x9

1/10

 24 x 6

1/10

 24/10 x 6/10  22/5 x 3/5  22 x 3

1/5

 5 4x3

Solution 6

27 x 9  27 x 9

1/6

 33 x 9

1/6

 33/6 x 9/6  31/2 x 3/2  3 x 3

1/2

 3x 3  x 3x

Fix It In Exercises 127 and 128, identify the step the first error is made and fix it. 127. Use the properties of exponents to simplify –10004/3.

Solution Step 2 was incorrect. Step 1: 

 1000  3

4

Step 2:  10 4 Step 3: 10,000

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62


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

In Exercise 128, identify the step the first error is made and fix it. 3 128. Simplify the radical expression 5x 27 x  48x .

Solution Step 5 was incorrect.

 

  

Step 1: 5 x 9 3 x  16 3 x 2 x 2 Step 2: 5x 9 3x  16 x 3x

Step 3: 5 x  3  3 x  4 x 3 x Step 4: 15x 3x  4x 3x Step 5: 11x 3x

Applications 129. Hiking collage A square-shaped hiking collage of photos has an area of 120 square inches. What is the length of each of its sides?

Solution

s  120  4 30  2 30 inches 130. Volume The volume of a cube-shaped box is 2000 square inches. What is the length of each of its sides?

Solution

s  3 2000  3 1000 3 2  103 2 inches Discovery and Writing We often can multiply and divide radicals with different indices. For example, to multiply

3 by 3 5, we first write each radical as a sixth root 3 = 3 1/2 = 33/6 = 6 33 = 6 27 3

5 = 5 1/3 = 5 2/6 = 6 5 2 = 6 25

and then multiply the sixth roots. 3 3 5 = 6 27 6 25 = 6  27  25  = 6 625

Division is similar. Use this idea to write each of the following expressions as a single radical. 131.

23 2 Solution 6

6

2 3 2  21/2  21/3  23/6  22/6  23 22  6 8 6 4  6 32

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63


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

33 5

132.

Solution

3 3 5  31/251/3  33/652/6  6 33 6 52  6 27 6 25  6 675 133.

4

3 2

Solution 4

3 2

3

134.

4 4 3 3 4 3 4 4 4 12 4 12       4 2 22/4 4 22 4 4 4 4 4 4 16

31/4 2

31/4

 1/2

2 5

Solution 3

2 5

21/3 5

 1/2

6 2 6 6 6 6 2 4 4 6 125 500 500       3/6 6 6 6 6 6 3 5 5 125 125 125 15, 625 5

22/6

135. Explain why a1/n is undefined if n is even and a represents a negative number.

Solution If a1/ n  x, then x n  a. However, if n is even, xn cannot be negative. 136. For what values of x does

4

x4 = x? Explain.

Solution 4

x 4  x . Since x  x if x  0, then 4 x 4  x if x  0.

137. Explain what is meant by rationalizing the denominator of a radical expression.

Solution To rationalize a denominator means to write an equivalent fraction with a denominator equal to a rational number. 138. If all of the radicals involved represent real numbers and y  0, explain why

n

n x x  n y y

Solution x x n   y y

1/ n

x 1/ n

n

x

y

n

y

 1/ n

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64


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

139. If all of the radicals involved represent real numbers and there is no division by 0, explain why

x   y

 m/ n

n

ym xm

Solution

x   y

 m/ n

x  m/ n

x  m/ n

x m/ n y m/ n

y m/ n

y

y

x m/ n y

x m/ n

  m/ n

  m/ n

 m/ n

y   x  m

1/ n

 ym    1/ n  x m  m  

1/ n

n

ym xm

140. The definition of x m / n requires that n x be a real number. Explain why this is important. (Hint: Consider what happens when n is even, m is odd, and x is negative.)

Solution Consider the case when n is even, m is odd and x is negative. Then

    x  . Thus,

x m/ n  x 1/ n

m

n

m

n

x must be a real number for the expression to be

defined.

Critical Thinking In Exercises 141–148, match each expression on the left with an equivalent expression on the right. Assume all variables represent positive numbers. 141.

 16 

1/4

142.  1024 

1/10

a.

2

b.

x 87 x

143. 0111/19

c. 2

144.  1

d. 0

12/19

87

x 86 x

145.

87

1

e.

146.

87

x88

f.

1

g.

undefined

147.

148.

1 87

x

3 3

512

h. –1

Solution 141.

 16 

142.

 1024

1/4

is undefined. g

1/10

 

 210

1/10

 2. c

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

65


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

143. 0111/19  0. d 144.

 1

   1. f

12/19

 19 1

145.

87

1  1. h

146.

87

x88 

87

1

1

147.

148.

87

x

3 3

87

12

x87 87 x  x 87 x. b 

x

87

x 86

87

x 86

87

x 86 . e x

512  3 8  2. a

EXERCISES R.4 Getting Ready

Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Combine like terms and write in descending powers of x. 7  3 x  2 x 2  5 x 2  4 x  5

Solution

7  3 x  2 x 2  5 x 2  4 x  5  2 x 2  5 x 2   3 x  4 x    7  5   3 x 2  x  2

2. Remove parentheses and write in descending powers of  3 y 2  2 y 3  5  y

Solution

 3 y 2  2 y 3  5  y  3 y 2  2 y 3   5     y   3 y  2 y  5  y 2

3

 2y3  3y2  y  5

3. Use the Distributive Property to multiply. 3 x 4 x 2  7 x  2

Solution

3 x 4 x 2  7 x  2  3 x 4 x 2   3 x  7 x    3 x  2   12 x 3  21x 2  6 x

4. Use the Distributive Property to multiply. 5 yz 2 yz 2  7 yz  2

Solution

5 yz 2 yz 2  7 yz  2  5 yz 2 yz 2  5 yz 2  7 yz   5 yz 2  2   5 y z  35 y 2 z 3  10 yz 2 2

4

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66


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

5. Use the Distributive Property to multiply.

Solution

5 1  5  5  1  5

5 1 5

 5   5  25  5  5

6. Identify the conjugate of 3  11.

Solution The conjugate of 3  11 is 3  11 , since a  b and a  b are conjugates.

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A __________ is a real number or the product of a real number and one or more ________.

Solution monomial, variables 8. The _________ of a monomial is the sum of the exponents of its _________.

Solution degree, variables 9. A _________ is a polynomial with three terms.

Solution trinomial 10. A _________ is a polynomial with two terms.

Solution binomial 11. A monomial is a polynomial with _________ term.

Solution one 12. The constant 0 is called the _________ polynomial.

Solution zero 13 Terms with the same variables with the same exponents are called _________ terms.

Solution like

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67


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

14. The ________ of a polynomial is the same as the degree of its term of highest degree.

Solution degree 15. To combine like terms, we add their _________ and keep the same ________ and the same exponents.

Solution coefficients, variables 16. The conjugate of 3 x  2 is _________.

Solution

3 x 2 Determine whether the given expression is a polynomial. If so, tell whether it is a monomial, a binomial, or a trinomial, and give its degree. 17. x 2  3 x  4

Solution yes, trinomial, 2nd degree 18. 5xy  x

3

Solution yes, binomial, 3rd degree 3 1/2 19. x  y

Solution no 20. x

3

 5 y 2

Solution no 2 3 21. 4x  5x

Solution yes, binomial, 3rd degree 2

22. x y

3

Solution yes, monomial, 5th degree 23.

15 Solution yes, monomial, 0th degree

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68


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

24.

5 x  5 x 5

Solution no 25. 0

Solution yes, monomial, no degree 3 2 26. 3 y  4 y  2 y

Solution yes, none, 3rd degree Practice Perform the operations and simplify. 27.

 x  3x   5x  8x  3

2

3

Solution

 x  3x   5x  8x   x  3x  5x  8x  x  5x  3x  8x  6x  3x  8x 3

2

3

3

 

28. 2 x 4  5 x 3  7 x 3  x 4  2 x

2

3

3

3

2

3

2

Solution

 2 x  5x    7 x  x  2x   2 x  5 x  7 x  x  2x 4

3

3

4

4

3

3

4

 2 x 4  x 4  5x 3  7 x 3  2 x  x 4  2 x 3  2x 29.

 y  2 y  7   y  2 y  7 5

3

5

3

Solution

 y  2 y  7   y  2 y  7  y  2 y  7  y  2 y  7 5

3

5

3

5

3

5

3

 y 5  y 5  2 y 3  2 y 3  7  7  4 y 3  14

 

30. 3t 7  7t 3  3  7t 7  3t 3  7

Solution

3t  7t  3   7t  3t  7  3t  7t  3  7t  3t  7 7

3

7

3

7

3

7

3

 3t 7  7t 7  7t 3  3t 3  3  7  4t 7  4t 3  4

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69


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 

 

31. 2 x 2  3 x  1  3 x 2  2 x  4  4

Solution

 

 

2 x 2  3x  1  3 x 2  2x  4  4  2 x 2  2  3x   2  1  3 x 2  3  2x   3  4   4  2x  6 x  2  3x  6 x  12  4 2

2

 2x 2  3x 2  6 x  6 x  2  12  4   x 2  14

 

 

32. 5 x 3  8 x  3  2 3 x 2  5 x  7

Solution

 

 

5 x 3  8x  3  2 3x 2  5x  7  5 x 3  5  8x   5  3  2 3 x 2  2  5x   7  5x  40 x  15  6 x  10 x  7 3

2

 5x 3  6 x 2  40 x  10 x  15  7  5x 3  6 x 2  30 x  8

 

 

33. 8 t 2  2t  5  4 t 2  3t  2  6 2t 2  8

Solution

8 t 2  2t  5  4 t 2  3t  2  6 2t 2  8

   8  2t   8 5  4 t   4  3t   4  2  6 2t   6  8

8 t

2

2

2

 8t 2  16t  40  4t 2  12t  8  12t 2  48  8t 2  4t 2  12t 2  16t  12t  40  8  48  28t  96

 

 

 

 

34. 3 x 3  x  2 x 2  x  3 x 3  2 x

Solution

 

 

 

3 x 3  x  2 x 2  x  3 x 3  2 x  3 x 3  3   x   2 x 2  2  x   3 x 3  3  2 x   3 x  3 x  2 x  2x  3x  6 x 3

2

3

 3 x 3  3 x 3  2x 2  3x  2x  6 x  2 x 2  x

35. y y 2  1  y 2  y  2   y  2 y  2 

Solution

 

y y 2  1  y 2  y  2   y  2 y  2   y y 2  y  1  y 2  y   y 2  2   y  2 y   y  2   y  y  y3  2y2  2y2  2y 3

 y 3  y 3  2 y 2  2 y 2  y  2 y  4 y 2  y

36. 4a2  a  1  3a a2  4  a2  a  2 

Solution

 4a2  a  1  3a a2  4  a2  a  2 

 

 4a  a   4a2  1  3a a2  3a  4   a2  a   a2  2  2

 4a  4a  3a  12a  a3  2a2 3

2

3

 4a3  3a3  a3  4a2  2a2  12a  2a3  6a2  12a

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70


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

37. xy  x  4 y   y x 2  3 xy  xy  2 x  3 y 

Solution

xy  x  4 y   y x 2  3 xy  xy  2 x  3 y 

 

 xy  x   xy  4 y   y x 2  y  3 xy   xy  2 x   xy  3 y   x y  4 xy  x y  3 xy 2  2 x 2 y  3 xy 2 2

2

2

 x 2 y  x 2 y  2 x 2 y  4 xy 2  3 xy 2  3 xy 2  2 x 2 y  4 xy 2

38. 3mn m  2n  6m 3mn  1  2n 4mn  1

Solution

3mn  m  2n   6m  3mn  1  2n  4mn  1

 3mn  m   3mn  2n   6m  3mn   6m  1  2n  4mn   2n  1  3m2 n  6mn2  18m2 n  6m  8mn2  2n  3m2 n  18m2 n  6mn2  8mn2  6m  2n  15m2 n  2mn2  6m  2n

39. 2 x 2 y 3 4 xy 4

Solution

2 x 2 y 3 4 xy 4  2  4  x 2 xy 3 y 4  8 x 3 y 7

40. 15a3 b 2a2 b3

Solution

 

15a 3 b 2a 2 b3  15  2  a 3a 2 bb3  30a 5 b4

 mn  41. 3m2 n 2mn2     12 

Solution  mn   1  2 6 4 4 m 4 n4 2 3m2 n 2mn2   m n     3  2     m mmnn n  12 2  12   12 

42. 

3r 2 s 3  2r 2 s   15rs 2     5  3   2 

Solution 3r 2 s 3  2r 2 s   15rs 2   3   2   15  2 2 3 2 5 6             r r rs ss  3r s 5  3   2   5   3   2 

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71


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

43. 4rs r 2  s2

Solution

 

 

4rs r 2  s 2  4rs r 2  4rs s 2  4r 3 s  4rs 3

44. 6u2v 2uv 2  y

Solution

 

6u2v 2uv 2  y  6u2v 2uv 2  6u2v   y   12u3v 3  6u2vy

45. 6ab2c 2ac  3bc 2  4ab2c

Solution

 

6ab2c 2ac  3bc 2  4ab2c  6ab2c  2ac   6ab2c 3bc 2  6ab2c 4ab2c  12a b c  18ab c  24a b c 2

46. 

mn2 4mn  6m2  8 2

2

2

3

3

2

4

2

Solution 

mn2 mn2 mn2 mn2 4mn  6m2  8   4mn   6m2    8 2 2 2 2  2m2 n3  3m3 n2  4mn2



47. a  2 a  2

Solution

a  2a  2  a  2a  2a  4 2

48. y  5

 y  5

 a2  4a  4

Solution  y  5  y  5   y 2  5 y  5 y  25  y 2  10 y  25

49.  a  6 

2

Solution

a  6  a  6a  6 2

 a2  6a  6a  36  a2  12a  36

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72


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

50.  t  9 

2

Solution

t  9  t  9t  9 2

 t 2  9t  9t  81  t 2  18t  81 51.

 x  4 x  4  Solution

 x  4  x  4   x  4 x  4 x  16 2

 x 2  16



52. z  7 z  7

Solution

 z  7  z  7   z  7z  7 z  49 2

 z 2  49



53. x  3 x  5

Solution

 x  3 x  5  x  5 x  3x  15 2

 x 2  2 x  15



54. z  4 z  6

Solution

 z  4  z  6  z  6z  4z  24 2

 z 2  2z  24



55. u  2 3u  2

Solution

u  2 3u  2  3u  2u  6u  4 2

 3u2  4u  4



56. 4 x  1 2 x  3

Solution

 4 x  1 2x  3  8x  12 x  2 x  3 2

 8 x 2  10 x  3

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73


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



57. 5 x  1 2 x  3

Solution

5x  1 2x  3  10 x  15x  2x  3 2

 10 x 2  13 x  3



58. 4 x  1 2x  7

Solution

 4 x  1 2x  7   8x  28x  2x  7 2

 8 x 2  30 x  7 59.  3a  2b 

2

Solution

 3a  2b    3a  2b  3a  2b   9a  6ab  6ab  4b  9a  12ab  4b 2

2



60. 4a  5b 4a  5b

2

2

2

Solution

 4a  5b 4a  5b  16a  20ab  20ab  25b  16a  25b 2

61.

2

2

2

 3m  4n 3m  4n Solution

 3m  4n 3m  4n  9m  12mn  12mn  16n  9m  16n 2

62.  4r  3s 

2

2

2

2

Solution

 4r  3s    4r  3s  4r  3s   16r  12rs  12rs  9s  16r  24rs  9s 2

2



63. 2 y  4 x 3 y  2 x

2

2

2

Solution

 2 y  4 x  3 y  2x   6 y  4 xy  12xy  8x  6 y  16xy  8x 2



64. 2 x  3 y 3 x  y

2

2

2

Solution

 2x  3 y  3x  y   6x  2xy  9xy  3 y  6x  7 xy  3 y 2

2

2

2

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74


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

65.  9 x  y  x 2  3 y

Solution

9x  y   x  3 y   9x  27 xy  x y  3 y  9x  x y  27 xy  3 y 2

3

2

2

3

2

2

66. 8a2  b  a  2b 

Solution

8a  b  a  2b   8a  16a b  ab  2b 2

3

67.  5z  2t  z 2  t

2

2

Solution

5z  2t   z  t   5z  5tz  2tz  2t  5z  2tz  5tz  2t 2

68. y  2 x 2

3

2

2

3

2

2

 x  3 y  2

Solution

 y  2 x  x  3 y   x y  3 y  2x  6 x y  2 x  5x y  3 y 2

69.  3 x  1

2

2

2

4

2

4

2

2

3

Solution

 3x  1   3x  1 3x  1 3x  1   9 x  3 x  3 x  1  3 x  1   9 x  6 x  1  3 x  1  9 x  3 x   9 x  1  6 x  3 x   6 x  1  1  3 x   1  1 3

2 2

2

2

 27 x 3  9 x 2  18 x 2  6 x  3 x  1  27 x 3  27 x 2  9 x  1 70.  2 x  3 

3

Solution

 2x  3   2x  3 2x  3 2x  3   4 x  6 x  6 x  9  2 x  3   4 x  12 x  9  2 x  3   4 x  2 x   4 x  3   12 x  2 x   12 x  3   9  2 x   9  3  3

2 2

2

2

 8 x 3  12 x 2  24 x 2  36 x  18 x  27  8 x 3  36 x 2  54 x  27

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75


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

71.

 3x  1  2x  4 x  3 2

Solution

 3 x  1  2 x  4 x  3   3 x  2 x   3 x  4 x   3 x   3   1  2 x   1  4 x   1   3  2

2

2

 6 x 3  12 x 2  9 x  2 x 2  4 x  3  6 x 3  14 x 2  5 x  3

72.  2 x  5 x 2  3 x  2

Solution

 2 x  5   x  3 x  2   2 x  x   2 x  3 x   2 x  2   5  x   5  3 x   5  2  2

2

2

 2 x 3  6 x 2  4 x  5 x 2  15 x  10  2 x 3  11x 2  19 x  10

73.  3 x  2 y  2 x 2  3 xy  4 y 2

Solution

 3x  2 y   2x  3xy  4 y  2

2

 3 x 2 x 2  3 x  3 xy   3 x 4 y 2  2 y 2 x 2  2 y  3 xy   2 y 4 y 2

 6 x  9 x y  12 xy  4 x y  6 xy  8 y  6 x  5 x y  6 xy  8 y 3 3

2

74.  4r  3s  2r 2  4rs  2s2

2

2

2

3

3

2

2

Solution

 4r  3s   2r  4rs  2s  2

2

 

 

 4r 2r 2  4r  4rs   4r 2s 2  3s 2r 2  3s  4rs   3s 2s 2

 8r  16r s  8rs  6r s  12rs  6s  8r  10r s  20rs  6s 3 3

2

2

2

2

3

3

2

2

Multiply the expressions as you would multiply polynomials.

75. 5  6 x

5  6 x 

Solution

5  6x 5  6x   5 5  5   6x   5  6x   6x 6x  25  5 6 x  5 6 x  36 x 2  25  6 x



76. 2 x  6 7 x  2

Solution

2x  6 7 x  2   2x(7 x )  2x 2  7 x 6  6 2  14 x 2  2 x 2  7 x 6  12  14 x 2  2 x 2  7 x 6  4  3  14 x 2  2 x 2  7 x 6  2 3

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76


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

77. 3  x

2

Solution

3  x   3  x 3  x   9  3 x  3 x  x  9  6 x  x 2

78. 5 y  2 x

2

2

Solution

5 y  2 x   5 y  2 x 5 y  2 x   25 y   5 y   2 x    5 y   2 x    2 x  2 x  2

2

 25 y 2  10 y x  10 y x  4 x 2  25 y 2  20 y x  4 x 79.

 5  3x 2  5x  Solution

 5  3x 2  5x   2 5  5x  6x  3 5x  3 5x  x  2 5 2

80.

2

 2  x 3  2x  Solution

 2  x 3  2x   3 2  2x  3x  2x  2x  2x  5x  3 2 2

81. 2 y n 3 y n  y  n

Solution

2

2

 

 

2 y n 3 y n  y  n  2 y n 3 y n  2 y n y  n  6 y n n  2 y n

82. 3a  n 2a n  3a n  1

   n

 6 y 2n  2 y 0  6 y 2n  2

Solution

3a  n 2a n  3a n  1  3a  n 2a n  3a  n 3a n  1  6a  n  n  9a  n  n  1  6a0  9a 1  6 

83. 5 x 2n y n 2 x 2 n y  n  3 x 2 n y n

Solution

 

 5 x 2 n y n 2 x 2 n y  n  3 x 2 n y n   5 x 2 n y n 2 x 2 n y  n  5 x 2 n y n 3 x 2 n y n  10 x

2n 2n

y

n    n

 15 x

 10 x y  15 x y 4n

0

0

2n

2 n   2 n 

 10 x

y

9 a

nn

4n

 15 y 2 n

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77


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

84. 2a3n b2n 5a 3n b  ab2 n

Solution

2a3n b2n 5a 3n b  ab2n  2a3n b2n 5a 3n b  2a3n b2n ab2n  10a

3 n   3 n

b

2n 1

 2a

3n 1

b

2 n   2 n

 10a0 b2n  1  2a3n  1b0  10b2n 1  2a3n  1



85. x n  3 x n  4

Solution

 x  3 x  4   x x  4 x  3x  12  x  x  12 n

n

n



86. a n  5 an  3

n

n

n

2n

n

Solution

a  5a  3  a a  3a  5a  15  a  8a  15 n

n

n



87. 2r n  7 3r n  2

n

n

2n

n

n

Solution

 2r  7  3r  2   2r 3r   2r  2   7  3r   14 n

n

n

n

n

n

 6r 2 n  4r n  21r n  14  6r 2 n  25r n  14



88. 4 z n  3 3z n  1

Solution

 4 z  3 3z  1  4 z 3z   4 z  1  3  3z   3 n

n

n

n

n

n

 12 z 2 n  4 z n  9 z n  3  12 z 2 n  13 z n  3

89. x 1/2 x 1/2 y  xy 1/2

Solution

 

x 1/2 x 1/2 y  xy 1/2  x 1/2 x 1/2 y  x 1/2 xy 1/2  x 2/2 y  x 3/2 y 1/2  xy  x 3/2 y 1/2

90. ab1/2 a 1/2 b1/2  b1/2

Solution

 

ab1/2 a 1/2 b1/2  b1/2  ab1/2a 1/2 b1/2  ab1/2 b1/2  a3/2 b2/2  ab2/2  a3/2 b  ab

91.

a

1/2

 b 1/2

Solution

a

1/2

 b 1/2

a

1/2

 b 1/2

a

1/2

 b 1/2  a 1/2a 1/2  a 1/2 b 1/2  a 1/2 b 1/2  b 1/2 b 1/2

 a 2/2  b2/2  a  b

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78


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

92. x 3/2  y 1/2

2

Solution

x

3/2

 y 1/2

  x 2

3/2

 y 1/2

 x

 y 1/2  x 3/2 x 3/2  x 3/2 y 1/2  x 3/2 y 1/2  y 1/2 y 1/2

3/2

 x 6/2  2 x 3/2 y 1/2  y 2/2  x 3  2 x 3/2 y 1/2  y

Rationalize each denominator. 93.

2 31

Solution 2 31

94.

31

2

2

52

1 52

 5  2  5  2  5  2  5  2 5 2 5 2  5  2 5  4 1 1

5 2

2

2

7 2

3x 7 2

3x 7 2

7 2 7 2

 7  2  3 x  7  2  3 x  7  2  x 7  2   74 3 7  2  

3x

2

2

14 y 2 3 Solution 14 y 2 3

97.

1

3x

Solution

96.

2

1

Solution

95.

 3  1  2  3  1  2  3  1  3  1 2 31 31  3  1 3  1 2

14 y 2 3

23 23

 2  3  14 y  2  3  2 y 2  3   29  2  3

14 y

2

2

x x 3

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79


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution x x 3

98.

x

x 3

x 3 x 3

x x 3 x2 

 3

  x x  3

2

x2  3

y 2y  7

Solution

  y 2 y  7  2 y  7 2 y  7 2 y  7 2 y   7   4y  7 y

99.

2y  7

y 2y  7

2

2

2

y 2

   

   

2  3 1 3    1 3 1 3 1 3

2 6 3

y 2 y 2

y 2 y 2 y 2 y 2 y2  2y 2  2    2 y2  2 y 2 y 2 y2  2

x 3 x 3 Solution x 3 x 3

101.

y 2

Solution

100.

y

x 3 x 3 x 3 x 3 x2  2x 3  3    2 x2  3 x 3 x 3 x2  3

2 3 1 3 Solution 2 3

12 

 3

2

 3  2  6  3  3 2

13

  

2  6  3 3 2  2  6  3 3

2 3 3 2  6 2

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80


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

102.

3 2 1 2

Solution

3 2 1 2

3 6 2

3  2 1 2   1 2 1 2

12 

 2

2

 2  3  6  2  2 2

12

3 6 2 2 1   3 6 22 

 6  2  3 2 103.

x

y

x

y

Solution x x

104.

y y

x x

y

y

x x

y y

x 2  xy  xy 

 x  y 2

y2

2

x  2 xy  y xy

2x  y 2x  y Solution 2x  y 2x  y

2x  y 2x  y

2x  y 2x  y

4 x 2  y 2x  y 2x  y 2

 2x   y 2

2

2x  2 y 2x  y 2 2x  y 2

Rationalize each numerator. 105.

21 2 Solution

    2

21  2

106.

2  12 21 21 21     2 21 2 21 2 21 2

1

  2  1

x 3 3 Solution

x 3 x 3 x 3  x 3 x 9     3 3 x  3 3  x  3 3  x  3 2

2

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81


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

107.

y 3 y 3 Solution

y 3 y 3 108.

y2 

  3

2

y 3 y 3 y2  3    y  3 y  3 y2  y 3  y 3  9 y2  2y 3  3

a b a b Solution

 a   b 2

a b a b 109.

x3 3

a b

a b

a b a b

2

a  ab  ab  b 2

2

ab

a  2 ab  b

x

Solution

 x  3   x   x3 x 3 x  3  x  2

x3 x  3

x3 x  3

x 3 x

3

 x 3  x 3

110.

2

x3 x

3

 x 3  x

1 x3 x

2h  2 h Solution 2h  2  h

     2

2h  2 2h  2 2h  2   h 2h  2 h 2h  2  

2

2h2

h

 2  h  2 h

h

 2  h  2

1 2h  2

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82


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Perform each division and write all answers without using negative exponents. 111.

36a2 b3 18ab6

Solution 36a2 b3 2a  2a2  1b3 6  2a 1b 3  3 6 18ab b 112.

45r 2 s5t 3 27r 6 s 2t 8

Solution 45r 2 s5t 3 5 2 6 5 2 3 8 5 4 3 5 5s 3   r s t   r s t   3 3 27r 6 s 2t 8 3r 4 t 5 113.

16 x 6 y 4 z 9 24 x 9 y 6 z 0

Solution 16 x 6 y 4 z 9 24 x y z 9

114.

6



0

2 6 9 4 6 90 2 2z 9   x 3 y 2 z 9   3 2 x y z 3 3 3x y

32m6 n4 p2 26m6 n7 p2 Solution 32m6 n4 p2 26m6 n7 p2

115.

16 6 6 4 7 2 2 16 0 3 0 16 m n p  mn p  13 13 13n3

5 x 3 y 2  15 x 3 y 4 10 x 2 y 3 Solution

5x 3 y 2  15x 3 y 4 5x 3 y 2 15x 3 y 4   10 x 2 y 3 10 x 2 y 3 10 x 2 y 3 x 3xy   2y 2 116.

9m4 n9  6m3 n4 12m3 n3

Solution

9m4 n9  6m3 n4 9m4 n9 6m3 n4   12m3 n3 12m3 n3 12m3 n3 3mn6 n   4 2 117.

24 x 5 y 7  36 x 2 y 5  12 xy 60 x 5 y 4

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83


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 24 x 5 y 7  36 x 2 y 5  12 xy 60 x 5 y 4 118.

24 x 5 y 7

36 x 2 y 5

12 xy

60 x 5 y

60 x 5 y

60 x 5 y

 4

 4

 4

2y3 3y 1  3  4 3 5 5x 5x y

9a 3 b4  27a 2 b4  18a 2 b3 18a 2 b7

Solution 9a 3 b4  27a2 b4  18a2 b3 9a3 b4 27a 2 b4 18a 2 b3 a 3 1       4 2 7 2 7 2 7 2 7 3 3 18a b 18a b 18a b 18a b 2b 2b b Perform each division. If there is a nonzero remainder, write the answer in quotient +

remainder divisor

form. 119. x  3 3x 2  11x  6

Solution 3x 

2

x  3 3 x 2  11x  6 3x 2  9x 2x  6 2x  6 0 120. 3 x  2 3 x 2  11x  6

Solution x  3 3 x  2 3 x 2  11x  6 3x 2  2x 9x  6 9x  6 0 121. 2x  5 2x 2  19x  37

Solution x 

7  2 x25

2 x  5 2 x 2  19 x  37 2x 2  5x  14 x  37  14 x  35 2

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84


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

122. x  7 2 x 2  19 x  35

Solution 2x 

5

x  7 2 x  19 x  35 2

2 x 2  14 x  5 x  35  5 x  35 0

123.

2x 3  1 x1 Solution 2 x 2  2 x  2  X3 1 x  1 2x 3  0x 2  0x  2x  2x 3

1

2

2x 2  0 x 

1

2x  2x 2

2x 

1

2x 

2 3

124.

2 x 3  9x 2  13x  20 2x  7 Solution x 2  x  3  2 x1 7 2 x  7 2 x 3  9 x 2  13 x  20 2x 3  7 x 2  2 x 2  13 x  20  2x 2  7 x 6 x  20 6 x  21 1

125. x 2  x  1 x 3  2 x 2  4 x  3

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85


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution x

3

x  x  1 x  2x 2  4 x  3 2

3

x3  x2  x  3x 2  3x  3  3x 2  3x  3 0 126. x 2  3 x 3  2x 2  4 x  5

Solution x  2  x2x31 x 2  3 x 3  2x 2  4 x  5  3x

x3

 2x  x  5 2

 2x 2

6  x 1

127.

x 5  2x 3  3x 2  9 x3  2

Solution

x2 

 x2 5 x3 2

2 

x 3  2 x5  0x 4  2x 3  3x 2  0x  9  2x 2

x5

 2x 3  x 2  0x  9  2x 3

4  x

128.

2

5

x 5  2x 3  3x 2  9 x3  3

Solution

x2 

2

3 x3 3

x 3  3 x 5  0 x 4  2x 3  3x 2  0x  9  3x 2

x5  2x 3

 0x  9

 2x

6

3

3

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86


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

129.

x 5  32 x 2 Solution x 4  2 x 3  4 x 2  8 x  16 x  2 x 5  0 x 4  0 x 3  0 x 2  0 x  32 x 5  2x 4 2x 4  0x 3 2x 4  4 x 3 4x 3  0x 2 4 x 3  8x 2 8x 2  0 x 8 x 2  16 x 16 x  32 16 x  32 0

130.

x4  1 x1 Solution

x3  x2  x 

1

x  1 x 4  0x 3  0x 2  0x  1 x4  x3  x 3  0x 2  x3  x2 x 2  0x x2  x  x1  x1 0 131. 11x  10  6 x 2 36 x 4  121x 2  120  72x 3  142x

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87


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 6 x 2  x  12 6 x 2  11x  10 36 x 4  72 x 3  121x 2  142 x  120 36 x 4  66 x 3  60 x 2 6 x 3  61x 2  142 x 6 x 3  11x 2  10 x  72 x 2  132 x  120  72 x 2  132 x  120 0 132. x  6 x 2  12 121x 2  72 x 3  142 x  120  36 x 4

Solution 6 x 2  11x  10 6 x 2  x  12 36 x 4  72 x 3 36 x 4  6 x 3

 121x 2  142 x  120  72 x 2

66 x 3

 49 x 2  142 x

66 x 3

 11x 2  132 x  60 x 2  10 x  120  60 x 2  10 x  120 0

Fix It In Exercises 133 and 134, identify the step the first error is made and fix it. 133. Use the Special Product Formula  x  y   x 2  2 xy  y 2 to square the given binomial 2

difference.  4 x  7 y 

2

Solution Step 2 was incorrect. Step 1: Square 4x and get 16 x 2 .

  

Step 2: Multiply 2 4 x 7 y and get 56xy Step 3: Square 7y and get 49y

2

2 2 Step 4: Combine the results of Steps 1, 2, and 3, and get 16x  56xy  49 y

134. Rationalize the denominator.

4 2 3

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88


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution Step 4 was incorrect. Step 1:

Step 2:

5 3    5  3  5  3  4

20  4 3 25  5 3  5 3  3

Step 3:

20  4 3 22

Step 4:

10  4 3 11

Applications 135. Geometry Find an expression that represents the area of the brick wall.

Solution

Area  length  width   x  5  x  2  ft 2  x 2  2 x  5 x  10 ft 2  x 2  3 x  10 ft 2

136. Geometry The area of the triangle shown in the illustration is represented as

 x  3 x  40  square feet. Find an expression that represents its height. 2

Solution

2 x  10 1  base  height 2 x  8 2 x 2  6 x  80 1 x 2  3 x  40   x  8   height 2 x 2  16 x 2  10 x  80 2 x 2  3 x  40   x  8   height  10 x  80 2 x 2  6 x  80   x  8   height 0 2 x 2  6 x  80  height The height is  2 x  10 ft. x 8 Area =

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89


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

137. Gift boxes The corners of a 12 in.-by-12 in. piece of cardboard are folded inward and glued to make a box. Write a polynomial that represents the volume of the resulting box. Crease here and fold inward

Solution Volume  l  w  h

  12  2 x  12  2 x  x in.3

    144 x  48 x  4 x  in.   4 x  48 x  144 x  in.

 144  48 x  4 x 2 x in.3 2

3

3

2

3 3

138. Travel Complete the following table, which shows the rate (mph), time traveled (hr), and distance traveled (mi) by a family on vacation.

r 3x + 4

t

=

d 3x2 + 19x + 20

Solution t

d 3 x 2  19 x  20  3x  4 r x  5

3 x  4 3 x 2  19 x  20 3x 2  4 x 15 x  20 15 x  20 0 t  x 5

Discovery and Writing 139. Show that a trinomial can be squared by using the formula

a  b  c   a  b  c  2ab  2bc  2ac. 2

2

2

2

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90


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

a  b  c   a  b  c a  b  c   a a  b  c   b a  b  c   c a  b  c  2

 a2  ab  ac  ab  b2  bc  ac  bc  c2  a2  b2  c2  2ab  2bc  2ac 140. Show that  a  b  c  d   a2  b2  c 2  d 2  2ab  2ac  2ad  2bc  2bd  2cd . 2

Solution

a  b  c  d   a  b  c  d a  b  c  d   a a  b  c  d   b a  b  c  d   c a  b  c  d   d a  b  c  d  2

 a2  ab  ac  ad  ab  b2  bc  bd  ac  bc  c2  cd  ad  bd  cd  d 2  a2  b2  c2  d 2  2ab  2ac  2ad  2bc  2bd  2cd 141. Explain the FOIL method.

Solution Answers may vary. 142. Explain how to rationalize the numerator of

X 2 . X

Solution Multiply the numerator and denominator by the conjugate of the numerator

 x  2 .

143. Explain why  a  b   a 2  b2 . 2

Solution Check the formula with a  1 and b  2 . 144. Explain why

a2  b2  a2  b2 .

Solution Check the formula with a  3 and b  4 . Critical Thinking In Exercises 145–150, determine if the statement is true or false. If the statement is false, then correct it and make it true. 145. All polynomials are trinomials.

Solution False. Some polynomials are trinomials. 146. All binomials are polynomials.

Solution True.

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91


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

147.  12 x  5 y   144 x 2  120 xy  25 y 2 2

Solution True.  12 x  5 y    12 x  5 y  12 x  5 y   144 x 2  60 xy  60 xy  25 y 2 2

 144 x 2  120 xy  25 y 2

148.  6 x  y   36 x 2  y 2 2

Solution False.  6 x  y    6 x  y  6 x  y   36 x 2  6 xy  6 xy  y 2  36 x 2  12 xy  y 2 2



149. x 1/3  6 4 x 1/3  7  4 x 1/9  17 x 1/3  42

Solution



False. x 1/3  6 4 x 1/3  7  4 x 2/3  7 x 1/3  24 x 1/3  42  4 x 2/3  17 x 1/3  42



150. x 3  5 x 3  5  x 9  25

Solution



False. x 3  5 x 3  5  x 6  5 x 3  5 x 3  25  x 6  25  x16  25

x

0 0 2 +

2 x

In Exercises 151 and 152, the revenue associated with selling x units of a product is dollars, and the cost associated with producing x units of the product is –200x + 500 dollars.

151. Determine the polynomial that represents the profit in dollars of making x units of the product.

Solution

Profit  Revenue  Cost  x 2  200 x   200 x  500   x 2  200 x  200 x  500  x 2  400 x  500

152. If 100 units are produced and sold, would the profit exceed $50,000?

Solution Use the answer to #139. x 2  400 x  500   100   400  100   500  49, 500  False. 2

EXERCISES R.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

The prime factorization of three terms are shown. Find their greatest common factor.

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92


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

12x3 y2 = 2  2  3 x  x  x  y  y 18xy 4  2  3  3  x  y  y  y  y

30x2 y 3  2  3  5  x  x  y  y  y Solution All terms have 2, 3, x, and y 2 in common. The greatest common factor of all 3 terms 2 2 is 2  3  x  y  6xy

2. Find the greatest common factor of 10 x 5  y  2  , 20 x 4  y  2  , and 15x 4  y  2  . 2

3

2

Solution The greatest common factor of 10, 20, and 15 is 5. The greatest common factor of

x5 , x4 , and x4 is x 4 . The greatest common factor of  y  2  ,  y  2  and  y  2  is 2

3

2

 y  2  . The greatest common factor of all 3 terms is 5 x  y  2 . 2

4

2

3. Fill in the boxes to complete each factorization.

a.

8 x 3  6 x 2  10 x  2 x 4 x 2    5

b.

8 x  11  1  8 x 11

Solution

a.

8 x 3  6 x 2  10 x  2 x 4 x 2  3 x  5

b.

8 x  11  1  8 x  11

4. Fill in the boxes to complete each factorization. a.

x 2  10 x  9   x  9  x   

b.

12 x 2  5 x  2   3 x  2   1

Solution a.

x 2  10 x  9   x  9 x  1

b.

12 x 2  5 x  2   3 x  2 4 x  1

5. Fill in each set of parentheses. a.

49 x 2  100z 4      

b.

125 x 3  8z 3      

c.

27 y 6  64 z 3      

2

3

3

2

3

3

Solution a.

49 x 2  100z 4   7 x   10z 2 2

2

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93


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

b.

125 x 3  8z 3   5 x    2z 

c.

27 y 6  64z 3  3 y 2

3

6. Fill in the box. x  7

3

   4z  3

3

  x  7  x  7 2/3

Solution

 x  7

2/3

 x  7

5/3

 x7

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. When polynomials are multiplied together, each polynomial is a __________ of the product.

Solution factor 8. If a polynomial cannot be factored using __________ coefficients, it is called a _________ polynomial.

Solution integer, prime Complete each factoring formula. 9.

ax  bx  ________

Solution

ax  bx  x  a  b 

2 2 10. x  y  ________

Solution

x 2  y 2   x  y  x  y 

2 2 11. x  2xy  y  ________

Solution x 2  2 xy  y 2   x  y  x  y    x  y 

2

2 2 12. x  2xy  y  _________

Solution x 2  2 xy  y 2   x  y  x  y    x  y 

2

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94


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3 3 13. x  y  ________

Solution

x 3  y 3   x  y  x 2  xy  y 2

3 3 14. x  y  ________.

Solution

x 3  y 3   x  y  x 2  xy  y 2

Practice In each expression, factor out the greatest common monomial. 15. 3x  6 _________.

Solution

3x  6  3  x  2

16. 5 y  15 _________.

Solution

5 y  15  5  y  3

17. 8 x 2  4 x 3 _________.

Solution

8x 2  4 x 3  4 x 2  2  x 

3 2 18. 9 y  6 y _________.

Solution

9 y 3  6 y 2  3 y 2  3 y  2

2 2 3 2 19. 7 x y  14x y ________.

Solution

7 x 2 y 2  14 x 3 y 2  7 x 2 y 2  1  2 x 

20. 25 y z  15 yz ________. 2

2

Solution

25 y 2 z  15 yz 2  5 yz  5 y  3z 

In each expression, factor by grouping.

21. a x  y  b x  y

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95


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

a  x  y   b  x  y    x  y  a  b 

22. b x  y  a x  y

Solution

b  x  y   a  x  y    x  y  b  a 

23. 4a  b  12a 2  3ab

Solution

4a  b  12a2  3ab  4a  b  3a  4a  b   1  4a  b   3a  4a  b    4a  b  1  3a 

2 24. x  4x  xy  4 y

Solution

x 2  4 x  xy  4 y  x  x  4   y  x  4    x  4  x  y 

In each expression, factor the difference of two squares. 25. 4 x 2  9

Solution 4 x 2  9   2 x   32   2 x  3  2 x  3  2

26. 36 z 2  49

Solution 36 z 2  49   6z   7 2   6 z  7  6 z  7  2

27. 4  9r 2

Solution 4  9r 2  22   3r    2  3r  2  3r  2

28. 16  49x 2

Solution 16  49 x 2  42   7 x    4  7 x  4  7 x  2

29. 81x 4  1

Solution

81x 4  1  9 x 2

  1  9 x  19 x  1  9 x  1 3 x  1 3 x  1 2

2

2

2

2

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96


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

30. 81  x 4

Solution

   9  x 9  x   9  x   3  x  3  x 

81  x 4  92  x 2 31.

2

2

2

2

 x  z   25 2

Solution

 x  z   25   x  z   5   x  z  5 x  z  5 2

2

2

32.  x  y   9 2

Solution

x  y  9  x  y  3   x  y  3 x  y  3 2

2

2

In each expression, factor the trinomial. 33. x 2  8 x  16

Solution x 2  8 x  16   x  4  x  4    x  4 

2

34. a 2  12a  36

Solution a2  12a  36   a  6  a  6    a  6 

2

35. b2  10b  25

Solution b2  10b  25   b  5  b  5    b  5 

2

2 36. y  14 y  49

Solution y 2  14 y  49   y  7  y  7    y  7 

2

37. m2  4mn  4n2

Solution

m2  4mn  4n2   m  2n m  2n

  m  2n

2

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97


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

38. r 2  8rs  16 s 2

Solution

r 2  8rs  16s2   r  4s  r  4s    r  4s 

2

2 2 39. 12x  xy  6 y

Solution

12 x 2  xy  6 y 2   4 x  3 y  3x  2 y 

2 2 40. 8x  10xy  3 y

Solution

8 x 2  10 xy  3 y 2   4 x  y  2 x  3 y 

In each expression, factor the trinomial by grouping. 41. x 2  10 x  21 Solution x 2  10 x  21 : a  1, b  10, c  21

key number  ac  1  21  21

x 2  10 x  21  x 2  7 x  3 x  21

 x  x  7  3  x  7   x  7  x  3 

42. x 2  7 x  10

Solution x 2  7 x  10 : a  1, b  7, c  10

key number  ac  1  10   10

x 2  7 x  10  x 2  5 x  2 x  10

 x  x  5  2  x  5   x  5  x  2 

43. x 2  4 x  12

Solution x 2  4 x  12 : a  1, b  4, c  12

key number  ac  1  12   12 x 2  4 x  12  x 2  6 x  2 x  12

 x  x  6  2  x  6   x  6  x  2 

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98


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

44. x 2  2 x  63

Solution x 2  2 x  63 : a  1, b  2, c  63

key number  ac  1  63   63 x 2  2 x  63  x 2  9 x  7 x  63

 x  x  9  7  x  9   x  9  x  7 

45. 6p  7 p  3 2

Solution 6 p2  7 p  3 : a  6, b  7, c  3

key number  ac  6  3   18 6 p2  7 p  3  6 p2  9 p  2 p  3

 3p  2p  3    2p  3    2 p  3  3 p  1

46. 4q  19q  12 2

Solution 4q2  19q  12 : a  4, b  19, c  12

key number  ac  4  12   48

4q2  19q  12  4q2  3q  16q  12

 q  4q  3   4  4q  3    4q  3  q  4 

In each expression, factor the sum of two cubes. 47. t 3  343 Solution

t 3  343  t 3  73   t  7  [t 2   t  7   72 ]   t  7  t 2  7t  49

48. r 3  8s 3

Solution

3 2 r 3  8s 3  r 3   2s    r  2s  r 2   r  2s    2s     r  2s  r 2  2rs  4s2  

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99


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3 3 49. 125 y  216z

Solution 3 3 2 2 125 y 3  216z 3   5 y    6z    5 y  6z   5 y    5 y  6z    6z    

  5 y  6z  25 y 2  30 yz  36z 2

3 3 50. 27 y  1000z

Solution 3 3 2 2 27 y 3  1000z 3   3 y    10z    3 y  10z   3 y    3 y  10z    10z    

  3 y  10z  9 y 2  30 yz  100z 2

In each expression, factor the difference of two cubes. 51. 8 z 3  27

Solution

3 2 8z 3  27   2z   33   2z  3   2z    2z  3   32    2z  3  4 z 2  6z  9  

52. 125a 3  64

Solution

3 2 125a3  64   5a   43  (5a  4)  5a    5a  4   42    5a  4  25a2  20a  16  

3 3 53. 343y  z

Solution

343 y 3  z 3   7 y   z 3   7 y  z  (7 y )2   7 y  z   z 2    7 y  z  49 y 2  7 yz  z 2 3

3 3 54. 27 y  512z

Solution 3 3 2 2 27 y 3  512z 3   3 y    8z    3 y  8z   3 y    3 y  8z    8z     2 2   3 y  8z  9 y  24 yz  64z

Factor each expression completely. If an expression is prime, so indicate. 55. 3a 2 bc  6ab 2c  9abc 2

Solution

3a2 bc  6ab2c  9abc2  3abc  a  2b  3c 

3 3 3 2 2 2 56. 5x y z  25x y z  125xyz

Solution

5 x 3 y 3 z 3  25 x 2 y 2 z 2  125 xyz  5 xyz x 2 y 2 z 2  5 xyz  25

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100


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

57. 3 x 3  3 x 2  x  1

Solution

3 x 3  3 x 2  x  1  3 x 2  x  1  1  x  1   x  1 3 x 2  1

58. 4 x  6 xy  9 y  6

Solution

4 x  6 xy  9 y  6  2 x  2  3 y   3  3 y  2   3 y  2 2 x  3

59. 2txy  2ctx  3ty  3ct

Solution

2txy  2ctx  3ty  3ct  t  2 xy  2cx  3 y  3c   t 2 x  y  c   3  y  c    t  y  c  2 x  3 

60. 2ax  4ay  bx  2by

Solution

2ax  4ay  bx  2by  2a  x  2 y   b  x  2 y    x  2 y  2a  b 

61. ax  bx  ay  by  az  bz

Solution

ax  bx  ay  by  az  bz  x  a  b   y  a  b   z  a  b    a  b  x  y  z 

2 3 2 2 62. 6x y  18xy  3x y  9x

Solution

 

6 x 2 y 3  18 xy  3 x 2 y 2  9 x  3x 2 xy 3  6 y  xy 2  3  3x 2 y xy 2  3  1 xy 2  3    2  3 x xy  3  2 y  1 63. x 2   y  z 

2

Solution

x 2   y  z    x   y  z    x   y  z     x  y  z  x  y  z  2

64. z 2   y  3 

2

Solution

z 2   y  3   z   y  3   z   y  3    z  y  3 z  y  3 2

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101


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

65.  x  y    x  y  2

2

Solution

 x  y    x  y    x  y    x  y   x  y    x  y    x  y  x  y  x  y  x  y    2 x  2 y   4 xy 2

2

66.  2a  3    2a  3  2

2

Solution

 2a  3   2a  3   2a  3   2a  3  2a  3   2a  3   2a  3  2a  3 2a  3  2a  3   4a  6   24a 2

2

4 4 67. x  y

Solution

    y    x  y  x  y    x  y   x  y  x  y  2

x4  y 4  x2

2

2

2

2

2

2

2

2

68. z 4  81

Solution

   9   z  9 z  9   z  9 z  3    z  9  z  3 z  3

z 4  81  z 2

2

2

2

2

2

2

2

2

69. 3 x 2  12

Solution

3 x 2  12  3 x 2  4  3  x  2  x  2 

70. 3x y  3xy 3

Solution

3 x 3 y  3 xy  3 xy x 2  1

 3 xy  x  1 x  1 2 71. 18xy  8x

Solution

18 xy 2  8 x  2 x 9 y 2  4

 2 x  3 y  2  3 y  2 

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102


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

72. 27 x 2  12

Solution

27 x 2  12  3 9 x 2  4

 3  3 x  2  3 x  2 

73. x 2  2 x  15

Solution

x2  2x  15  prime 74. x 2  x  2

Solution

x2  x  2  prime 75. 15  2a  24a 2

Solution

15  2a  24a2  24a2  2a  15

  6a  5 4a  3

76.  32  68 x  9 x 2

Solution

32  68 x  9 x 2  9 x 2  68 x  32   9 x  4  x  8

2 2 77. 6x  29xy  35 y

Solution

6 x 2  29 xy  35 y 2   3 x  7 y  2 x  5 y 

2 2 78. 10x  17 xy  6 y

Solution

10 x 2  17 xy  6 y 2   5 x  6 y  2 x  y 

79. 12p  58pq  70q 2

2

Solution

12 p2  58pq  70q2  2 6 p2  29pq  35q2  2  6 p  35q  p  q 

2 2 80. 3x  6xy  9 y

Solution

3 x 2  6 xy  9 y 2  3 x 2  2 xy  3 y 2  3  x  3 y  x  y 

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103


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

81. 6m2  47 mn  35n2

Solution

6m2  47mn  35n2   6m2  47mn  35n2    6m  5n  m  7n 

82.  14 r 2  11rs  15 s 2

Solution

14r 2  11rs  15s2   14r 2  11rs  15s2    7r  5s  2r  3s 

83. 6 x 3  23 x 2  35 x

Solution

6 x 3  23 x 2  35 x   x 6 x 2  23 x  35   x  6 x  7  x  5 

3 2 84.  y  y  90 y

Solution

 y 3  y 2  90 y   y y 2  y  90   y  y  10  y  9 

85. 6 x 4  11x 3  35 x 2

Solution

6 x 4  11x 3  35 x 2  x 2 6 x 2  11x  35  x 2  2 x  7  3 x  5 

86. 12 x  17 x 2  7 x 3

Solution

12 x  17 x 2  7 x 3  7 x 3  17 x 2  12 x   x 7 x 2  17 x  12   x  x  3  7 x  4 

87. x 4  2 x 2  15

Solution



x 4  2 x 2  15  x 2  5 x 2  3

88. x 4  x 2  6

Solution



x4  x2  6  x2  3 x2  2

89. a 2 n  2a n  3

Solution



a2n  2an  3  an  3 an  1

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104


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

90. a 2 n  6a n  8

Solution



a2n  6an  8  an  4 an  2

91. 6 x 2 n  7 x n  2

Solution





6 x 2n  7 x n  2  3 x n  2 2 x n  1

92. 9 x 2 n  9 x n  2

Solution

9 x 2n  9 x n  2  3 x n  2 3 x n  1

93. 4x

2n

 9 y 2n

Solution

   3 y    2 x  3 y  2 x  3 y 

4 x 2n  9 y 2n  2 x n

2

n

n

n

2

n

n

94. 8 x 2 n  2 x n  3

Solution



8 x 2n  2 x n  3  4 x n  3 2 x n  1

95. 10 y

2n

 11 y n  6

Solution



10 y 2n  11 y n  6  5 y n  2 2 y n  3

96. 16 y

4n

 25 y 2n

Solution

16 y 4 n  25 y 2n  y 2n 16 y 2 n  25

 y 2n  4 y n  52    n 2n  y 4y  5 4yn  5

2



97. 2 x 3  2000

Solution

2 x 3  2000  2 x 3  1000  2 x 3  103  2  x  10  x 2  10 x  100

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105


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3 98. 3 y  648

Solution

3 y 3  648  3 y 3  216  3 y 3  63  3  y  6  y 2  6 y  36

99.  x  y   64 3

Solution

 x  y   64   x  y   4   x  y   4  x  y   4  x  y   4    x  y  4   x  2xy  y  4 x  4 y  16 3

3

2

3

2

2

2

100.  x  y   27 3

Solution

 x  y   27   x  y   3   x  y   3  x  y   3  x  y   3    x  y  3  x  2 xy  y  3x  3 y  9 3

3

2

3

2

2

2

6 6 101. 64a  y

Solution

    y   8a  y 8a  y    2a  y   4a  2ay  y   2a  y   4a  2ay  y    2a  y  2a  y   4a  2ay  y  4a  2ay  y 

64a6  y 6  8a3

2

3

2

3

3

3

3

2

2

2

2

2

2

2

2

102. a 6  b6

Solution

    b   a  b (a )  a b  (b )   a  b a  a b  b 

a6  b6  a2

3

2

3

2

2

2 2

2

2

2 2

2

2

4

2

2

4

103. a 3  b3  a  b

Solution

a3  b3  a  b   a  b  a2  ab  b2   a  b  1   a  b  a2  ab  b2  1

104. a2  y 2  5  a  y 

Solution

a  y   5 a  y   a  y a  y   5 a  y   a  y a  y  5 2

2

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106


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

6 6 105. 64x  y

Solution

64 x 6  y 6  4 x 2

   y    4 x  y (4 x )  4 x y  ( y )    4 x  y  16 x  4 x y  y  3

2

3

2

2

2 2

2

2

4

2

2

2

2

2 2

4

2 2 106. z  6z  9  225 y

Solution

z 2  6z  9  225 y 2   z  3 z  3  225 y 2   z  3   15 y  2

2

  z  3  15 y  z  3  15 y 

2 2 107. x  6x  9  144 y

Solution

x 2  6 x  9  144 y 2   x  3 x  3  144 y 2   x  3   12 y  2

2

  x  3  12 y  x  3  12 y 

2 2 108. x  2x  9 y  1

Solution

x 2  2 x  9 y 2  1  x 2  2 x  1  9 y 2   x  1 x  1  9 y 2   x  1   3 y    x  1  3 y  x  1  3 y  2

2

109.  a  b   3  a  b   10 2

Solution

a  b  3 a  b  10  a  b  5 a  b  2  a  b  5a  b  2 2

110. 2  a  b   5  a  b   3 2

Solution

2  a  b   5  a  b   3  2  a  b   1  a  b   3   2a  2b  1 a  b  3 2

111.

x6  7x3  8

Solution



x 6  7 x 3  8  x 3  8 x 3  1   x  2  x 2  2 x  4  x  1 x 2  x  1

112. x 6  13 x 4  36 x 2

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107


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution



x 6  13 x 4  36 x 2  x 2 x 4  13 x 2  36  x 2 x 2  9 x 2  4  x 2  x  3  x  3  x  2  x  2 

113. x 4  x 2  1

Solution

x 4  x 2  1  x 4  2x 2  1  x 2

     x  1  x   x  1  x  x  1  x    x  x  1 x  x  1  x2  1 x2  1  x2 2

2

2

2

2

2

2

114. x 4  3 x 2  4

Solution

x 4  3x 2  4  x 4  4 x 2  4  x 2

     x  2  x   x  2  x  x  2  x    x  x  2  x  x  2   x2  2 x2  2  x2 2

2

2

2

2

2

2

115. x 4  7 x 2  16

Solution

x 4  7 x 2  16  x 4  8 x 2  16  x 2

     x  4  x   x  4  x  x  4  x    x  x  4  x  x  4   x2  4 x2  4  x2 2

2

2

2

2

2

2

4 2 116. y  2 y  9

Solution

y4  2y2  9  y4  6y2  9  4y2

  y  3  4 y   y  3   2 y    y  3  2 y  y  3  2 y    y  2 y  3  y  2 y  3   y2  3 2

2

2

2

2

2

2

2

2

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108


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

117. 4a 4  1  3a 2

Solution



 a   2a  1  a  2a  1  a    2a  a  1 2a  a  1

4a4  1  3a2  4a4  4a2  1  a2  2a2  1 2a2  1  a2  2a2  1

2

2

2

2

2

2

118. x 4  25  6 x 2

Solution



  2x    x  5  2 x  x  5  2 x    x  2 x  5  x  2 x  2 

x 4  25  6 x 2  x 4  10 x 2  25  4 x 2  x 2  5 x 2  5  4 x 2  x 2  5

2

2

2

2

2

2

Factor each expression by grouping three terms and two terms. 2 119. x  x  6  xy  2 y

Solution

x 2  x  6  xy  2 y   x  3 x  2  y  x  2   x  2 x  3  y 

2 120. 2x  5x  2  xy  2 y

Solution

2x 2  5 x  2  xy  2 y   2x  1 x  2  y  x  2   x  2 2x  1  y 

121. a 4  2a 3  a 2  a  1

Solution

a4  2a3  a2  a  1  a2 a2  2a  1  a  1  a2  a  1 a  1  1  a  1   a  1 a2  a  1  1

 (a  1) a3  a2  1 122. a 4  a 3  2a 2  a  1

Solution

a4  a3  2a2  a  1  a2 a2  a  2  a  1  a2  a  2 a  1  1  a  1

  a  1 [a (a  2)  1] 2

 (a  1) a3  2a2  1

Factor the indicated monomial from the given expression. 123. 3 x  2; 2

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109


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

3x  2  2  32x  22   2  32 x  1

124. 5 x  3; 5

Solution

5 x  3  5  55x  53   5  x  53 

125. x 2  2x  4;2

Solution

   2  x  x  2 2

x 2  2 x  4  2 x2  22x  42 1 2

2

126. 3 x 2  2 x  5;3

Solution

   3x  x   2

3 x 2  2 x  5  3 33x  23x  53 2

2 3

5 3

127. a  b; a

Solution

a  b  a  aa  ab   a  3  ab 

128. a  b; b

Solution

a  b  b  ab  bb   b  ab  1

129. x  x 1/2 ; x 1/2

Solution

  x x

x  x 1/2  x 1/2 x 1 1/2  x 1/2 1/2 1/2

1/2

1

130. x 3/2  x 1/2 ; x 1/2

Solution

x 3/2  x 1/2  x 1/2 x 3/2  1/2  x 1/2  1/2 x

1/2

 x  1

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110


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

131. 2x  2 y ; 2

Solution

 2x 2y    2x  2 y  2   2  2    2 132.

 2x  y 

3a  3b; 3 Solution

 3a 3b    3a  3b  3   3 3  

 3 a  3b

133. ab3/2  a3/2b; ab

Solution  ab3/2 a3/2 b  ab3/2  a3/2 b  ab    ab   ab  ab b1/2  a 1/2

134. ab2  b; b1

Solution  ab2 b  ab2  b  b 1  1  1  b   b 1 3  b ab  b2

Factor completely and simplify each algebraic expression. Write answers using positive exponents. 135. 2 x 4 /5  8 x 2/5

Solution

2 x 4/5  8 x 2/5  2 x 2/5 x 2/5  4

136. 9 x 3/7  18 x 1/7

Solution

9 x 3/7  18 x 1/7  9 x 1/7 x 2/7  2

137. x 6  3 x 2

Solution x 6  3 x 2  x 6 

3 x8 3 x8  3  2  2  2 x x x x2

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111


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

138. 5 x 3  10 x 7

Solution 5 x  10 x 3

139.  x  1

1/2

7

10

 5x  3

  x  1

x

7

5 x 10 x

7

10 x

7

5 x 10  10 x

7

5 x 10  2 x

7

3/2

Solution

 x  1

1/2

140.  x  6 

  x  1

2/3

  x  1

3/2

1/2

[1   x  1]   x  x  1

  x  6

 1   x  6    x  6  

1/2

  x  6

5/3

Solution

 x  6

2/3

141. 2  x  1

  x  6

5/3

3/5

 4 x  x  1

2/3

2/3

8/5

Solution

2  x  1

142. x 2  5

3/5

1/5

 4 x  x  1

 x2  5

 x  7

 2  x  1

8/5

8/5

2   x  1

 x  1  2 x    

 x  1

8/5

2  x  1

 x  1

8/5

4/5

Solution

 x  5 2

1/5

 x2  5

4/5

 x2  5

4/5

 x 2  5  1   

x2  6

 x  5 2

4/5

Fix It In Exercises 143 and 144, identify the step the first error is made and fix it. 143. Factor completely: 2x2 − 2xy − 8x + 8y

Solution Step 2 was incorrect.

Step 1: 2 x 2  xy  4 x  4 y

Step 2: 2  x  x  y   4  x  y  

Step 3: 2 x  y

 x  4

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112


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

144. Factor completely: 3m3 – 81n6

Solution Step 3 was incorrect.

Step 1: 3 m3  27n6

Step 2: 3[(m)3  (3n2 )3 ]

Step 3: 3 m  3n2

 m  3mn  9n  2

2

4

Applications 145. Candy To find the amount of chocolate used in the outer coating of one of the malted-milk balls shown, we can find the volume V of the chocolate shell using the formula v  43  r1  43  r2 . Factor the expression on the right side of the formula. 3

3

Solution

4 3 4 3 r  r 3 1 3 2 4   r13  r23 3 4    r1  r2  r12  r1r2  r22 3

v 

 

146. Movie stunts The formula that gives the distances a stuntwoman is above the ground t seconds after she falls over the side of a 144-foot tall building is s  144  16t 2 . Factor the right side.

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113


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

f  144  16t 2

 16 9  t 2

 16  3  t  3  t  Discovery and Writing 147. Explain how to factor the difference of two squares.

Solution Answers may vary. 148. Explain how to factor the difference of two cubes.

Solution Answers may vary. 149. Explain how to factor by grouping.

Solution Answers may vary. 150. Explain what is meant by factor completely.

Solution Answers may vary.

Critical Thinking In Exercises 151–156, determine if the statement is true or false. If the statement is false, then correct it and make it true. 22 44 44 22 22 22 151. The GCF of 22x y  44x y is 22x y .

Solution True.

2

152. 25 x 200 z 200  36 factors completely as 5 x 100 z 100  6 .

Solution False. 25 x 200 z 200  36 is the sum of two squares and is thus prime. 3 3 3 153. p q r  64 factors completely as  pqr  4  . 3

Solution

False. p3q3 r 3  64  pqr

  4   pqr  4  p q r  4pqr  16 3

3

2

2 2

2 2 3 3 154. 9x  15xy  25 y is a factor of 27 x  125 y .

Solution

   5 y    3x  5 y  9x  15xy  25 y 

True. 27 x 3  125 y 3  3 x

3

3

2

2

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114


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



155.  2 x  5 y    7 z  9w  factors completely as 2 x  5 y  7 z  9w 2 x  5 y  7 z  9w . 2

2

Solution True. 156. The polynomial x 2  kx  12 can be factored for integer values k  7, 8, and 13 only.

Solution False. It can be factored for these values of k: 7, 8, 12,  7,  8,  12.

EXERCISES R.6 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

a. Add:

2 5  3 9

b. Subtract:

3 3  5 4

 7  8  c. Multiply:      4   21 

d. Divide:

9  27     2  4 

Solution 2 5 6 5 11 a.     3 9 9 9 9 b.

3 3 12 15 3     5 4 20 20 20

c.

 7  8  7 8 7  8 2 2         21  3 3 4 21 4 21 4   

d.

9  27  9  4  9  4  9  4  2  2        2  4  2  27  2  27  3 2  27  3 

2. Simplify each expression. a.

 x  8 x  5  x  3 x  8 a  a  11  11  a  11  a  4

b.

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115


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution a.

b.

 x  8 x  5   x  8   x  5  x  5  x  3 x  8  x  3  x  8  x  3 a4  a  11

 11  a  11  a 

a4  a  11

  a  11 11  a 



a4  a  11

a  11 11  a 



a4 a  11

 a  11   11  a



a4 a  11

3. Determine whether each expression is equivalent. 7 7 a. and  x 6 x 6 b.

  x  9 x  9 and 3x  8 3x  8

c.

y  5z 5z  y and  7z  y y  7z

Solution a. yes, equivalent since

a a  b b

b. yes, equivalent since  x  9   x  9

c. no, not equivalent 4. Factor each polynomial a. 8x4 – 16x3 b. 4z2 – 121 c. 5y2 + 23y – 10 d. 8b3 – 125

Solution a.

8 x 4  16 x 3  8x 3  x  2

b.

4 z 2  121   2z   112   2z  11 2z  11

c.

5 y 2  23 y  10   5 y  2 y  5

d.

8b3  125   2b   53   2b  5 4b2  10b  25

2

3

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116


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

5. Identify the number that makes the denominator of

5 zero? x 2

Solution –2 would make the denominator 0 6. The denominators of several fractions are given. Identify the LCD. a. 8, 16, 24 b. 5a, 6a2 c. x, 2x – 3, 2x – 7 d. 2x – 3, (2x – 3)3

Solution a. The LCD of 8, 16, and 24 is 48 b. The LCD of 5a and 6a 2 is 30a 2



c. The LCD of x , 2x  3 , and 2 x  7 is x 2 x  3 2 x  7 d. The LCD of 2x  3 and  2 x  3  is  2 x  3  3

3

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. In the fraction

a , a is called the _________. b

Solution numerator 8. In the fraction

a , b is called the ________. b

Solution denominator 9.

a c  if and only if _________. b d Solution ad  bc

10. The denominator of a fraction can never be _________.

Solution Zero

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117


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Complete each formula. 11.

a c   _________ b d

Solution

ac bd 12.

a c   _________ b d

Solution

ad bc 13.

a c   ________ b b

Solution

ac b 14.

a c   ________ b b

Solution

ac b Determine whether the fractions are equal. Assume that no denominators are 0. 15.

8 x 16 x , 3y 6y

Solution 8 x ? 16 x  3y 6y ? 8 x  6 y  3 y  16 x 48 xy  48 xy EQUAL

16.

3x 2 12 y 2 , 4 y 2 16 x 2

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118


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 3 x 2 ? 12 y 2  4 y 2 16 x 2 ? 3 x 2  16 x 2  4 y 2  12 y 2 48 x 4  48 y 4 NOT EQUAL

17.

25 xyz 50a2 bc , 12ab2c 24 xyz

Solution 25 xyz ? 50a2 bc  12ab2c 24 xyz ? 25 xyz  24 xyz  12ab2c  50a2 bc 600 x 2 y 2 z 2  600a3 b3c 2 NOT EQUAL

18.

15rs 2 37.5a 3 , 4rs 2 10a 3

Solution 15rs 2 ? 37.5a 3  4rs 2 10a 3 ? 15rs 2  10a 3  4rs 2  37.5a 3 150rs 2a 3  150rs 2a 3 EQUAL

In Exercises 19–26, identify the restricted numbers of the rational expression. 19.

11x x 5

Solution x  5 since 5  5  0 20.

13 x  18

Solution x  18 since  18  18  0 21.

4 x 2  169

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119


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

x 2  169   x  13 x  13 , x   13, 13

22.

since  13  13  0 and 13  13  0

8x x  144 2

Solution

x 2  144   x  12 x  12 , x   12, 12 since  12  12  0 and 12  12  0

23.

x1 x  4 x  21 2

Solution

x 2  4 x  21   x  7  x  3 , x   7, 3

24.

since  7  7  0 and 3  3  0

4x x 2  2 x  63

Solution

x 2  2 x  63   x  9 x  7  , x   9, 7 since  9  9  0 and 7  7  0

25.

5 2x  9x  5 2

Solution

2 x 2  9 x  5   x  5  2x  1 , x  5,

26.

1  1 since  5  5  0 and 2    1  0 2 2

x2 2x2  6x

Solution

2x 2  6 x  2 x  x  3 , x  0,  3 , since 2  0   0 and 3  3  0

Practice Simplify each rational expression. Assume that no denominators are 0. 27.

7a2 b 21ab2

Solution 7a2 b a  7ab a 7ab a     2 3b  7ab 3b 7ab 3b 21ab 28.

35p3q2 49p4q

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120


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 35p3q2 5q  7 p3q 5q 7 p3q 5q     49p4q 7 p  7 p3q 7 p 7 p3q 7 p

Perform the operations and simplify, whenever possible. Assume that no denominators are 0. 29.

4x 2  7 5a

Solution 4x 2 4x  2 8x    7 5a 7  5a 35a 30. 

5y 4  2z y 2

Solution

5 y 4 5 y  4 20 y 10  2    2 2 yz 2z y 2z  y 2y z 31.

8m 3m  5n 10n

Solution 8m 3m 8m 10n 80mn 16      5n 10n 5n 3m 15mn 3 32.

15 p 5 p  8q 16q2

Solution 15p 5 p 15p 16q2 240 pq2      6q 8q 16q2 8q 5 p 40 pq 33.

3z 2z  5c 5c

Solution 3z 2z 3z  2z 5z z     5c 5c 5c 5c c 34.

7a 3a  4b 4 b

Solution 7a 3a 7a  3a 4a a     4b 4 b 4b 4b b

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121


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

35.

15 x 2 y x2 y  7a2 b3 7a2 b3

Solution 15 x 2 y x2 y 14 x 2 y 2 x 2 y    2 3 2 3 2 3 7a b 7a b 7a2 b3 a b 36.

8rst 2 7 rst 2  15m4 t 2 15m4 t 2

Solution 8rst 2 7 rst 2 15rst 2 rs    15m4 t 2 15m4 t 2 15m4t 2 m4 Simplify each fraction. Assume that no denominators are 0. 2x  4 37. 2 x 4 Solution 2x  4 x 4 2

38.

2  x  2

 x  2 x  2

2 x2

x 2  16 x 2  8 x  16

Solution

 x  4  x  4   x  4 x 2  16  2 x  8 x  16  x  4  x  4  x  4

39.

4  x2 x 2  5x  6

Solution 4  x2 x  5x  6 2

40.

 2  x  2  x    x  2  x  3 x  2 x  3

25  x 2 x 2  10 x  25

Solution 25  x 2 x  10 x  25 2

41.

5  x 5  x    x  5  x  5 x  5 x  5

6 x 3  x 2  12 x 4 x 3  4x 2  3x

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122


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

 

 

x 6 x 2  x  12 x  2 x  3  3 x  4  3 x  4 6 x 3  x 2  12 x    3 2 2 2x  1 4 x  4 x  3x x 4 x  4 x  3 x  2 x  3  2 x  1

42.

6x 4  5x 3  6x 2 2 x 3  7 x 2  15 x

Solution

 

2 2 x 2  2 x  3 3 x  2 x  2 x  3 3 x  2 6x 4  5x 3  6x 2 x 6x  5x  6    2 x 3  7 x 2  15 x x 2 x 2  7 x  15 x  2 x  3 x  5  2x  3 x  5

43.

x3  8 x 2  ax  2 x  2a

Solution

 x  2  x  2x  4  x  2x  4    x a x  ax  2 x  2a x  x  a   2  x  a   x  a  x  2 x3  8

2

x 3  23

2

2

44.

xy  2 x  3 y  6 x 3  27

Solution xy  2 x  3 y  6 x  27 3

x  y  2  3  y  2 x 3 3

3

 y  2 x  3  y  2  x  3  x  3x  9 x  3 x  9 2

2

Perform the operations and simplify, whenever possible. Assume that no denominators are 0. 45.

x2  1 x2  2 x x  2x  1

Solution

x  x  1  x  1 x  1  x2  1 x2 x2  2   x x x  2x  1  x  1 x  1 x  1

46.

y2  2y  1 y 2  2 y y  y 2

Solution

 y  1 y  1  y  2  y  1 y2  2y  1 y 2  2  y y y  y 2  y  2 y  1 y

47.

3x 2  7 x  2 x 2  x  2 x 2  2x 3x  x

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123


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 3x 2  7 x  2 x  2x 2

48.

x2  x

3x  x 2

 3x  1 x  2  x  x  1  x  1 x x  x  2 x  3 x  1

x 2  x 2x 2  x  3  2x 2  3x x2  1

Solution

x  x  1  2 x  3  x  1 x 2  x 2x 2  x  3    1 2 2 2x  3x x 1 x  2 x  3   x  1 x  1

49.

x2  x x2  1  x1 x2 Solution

x 2  x x 2  1 x  x  1  x  1 x  1 x  x  1     x1 x2 x1 x2 x2 50.

x 2  5x  6 x  2  x2  6x  9 x2  4

Solution x 2  5x  6

x2

x 2  6x  9 x 2  4

51.

2

 x  2 x  3  x  2  x2  x  3 x  3  x  2 x  2  x  3 x  2

2 x 2  32 x 2  16  8 2

Solution

2 x 2  16 2 x 2  32 x 2  16 2 x 2  32 2 2 1    2   2  8 2 8 8 x  16 x  16 2

52.

x2  x  6 x2  4  2 2 x  6x  9 x  9

Solution

x2  x  6 x2  4 x 2  x  6 x 2  9  x  3 x  2   x  3  x  3       x 2  6 x  9 x 2  9 x 2  6 x  9 x 2  4  x  3 x  3  x  2  x  2 

 x  3   x  3 x  2 2

53.

z 2  z  20 z 2  25  z 5 z2  4

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124


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution z 2  z  20 z2  4

 z  5 z  4   z  5 z 2  25 z 2  z  20 z  5   2  2 z 5 z 4 z  25  z  2  z  2   z  5  z  5  

54.

z4

 z  2 z  2

ax  bx  a  b x2  1  2 2 2 a  2ab  b x  2x  1

Solution

ax  bx  a  b x2  1 ax  bx  a  b x 2  2 x  1    a2  2ab  b2 x 2  2x  1 a2  2ab  b2 x2  1 x  a  b   1  a  b   x  1 x  1   a  ba  b  x  1 x  1 

55.

a  b x  1   x  1 x  1  x  1 a  ba  b  x  1 x  1 a  b

3 x 2  5 x  2 6 x 2  13 x  5  x 3  2x 2 2x 3  5x 2

Solution 3 x 2  5 x  2 6 x 2  13 x  5 3 x 2  5 x  2 2x 3  5x 2    2x 3  5x 2 6 x 2  13 x  5 x 3  2x 2 x 3  2x 2  3x  1 x  2  x 2  2x  5  1  x 2  x  2  3 x  1 2 x  5

56.

x 2  13 x  12 2x 2  x  3  8 x 2  6 x  5 8 x 2  14 x  5

Solution x 2  13 x  12 2x 2  x  3 x 2  13 x  12 8 x 2  14 x  5    2 2 8 x  6 x  5 8 x  14 x  5 8 x 2  6 x  5 2 x 2  x  3  x  12 x  1   4 x  5 2 x  1   x  12 2 x  1   4 x  5 2x  1  2 x  3 x  1  2 x  1 2x  3

57.

x 2  7 x  12 x 2  3 x  10 x 3  4 x 2  3 x   2 x 3  x 2  6x x 2  2x  3 x  x  20

Solution x 2  7 x  12 x 2  3 x  10 x 3  4 x 2  3 x   2 x 3  x 2  6x x 2  2x  3 x  x  20  x  3 x  4    x  5 x  2  x  x  3 x  1  1  x  x  3  x  2   x  3  x  1  x  5  x  4 

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125


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

58.

x  x  2  3

x  x  1  2

x  x  7   3  x  1 x  x  7   3  x  1

Solution

x  x  2  3

x  x  1  2

x  x  7   3  x  1 x  x  7   3  x  1

 

x 2  2x  3 x2  x  2  x 2  7 x  3x  3 x 2  7 x  3x  3 x 2  2x  3

x2  x  2

x2  4x  3 x2  4x  3  x  3 x  1   x  2 x  1  x  2   x  3 x  1  x  3 x  1 x  3 59.

x2  2x  3 3x  8 x2  6x  5   2 21x  50 x  16 x  3 7 x 2  33 x  10

Solution x 2  2x  3 x 2  6x  5 x 2  2x  3 3x  8 3 x  8 7 x 2  33 x  10      x 2  6x  5 21x 2  50 x  16 x  3 7 x 2  33 x  10 21x 2  50 x  16 x  3  x  3 x  1  3x  8   7 x  2 x  5   7 x  2 3x  8 x  3  x  5 x  1 

60.

x 5 x 5

 x2  4x  3 x2  x  6      x 2  4  x 2  2 x x 2  3 x  9 

x 3  27

Solution x 3  27  x 2  4 x  3 x 2  x  6  x 3  27  x 2  4 x  3 x 2  3 x  9    2   2  2  x 2  4  x 2  2 x x  3 x  9  x  4  x 2  2 x x  x  6  

x 3  27   x  3  x  1 x 2  3 x  9     x 2  4  x  x  2   x  3 x  2 

 x  1  x  3x  9   x 4 x  x  2  x  2   x  3  x  3x  9 x  x  2 x  2    x  2 x  2  x  1  x  3x  9 x  x  3  2

x 3  27 2

2

2

x1

61.

3 x2  x3 x3

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126


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 3 x 2 3 x 2 x 5    x3 x3 x3 x3 62.

3 x2  x1 x1 Solution

3 x 2 3 x 2 x 5    x1 x1 x1 x1 63.

4x 4  x1 x1 Solution

4x 4 4 x  4 4  x  1    4 x1 x1 x1 x1

64.

6x 3  x 2 x 2 Solution

6x 3 6 x  3 3  2 x  1    x 2 x 2 x 2 x 2

65.

2 1  5 x x 5

Solution 2 1 2 1 1     5 x x 5 x 5 x 5 x 5 66.

3 2  x 6 6 x

Solution 3 2 3 2 5     x 6 6 x x 6 x 6 x 6 67.

3 2  x1 x1 Solution

3  x  1 2  x  1 3 2 3x  3 2x  2      x  1 x  1  x  1 x  1  x  1 x  1  x  1 x  1  x  1 x  1 

5x  1

 x  1 x  1

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127


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

68.

3 x  x 4 x 4 Solution

3  x  4 x  x  4 3 x 3 x  12 x2  4x      x  4 x  4  x  4  x  4   x  4  x  4   x  4  x  4   x  4  x  4  

69.

x 2  7 x  12

 x  4  x  4 

a3 a  2 a  7a  12 a  16 2

Solution a3 a a3 a    a2  7a  12 a2  16  a  3  a  4   a  4  a  4   

a 1  a  4  a  4  a  4  1 a  4 

a

a  4 a  4  a  4 a  4  a4

a

a  4 a  4  a  4 a  4  2 a  2 2a  4   a  4 a  4  a  4 a  4  70.

a 2  2 a  a  2 a  5a  4 2

Solution a 2 a 2    a2  a  2 a2  5a  4  a  2  a  1  a  4  a  1   

71.

a a  4 

2 a  2

a  2a  1a  4  a  4 a  1a  2 a2  4a

2a  4

a  2a  1a  4  a  2a  1a  4  a2  2a  4

a  2a  1a  4 

x 1  x  2 x 4 2

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128


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

1  x  2 1 1 x x x      x 2  4 x  2  x  2  x  2  x  2  x  2  x  2   x  2  x  2 

72.

x

x 2

 x  2 x  2  x  2 x  2

 x  2 x  2

2

b2 4  b  4 b2  2b 2

Solution

b2  b  4  b  2 b2 4 b2 4      b2  4 b2  2b  b  2  b  2  b  b  2  b  b  2  b  2  b  b  2  b  2  b3

b  b  2  b  2 

4b  8

b  b  2  b  2 

b  4b  8 3

73.

b  b  2  b  2 

3x  2 x  2 x  2x  1 x  1 2

Solution x x 3x  2 3x  2  2   2 x  2 x  1 x  1  x  1 x  1  x  1 x  1   

74.

x  x  1  3x  2 x  1   x  1 x  1 x  1  x  1 x  1 x  1 3x 2  5x  2

x2  x

 x  1 x  1 x  1  x  1 x  1 x  1 2x 2  6x  2

2 x 2  3x  1

 x  1 x  1 x  1  x  1  x  1 2

2t t1  t 2  25 t 2  5t

Solution

2t  t  t  1t  5 2t t1 2t t1      t 2  25 t 2  5t  t  5  t  5  t  t  5  t  t  5  t  5  t  t  5  t  5   

2t 2

t  t  5  t  5 

t 2  4t  5

t  t  5  t  5 

t  4t  5 2

t  t  5  t  5 

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129


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

75.

2 y 1 2

3

1 y1

Solution 2 1 2 3 1 3    2 y  1 1 y 1 y 1  y  1 y  1 

2

 y  1 y  1 2

3  y  1 y  1 1  y  1 y  1 3y  3 2

1  y  1

 y  1 y  1

y 1

 y  1 y  1  y  1 y  1  y  1 y  1  3 y  2 y  1  3 y  2 3y  y  2    y  1 y  1  y  1 y  1 y  1 2

76. 2 

4 1  t 4 t 2 2

Solution 4 1 2 4 1 2 2     t  4 t  2 1  t  2 t  1 t  2 

2  t  2  t  2  1  t  2 t  2

2t  8 2

4

1 t  2

t  2t  2 t  2t  2 4

t 2

t  2t  2 t  2t  2 t  2t  2  2t  3t  2  2t  3 2t  t  6   t  2t  2 t  2t  2 t  2 2

77.

1 3 3x  2   x  2 x  2 x2  4

Solution

1 3 3x  2 1 3 3x  2   2    x  2 x  2 x  4 x  2 x  2  x  2 x  2   

1  x  2

3  x  2

3x  2

 x  2 x  2  x  2 x  2  x  2 x  2 x2

3x  6

3x  2

 x  2 x  2  x  2 x  2  x  2 x  2 x 2

 x  2 x  2

1 x2

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130


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

78.

3  3 x  1 x 5   x 3 x 3 x2  9

Solution

3  3 x  1 x x 5 5 9x  3      2 x 3 x 3 x  3 x  3  x  3  x  3  x 9 

x  x  3

5  x  3

9x  3

 x  3 x  3  x  3 x  3  x  3 x  3 x 2  3x

5 x  15

9x  3

 x  3 x  3  x  3 x  3  x  3 x  3  x  3 x  4   x  4 x  7 x  12    x  3 x  3  x  3 x  3 x  3 2

 1 1  x 3 79.     x  2 x  3  2x

Solution

1  x  3 1  x  2  x  3  1 1  x 3          x  2  x  3   x  3  x  2   2 x  x  2 x  3  2x     x3 x 3 x 2      x  2  x  3   x  2  x  3   2 x   x 3 2x  5 2x  5     x  2 x  3 2x 2 x  x  2

 1 1  1 80.       x 1 x 2 x 2  

Solution

 1  x  2 1  x  1  x  2  1 1  1       1  x  1 x  2  x  2   x  1 x  2   x  2  x  1    x 2 x 2 x1      x  1 x  2   x  1 x  2   1   3 x 2 3     x  1 x  2 1 x  1

81.

3x x 3x  1   x  4 x  4 16  x 2

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131


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 3x 3x  1 3x 3x  1 x x      2 x  4 x  4 16  x x  4 x  4  4  x  4  x     

82.

3x 3x  1 x   x  4 x  4  x  4  x  4  3x  x  4

x  x  4

3x  1

 x  4  x  4   x  4 x  4   x  4 x  4 3 x 2  12 x

x2  4x

3x  1

 x  4  x  4   x  4 x  4   x  4 x  4 2 x 2  19 x  1

 x  4  x  4 

7x 3x 3x  1   2 x  5 5  x x  25

Solution 7x 3x 3x  1 7x 3 x 3x  1      x  5 5  x x 2  25 x  5 x  5  x  5  x  5

  

83.

4x 3x  1  x  5  x  5  x  5  4 x  x  5

3x  1

 x  5 x  5  x  5 x  5 4 x 2  20 x

3x  1

4 x 2  23 x  1

 x  5 x  5  x  5 x  5  x  5 x  5

1 2 1  2  2 x  3x  2 x  4x  3 x  5x  6 2

Solution 1 2 1   x 2  3x  2 x 2  4 x  3 x 2  5x  6 1 2 1     x  2 x  1  x  3 x  1  x  2 x  3   

1  x  3

2  x  2

1  x  1

 x  2 x  1 x  3  x  3 x  1 x  2  x  2 x  3 x  1 x3

2x  4

x1

 x  2 x  1 x  3  x  2 x  1 x  3  x  2 x  1 x  3 x  3  2x  4  x  1

 x  2 x  1 x  3

0

 x  2 x  1 x  3

0

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132


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

84.

2 2 2z  2 y   x  y x  z  y  x  z  x  Solution

2 2 2z  2 y 2 2 2z  2 y      x  y x  z  y  x  z  x  y  x z  x  y  x  z  x    

85.

2z  x

2  y  x 

2z  2 y

 y  x  z  x   z  x  y  x   y  x  z  x  2z  2 x

2 y  2 x

2z  2 y

 y  x  z  x   y  x  z  z   y  x  z  x  2z  2 x  2 y  2 x  2z  2 y

 y  x  z  x 

0

 y  x  z  x 

0

3x  2 4 x 2  2 3 x 2  25   2 x 2  x  20 x 2  25 x  16

Solution 3x  2 x 2  x  20

4x2  2

3 x 2  25

x 2  25 x 2  16 3x  2 4x2  2 3 x 2  25     x  5 x  4   x  5 x  5  x  4  x  4 

 3x  2 x  5 x  4    4 x  2  x  4  x  4     x  5 x  4  x  5 x  4   x  5 x  5 x  4  x  4  3x  25  x  5 x  5   x  4  x  4  x  5 x  5 2

2

3 x 3  5 x 2  58 x  40

4 x 4  62 x 2  32

 x  5 x  4  x  5 x  4   x  5 x  4  x  5 x  4  

3 x  5 x  58 x  40 3

2

 x  5 x  4  x  5 x  4 

2

 x  5 x  4  x  5 x  4   x  5 x  4  x  5 x  4  

3 x 4  100 x 2  625

4 x  62 x  32 4

3 x 4  100 x 2  625

 x  5 x  4  x  5 x  4 

3 x  5 x  58 x  40  4 x  62 x  32  3 x  100 x 2  625 3

2

4

2

4

 x  5 x  4  x  5 x  4 

 x  3 x  43 x 2  58 x  697 4

3

 x  5 x  4  x  5 x  4 

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133


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

86.

3x  2 x4 1   2 8 x  10 x  3 6 x  11x  3 4 x  1 2

Solution 3x  2 x4 1   8 x 2  10 x  3 6 x 2  11x  3 4 x  1 3x  2 x4 1     4 x  1 2 x  3  3 x  1 2x  3 4 x  1   

 3 x  2 3x  1   x  4  4 x  1  1  2x  3 3 x  1  4 x  1 2 x  3 3 x  1  3x  1 2x  3 4 x  1  4 x  1 2 x  3 3 x  1 9x 2  3x  2

4 x 2  17 x  4

6 x 2  11x  3

 4 x  1 2 x  3 3 x  1  4 x  1 2 x  3 3x  1  4 x  1 2 x  3 3 x  1 9 x 2  3 x  2  4 x 2  17 x  4  6 x 2  11x  3

 4 x  1 2 x  3 3 x  1

7 x 2  31x  1

 4 x  1 2x  3 3 x  1

Simplify each complex fraction. Assume that no denominators are 0.

3a 87. b 6ac b2 Solution 3a b 6ac b2

3a 6ac 3a b2 b  2    b b 6 ac 2 c b

3t 2 88. 9 x t 18 x Solution 3t 2 9x t 18 x

89.

3t 2 t 3t 2 18 x     6t 9 x 18 x 9 x t

3a2 b ab 27 Solution 3a2 b 3a2 b ab 3a2 b 27      81a ab 1 27 1 ab 27

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134


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3u2v 90. 4t 3uv

Solution 3 u2 v 4t

3uv

3u2v 3uv 3u2v 1 u     4t 1 4t 3uv 4t

xy 91. ab yx ab

Solution x y ab y x ab

x  y y  x x  y ab     1 ab ab ab y  x

x 2  5x  6 2x 2 y 92. x2  9 2x 2 y

Solution x 2 5  6 2 x2 y

x2 9 2 x2 y

x 2  5x  6 x 2  9 x 2  5x  6 2x 2 y    2 2x 2 y 2x 2 y 2x 2 y x 9 

 x  3 x  2  2

2x y

2x 2 y

 x  3 x  3

x 2 x3

1 1  x y 93. xy

Solution 1 x

 y1 xy

94.

xy

    xy    xy    y  x 1 x

1 y

1 x

xy  xy 

1 y

x2 y 2

x2 y 2

xy 11 11  x y

Solution

xy  xy  xy x2 y 2   11  11y xy 11x  11y xy  11x   xy x

  11 y

x2 y 2 x2 y 2  11 y  11x 11  y  x 

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135


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

1 1  x y 95. 1 1  x y Solution 1 x

 y1

1 x

 y

 1

    xy    xy   xy    xy    xy  

xy

1 x

1 y

1 x

1 y

1 x

1 y

1 x

1 y

yx yx

1 1  x y 96. 1 1  x y Solution 1 x

 y1

1 x

 y1

    xy    xy    xy    xy    xy   xy

1 x

1 y

1 x

1 y

1 x

1 y

1 x

1 y

yx yx

3a 4a2  x 97. b 1 1  b ax Solution 3a b 1 b

 4 ax  ax1

1 98.

2

abx

    abx    abx    3a x  4a b  a 3 x  4ab  3a b

4 a2 x

4 a2 x

3a b

abx  b1  ax1 

2

abx  b1   abx  ax1 

3

2

ax  b

ax  b

x y

x2 1 y2 Solution

1  xy 2

x y2

1

y 2 1  xy y

2

  y  1  y    y  xy  2

2

x y

  1 y    y  1 2

x y2

2

2

x y2

2

2

x y 2

2

y  y  x

 x  y  x  y 

y xy

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136


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

6 x 99. 6 x 5 x x  1

Solution x  1  6x

 6

x 5 x

100.

x  x  1  6x 

x  x  5  6x 

x  x   x  1  x  6x 

x  x   x  5   x  6x 

x2  x  6 x  5x  6 2

 x  3 x  2  x  2  x  2 x  3 x  2

2z 3 1 z Solution

z  2z  2z 2z 2   1  3z z  1  3z  z  1  z  3z  

101.

2z 2 z3

3 xy 1 1 xy

Solution

xy  3 xy  3 xy 3x 2 y 2   1  xy1 xy 1  1 xy  1  xy

xy

102.

x 3 

1 xy

3x 2 y 2 xy  1

1 x

1 x3 x

Solution x  3  x1  x1  x  3

103.

 

x  x  3  x1 

x   x1  x  3 

x 2  3x  1 1  x 2  3 x

x 2  3x  1

 x 2  3x  1

 1

3x x

1 x

Solution

x 3x  3x 3x 2   x  x1 x  x  x1  x 2  1

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137


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

104.

2x 2  4 4x 2 5 Solution

2 2 2x 2  4 5 2x  4 10 x 2  20 2 5 x  10 5 x 2  10     2  45x 10  4 x 5  2x 5  2  45x  2 5  2 x 

x 2  x  2 x 1 105. 3 x  x2 x1

Solution x x 2 3 x 2

 x2 1  xx 1

 x  2 x  1  x  2 x  1

x x 2

 x2 1 

 x  2 x  1    x  2 x  1    x  2 x  1    x  2 x  1   x  1 x    x  2 2   x  1 3   x  2 x 

3 x 2

 xx 1

x x 2

2 x 1

3 x 2

x x 1

x 2  x  2x  4 3x  3  x 2  2x

x 2  3x  4 x 2  5x  3

2x 1  106. x  3 x  2 3 x  x 3 x 2

Solution 2x x 3 3 x 3

 x 1 2  x x 2

 x  3 x  2  x  3 x  2

2x x 3

 x 1 2 

 x  3 x  2    x  3 x  2    x  3 x  2    x  3 x  2   x  2 2x    x  3 1   x  2 3   x  3 x 

3 x 3

 x x2

2x x 3

1 x 2

3 x 3

x x 2

2x 2  4 x  x  3 3x  6  x 2  3x 2x 2  3x  3 2x 2  3x  3     x 2  6x  6 x 2  6x  6 

Write each expression without using negative exponents, and simplify the resulting complex fraction. Assume that no denominators are 0. 1 107. 1  x 1 Solution 1 1 x

 1

x  1 1 x   1 1 1 x x 1  x  x  1

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138


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

108.

y 1 x 1  y 1

Solution

 

1 xy y1 y 1 x y    1 1 1 1  x y x 1  y 1 xy x  y x y

3  x  2   2  x  1 1

109.

 x  2

1

1

Solution 3  x  2   2  x  1 1

 x  2

3  2  x  2 x  1 x 32  x2 1   x 2 1 x 1   x  2 x  1 x 1 2  x 2

1

1

 

 x  2 x  1    x  2 x  1  3 x 2

2 x 1

x1 x  1 3  x  2      2

x1 3x  3  2x  4 5x  1   x1 x1

2 x  x  3  3  x  2 1

110.

 x  3  x  2 1

1

1

Solution 2 x  x  3  3  x  2 1

 x  3  x  2 1

1

1

2x  3  x  3 x  2 x2x3  x32   x 3 1 x 2   x  3 x  2  x  3 x  2  x 31 x 2

 x  2 2x    x  3 3

1  2x 2  4 x  3x  9  2x 2  x  9

x

111.

1

1 3 x 1

Solution x x   1  3 x11 1  31 x

x 1

x  1

x  3x 

x 3x 3x   1  3x 3  1  3x  3  x

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139


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

112.

ab 3 2  1 2a Solution ab ab   2  2a31 2  32 a

ab 2

a  3

a  a2 

ab 2ab 2ab   2  32a 2  2  32a  4  3a

1

113.

1

1

1

1 x

Solution

 x  1 1  x  1  x  1 1 1 1    1 x  1 1  1 1 x  x  1 1  xx 1  x  1  x 2x  1 1  x 1 1 x  x 1  x1 y

114.

2

2 2

2 y

Solution y y   2 2  2  2 2  y  2 y

 

y 2  2y

 2 y  2 y   2 y 2  2 y y  2y  2 y  2 2  22y y2  2 y  2 2  2 y 2 2y  2 2 y  y  1  4y  4  2y 2 y  y  1 y  y  1   3y  2 2  3 y  2

Fix It In Exercises 115 and 116, identify the step the first error is made and fix it. x3 + 8 x2 -2x + 4 115. Divide: ÷ 6x + 6 4x2 -4 Solution Step 4 was incorrect. Step 1:

x3 + 8 4x 2 - 4 × 2 6x + 6 x - 2x + 4

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140


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 x  2  x  2x  4  4  x  1 Step 2:  x  2x  4 6  x  1 2

2

2

Step 3:

Step 4:

2  x  2  x  1 x  1 3  x  1

2  x  2  x  1 3

116. Subtract :

4x  2 3x  5  x7 x7

Solution Step 5 was incorrect. Step 1:

 4 x  2 x  7    3 x  5 x  7   x  7  x  7   x  7  x  7 

Step 2:

 4 x  2 x  7    3x  5 x  7   x  7  x  7 

Step 3:

Step 4:

Step 5:

Step 6:

4 x 2  28 x  2 x  14  3 x 2  21x  5 x  35

 x  7  x  7 

4 x 2  26 x  14  3 x 2  26 x  35

 x  7  x  7 

4 x 2  26 x  14  3 x 2  26 x  35

 x  7  x  7 

x 2  52 x  49

 x  7  x  7 

Applications 117. Engineering The stiffness k of the shaft shown in the illustration is given by the following formula where k1 and k2 are the individual stiffnesses of each section. Simplify the complex fraction on the right side of the formula. 1 k 1 1  k1 k2

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141


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution 1 k1

k1k2  1 1  1 k kk 1  1 1 2

2

k1

k2

k1k2

k1k2

 kk  kk   k k 1 2

1 k1

1 k2

1 2

2

1

118. Electronics The combined resistance R of three resistors with resistances of R1, R2, and R3 is given by the following formula. Simplify the complex fraction on the right side of the formula. 1 R 1 1 1   R1 R2 R3

Solution 1 1 R1

1 R2

1 R3

R1R2 R3  1 R1R2R3

R1R2R3

    RR R  RR R  RR R   1 R1

1 R2

1 R3

1

2

3

1 R1

1

2

3

1 R2

1

2

3

1 R3

R1R2 R3 R2 R3  R1R3  R1R2

Discovery and Writing 119. Explain what a rational expression is.

Solution Answers may vary. 120. Describe how to simplify a rational expression.

Solution Answers may vary. 121. Explain why the denominator of a rational expression cannot be 0.

Solution Answers may vary. 122. Describe how to add or subtract rational expressions.

Solution Answers may vary. 123. Explain how to multiply rational expressions.

Solution Answers may vary. 124. Explain how to divide rational expressions.

Solution Answers may vary.

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142


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

125. Explain why the formula

a c ad  bc is valid.   b d bd

Solution a c a d c b ad bc ad  bc         . b d b d d b bd bd bd 126. Explain why the formula

a c a d is valid.    b d b c

Solution Let x  ab  dc . Then x  dc  ab , and x  dc  dc  ab  dc  Thus, ab  dc  x  ab  dc . Critical Thinking In Exercises 127–130, determine if the statement is true or false. If the statement is false, then correct it and make it true. 127. The numerator of a rational expression can never be 0.

Solution False. The denominator can never equal 0. 128. The denominator of a rational expression can never be 0.

Solution True. 129.

x7  1 for all values of x. x7 Solution False. xx 77  1 for all values of x except x  7 .

130.

x 7  1 for all values of x. 7x Solution False. 7xx7  1 for all values of x except x = 7.

In Exercises 131–134, determine if the statement is true or false. If the statement is false, then correct it and make it true. Assume no denominators are 0.

x  y  1 131.   y  x 3 3

Solution

 x  y     x  y     x  y    1  1. True.  1 [  x  y ]  y  x  1  x  y  3 3

3

3

3

3

3

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143


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

132.

25  x  1 x 25

Solution 25  x 25 x x False.    1 . 25 25 25 25 133.

5 5 10   x y xy

Solution

False.

134. 10 

5 5 5 y 5 x 5 y  5x       . x y x y y x xy

1 9  x x

Solution

False. 10 

1 10 x 1 10 x  1 .     x 1 x x x

135. Domain The set of all real numbers for which an algebraic expression is defined is 3x  8 called the domain. What is the domain of the rational expression 3 ? x 1

Solution The domain is the set of all real numbers except x = 6. 136. Domain What is the domain of the rational expression

3 ? Refer to Exercise 135. x 3 2

Solution The domain is the set of all real numbers.

CHAPTER REVIEW SOLUTIONS 1.

8 , 6 , 5 , , 3 , 12

Consider the set {6, 

, 0 , 3

Exercises

︸. List the numbers in this set that are

natural numbers.

Solution natural: 3, 6, 8 2. whole numbers.

Solution whole: 0, 3, 6, 8

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144


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3. integers.

Solution integers: –6, –3, 0, 3, 6, 8 4. rational numbers.

Solution rational: 6,  3, 0, 21 , 3, 6, 8 5. irrational numbers.

Solution irrational:  , 5 6. real numbers.

Solution real: 6,  3, 0, 21 , 3,  , 5, 6, 8 Consider the set {6,  3, 0, 21 , 3, π , 5, 6, 8}. List the numbers in this set that are 7. prime numbers.

Solution prime: 3 8. composite numbers.

Solution composite: 6, 8 9. even integers.

Solution even integers: –6, 0, 6, 8 10. odd integers.

Solution odd integers: –3, 3 Determine which property of real numbers justifies each statement. 11.

a  b   2  a   b  2 Solution Associative Property of Addition

12. a  7  7  a

Solution Commutative Property of Addition

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145


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

  

13. 4 2 x  4  2 x

Solution Associative Property of Multiplication

14. 3 a  b  3a  3b

Solution Distributive Property 15.

5a  7  7 5a  Solution Commutative Property of Multiplication

16.

 2x  y   z   y  2x   z Solution Commutative Property of Addition

 

17.  6  6

Solution Double Negative Rule Graph each subset of the real numbers: 18. the prime numbers between 10 and 20

Solution

19. the even integers from 6 to 14

Solution

Graph each interval on the number line. 20. 3  x  5

Solution

3  x  5

21. x  0 or x  1

Solution x  0 or x  1

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146


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

22. ( 2, 4]

Solution (  2, 4]

 

23. , 2  5, 

Solution

 , 2   5,    , 2   5,  

__________________________________

 , 2    5,  

24. ,  4  [6, )

Solution

 ,  4  [6, )

Write each expression without absolute value symbols. 25. 6

Solution

Since 6  0, 6  6. 26. 25

Solution

Since  25  0, 25    25  25.

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147


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

27. 1  2

Solution

Since 1  2  0, 1  2   1  2  2  1. 31

28.

Solution

Since 3  1  0,

3  1  3  1.

29. On a number line, find the distance between points with coordinates of –5 and 7.

Solution

distance  7   5   12  12

Write each expression without using exponents. 30. 5a 3

Solution  5a 3  5  a  a  a

31.

 5a 

2

Solution

 5a    5a  5a  2

Write each expression using exponents. 32. 3  t  t  t

Solution 3ttt  3t 3

 

33. 2b 3b

Solution

 2b 3b   2 3 bb  6b

2

Simplify each expression. 34. n2 n4

Solution n 2 n 4  n 2  4  n6

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148


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 

35. p 3

2

Solution

p   p 3

2

36. x 3 y 2

32

 p6

4

Solution

x y   x   y   x y 3

4

2

3

4

2

4

12

8

3

 a4  37.  2  b 

Solution

   

3

3 a4  a4  a 12   2   3 b6 b  b2

38. m 3 n0

2

Solution

 m n    m  1  m  m1 3

0

2

3

2

6

6

 p2q2  39.    2 

3

Solution

   

3

3 3 q2  p 2 q 2   q2  q6      2   3 8p6  2   2p  2p2

40.

a5 a8

Solution a5 1  a 5  8  a 3  3 8 a a

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149


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 a2  41.  3  b 

2

Solution

 a2   3  b 

2

2

 b3  b6   2   4 a a 

 3x 2 y 2  42.  2 2   x y 

2

Solution

 3x 2 y 2   2 2   x y   a3b2  43.  3    ab 

2

2

2

2

 x2 y 2   x2 y 2 y 2   y4  y8   2 2         2   3  9  3x y   3x   

2

Solution

 a3b2   3    ab 

2

 3x 3 y  44.  3   xy 

2

2

2

 ab3   aa3   a4  a8   3 2    2 3    5   10 b a b  b b  b 

2

Solution

 3x 3 y    3  xy 

2

 2m2 n0  45.    2 1   4m n 

2

2

 xy 3   y2  y4        3   3x 2  9x 4  3 x y    3

Solution

 2m2 n0    2 1    4m n 

3

3

3

3

 4m2 n1   2m2 m2   2m4  8m12               2 0    n  n1n0  n3  2m n   

46. If x  3 and y  3, evaluate  x 2  xy 2 .

Solution  x 2  xy 2    3    3  3     9    3  9   9   27   9  27  18 2

2

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150


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Write each number in scientific notation. 47. 6750

Solution 6750  6.750  103

48. 0.00023

Solution 0.00023  2.3  10 4

Write each number in standard notation. 49. 4.8  102

Solution 4.8  10 2  480

50. 0.25  10 3

Solution 0.25  10 3  0.00025

51. Use scientific notation to simplify

 45,000 350,000 . 0.000105

Solution

 45, 000 350, 000   4.5  10  3.5  10   4.5  3.5  10  10  15.75  10 4

1.05  104

0.000105

5

4

1.05  104

5

9

1.05  104  15  1013  1.5  1014

Simplify each expression, if possible. 52. 1211/2

Solution

 

1211/2  112

 27  53.    125 

1/2

 11

1/3

Solution

 27     125 

1/3

 3 3       5    

1/3

3 5

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151


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

54. 32 x 5

1/5

Solution

 32 x  5

55. 81a4

 

1/5

 321/5 x 5

1/5

 2x

1/4

Solution

81a  4

 

1/4

 811/4 a 4

56. 1000 x 6

1/4

3a

1/3

Solution

 1000x  6

57. 25 x 2

1/3

  1000 

1/3

x  6

1/3

 10 x 2

1/2

Solution

 25x  2

1/2

  25

1/2

x  2

1/2

 not a real number

58. x 12 y 2

1/2

Solution

x y  12

 x 12  59.  4  y 

2

1/2

  y 

 x 12

1/2

2

1/2

 x6 y

1/2

Solution

 x 12   4  y 

1/2

 y4    12  x 

 c2/3c5/3  60.   2/3  c 

1/2

y2 x6

1/3

Solution

 c2/3c5/3    2/3  c 

1/3

 c7/3    2/3  c 

1/3

 c9/3

1/3

 

 c3

1/3

 c

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152


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 a1/4a3/4  61.   9/2  a 

1/2

Solution

 a1/4a3/4    9/2  a 

1/2

 a9/2    1/4 3/4  a a 

1/2

 a9/2    2/4  a 

1/2

 a9/2    1/2  a 

1/2

 a8/2

1/2

 

 a4

1/2

 a2

Simplify each expression. 62. 642/3

Solution

642/3  64 1/3

  4  16 2

2

63. 32 3/5

Solution 1

323/5 

 16  64.    81 

3/5

32

1

32  1/5

3

1

3

2

1 8

3/4

Solution

 16     81 

3/4

 32  65.    243 

 

 

3

161/4 163/4 23 8  3/4    3 3 27 81 3 811/4

2/5

Solution

 32     243 

 8 66.    27 

2/5

322/5 2/5

243

 32   2  4   243  3 9 1/5

1/5

2

2

2

2

2/3

Solution

 8     27 

2/3

 27     8 

2/3

   

2

27 1/3 272/3 32 9  2/3    2 8 22 4 81/3

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153


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 16  67.    625 

3/4

Solution

 16     625 

3/4

68. 216 x 3

 625     16 

3/4

6253/4 163/4

625   5  125   16  2 8 3

1/4

1/4

3

3

3

2/3

Solution

 216 x  3

69.

2/3

  216 

2/3

x  3

2/3

 36 x 2

pa/2 pa/3 pa/6

Solution pa /2 pa /3 p3a /6 p2a /6 p5a /6    p4a /6  p2a /3 pa /6 pa /6 pa /6 Simplify each expression.

36

70.

Solution

36  6 71.  49

Solution

 49  7 9 25

72.

Solution

9  25 73. 3

9 25

3 5

27 125

Solution 3

3 27 27 3   3 125 125 5

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154


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

x2 y 4

74.

Solution x2 y 4 

x2

y4

 x y2

75.

3

x3

Solution 3

76. 4

x3  x m8 n4 p16

Solution 4

77. 5

m8 n4 p16

4

m8 4 n4 4

p16

m2 n p4

a15b10 c5

Solution 5

5 a15 b10 a15 5 b10 a3 b2   5 5 c c5 c

Simplify and combine terms.

50  8

78.

Solution

50  8  25 2  4 2  5 2  2 2  7 2 12  3  27

79.

Solution

12  3  27  4 3  3  9 3  2 3  3  3 3  3 3  3 3  0 80.

3

24 x 4  3 3x 4

Solution 3

3

24 x 4  3 3x 4  3 8x 3 3 3x  x 3 3 3x  2x 3 3x  x 3 3x  x 3 3x

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155


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Rationalize each denominator.

7

81.

5 Solution

7 5 82.

7

5

5

5

35 5

8 8 Solution

8 8 83.

8

8

2

2

8 2 16

8 2 2 2 4

1 3

2

Solution

1 3

84.

2

1 3

2

3 3

4 4

3

4

3

8

3

4 2

2 3

25

Solution

2 3

25

2 3

25

3

5

3

5

23 5 3

125

23 5 5

Rationalize each numerator. 85.

2 5

Solution

2 2 2 2    5 5 2 5 2 86.

5 5 Solution

5 5 5 5 1     5 5 5 5 5 5

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156


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

87.

2x 3

Solution

2x 2x 2x 2x    3 3 2x 3 2x 88.

33 7 x 2 Solution

3 3 7 x 3 3 7 x 3 49 x 2 3 3 343 x 3 21x     3 3 3 2 2 2 2 49 x 2 49 x 2 49 x 2 Give the degree of each polynomial and tell whether the polynomial is a monomial, a binomial, or a trinomial. 89. x 3  8

Solution 3rd degree, binomial 90. 8 x  8 x 2  8

Solution 2nd degree, trinomial 91.

3x 2 Solution 2nd degree, monomial

92. 4 x 4  12 x 2  1

Solution 4th degree, trinomial Perform the operations and simplify.

93. 2 x  3  3 x  4

Solution

2  x  3  3  x  4   2 x  6  3 x  12  5 x  6

94. 3 x 2 x  1  2 x x  3  x 2 x  2

Solution

3 x 2  x  1  2 x  x  3  x 2  x  2  3 x 3  3 x 2  2 x 2  6 x  x 3  2 x 2  2 x 3  7 x 2  6 x

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157


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra



95. 3 x  2 3 x  2

Solution

 3x  2 3x  2  9x  6x  6x  4  9x  12x  4 2



96. 3x  y 2x  3 y

2

Solution

 3x  y  2x  3 y   6x  9xy  2xy  3 y  6x  7 xy  3 y 2



97. 4a  2b 2a  3b

2

2

2

Solution

 4a  2b 2a  3b  8a  12ab  4ab  6b  8a  8ab  6b 2

98.  z  3  3z 2  z  1

2

2

2

Solution

 z  3  3z  z  1  3z  z  z  9z  3z  3  3z  10z  2z  3 2



3

2

2

3

2

99. an  2 an  1

Solution

a  2a  1  a  a  2a  2  a  a  2 n

100.

n

 2  x

2n

n

n

2n

n

2

Solution

 2  x    2  x  2  x    2   x 2  x 2  x  2  2x 2  x 2

101.

2

2

2

 2  1 3  1 Solution

 2  1 3  1  6  2  3  1

102.

 3  2 9  2 3  4 3

3

3

Solution

 3  2 9  2 3  4  27  2 9  4 3  2 9  4 3  8  3  8  5 3

3

3

3

3

3

3

3

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158


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Rationalize each denominator. 103.

2 31 Solution

2 31

104.

 3  1  2  3  1  2  3  1  3  1 2 31 31  3  1 3  1 2

31

2

2

3 2

2 3 2

2 3 2

 3  2   2  3  2   2  3  2   2 3  2   32 1 3 2 3  2    

3 2

2

2

2

2x x 2

Solution

2x x 2

106.

2

2

Solution

105.

2x

x

y

x

y

x 2

x 2 x 2

 x  2  2 x  x  2 x 4  x 2

2x

2

2

Solution x x

y y

x x

y y

x x

y y

x 2  xy  xy  y

 x  y 2

2

x  2 xy  y xy

Rationalize each numerator. 107.

x 2 5 Solution

x 2 x 2 x 2  x 2 x 4     5 5 x  2 5  x  2 5  x  2 2

2

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159


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

108.

1 a a

Solution

 a  1 a 1 a 1 a 1 a    a a 1  a a 1  a  a 1  a  12 

2

Perform each division. 109.

3x2 y 2 6x3 y

Solution 3x 2 y 2 y  3 2 x 6x y 110.

4a2 b3  6ab4 2b2

Solution

4a2 b3  6ab4 2b2

4a2 b3

6ab4

2b2 2b2 2  2a b  3ab2

111. 2x  3 2x 3  7 x 2  8x  3

Solution

x 2  2x 

1

2x  3 2x  7 x  8x  3 3

2

2x 3  3x 2 4 x 2  8x 4x 2  6x 2x  3 2x  3 0 112. x 2  1 x 5  x 3  2x  3x 2  3

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160


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution x3 

2x  3 

6 x2  1

x 2  1 x 5  0x 4  x 3  3x 2  2x  3  x3

x5

2x 3  3x 2  2x  2x

2x 3  3x  3x

2

3

2

3 6

Factor each expression completely, if possible. 113. 3t 3  3t

Solution

3t 3  3t  3t t 2  1  3t  t  1 t  1

114. 5r 3  5

Solution

5r 3  5  5 r 3  1  5 r 3  13  5  r  1 r 2  r  1

115. 6 x 2  7 x  24

Solution

6 x 2  7 x  24   3x  8 2 x  3

116. 3a 2  ax  3a  x

Solution

3a2  ax  3a  x  a  3a  x   1  3a  x    3a  x  a  1

117. 8 x 3  125

Solution

3 2 8 x 3  125   2 x   53   2 x  5   2 x    2 x  5   52    2 x  5  4 x 2  10 x  25  

118. 6 x 2  20 x  16

Solution

6 x 2  20 x  16  2 3 x 2  10 x  8  2  3 x  2  x  4 

119. x 2  6 x  9  t 2

Solution x 2  6 x  9  t 2   x  3  x  3   t 2   x  3   t 2   x  3  t  x  3  t  2

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161


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

120. 3 x 2  1  5 x

Solution

3x 2  1  5x  3x 2  5x  1  prime 121. 8 z 3  343

Solution

3 2 8z 3  343   2z   7 3   2z  7   2z    2z  7   7 2    2z  7  4 z 2  14 z  49  

122. 1  14 b  49b 2

Solution 1  14b  49b2  49b2  14b  1   7b  1 7b  1   7b  1

2

123. 121z 2  4  44 z

Solution 121z 2  4  44 z  121z 2  44 z  4   11z  2  11z  2    11z  2 

2

124. 64 y 3  1000

Solution

3 64 y 3  1000  8 8 y 3  125  8  2 y   53   8  2 y  5  4 y 2  10 y  25  

125. 2 xy  4 zx  wy  2 zw

Solution

2xy  4zx  wy  2zw  2x  y  2z   w  y  2z    y  2z  2x  w 

126. x 8  x 4  1

Solution

     x  1   x    x  1  x  x  1  x    x  2 x  1  x  x  x  1   x  1 x  1  x   x  x  1     x  1  x  x  1  x  x  x  1

x8  x 4  1  x8  2x 4  1  x 4  x 4  1 x 4  1  x 4 2

4

2

4

2

4

2

4

2

2

2

2

2

2

4

2

2

2

4

2

4

2

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162


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Factor completely and simplify each algebraic expression. Write answers using positive exponents. 127. 7 x 5  14 x 3

Solution

7 x  14 x 5

128. 2  x  1

3

2/5

8 14 7 x8 14 7 x 8  14 7 x  2  7x  3  3  3   x x x x3 x3 5

 4 x  x  1

7/5

Solution

2  x  1

2/5

 4 x  x  1

7/5

 2  x  1

7/5

 2  x  1

7/5

2  x  1

 x  1

 x  1  2 x   

  x  1

7/5

In Exercises 129 and 130, identify the restricted numbers of the rational expression. 129.

11x x 2  225

Solution

x 2  225   x  15 x  15 , x  15, 15, since 15  15  0 and 15  15  0

130.

x1 x  3 x  40 2

Solution

x 2  3 x  40   x  8 x  5 , x  8, 5 , since 8  8  0 and 5  5  0

Simplify each rational expression. 131.

2 x x  4x  4 2

Solution 2 x x  4x  4 2

132.

  x  2

 x  2 x  2

1 x 2

a2  9 a  6a  9 2

Solution

a  3a  3  a  3 a2  9  2 a  6a  9  a  3  a  3  a  3

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163


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Perform each operation and simplify. Assume that no denominators are 0. 133.

x 2  4 x  4 x 2  5x  6  x2 x 2 Solution

x 2  4 x  4 x 2  5 x  6  x  2  x  2   x  2  x  3       x  2  x  3  x2 x 2 x2 x 2

134.

2 y 2  11 y  15 y 2  2 y  8  2 y2  6y  8 y  y 6 Solution

2 y 2  11 y  15 y 2  2 y  8  2 y  5  y  3   y  4  y  2  2 y  5  2    y2  6y  8 y  y 6  y  4  y  2  y  3 y  2 y  2

135.

2t 2  t  3 10t  15  2 2 3t  7t  4 3t  t  4

Solution 2t 2  t  3 10t  15 2t 2  t  3 3t 2  t  4    3t 2  7t  4 3t 2  t  4 3t 2  7t  4 10t  15  2t  3t  1   3t  4 t  1  t  1  5  3t  4 t  1 5  2t  3

136.

p2  7 p  12 p3  8 p2  4 p

p2  9 p2

Solution

 p  3 p  4  p2  7 p  12 p2  9 p2  7 p  12 p2 p2     p3  8p2  4 p p2 p3  8p2  4 p p2  9 p p2  8p  4  p  3  p  3 

137.

p  p  4

 p  8p  4   p  3  2

x2  x  6 x2  x  6 x2  4   x2  x  6 x 2  x  2 x2  5x  6

Solution x2  x  6 x2  x  6 x2  4  2  2 2 x  x  6 x  x  2 x  5x  6 x2  x  6 x 2  x  6 x 2  5x  6  2   x  x  6 x2  x  2 x2  4  x  3 x  2   x  3 x  2   x  2 x  3   x  3 x  2 x  3  2  x  3 x  2  x  2 x  1  x  2 x  2  x  2  x  1

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164


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

 2x  6 2x 2  2x  4  x 2  x  2  138.   x 2  25  x 2  2x  15  x 5 Solution  2x  6 2x 2  2x  4  x 2  x  2 x 2  25 x2  x  2 2x  6       x  5 2 x 2  x  2 x 2  2 x  15 x 2  25  x 2  2 x  15  x 5 

139.

2  x  3 x 5

 x  5 x  5   x  2 x  1  1 2  x  2  x  1  x  5  x  3 

2 3x  x 4 x 5

Solution

2  x  5 3x  x  4 2 3x 2 x  10 3 x 2  12 x      x  4 x  5  x  4  x  5   x  5  x  4   x  4  x  5   x  4  x  5  

140.

3 x 2  10 x  10

 x  4  x  5

5x 3x  7 2x  1   x 2 x2 x2

Solution

5 x  x  2   x  6 x  2 5x 3x  7 2x  1 5x x  6       x 2 x 2 x2 x 2 x2  x  2 x  2  x  2 x  2  

141.

5 x 2  10 x

 x 2  4 x  12

 x  2 x  2  x  2 x  2 4 x 2  6 x  12

 x  2 x  2

2 2x 2  3x  6

 x  2 x  2

x x x   x 1 x 2 x 3

Solution x x x   x 1 x 2 x 3 x  x  2 x  3 x  x  1 x  3 x  x  1 x  2     x  1 x  2 x  3  x  2 x  1 x  3  x  3 x  1 x  2

 

x 3  5x 2  6x

x 3  4 x 2  3x

x 3  3x 2  2x

 x  1 x  2 x  3  x  1 x  2 x  3  x  1 x  2 x  3 3 x 3  12 x 2  11x

x 3 x 2  12 x  11

 x  1 x  2 x  3  x  1 x  2 x  3

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165


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

142.

x 3x  7 2x  1   x1 x2 x2

Solution 3 x  7 2 x  1 x x 3x  7 2x  1      x1 x2 x2 x1 x2 x2 x  6 x   x1 x2 x  x  2   x  6 x  1    x  1 x  2  x  2 x  1 

143.

3  x  1 x

Solution 3  x  1 x

5 x2  3 x

2

5 x2  3 x2

x2  2x

x2  7x  6

5 x  6

 x  1 x  2  x  1 x  2  x  1 x  2

 x

x1

  x  3x  x  1 x  1  5  x  3  x  1  x  x  2

3x 3  6x 2  3x x  x  1 2

2

x 2  x  1

x 2  x  1

x1

144.

x 2  x  1

5 x 3  5 x 2  15 x  15 x  x  1 2

x3

x  x  1 2

 x 3  x 2  12 x  15 x 2  x  1

3x x2  4x  3 x2  x  6  2  x  1 x  3x  2 x2  4

Solution

 x  3 x  1   x  3 x  2 3x 3x x2  4x  3 x2  x  6  2    2 x  1 x  3x  2 x  1  x  1 x  2   x  2  x  2  x 4 

3x 3x x3 x3    x1 x2 x2 x1

Simplify each complex fraction. Assume that no denominators are 0. 5x 145. 2 2 3x 8

Solution 5x 2 3x 8

 2

5x 3x 2 5x 8 20    2  2 8 2 3x 3x

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166


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

3x y 146. 6x y2 Solution 3x y

 6x y2

3x 6x 3x y 2 y  2    y y 6x 2 y

1 1  x y 147. xy Solution 1 1  xy x1  y1 xy  x1   xy y1 yx x y    xy xy  x  y  xy  x  y  xy  x  y 

148.

 

x 1  y  1 y 1  x 1 Solution 1 xy  y1 x 1  y 1 x   1 1 1 1 x y x xy y

    xy    xy    y  x    xy    xy   x  y 1 x

1 y

1 x

1 y

1 y

1 x

1 y

1 x

CHAPTER TEST SOLUTIONS Consider the set {7,  23 , 0, 1, 3, 10, 4 }. 1.

List the numbers in the set that are odd integers.

Solution odd integers: –7, 1, 3 2. List the numbers in the set that are prime numbers.

Solution prime numbers: 3 Determine which property justifies each statement. 3.

a  b  c   b  a   c Solution Commutative Property of Addition

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167


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

4.

a  b  c   ab  ac Solution Distributive Property

Graph each interval on a number line. 5.

4  x  2

Solution

4  x  2  6.

  ,  3  [6, ) Solution

  ,  3  [6, )  Write each expression without using absolute value symbols. 7.

17 Solution

Since  17  0, 17    17   17

8.

x  7 , when x  0 Solution

If x  0, then x  7  0. Then x  7    x  7  .

Find the distance on a number line between points with the following coordinates. 9.

4 and 12 Solution

distance  12   4   16  16

10. 20 and  12

Solution

distance  12   20   8  8

Simplify each expression. Assume that all variables represent positive numbers, and write all answers without using negative exponents. 11. x 4 x 5 x 2

Solution x 4 x 5 x 2  x 4  5  2  x 11

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168


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

12.

r 2r 3s r 4 s2

Solution r 2r 3s r 5s r   4 2 4 2 s r s r s

a a  13. 1

2

2

a 3

Solution

a a   a   a  a 1

2

2

a3

1

2

a3

2

a3

6

 x0 x 2  14.  2   x 

Solution 6

6

6  x0 x 2   x2  4  x 24  2    2   x x x    

 

Write each number in scientific notation. 15. 450,000

Solution

450,000  4.5  105 16. 0.000345

Solution

0.000345  3.45  104

Write each number in standard notation. 17. 3.7  103

Solution

3.7  103  3, 700 18. 1.2  103

Solution

1.2  103  0.0012

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169


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Simplify each expression. Assume that all variables represent positive numbers, and write all answers without using negative exponents. 19.

 25a  4

1/2

Solution

 25a  4

 36  20.    81 

1/2

 

 251/2 a 4

1/2

 5a 2

3/2

Solution

 36     81 

3/2

 8t 6  21.  9    27s 

363/2 813/2

 36   216  8  81  729 27 1/2

1/2

3

3

2/3

Solution

 8t 6   27 s9    22.

3

2/3

 27 s9     6   8t 

2/3

   8 t  272/3 s9 2/3

6

2/3

2/3

 27  s  9s  8  t 4t 1/3

1/3

2

2

6

4

6

4

27a6

Solution 3

23.

27a6  3 27 3 a6  3a2 12  27

Solution

12  27  4 3  9 3 2 33 3 5 3 3

24. 2 3x 4  3x 3 24 x

Solution 2 3 3 x 4  3 x 3 24 x  2 x 3 3 3 x  3 x 3 8 3 3 x  2 x 3 3 x  3 x  2  3 3 x  2 x 3 3 x  6 x 3 3 x 3

 4 x 3 3 x

25. Rationalize the denominator:

x x 2

.

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170


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

x x 2

x

x 2

 x  2  x  x  2 x 2  x 2 x 4

x 2

x

2

26. Rationalize the numerator:

2

x

y

x

y

Solution

.

 x  y 2

x

y

x

y

x

y

x

y

x

y

x

y

2

x  xy  xy  2

y

2

xy x  2 xy  y

Perform each operation.

 

27. a 2  3  2a 2  4

Solution

a  3  2a  4  a  3  2a  4 2

2

2

2

 a2  7

28. 3a 3 b2

 2a b  3

4

Solution

 3a b  2a b   6a b 3

2

3

4



29. 3 x  4 2 x  7

6

6

Solution

 3 x  4  2 x  7   6 x  21x  8x  28 2

 6 x 2  13 x  28



30. a n  2 a n  3

Solution

a  2a  3  a  3a  2a  6 n

n

2n

n

n

 a2n  an  6 31.

 x  4  x  4  2

2

Solution

 x  4  x  4   x  4 x  4 x  16  x  16 2

2

4

2

2

4

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171


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

32. x 2  x  2  2 x  3 

Solution

 x  x  2 2x  3  2x  3x  2x  3x  4 x  6  2x  5x  7 x  6 2

3

2

2

3

2

33. x  3 6 x 2  x  23

Solution 6 x  19

 x343

x  3 6x2 

x  23

6 x  18 x 2

19 x  23 19 x  57 34 34. 2 x  1 2 x 3  3 x 2  1

Solution x 2  2x 

1

2x  1 2x  3x  0x  1 3

2

2x 3  x 2 4x2  0x 4x 2  2x 2x  1 2x  1 0

Factor each polynomial completely. 35. 3 x  6 y

Solution

3x  6 y  3  x  2 y 

36. x 2  100

Solution

x 2  100  x 2  102   x  10  x  10 

37. 45 x 2  20 y 2

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172


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution

45 x 2  20 y 2  5 9 x 2  4 y 2

 5  3 x  2 y  3 x  2 y 

38. 10t 2  19tw  6w 2

Solution

10t 2  19tw  6w 2   5t  2w  2t  3w 

39. 64m3  125n3

Solution

64m3  125n3   4m  5n  16m2  20mn  25n2

40. 3a3  648

Solution

3a 3  648  3 a 3  216  3  a  6  a 2  6a  36

41. x 4  x 2  12

Solution

 x  3    x  2  x  2   x  3 

x 4  x 2  12  x 2  4

2

2

42. 6 x 4  11x 2  10

Solution



6 x 4  11x 2  10  3 x 2  2 2 x 2  5

Simplify each rational expression. Assume no denominators are 0. 43.

44 p3q6 33p4q2

Solution 44 p3q6

4q4  3p 33p4q2

44.

49  x 2 x  14 x  49 2

Solution 49  x 2 x  14 x  49 2

  x  7  x  7 

 x  7  x  7 



x 7 x7

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173


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Perform each operation and simplify if possible. Assume that no denominators are 0. 45.

x 2  x2 x2

Solution 2 x x2   1 x2 x2 x2 46.

x x  x1 x1

Solution

x  x  1 x  x  1 x x x2  x  x2  x 2 x      x  1 x  1  x  1 x  1  x  1 x  1  x  1 x  1  x  1 x  1

47.

x 2  x  20 x 2  25  x 5 x 2  16

Solution x 2  x  20 x 2  25  x  5  x  4   x  5  x  5   x  5      x 5 x 5 x4 x 2  16  x  4  x  4 

48.

2

x2 x2  4  x1 x  2x  1 2

Solution

x2 x2  4 x2 x1 1     x  2x  1 x  1 x  1 x  1 x  2 x  2 x  1         x  2 2

Simplify each complex fraction. Assume that no denominators are 0. 1 1  49. a b 1 b

Solution 1 a

 b1 1 b

50.

ab  a1  b1  ab   1 b

ab  a1   ab  b1  a

ba a

x 1 x

1

 y 1

Solution

1 xy  x1  x 1 y x    1 1 1 1 1  x 1  y 1 xy x  y xy  x   xy x y

  1 y

y yx

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174


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

GROUP ACTIVITY SOLUTIONS Anesthesiology Real-World Example of a Rational Expression For surgeries and serious injuries, anesthesiologist are responsible for administering anesthesia. How anesthesia is administered and monitored ensures safety for patients. There is a mathematic model, a rational expression, that describes how medicine concentration varies over time in the human body. The model is determined by various factors including the patient’s weight, types of drugs used, and quantity of drugs administered. The right concentration of anesthesia is required to ensure the patient remains unconscious, and safe, during a procedure.

Group Activity The rational expression shown below models a patient’s bloodstream concentration of anesthesia over time. 2.7t 0.4t 2  1.2

t represents minutes since the dosage was given

If we input numbers for time (in minutes), the values we obtain will be in concentration of the anesthesia (in milligrams per liter or mg/L). a. Complete the table shown for the times given. Round the concentration values to three decimal places.

Minutes (t) since dose

Concentration in mg/L

1

1.688

2

1.929

5

1.205

10

0.655

20

0.335

30

0.224

40

0.168

50

0.135

60

0.112

70

0.096

80

0.084

90

0.075

b. If a concentration of anesthesia is below 0.112 mg/L, it is no longer effective. Based on the data in the table, approximately when will the patient wake up? c. If the surgery or procedure lasts for 1.5 hours, will additional anesthesia be required? d. If the concentration is over 2 mg/L in the bloodstream, then it could be very dangerous for the patient. Based on the data in the table, does that ever occur?

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175


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter R: A Review of Basic Algebra

Solution a. t  1

1.688 mg/L t  2 1.929 mg/L t  5 1.205 mg/L t  10 0.655 mg/L t  20 0.335 mg/L t  30 0.224 mg/L t  40 0.168 mg/L t  50 0.135 mg/L t  60 0.112 mg/L t  70 0.096 mg/L t  80 0.084 mg/L t  90 0.075 mg/L

b. Below 0.112 mg/L, the patient will wake up after 60 minutes. c. Yes, more anesthesia will be required. d. No, the concentration will never reach 2 mg/L.

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176


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER 1: EQUATIONS AND I NEQUALITIES

TABLE OF CONTENTS End of Section Exercise Solutions .................................................................................. 177 Exercises 1.1 .............................................................................................................................. 177 Exercises 1.2 ............................................................................................................................. 210 Exercises 1.3 ............................................................................................................................ 238 Exercises 1.4 ............................................................................................................................. 261 Exercises 1.5 ............................................................................................................................ 302 Exercises 1.6 ............................................................................................................................ 329 Exercises 1.7 ............................................................................................................................ 360 Exercises 1.8 ............................................................................................................................ 406 Chapter Review Solutions................................................................................................ 437 Chapter Test Solutions .................................................................................................... 472 Cumulative Review Exercises.......................................................................................... 483 Group Activity Solutions .................................................................................................. 497

END OF SECTION EXERCISE SOLUTIONS EXERCISES 1.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Remove parentheses and simplify. – 4 ( 2 x  1) – 2 ( x – 5 )  3 x

Solution – 4 ( 2 x  1) – 2 ( x – 5 )  3 x  – 8 x – 4 – 2 x  10  3 x  – 7 x  6

2. Factor completely and simplify. 5 x 2 – 17 x  6

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177


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

5x 2 – 17 x  6  (5x – 2)( x – 3) 3. Substitute –6 for x in the equation 3 x – 2 ( x  1)  2 x  4 . Is the equation that results true or false? Solution 3(–6) – 2(–6  1)  2( 6)  4 –18  12 – 2  –12  4 –8  –8, which is a true statement.

4. Identify the LCD of

3 2x x  1 , , and . 4 5 2

Solution The LCD of 5, 2, and 4 is 20. 5. Identify the LCD of

3x  1 2 . and x  8 x 2  9x + 8

Solution x2 – 9x + 8 = (x – 8) (x – 1) The LCD between x – 8 and (x – 8) (x – 1) is (x – 8) (x – 1). 6. Multiply and simplify.

 x  3 x + 3 x 4 3  x 2 2  Solution    4   2   x  3 x  2    x  3 x  2   x  3   x  2       4   2    x  3  x  2    x  3  x  2    x  2   x  3    

 x  2 4 –  x – 3 2     4 x  8 – 2 x – 6  4 x  8 – 2 x  6

2x 

14

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If a number satisfies an equation, it is called a __________ or a __________ of the equation.

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178


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution root, solution 8. If an equation is true for all values of its variable, it is called an __________.

Solution identity 9. A contradiction is an equation that is true for __________ values of its variable.

Solution no 10. A __________ equation is true for some values of its variable and is not true for others.

Solution conditional 11. An equation of the form ax + b = 0 is called a __________ equation.

Solution linear 12. If an equation contains rational expressions, it is called a __________ equation.

Solution rational 13. A conditional linear equation has __________ root.

Solution one 14. The __________ of a fraction can never be 0.

Solution denominator Practice Find any restrictions on the values of x in each equation. 15. 2x  8  17

Solution 2x  5  17 no restrictions 16.

1 x  7  12 2 Solution

1 x  7  14 2 no restrictions

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179


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

17.

1  10 x Solution

1  12 x x  0 18.

4  9x x  2 Solution

3  9x x  2 x  2 19.

8 1  x  6 x  2

Solution

8 5  x  6 x  2 x  6  0 x  2  0 x  6

x  2

x  6, x  2 20.

x 3   x  3 x  4

Solution

x 4   x  3 x  4 x  3  0 x  4  0 x  3

x  4

x  3, x  4 21.

x 5x  2 x  3 x  16

Solution

1 5x  2 x  3 x  16 1 5x  x  3 ( x + 4)( x  4)

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180


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  3  0

x  4  0

x  4  0

x  3

x  4

x  4

x  3, x  4, x  4 22.

x + 5 x

2

 3x  4

5  2 x

Solution

1 5   2 x x 2  3x  4 1 5   2 x ( x  1)( x  4) x  1  0 x  4  0 x  0 x  1

x  4

x  1, x  4, x  0 Solve each equation, if possible. Classify each one as an identity, a conditional equation, or a contradiction. 23. 2 x  5  15

Solution 2 x  5  15 2 x  5  5  15  5 2 x  10 2x 10  2 2 x  5 conditional equation

24. 3 x  2  x  8

Solution 3x  2  x  8 3x  x  2  x  x  8 2x + 2  8 2x + 2  2  8  2 2x  6 2x 6  2 2 x  3 conditional equation

25. 2( n  2)  5  2 n

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181


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 2(n  2)  5  2n 2n  4  5  2n 2n  1  2n 2n  2n  1  2n  2n 1  0 no solution contradiction

26. 3( m  2 )  2( m  3 )  m

Solution 3(m  2)  2(m  3)  m

3m  6  2m  6  m 3m  6  3m  6 all real numbers identity 27.

x + 7  7 2 Solution x + 7  7 2 x + 7 2   2(7) 2 x + 7  14 x + 7  7  14  7 x  7 conditional equation

28.

x  7  14 2 Solution x  7  14 2 x  7  7  14  7 2 x  21 2 x 2   2(21) 2 x  42 conditional equation

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182


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

29. 2( a + 1)  3( a  2)  a

Solution 2(a + 1)  3(a  2)  a 2a + 2  3a  6  a 2a + 2  2a  6 2a  2a + 2  2a  2a  6 2  6 no solution; contradiction

30. x2  ( x  4)( x  4)  16

Solution

x 2  ( x  4)( x  4)  16 x 2  x 2  16  16 x2  x2 all real numbers; identity 6 x  18 2

31. 3( x  3) 

Solution

6 x  18 2 6 x  18 3x  9  2 6 x  18 2(3x  9)  2  2 6 x  18  6 x  18 all real numbers; identity 3( x  3) 

32. x( x  2)  ( x  1)2

Solution

x( x  2)  ( x  1)2 x 2  2 x  ( x  1)( x  1) x 2  2x  x 2  2x  1 x 2  x 2  2x  x 2  x 2  2x  1 2x  2x  1 2x  2x  2x  2x  1 0  1 no solution; contradiction

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183


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

33.

3  1 b  3

Solution

3  1 b  3 3  (b  3)(1) (b  3)  b  3 3  b  3 3  3  b  3  3 6  b conditional equation 34. x2  8x  15 = ( x  3)( x  5)

Solution x 2  8 x  15 = ( x  3)( x  5) x 2  8 x  15 = x 2  2 x  15 x 2  x 2  8 x  15 = x 2  x 2  2 x  15 8 x  15 = 2 x  15 8 x  8 x  15 = 2 x  8 x  15 15 = 10 x  15 15  15  10 x  15  15 30  10 x 30 10 x  10 10 3  x

conditional equation 35. 2x2  5x  3 = (2x  1)( x  3)

Solution

2x 2  5x  3 = (2x  1)( x  3) 2x 2  5x  3 = 2x 2 + 5x  3 all real numbers; identity

 19  36. 2x 2  5x  3 = 2x  x   2 

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184


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution  19  2x 2  5x  3 = 2x  x   2  2 x 2  5 x  3 = 2 x 2  19 x 2 x 2  2 x 2  5 x  3 = 2 x 2  2 x 2  19 x 5 x  3 = 19 x 5 x  5 x  3 = 19 x  5 x 3  14 x 14 x 3  14 14 3   x 14 conditional equation

Solve each linear equation. 37. 4 x  11  9

Solution 4 x  11  9 4 x  20 x  5

38. 8 y + 16   8

Solution 8 y + 16 = 8 8 y  24 y  3

39. 16 = 3z  2

Solution 16 = 3z  2 18 = 3z 6 = z

40. 20 = 5 x + 10

Solution 20 = 5 x + 10 30 = 5 x 6 = x

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185


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

41. 2 x + 7 = 10  x

Solution 2 x + 7 = 10  x

3 x + 7 = 10 3x = 3 x = 1 42. 9a  3 = 15 + 3a

Solution 9a  3 = 15 + 3a 6a  3 = 15 6a = 18 a = 3 43. 4y – 1 = –2y + 19

Solution 4 y  1 = 2 y + 19

6 y  1 = 19 6 y = 20 y =

10 3

44. –2 – 7x = 1 + 2x

Solution 2  7 x = 1 + 2x

2  9x = 1 9x = 3 x = 

1 3

45. 5(2y – 9) = 3y – 4

Solution 5(2 y  9) = 3 y  4

10 y  45 = 3 y  4 7 y  45 = 4 7 y = 41 y =

41 7

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186


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

46. 13x + 1 = –(5 – x)

Solution 13x + 1 = (5  x )

13x + 1 = 5  x 12x + 1 = 5 12x = 6 x = 

1 2

47. 5( x  2) = 2( x  4)

Solution 5( x  2) = 2( x  4) 5 x  10  2 x  8 3 x  10  8 x  6

48. 5(r  4) = 5(r  4)

Solution 5(r  4) = 5(r  4)

5r  20 = 5r  20 10r  20 = 20 10r = 40 r = 4 49. 7(2x  5)  6( x  8) = 7

Solution 7(2x  5)  6( x  8) = 7

14 x  35  6x  48 = 7 8x  13 = 7 8x = 20 x =

20 5  8 2

50. 6( x  5)  4( x  2) = 1 Solution 6( x  5)  4( x  2) = 1

6 x  30  4 x  8 = 1 2x  38 = 1 2x = 37 x =

37 2

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187


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

51. 3( x  2)  ( x + 5) = 7 + 4( x  3)

Solution 3( x  2)  ( x + 5) = 7 + 4( x  3) 3 x  6  x  5 = 7 + 4 x  12 2 x  11 = 4x  5 2 x  11 = 5 2 x = 6 x  3 52. 8 – 2( y  1)  3( y  6)  ( y  2)

Solution 8 – 2( y  1)  3( y  6)  ( y  2)

8 – 2 y  2  3 y  18  y  2 6 – 2 y = 2 y  20 6  4 y = 20 4 y = 14 y = 

7 2

53. (t  1)(t  1)  (t  2)(t  3)  4

Solution (t  1)(t  1)  (t  2)(t  3)  4 t2  1  t2  t  6  4 1  t  2 t  1  2 t  1 54. ( x  2)( x  3) = ( x  3)( x  4)

Solution ( x  2)( x  3) = ( x  3)( x  4) x 2  5 x  6 = x 2  7 x  12 5 x  6 = 7 x  12 12 x  6 = 12 12 x = 6 x = 

6 1   12 2

Solve the linear equations containing fractions. 55.

5 z  8 = 7 3

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188


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 5 z  8 = 7 3 5 z = 15 3 5 3  z = 3(15) 3 5z = 45 z = 9

56.

4 y + 12 = 4 3

Solution

4 y + 12 = 4 3 4 y = 16 3 4 3  y = 3( 16) 3 4 y = 48 y = 12 57.

z + 2 = 4 5

Solution z + 2 = 4 5 z = 2 5 z 5  = 5(2) 5 z  10 58.

3p  p = 4 7 Solution 3p  p = 4 7  3p  7  p  = 7( 4)  7 

3p  7 p = 28 4 p = 28 p = 7

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189


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

59.

7 15 x + 5 = x + 2 2 Solution 7 15 x + 5 = x + 2 2 7   15  2 x + 5  = 2 x +  2 2  

7 x + 10  2 x + 15 5 x + 10  15 5x  5 x  1 60.

x 1 1 x  =  2 5 2 3

Solution x 1 1 x   = 2 5 2 3 x 1 1 x 30   = 30   5 3 2 2

(multiply by common denominator)

15 x  6 = 15 + 10 x 5 x  6 = 15 5 x = 21 x =

61.

21 5

3x  2 7 = 2x + 3 3

Solution 3x  2 7 = 2x + 3 3  3x  2 7 3  = 3 2 x +  3 3 

3x  2  6x  7 3 x  2  7 3x  9 x  3

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190


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

62. 2 x 

7 x 4x + 3 + = 6 6 6

Solution 7 x 4x + 3 + = 6 6 6  7 4x + 3 x 6 2 x  +  = 6  6 6 6  2x 

12 x  7  x  4 x  3 13 x  7  4 x  3 9x  7  3 9 x  10 x 

63.

10 9

3x  1 1 = 20 2

Solution 3x  1 1 = 20 2 3x  1 1 20  = 20  20 2 3 x  1  10

3x = 9 x = 3 64. 2(2 x  1) 

3x 3(4  x ) = 2 2

Solution 3(4  x ) 3x = 2 2  3(4  x ) 3x  2 2(2 x  1)   = 2  2 2  2(2 x  1) 

4(2 x  1)  3 x  3(4  x ) 8 x  4  3 x  12  3 x 5 x  4  12  3 x 8 x  4  12 8 x  16 x  2 65.

3  x x  7 + = 4x  1 3 2

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191


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution x  7 3  x + = 4x  1 3 2 3  x x  7 6 +  = 6(4 x  1) 2   3 2(3  x )  3( x  7)  24 x  6 6  2 x  3 x  21  24 x  6 5 x  27  24 x  6 19 x  27  6 19 x  21 x 

66.

21 19

3 (3x  2)  10 x  4 = 0 2 Solution 3 (3 x  2)  10 x  4 = 0 2 3  2  (3 x  2)  10 x  4 = 2(0) 2  3(3 x  2)  20 x  8  0 9 x  6  20 x  8  0

11x  14  0 11x  14 x  

67.

14 11

a(a  3)  5 (a  1)2 = 7 7

Solution (a  1)2 a(a  3)  5 = 7 7 (a  1)2   a(a  3)  5 7   = 7 7 7     a(a  3)  5  (a  1)(a  1) a2  3a  5  a2  2a  1 3a  5  2a  1 5  a  1 4  a

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192


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

68.

( y  2)2 y2 = y  2  3 3

Solution ( y  2)2 y2 = y  2  3 3  ( y  2)2   y2  3   = 3 y  2  3 3     ( y  2)2  3 y  6  y 2 y2  4y  4  y2  3y  6 4y  4  3y  6 y  4  6 y  2

Solve each rational equation. Check for false or extraneous solutions. 69.

4 2 6  = x 5 x

Solution 4 2 6  = x x 5 4 2 6 5 x   = 5 x  x 5 x

20  2 x  30 2 x  10 x  5 70.

3 1 4  = x 2 x Solution 3 1 4  = x x 2 3 1 4 2 x   = 2 x  x x 2   6  x = 8 x = 2

71.

1 1 2 1  =   4x 3 3x 2

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193


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 1 1 2 1   =  4x 3 3x 2  1  2 1 1   = 12 x     12 x  3 2  4x  3x

(multiply by common denominator)

3  4 x  8  6 x 3  2 x  8 2 x  11 x  72.

11 2

2 1 2 3  =  5x 3 x 5

Solution 2 1 2 3   = x 5x 3 5  2 2 1 3   = 15 x    (multiply by common denominator) 15 x  3 5  5x x

6  5 x  30  9 x 6  14 x  30 14 x  24 x  73.

12 7

2 1 1  = x  1 3 x  1

Solution

2 1 1  = x  1 x  1 3  2 1 1   = 3( x  1)  3( x  1) x  1 3 x  1 6  1( x  1)  3(1) 6  x  1  3 x  7  3 x  4 74.

3 1 3  = x  2 x x  2

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194


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3 1 3  = x  2 x x  2  3 1 3 x( x  2)   = x ( x  2)  x x  2 x  2 3 x + 1( x  2)  3 x 3x + x  2  3x 4 x  2  3x x  2 The answer does not check. Þ no solution 75.

9t  6 7 = t (t  3) t  3

Solution

9t  6 7 = t(t  3) t  3  9t  6  7 t(t  3)   = t (t  3)  ( 3) 3   t t t   9t  6  7t 2t  6  0 2t  6 t  3 The answer does not check. Þ no solution 76. x 

2( 2 x  1) 3x 2 = 3x  5 3x  5

Solution 2( 2 x  1) 3x2 = 3x  5 3x  5  2( 2 x  1)  3x2   (3 x  5)  x  = (3 x 5)  3x  5  3x  5  x 

x (3 x  5)  2( 2 x  1) = 3 x 2 3x2  5x  4 x  2 = 3x2 x  2 = 0 x = 2

77.

2 4 = (a  7)(a  2) (a  3)(a  2)

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195


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

2 4 = (a  7)(a  2) (a  3)(a  2) 2 4 (a  7)(a  2)(a  3)   (a  7)(a  2)(a  3)  (a  7)(a  2) (a  3)(a  2) 2(a  3)  4(a  7) 2a  6  4a  28 2a  34 a  17 78.

2 1 1  = 2 n  2 n  1 n  n  2

Solution 2 1 1  = 2 n  2 n  1 n  n  2 2 1 1 =  ( n  2)(n  1) n  2 n  1  2 1  1 ( n  2)( n  1)   = ( n  2)(n  1)  ( n  2)( n  1) n  1 n  2 2( n  1)  1( n  2)  1 2n  2  n  2  1 3n  1 n 

79.

3 x

2

 16

1 3

2 3 + x  4 x  4

Solution 3

2 3 + x  4 x  4 x  16 3 2 3 +  x  4 x  4 ( x + 4)( x  4)    2 3 3  ( x + 4)( x  4)  +   ( x  4)( x + 4)   x  4 ( x + 4)( x  4)  x  4 3  2( x + 4)  3( x  4) 2

3  2 x  8  3 x  12 3  5x  4 7  5x 7  x 5

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196


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

80.

2 5 x   2 x  6 x  6 x  36

Solution 2 5 x   2 x  6 x  6 x  36 2 5 x   x  6 x  6 ( x  6)( x + 6)  2   5  x ( x + 6)( x  6)    ( x + 6)( x  6)   ( x  6)( x + 6   x  6 x  6 2( x + 6)  5( x  6)  x 2 x + 12  5 x  30  x 2 x + 12  4 x  30 2 x + 12  30 2 x  42 x  21

81.

2x  3 x

2

 5x  6

3x  2 x

2

 x  6

5x  2 x2  4

Solution 2x  3 2

3x  2 2

5x  2

x  5x  6 x  x  6 x2  4 2x  3 3x  2 5x  2   ( x  3)( x  2) ( x  3)( x  2) ( x  2)( x  2) ( x  2)(2 x  3)  ( x  2)(3 x  2)  ( x  3)(5 x  2) 2 x 2  x  6  3 x 2  4 x  4  5 x 2  13 x  6

{multiply by common denominator}

5 x 2  3 x  10  5 x 2  13 x  6 3 x  10  13 x  6 10 x  4 x  

82.

3x x

2

 x

2x x

2

 5x

4 2   10 5

x  2 x

2

 6x  5

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197


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 3x

2x

2

2

 x x  5x 3x 2x   x( x  1) x ( x  5) 3 2   x  1 x  5 3( x  5)  2( x  1)  x

x  2 2

 6x  5 x  2 ( x  5)( x  1) x  2 ( x  5)( x  1) {multiply by common denominator} x  2 x

3 x  15  2 x  2  x  2 x  13  x  2 13  2  no solution 83.

3x  5 x

3

 8

3 x

2

 4

2(3 x  2) ( x  2)( x 2  2 x  4)

Solution 3x  5

3

2(3 x  2)

( x  2)( x 2  2 x  4) x  4 3x  5 3 2(3 x  2)   2 ( x  2)( x  2) ( x  2)( x  2 x  4 ( x  2)( x 2  2 x  4) x

3

 8

2

( x  2)(3 x  5)  ( x 2  2 x  4)(3)  2( x  2)(3 x  2) {multiply by common denominator} 3 x 2  x  10  3 x 2  6 x  12  6 x 2  8 x  8 6x 2  7 x  2  6 x 2  8x  8 15 x  10 x 

84.

10 2  15 3

1 3n  4 1   2 n  8 5 n  2 5n  42n  16

Solution 1 3n  4 1   2 n  8 n  2 5 5n  42n  16 1 3n  4 1   n  8 (5n  2)(n  8) 5n  2 (5n  2)(1)  (3n  4)  n  8 {multiply by common denominator}

5n  2  3n  4  n  8 2n  6  n  8 n  2 85.

1 2(3n  1) 1   2 11  n 7 n  3 7n  74n  33

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198


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 1 2(3n  1) 1   2 11  n 7n  3 7n  74n  33 2(3n  1) 1 1   2 7n  3 n  11 7n  74n  33 6n  2 1 1   (7n  3)(n  11) 7n  3 n  11 (7n  3)  6n  2  (n  11)1

{multiply by common denominator}

7n  3  6n  2  n  11 n  5  n  11 2n  6 n  3 86.

4 a

2

 13a  48

Solution 4 2

2 a

2

1

 18a  32

2

a

2

2

 a  6

1 2

a  18a  32 a  a  6  13a  48 4 2 1   (a  16)(a  3) (a  16)(a  2) (a  3)(a  2) 4(a  2)  2(a  3)  1(a  16) {multiply by common denominator} a

4a  8  2a  6  a  16 2a  14  a  16 a  2 87.

5 2 6 1    2 y  4 y  2 y  2 y  6y  8 Solution 5 2 6 1    2 y  4 y  2 y  2 y  6y  8 5 4 1   ( y  2)( y  4) y  4 y  2 5( y  2)  4( y  4)  1 {multiply by common denominator} 5 y  10  4 y  16  1 5 y  10  4 y  15 y  5

88.

6 3 1   2a  6 3  3a a2  4a  3

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199


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 6 3   2a  6 3  3a 6 3   2(a  3) 3(1  a) 3 1   a  3 a  1 3(a  1)  1(a  3) 

1 2

 4a  3 1 (a  3)(a  1) 1 (a  3)(a  1) 1 {multiply by common denominator} a

3a  3  a  3  1 4a  6  1 4a  7 a 

89.

7 4

3y 2y 8   6  3y 2y  4 4  y2 Solution 3y 2y 8   6  3y 2y  4 4  y2 3y 2y 8   3(2  y) 2( y  2) (2  y )(2  y ) y y 8   2  y 2  y (2  y )(2  y ) y (2  y )  y (2  y )  8 {multiply by common denominator} 2y  y2  2y  y2  8 4y  8 y  2  The solution does not check, so the equation has no solution.

90.

3  2a a

2

 6  5a

2  3a a

2

 6  a

5a  2 a2  4

Solution 3  2a 2

2  3a 2

5a  2

a  6  5a a  6  a a2  4 2a  3 3a  2 5a  2   (a  2)(a  3) (a  3)(a  2) (a  2)(a  2) (a  2)(2a  3)  (a  2)(3a  2)  (a  3)(5a  2) {multiply by common denominator} 2a2  a  6  3a2  4a  4  5a2  13a  6 10a  4 a 

4 2   10 5

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200


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

91.

a 3a  2  1   2 a  2 a  4a  4

Solution a 3a  2  1  2 a  2 a  4a  4 3a  2 a  1   (a  2)(a  2) a  2  a    3a  2 (a  2)(a  2)   1  (a  2)(a  2)    a  2   (a  2)(a  2)  a(a  2)  (a  2)(a  2)  (3a  2) a2  2a  (a2  4a  4)  3a  2 a2  2a  a2  4a  4  3a  2 2a  4  3a  2 a  2 92.

x  1 x  2 1  2x   x  3 x  3 3  x

Solution x  1 x  2 1  2x   x  3 x  3 3  x x  1 x  2 2x  1   x  3 x  3 x  3 ( x  3)( x  1)  ( x  3)( x  2)  ( x  3)(2 x  1) {miltiply by common denominator} x 2  4 x  3  x 2  x  6  2x 2  5x  3 2x 2  3x  3  2x 2  5x  3 8 x  0 x 

0  0 8

Solve each formula for the specified variable. 93. f  ma; m

Solution f  ma f ma  a a f  m a

94. P  2l  2w; w

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201


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution P  2l  2w

P  2l  2w P  2l 2w  2 2 P  2l  w 2 95. ax + b = 0; x

Solution ax  b  0

ax  b x   96. V 

b a

1 2 r h; h 3

Solution 1 2 r h 3 1 3V  3  r 2 h 3 3V  r 2 h

V 

3V r 2 3V r 2

97. V 

r 2 h r 2

 h 1 2 r h; r 2 3

Solution 1 V  r 2 h 3 1 3V  3  r 2 h 3 3V  r 2 h r 2 h 3V  h h 3V  r2 h

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202


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

98. z 

x  

;

Solution z  z 

x  

x  

 

z  x  

  z  x   x  z 99. Pn  L 

si ;s f

Solution

si f si Pn  L  f Pn  L 

f (Pn  L)  f 

si f

f (Pn  L)  si f (Pn  L) si  i i f (Pn  L)  s i 100. Pn  L 

si ;f f

Solution Pn  L  Pn  L 

si f

si f

f (Pn  L)  f 

si f

f (Pn  L)  si f (Pn  L) si  Pn  L Pn  L f 

si Pn  L

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203


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

mMg

101. F 

r2

;m

Solution mMg F  r2 mMg Fr 2   r2 r2 Fr 2  mMg Fr 2 mMg  Mg Mg Fr 2  m Mg

102.

1 1 1   ;f f p q

Solution 1 1 1   f p q fpq 

1 1 1  fpq    f q p

pq  fq  fp pq  f (q  p) pq f (q  p)  q  p q  p pq  f q  p 103.

x y   1; y a b

Solution x y   1 a b y x  1  b a  y x  b 1   b  b a 

 x y  b 1   a  104.

x y   1; a a b

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204


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution x y   1 a b x y ab    ab  1 a b   bx  ay  ab bx  ab  ay bx  a( b  y ) bx a( b  y )  b  y b  y bx  a b  y

105.

1 1 1   ;r r r1 r2

Solution 1 1 1   r r1 r2

1 1 1  rr1r2    r r2   r1 r1r2  rr2  rr1

rr1r2 

r1r2  r (r2  r1 ) r1r2 r (r2  r1 )  r2  r1 r2  r1 r1r2  r r2  r1 106.

1 1 1   ;r r r1 r2 1

Solution 1 1 1   r r1 r2

rr1r2 

1 1 1  rr1r2    r r2   r1

r1r2  rr2  rr1 r1r2  rr1  rr2 r1 (r2  r )  rr2 r1 (r2  r ) rr2  r2  r r2  r r1 

rr2 r2  r

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205


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

107. n l  a  ( n  1)d;

Solution l  a  (n  1)d l  a  nd  d l  a  d  nd l  a  d nd  d d l  a  d  n d 108. l = a + (n – 1)d; d

Solution l  a  (n  1)d

l  a  (n  1)d (n  1)d l  a  n  1 n  1 l  a  d n  1 109. a  (n  2)

180 ;n n

Solution 180 n 180 an  (n  2)  n n an  (n  2)180 a  (n  2)

an  180n  360 360  180n  an 360  n(180  a) 360  n 180  a

110. S 

a  lr ;a 1  r

Solution a  lr 1  r a  lr (1  r ) S(1  r )  1  r S(1  r )  a  lr S 

S  Sr  lr  a

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206


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

111. R 

1 1 r1

1 r2

 r1

; r1

3

Solution R  R 

R 

1 1 r1

 r1

1 r2

3

r1r2 r3 (1)

r1r2 r3 r1  r1  r1 1

2

3

r1r2 r3 r2 r3  r1r3  r1r2

R(r2 r3  r1r3  r1r2 ) = r1r2 r3 Rr2 r3  Rr1r3  Rr1r2 = r1r2 r3 Rr1r3  Rr1r2  r1r2 r3 = Rr2 r3 r1 (Rr3  Rr2  r2 r3 ) = Rr2 r3 r1 =

112. R 

1 1 r1

 r1  r1 2

Rr2 r3 Rr3  Rr2  r2 r3

; r3

3

Solution

R  R 

R 

1 1 r1

1 r2

 r1

3

r1r2 r3 (1)

r1r2 r3 r1  r1  r1 1

2

3

r1r2 r3 r2r3  r1r3  r1r2

R(r2r3  r1r3  r1r2 ) = r1r2 r3 Rr2 r3  Rr1r3  Rr1r2 = r1r2 r3 Rr2 r3  Rr1r3  r1r2 r3 = Rr1r2 r3 (Rr2  Rr1  r1r2 ) = Rr1r2 r3 =

Rr1r2 Rr2  Rr1  r1r2

Fix It In exercises 113 and 114, identify the step the first error is made and fix it. 113. Solve the linear equation for x:

5(2 x  1)  8x  2( x  5)  7 x

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207


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Step 4 was incorrect. Step 1: 10 x  5  8 x  2 x  10  7 x Step 2: (10 x  8x )  5  (–2x  7 x )  10 Step 3: 2 x  5  5 x  10 Step 4: 3 x  5 Step 5: x  

5 3

114. Write the values of x that make the denominator zero and then solve the rational equation: 5 3 60   x  5 x  7 ( x  5)( x  7)

Solution Step 5 was incorrect. Recall from Step 1, that 5 and −7 would cause the denominators to be zero. Therefore, there is no solution because 5 is an extraneous solution. Discovery and Writing 115. Explain the difference between an identity and a contradiction. Give examples of each. Solution Answers may vary. 116. Share a strategy that can be used to identify the restrictions on a variable in a rational equation.

Solution Answers may vary. 117. Explain why a conditional linear equation always has exactly one root.

Solution Answers may vary. 118. Define an extraneous solution and explain how such a solution occurs.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 119. The equation 4 x  5( x  3)  9x  15 is a contradiction.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 4 x  5( x  3)  9 x  15

4 x  5 x  15  9 x  15 9 x  15  9 x  15 False. The equation is an identity. 120. The equation 4 x  5( x  3)  9x  15 is an identity.

Solution 4 x  5( x  3)  9 x  15 4 x  5 x  15  9 x  15 9 x  15  9 x  15 15  15 False. The equation is a contradiction.

121.

7, 4.5, and π would be included in the solution set for 2 x  8  (2 x  8). Solution 2 x  8  (2 x  8)  2 x  8  2 x  8 True. The equation is an identity, so all real numbers are solutions.

122. The equation x  188,424  x  188,425 has an infinite number of solutions.

Solution x  188,424  x  188,425 False. The equation is a contradiction, so it has no solution. 123. The solution set of

Solution 1

  1 x 3

1

  1 x 4

1

  1 x 3

1

  1 x 4

 7 is {7}.

 7

( x  3)  ( x  4)  7 2x  7  7 2 x  14 x  7 True.

124. If y 1 

1 and y2  1, then y1  y2 when x  2 x  

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution y1 

1  1  y2 x  

 1  (x   )     (x   )  1 x    1  x  

  1  x

False. y1  y2 when x    1.

EXERCISES 1.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Write an algebraic expression that represents the average of the four microbiology test scores 50, 75, 100 and x.

Solution

50  75  100  x 4 2. Rita watched 6 times as many movies on Netflix as Emma last year. If Emma watched x movies last year, write an expression for the number of movies Rita watched. Solution 6x 3. If the width of a rectangle is represented by x and the length is represented by 2 x  30, write a simplified expression representing its perimeter.

Solution Width = x Length = 2x + 30 P = 2 (Length)+ 2 (width) P = 2(2x + 30) + 2x P = 4x + 60 + 2x P = 6x +60 4. Jackson wins $50,000. If he invests x dollars of it in Amazon stock, write an expression for the amount remaining which he invests in an annuity.

Solution 50,000 – x

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

5. If the original price x of a North Face jacket is discounted 12%, write an expression for the selling price of the jacket.

Solution

Original  12% of original x  0.12x 6. If it takes Olivia x hours to complete a job, write an expression that represents the part of the job she completes in one hour.

Solution

1 x Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. To average n scores, __________ the scores and divide by n.

Solution add 8. The formula for the __________ of a rectangle is P = 2l + 2w.

Solution perimeter 9. The simple annual interest earned on an investment is the product of the interest rate and the __________ invested.

Solution amount 10. The number of units manufactured at which the cost on two machines is equal is called the __________.

Solution break point 11. Distance traveled is the product of the __________ and the __________.

Solution rate, time 12. 5% of 30 liters is __________ liters.

Solution 0.05(30) = 1.5

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Practice Solve each problem. 13. Algebra scores A student has completed all assignments in college algebra except for taking the comprehensive final exam. The student’s current test scores are shown in the table.

Test 1

Test 2

Test 3

Test 4

Online Homework

Final Exam

60

78

80

90

88

?

If the student’s online homework average for the semester is weighted as a test grade and the final exam grade is weighted as two test grades, what must the student score on the final exam to have an 80 average in the course?

Solution Let x = the score on the final exam. Since the final is weighted as two test grades, it counts as two test grades. Sum of scores  80 7 60  78  80  90  88  2 x  80 7 2 x  396  80 7 2 x  396  560 2 x  164 x  82 His score on the final exam needs to be 82. 14. Psychology scores Mandy has completed all assignments in psychology except for taking her comprehensive final exam. Her current scores are shown in the table.

Test 1

Test 2

Test 3

Test 4

Test 5

Final Exam

70

88

93

85

88

?

If the final exam score is weighted as a test grade and also replaces her lowest test score, what must she make on the final exam to have a 90 average in the course?

Solution Let x = the score on the final exam. Replace the lowest test score (70) with x also. Sum of scores  90 6 x  88  93  85  88  x  90 6 2 x  354  90 6 2 x  354  540 2 x  186 x  93 His score on the final exam needs to be 93.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

15. Test scores A student scored 5 points higher on his midterm and 13 points higher on his final than he did on his first exam. If his mean (average) score was 90, what was his score on the first exam?

Solution Let x = the score on the first exam. Then x + 5 = the score on the midterm, and x + 13 = the score on the final. Sum of scores  90 3 x  x  5  x  13  90 3 3 x  18  90 3 3 x  18  270 3 x  252 x  84 His score on the first exam was 84. 16. Test scores Rashida took four tests in chemistry class. On each successive test, her score improved by 3 points. If her mean score was 69.5, what did she score on the first test?

Solution Let x = the score on the first exam. Then her score on the following tests were x + 3, x + 6 and x + 9. Sum of scores  69.5 4 x  x  3  x  6  x  9  69.5 4 4 x  18  69.5 4 4 x  18  278 4 x  260 x  65 Her score on the first exam was 65%. 17. Teacher certification On the Illinois certification test for teachers specializing in learning disabilities, a teacher earned the scores shown in the accompanying table. What was the teacher’s score in program development? Human development with special needs

82

Assessment

90

Program development and instruction

?

Professional knowledge and legal issues

78

AVERAGE SCORE

86

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the program development score. Sum of scores  86 4 82  90  x  78  86 4 x  250  86 4 x  250  344 x  94 The program development score was 94.

18. Golfing Par on a golf course is 72. If a golfer shot rounds of 76, 68, and 70 in a tournament, what will she need to shoot on the final round to average par?

Solution Let x = the score on the final round. Sum of scores 4 76  68  70  x 4 x  214 4 x  214

 72  72  72  288

x  74 She needs to shoot 74 on the final round.

19. Replacing locks A locksmith at Pop-A-Lock charges $40 plus $28 for each lock installed. How many locks can be replaced for $236?

Solution Let x = the number of locks replaced. 40  28 

Number of locks

 236

40  28 x  236 28 x  196 x  7 7 locks can be changed for $236.

20. Delivering ads A University of Florida student earns $20 per day delivering advertising brochures door-to-door, plus $1.50 for each person he interviews. How many people did he interview on a day when he earned $56?

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the number of interviews. 20  0.75 

Number of interviews

 56

20  0.75 x  56 0.75 x  36 x  48 He interviewed 48 people.

21. Electronic LED billboard An electronic LED billboard in Times Square is 26 feet taller than it is wide. If its perimeter is 92 feet, find the dimensions of the billboard.

Solution Let x = the width Then x + 26 = the height.

Perimeter  92 2 x  2( x  26)  92 2 x  2 x  52  92 4 x  40 x  10 The dimensions are 10 ft by 36 ft. 22. Hockey rink A National Hockey League rink is 115 feet longer than it is wide. If the perimeter of the rink is 570 feet, find the dimensions of the rink?

Solution Let x = the width Then x + 115 = the height.

Perimeter  570 2 x  2( x  115)  570 2 x  2 x  230  570 4 x  340 x  85 The dimensions are 85 ft by 200 ft. 23. Debit card The width of your Visa bank debit card is 1.586 inches times its height. Find the dimensions of the debit card if its perimeter is 10.990 inches.

Solution Let h = height Let 1.586h = width Perimeter = 10.99

10.99 = 2h + 2(1.586h) 10.99 = 5.172h 2.125 inches = height 1.586(2.125) = 3.370 inches = width

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

24. International airport runway To accommodate the largest jets, an international airport runway is 1500 m more than 50 times its width. If the perimeter of the runway is 11,160 m, determine the length and width of the runway.

Solution Let w = width Let 50w + 1500 = length Perimeter = 11,160

11,160 = 2(50w + 1500) + 2w 11,160 = 100w + 3000 + 2w 11,160 = 102w + 3000 8160 = 102w 80 m = width 50(80) + 1500 = 5500 m = length 25. Width of a picture frame A picture frame with width x feet and height (x + 2) feet was built with 14 feet of framing material. Find x its width.

Solution

Perimeter  14 x  ( x  2)  x  ( x  2)  14 4 x  4  14 4 x  10 x 

5 1 1  2  The width is 2 feet. 2 2 2

26. Fencing a garden A gardener fences in a rectangular region that is formed by adjoining a rectangle of length 24 feet and width x feet to a square measuring x feet on each side. When he does, he needs twice as much fencing as he did just for the square region. How much fencing will he need?

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

Total Fence Length  2  Square Fence Length x  ( x  24)  x  ( x  24)  2  ( x  x  x  x ) 4 x  48  8 x 48  4 x x  12 The total fencing required is 4 x  48  4(12)  48  96 feet. 27. Swimming pool A rectangular swimming pool measures 12 meters by 6 meters and is surrounded by a 116 meter rectangular wooden fence. If the fence forms a border around the pool of uniform width, determine the width of the border.

Solution Let x = the width of the border.

Perimeter of fence  116 2(2 x  6)  2(2 x  12)  116 4 x  12  4 x  24  116 8 x  36  116 8 x  80 x  10 28. Aquarium A rectangular glass aquarium has a length of 15 yards, a width of 10 yards, and is placed inside a rectangular room for viewing at a museum. If the room has a perimeter of 146 yards and forms a walkway of uniform width that surrounds the aquarium, determine the width of the walkway.

Solution Let x = the width of the border.

Perimeter of room  146 2(2 x  10)  2(2 x  15)  146 4 x  20  4 x  30  146 8 x  50  146 8 x  96 x  12 The walkway has a width of 12 yards.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

29. Wading pool dimensions A larger swimming pool is created by adjoining a right triangular-shaped swimming pool, with legs 16 ft and 20 ft, to a rectangular-shaped wading pool that is x ft by 20 ft. Find the dimensions of the wading pool. (Hint: The area of a triangle  21 bh, and the area of a rectangle = lw.)

Solution

Total Area  2  Triangular Area 1 1 (16)(20)  2  (16)(20) 2 2 20 x  160  320

20 x 

20 x  160 x  8 The dimensions are 8 feet by 20 feet. 30. House construction A builder wants to install a triangular window with angles, x°, (x + 30)°, and (x + 30)°. What angles will he have to cut to make the window fit? (Hint: The sum of the angles in a triangle equals 180°.)

Solution

Sum of angles  180 x  x  30  x  30  180 3 x  60  180 3 x  120 x  40 The angles measure 40°, 70° and 70°.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

31. Length of a living room If a carpenter adjoins a rectangular-shaped porch, 12 ft by x ft, to rectangular-shaped living room, 12 ft by (x + 10) ft, the living area will be increased by 50%. Find the length of the living room.

Solution

New Area

Old Area

 0.50 

Old Area

12( x  10)  12 x  12( x  10)  0.50  12( x  10) 12 x  120  12 x  12 x  120  6 x  60 12 x  120  18 x  180 6 x  60 x  10 

The length of the living room is x  10  20 feet.

32. Depth of water in a trough A trapezoidal-shaped trough, with bases 8 inches and 12 inches, has a cross-sectional area of 54 square inches. If the depth of the trough is d inches, find its depth. (Hint: Area of a trapezoid  21 h( b1 + b2 ).)

Solution Area  54 1 d (12  8)  54 2 10d  54 d  5.4  The depth is 5.4 inches

33. Investment Jeffrey invested $16,000 in two accounts paying 4% and 6% annual interest. If the total interest earned in one year was $815, how much did he invest at each rate?

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the amount invested at 4%. Then 16000 – x = the amount invested at 6%.

Interest at 4%  Interest at 6%  Total interest 0.04 x  0.06(16000  x )  815 0.04 x  960  0.06 x  815 0.02 x  145 x  7250 $7250 was invested at 4% and $8750 was invested at 6%. 34. Investment An executive invests $22,000, some at 7% and the rest at 6% annual interest. If he receives an annual return of $1420, how much is invested at each rate?

Solution Let x = the amount invested at 7%. Then 22000 – x = the amount invested at 6%.

Interest at 7%  Interest at 6%  Total interest 0.07 x  0.06(22000  x )  1420 0.07 x  1320  0.06 x  1420 0.01x  100 x  10000 $10,000 was invested at 7% and $12,000 was invested at 6%. 35. Equity funds You invest part of $25,000 in a stock fund that earns 11% interest and the remainder in a stock fund that incurred at 4% loss. If the total interest earned in one year was $2,000, how much was invested in stock?

Solution Let x be the amount that earns 11%. Then 25,000 – x = the amount that incurs 4% loss. Interest gained at 11%  amount lost at 4%  Total interest 0.11x  0.04(25,000  x )  2000 0.11x  1000  0.04 x = 2000 0.15 x  1000  2000 0.15 x  3000 x  20,000 $20,000 was invested at 11% and $5000 was invested at a 4% loss.

36. Mutual funds You invest part of $45,000 in a mutual fund that earns 15% interest and the remainder in a mutual fund that incurred at 2% loss. If the total interest earned in one year was $5050, how much was invested in mutual fund?

Solution Let x be the amount that earns 15%. Then 45,000 – x = the amount that incurs 2% loss.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Interest gained at 15%  amount lost at 2%  Total interest 0.15 x  0.02(45,000  x )  5050 0.15 x  900  0.02 x = 5050 0.17 x  900  5050 0.17 x  5950 x  35,000 $35,000 was invested at 15% and $10,000 was invested at a 2% loss.

37. Financial planning After inheriting some money, a woman wants to invest enough to have an annual income of $5000. If she can invest $20,000 at 9% annual interest, how much more will she have to invest at 7% to achieve her goal? (See the table.)

Type

Rate

Amount

Income

9% investment

0.09

20,000

.09(20,000)

7% investment

0.07

x

.07x

Solution Let x = the amount invested at 7%.

Interest at 7%  Interest at 9%  Total interest 0.07 x  0.09(20000)  5000 0.07 x  1800  5000 0.07 x  3200 x  45714.29 She needs to invest $45,714.29 at 7% to reach her goal. 38. Investment A woman invests $37,000, part at 8% and the rest at 9 21 % annual interest. If the 9 21 % investment provides $452.50 more income than the 8% investment, how much is invested at each rate?

Solution Let x = the amount invested at 8%. Then 37,000 – x = the amount invested at 9 21 %.

Interest at 9 21 %  Interest at 8%  452.50 0.095(37,000  x )  0.08 x  452.50 3515  0.095 x  0.08 x  452.50 3062.50  0.175 x 17500  x $17,500 is invested at 8% and $19,500 is invested at 9 21 %. 39. Investment Equal amounts are invested at 6%, 7%, and 8% annual interest. If the three investments yield a total of $2037 annual interest, find the total investment.

Solution Let x = the amount invested at each rate.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Interest at 6%  Interest at 7%  Interest at 8%  Total Interest 0.06 x  0.07 x  0.08 x  2037 0.21x  2037 x  9700 $9,700 was invested at each rate, for a total investment of $29,100. 40. Investment Ali invested equal amounts of money at 5%, 7%, 9%, and 11% annual interest. If the four investments yielded a total of $1440 annual interest, find the total amount invested.

Solution Let x = the amount invested at each rate.

Interest at 5%  Interest at 7%  Interest at 9%  Interest at 11%  Total Interest 0.05 x  0.07 x  0.09 x  0.11x  1440 0.32 x  1440 x  4500 $4500 was invested at each rate, for a total investment of $18,000. 41. Ticket sales A full-price ticket for a college basketball game costs $2.50, and a student ticket costs $1.75. If 585 tickets were sold, and the total receipts were $1,217.25, how many tickets were student tickets?

Solution Let x = the number of full-price tickets sold. Then 585 – x = the number of student tickets sold. 2.50 

# of full-price

 1.75 

# of

 1217.25

student

2.50 x  1.75(585  x )  1217.25 2.50 x  1023.75  1.75 x  1217.25 0.75 x  193.50 x  258  There were 327 student tickets sold.

42. Ticket sales Of the 800 tickets sold to a movie, 480 were full-price tickets costing $7 each. If the gate receipts were $4960, what did a student ticket cost?

Solution Let x = the cost of a student ticket.

Cost of full-price

# of full-price

Cost of student

# of student

 4960

480(7)  x (800  480)  4960 3360  320 x  4960 320 x  1600 x  5  A student tickets cost $5. 43. Beachfront condo stay The cost per night to stay in a two-bedroom beachfront condo in Orange Beach, AL, is $377. This includes a 16% tax. What is the nightly cost?

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x be the original nightly cost in the condo.

Original nightly cost  tax  total nightly cost x + 0.16 x  377 1.16 x = 377 x  325 The original nightly cost in the condo was $325. 44. Beachfront condo stay The cost per night to stay in a three-bedroom beachfront condo in Myrtle Beach, SC, is $295. This includes a 18% tax. What is the nightly cost?

Solution Let x be the original nightly cost in the condo.

Original nightly cost  tax  total nightly cost x  0.18 x  295 1.18 x  295 x  250 The original nightly cost in the condo was $250. 45. Discount An iPad Air is on sale for $413.08. What was the original price of the iPad if it was discounted 8%?

Solution Let p = the original price.

Original price

 Discount 

New price

p  0.08p  413.08 0.92p  413.08 p  449 The original price was $449. 46. Discount After being discounted 20%, a weather radio sells for $63.96. Find the original price.

Solution Let p = the original price.

Original price

 Discount 

New price

p  0.20 p  63.96 0.80 p  63.96 p  79.95 The original price was $79.95.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

47. Markup A business owner increases the wholesale cost of a kayak by 70% and sells it for $365.50. Find the wholesale cost.

Solution Let w = the wholesale cost. wholesale cost

 Markup 

Selling price

w  0.70w  365.50 1.70w  365.50 w  215 The wholesale cost is $215.

48. Markup A merchant increases the wholesale cost of a surfboard by 30% to determine the selling price. If the surfboard sells for $588.90, find the wholesale cost.

Solution Let w = the wholesale cost. wholesale cost

 Markup 

Selling price

w  0.30w  588.90 1.30w  588.90 w  453 The wholesale cost is $453.

49. Break-point analysis A machine to mill a brass plate has a setup cost of $600 and a unit cost of $3 for each plate manufactured. A bigger machine has a setup cost of $800 but a unit cost of only $2 for each plate manufactured. Find the break point.

Solution Let x = # of plates for equal costs.

cost of 1st machine

cost of 2nd machine

600  3 x  800  2 x x  200 The break point is 200 plates. 50. Break-point analysis A machine to manufacture fasteners has a setup cost of $1200 and a unit cost of $0.005 for each fastener manufactured. A newer machine has a setup cost of $1500 but a unit cost of only $0.0015 for each fastener manufactured. Find the break point.

Solution Let x = # of fasteners for equal costs.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

cost of 1st machine

cost of 2nd machine

1200  0.005 x  1500  0.0015 x 0.0035 x  300 x  85714 The break point is about 85,714 fasteners.

51. Computer sales A computer store has fixed costs of $8925 per month and a unit cost of $850 for every computer it sells. If the store can sell all the computers it can get for $1275 each, how many must be sold for the store to break even? (Hint: The breakeven point occurs when costs equal income.)

Solution Let x = # of computer to break even.

Income  Expenses 1275 x  8925  850 x 425 x  8925 x  21 21 computers need to be sold to break even. 52. Restaurant management A restaurant has fixed costs of $137.50 per day and an average unit cost of $4.75 for each meal served. If a typical meal costs $6, how many customers must eat at the restaurant each day for the owner to break even?

Solution Let x = # of meals to break even.

Income  Expenses 6 x  137.50  4.75 x 1.25 x  137.50 x  110 More than 110 meals need to be sold to make a profit. 53. Roofing houses Kyle estimates that it will take him 7 days to roof his house. A professional roofer estimates that it will take him 4 days to roof the same house. How long will it take if they work together?

Solution Let x = days for both working together.

Man in 1 day

Roofer in 1 day

Total in 1 day

1 1 1   7 4 x 1  1 1 28x     28 x   4 7 x 4 x  7 x  28

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

11x  28 x 

28 6  2 11 11

They can roof the house in 2

6 days. 11

54. Sealing asphalt One crew can seal a parking lot in 8 hours and another in 10 hours. How long will it take to seal the parking lot if the two crews work together?

Solution Let x = hours for both working together.

Crew 1 in 1 hour

Crew 2 in 1 hour

Total in

1 hour

1 1 1   x 8 10 1  1 1  40 x     40 x   10  8 x 5 x  4 x  40 9 x  40 40 4  4 9 9

x 

They can seal the parking lot in 4

4 hours. 9

55. Mowing lawns Julie can mow a lawn with a lawn tractor in 2 hours, and her husband can mow the same lawn with a push mower in 4 hours. How long will it take to mow the lawn if they work together?

Solution Let x = hours for both working together.

Woman in 1 hour

Man in 1 hour

Total in 1 hour

1 1 1   2 4 x 1  1 1 4x     4x   2 4   x 2x  x  4 3x  4 x 

4 1  1 3 3

They can seal the parking lot in 1

1 hours. 3

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

56. Filling swimming pools A garden hose can fill a swimming pool in 3 days, and a larger hose can fill the pool in 2 days. How long will it take to fill the pool if both hoses are used?

Solution Let x = days for both houses to fill the pool. 1st hose

in 1 day

2nd hose in 1 day

Total in

1 day

1 1 1   3 2 x 1  1 1 6x     x   2 3 x 2x  3x  6 5x  6 x 

6 1  1 5 5

The pool can be filled in 1

1 days. 5

57. Filling swimming pools An empty swimming pool can be filled in 10 hours. When full, the pool can be drained in 19 hours. How long will it take to fill the empty pool if the drain is left open?

Solution Let x = hours for pool to fill with drain open.

Pipe in 1 hour

Drain in 1 hour

Total in 1 hour

1 1 1   10 19 x  1  1 1 190 x     190 x   19   10 x 19 x  10 x  190 9 x  190 x 

190 1  21 9 9

The pool can be filled in 21

1 hours. 9

58. Preparing seafood Kadek stuffs shrimp in his job as a seafood chef. He can stuff 1000 shrimp in 6 hours. When his sister helps him, they can stuff 1000 shrimp in 4 hours. If Kadek gets sick, how long will it take his sister to stuff 500 shrimp?

Solution Let x = hours for sister to stuff 1000 shrimp.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Sam in 1 hour

Sister in 1 hour

Total in

1 hour

1 1 1   6 x 4 1  1 1 24 x     24 x   x 6 4 4 x  24  6 x 24  2 x 12  x She can stuff 1,000 shrimp in 12 hours, so she can stuff 500 shrimp in 6 hours.

59. Diluting solutions How much water should be added to 20 ounces of a 15% solution of alcohol to dilute it to a 10% solution?

Solution Let x = the ounces of water added. Oz of alc. at start

Oz of alc. added

Oz of alc. at end

0.15 20  0 x   0.1024  x  3  2  0.1x 1  0.1x 1  x 0.1 10  x 10 oz of water should be added.

60. Increasing concentrations The beaker shown below contains a 2% saltwater solution. a. How much water must be boiled away to increase the concentration of the salt solution from 2% to 3%? b. Where on the beaker would the new water level be?

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the ml of water removed. ml of salt at start

ml of salt removed

ml of salt at end

0.02 300  0 x   0.03300  x  6  9  0.03 x 0.03 x  3 3 0.03 x  100 x 

a. 100 ml of water should be boiled away. b. The new level will be at the 200-ml mark. 61. Winterizing cars A car radiator has a 6-liter capacity. If the liquid in the radiator is 40% antifreeze, how much liquid must be replaced with pure antifreeze tobring the mixture up to a 50% solution?

Solution Let x = the liters of liquid replaced with pure antifreeze. Liters of a.f. at start

Liters of a.f. removed

Liters of a.f. replaced

Liters of a.f. at end

0.40 6  0.40 x  x  0.506 2.4  0.6 x  3 0.06 x  0.6 x  1  1 liter should be replaced with pure antifreeze. 62. Mixing milk If a bottle holding 3 liters of milk contains 3 21 % butterfat, how much skimmed milk must be added to dilute the milk to 2% butterfat?

Solution Let x = the liters of skimmed milk added. Liters of butterfat at start

Liters of butterfat added

Liters of butterfat at end

0.035 3  0 x   0.023  x  0.105  0  0.06  0.02 x 0.045  0.02 x 2.25  x 

2.25 liters of skimmed milk should be added.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

63. Preparing solutions A nurse has 1 liter of a solution that is 20% alcohol. How much pure alcohol must he add to bring the solution up to a 25% concentration?

Solution Let x = the liters of pure alcohol added. Liters of alcohol at start

Liters of

alcohol added

Liters of alcohol at end

0.20  1  x  0.25 1  x  0.20  x  0.25  0.25 x 0.75 x  0.05 x 

0.05 1 1   of a liters of pure alcohol 0.75 15 15 should be added.

64. Diluting solutions If there are 400 cubic centimeters of a chemical in 1 liter of solution, how many cubic centimeters of water must be added to dilute it to a 25% solution? (Hint: 1000 cc = 1 liter.)

Solution Let x = the cubic centimeters of water added.

Cubic centimeters of chemical at start

Cubic centimeters of chemical added

Cubic centimeters of chemical at end

400  0  0.25 1000  x  400  250  0.25 x 150  0.25 x 600  x  600 cubic centimeters of water should be added. 65. Cleaning swimming pools A swimming pool contains 15,000 gallons of water. How many gallons of chlorine must be added to “shock the pool” and bring the water to a 3 % 100

solution?

Solution Let x = the gallons of pure chlorine added.

Gallons of chlorine at start

Gallons of chlorine added

Gallons of chlorine at end

0 15000  x  0.0003 15000  x  x  4.5  0.0003 x 0.9997 x  4.5 x  4.5  About 4.5 gallons of pure chlorine should be added.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

66. Mixing fuels An automobile engine can run on a mixture of gasoline and a substitute fuel. If gas costs $3.50 per gallon and the substitute fuel costs $2 per gallon, what percent of a mixture must be substitute fuel to bring the cost down to $2.75 per gallon?

Solution Let x = the percentage of substitute fuel used. Then 1 – x = the percentage of gasoline used. Percentage of gasoline used

Cost per gallon of gasoline

Percentage of

fuel used

Cost per gallon of fuel

Cost per gallon of mixture

1  x 3.50  x 2  2.75 3.5  3.5 x  2 x  2.75 1.5 x  0.75 x  0.5  50%

The substitute fuel should be 25% of the mixture. 67. Evaporation How many liters of water must evaporate to turn 12 liters of a 24% salt solution into a 36% solution?

Solution Let x = the liters of water evaporated.

Liters of salt at start

Liters of salt evaporated

Liters of salt at end

0.24  12  0 x   0.36 12  x  2.88  0  4.32  0.36 x 0.36 x  1.44 x  4  4 liters of water should be evaporated. 68. Increasing concentrations A beaker contains 320 ml of a 5% saltwater solution. How much water should be boiled away to increase the concentration to 6%?

Solution Let x = the ml of water boiled away.

ml of salt at start

ml of salt removed

ml of salt at end

0.05 320  0 x   0.06320  x  16  0  19.2  0.06 x 0.06 x  3.2 x 

3.2 320 160 1 1    53  53 ml of water should be boiled away. 0.06 6 3 3 3

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

69. Lowering fat How many pounds of extra-lean hamburger that is 7% fat must be mixed with 30 pounds of hamburger that is 15% fat to obtain a mixture that is 10% fat?

Solution Let x = the pounds of extra-lean hamburger used. Pounds of fat in hamburger

Pounds of fat

in lean hamburger

Pounds of fat in mixture

0.15 30  0.07  x   0.1030  x  4.5  0.07 x  3  0.1x 1.5  0.03 x 50  x 50 pounds of the extra-lean hamburger should be used.

70. Dairy foods How many gallons of cream that is 22% butterfat must be mixed with milk that is 2% butterfat to get 20 gallons of milk containing 4% butterfat?

Solution Let x = the gallons of cream used. Then 20 – x = the gallons of milk used

Gallons of fat in cream

Gallons of fat in milk

Gallons of fat in mixture

0.22  x   0.0220  x   0.0420 0.22 x   0.4  0.02 x  0.8 0.2 x  0.4 x  2 2 gallons of cream should be used. 71. Mixing solutions How many gallons of a 5% alcohol solution must be mixed with 90 gallons of 1% solution to obtain a 2% solution?

Solution Let x = the gallons of 5% solution used. Gallons of alc. in 5% solution

Gallons of alc. in 1% solution

Gallons of alc. in 2% solution

0.05  x   0.0190  0.02 x  90 0.05 x  0.9  0.02 x  1.8 0.03 x  0.9 x  30 30 gallons of the 5% solution should be used.

72. Preparing medicines A doctor prescribes an ointment that is 2% hydrocortisone. A pharmacist has 1% and 5% concentrations in stock. How much of each should the pharmacist use to make a 1-ounce tube?

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the ounces of 1% cream used. Then 1 – x = the ounces of 5% cream used. Ounces of h.c. in 1% cream

Ounces of h.c. in 5% cream

Ounces of h.c. in final cream

0.01 x  0.05 1  x   0.02 1 0.01 x  0.05  0.05 x  0.02 0.04 x  0.03 x  0.75

0.75 ounces of the 1% cream should be used with 0.25 ounces of the 5% cream. 73. Feeding cattle A cattleman wants to mix 2400 pounds of cattle feed that is to be 14% protein. Barley (11.7% protein) will make up 25% of the mixture. The remaining 75% will be made up of oats (11.8% protein) and soybean meal (44.5% protein). How many pounds of each will he use?

Solution Since the mixture is to be 25% barley, there will be 0.25(2400) = 600 pounds of barley used. Thus, the other 1800 pounds will be either oats or soybean meal. Let x = the number of pounds of oats used. Then 1800 – x = the number of pounds of meal used.

Pounds of protein from barley

Pounds of protein from oats

Pounds of protein from soybean meal

Total pounds of protein

0.117 600  0.118 x  0.445 1800  x  0.142400 70.2  0.118 x  801  0.445 x  336 871.2  0.327 x  336 0.327 x  535.2 x  1637 The farmer should use 600 pounds of barley, 1,637 pounds of oats and 163 pounds of soybean meal. 74. Feeding cattle If the cattleman in Exercise 73 wants only 20% of the mixture to be barley, how many pounds of each should he use?

Solution Since the mixture is to be 20% barley, there will be 0.20(2400) = 480 pounds of barley used. Thus, the other 1920 pounds will be either oats or soybean meal. Let x = the number of pounds of oats used. Then 1920 – x = the number of pounds of meal used. Pounds of protein from barley

Pounds of protein from oats

Pounds of protein from soybean meal

Total pounds of protein

0.117 480  0.118 x  0.445 1920  x   0.142400

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

56.16  0.118 x  854.4  0.445 x  336 910.56  0.327 x  336 0.327 x  574.56 x  1757 The farmer should use 480 pounds of barley, 1,757 pounds of oats and 163 pounds of soybean meal. 75. Driving rates Javed drove to Daytona Beach, Florida, in 5 hours. When he returned, there was less traffic, and the trip took only 3 hours. If Javed averaged 26 mph faster on the return trip, how fast did he drive each way?

Solution Let r = his first rate. Then r + 26 = his return rate. Distance to city  Return distance 5r  3r  26 5r  3r  78 2r  78 r  39  He drove 39 mph going and 65 mph returning.

76. Distance problem Allison drove home at 60 mph, but her brother Austin, who left at the same time, could drive at only 48 mph. When Allison arrived, Austin still had 60 miles to go. How far did Allison drive?

Solution Let t = the time Allison and Austin travel.

Distance Allison travels

Distance Austin travels

 60

60t  48t  60 12t  60 t  5  They traveled for 5 hours, so Allison traveled 300 miles. 77. Distance problem Two cars leave Hinds Community College traveling in opposite directions. One car travels at 60 mph and the other at 64 mph. In how many hours will they be 310 miles apart?

Solution Let t = the time the cars travel.

Distance 1st car travels

Distance 2nd car travels

 Total Distance

60t  64t  310 124t  310 t  2.5  They will be 310 miles apart after 2.5 hours.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

78. Bank robbery Some bank robbers leave town, speeding at 70 mph. Ten minutes later, the police give chase, traveling at 78 mph. How long, after the robbery, will it take the police to overtake the robbers?

Solution Let t = the hours the robbers travel. Then t  Distance robbers travel

10 1  t   the hours the police travel. 60 6

Distance police

travel

 1 70t  78 t   6   70t  78t  13 8t  13 t 

13 5  1 8 8

The police will catch up 1

5 hours after the robbery. 8

79. Jogging problem Two cross-country runners are 440 yards apart and are running toward each other, one at 8 mph and the other at 10 mph. In how many seconds will they meet?

Solution Let t = the time the runners run. Distance 1st runs

Distance 2nd runs

Distance between them (in miles)

440 1760 1 18t  4 1 1 t  hour  60 minutes  65 minute  50 seconds 72 72

8t  10t 

They will meet after 50 seconds. 80. Driving rates One morning, Justin drove 5 hours before stopping to eat lunch. After lunch, he increased his speed by 10 mph. If he completed a 430-mile trip in 8 hours of driving time, how fast did he drive in the morning?

Solution Let r = the rate before lunch. Then r + 10 = the rate after lunch.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Distance before lunch

Distance after lunch

 Total Distance

5r  3r  10  430 5r  3r  30  430 8r  400 r  50  He drove 50 mph before lunch. 81. Boating problem A Johnson motorboat goes 5 miles upstream in the same time it requires to go 7 miles downstream. If the river flows at 2 mph, find the speed of the boat in still water.

Solution Let r = the speed of the boat in still water. Then the speed of the boat is r + 2 downstream and r – 2 upstream.

Time upstream  Time downstream

{Note: Time  Distance  Rate}

5 7  r  2 r  2 r  2r  2 r 5 2  r  2r  2 r 7 2 5r  2  7r  2 5r  10  7r  14 24  2r 12  r  The speed of the boat is 12 mph. 82. Wind velocity A plane can fly 340 mph in still air. If it can fly 200 miles downwind in the same amount of time it can fly 140 miles upwind, find the velocity of the wind.

Solution Let w = the speed of the wind. Then the speed of the plane is 340 + w downwind and 340 – w upwind.

Time upwind  Time downwind 140 200  340  w 340  w 140 340  w 340  w  340  w  340  w 340  w  340200 w 140340  w   200340  w  47,600  140w  68,000  200w 340w  20,400 w  60  The speed of the wind is 60 mph.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Use a calculator to help solve each problem. 83. Machine tool design 712.51 cubic millimeters of material was removed by drilling the blind hole as shown in the illustration. Find the depth of the hole. (Hint: The volume of a cylinder is given by V = πr2h.)

Solution

V  r 2 h 712.51  4.5 d 2

712.51 4.5

2

 d

11.2  d The hole is about 11.2 millimeters deep. 84. Architecture The Norman window with dimensions as shown is a rectangle topped by a semicircle. If the area of the window is 68.2 square feet, find its height h.

Solution Since the diameter of the semicircle is 6 feet, the radius of the semicircle is 3 feet. Area of rectangle

Area of semicircle

6h  3  21 3

2

Total area

 68.2

6h  18  4.5  68.2 6h  68.2  18  4.5 6h  72.0628 h  12 The height of the window is about 12 feet.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Discovery and Writing 85. Consider the strategy you use to solve investment and uniform motion problems. Describe any similarities you observe in these problem types.

Solution Answers may vary. 86. Which type of application was hardest for you to solve? Why? What strategy or approach works best for you when approaching solving this problem?

Solution Answers may vary. 87. Explain why the solution to an application problem should be checked in the original wording of the problem and not in the equation obtained from the words.

Solution Answers may vary.

EXERCISES 1.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Simplify the radical.

48

Solution

48 

16  3  4 3

2. Subtract and simplify.

8  2 50

Solution 8  2 50 

4  2  2 50

 2 2  2 25  2  2 2  2  5 2  2 2  10 2  8 2 3. Multiply. (3 + 2x)(2 – 7x)

Solution

(3  2x )(2  7 x)  6  21x  4x  14x2  14x2  17 x  6

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities



4. Multiply. 5  2 7 5  2 7

Solution

5  2 75  2 7  25  10 7  10 7  4  7  25  28  3

5. Rationalize the denominator and simplify.

4 6

Solution

4   6 

6 4 6 2 6    6 3 6 

6. Rationalize the denominator and simplify.

3  2 2 4 

2

Solution

3  2 24   4  2  4 

2 12  3 2  4 2  2  2 16  11 2     14 2 16  4 2  4 2  2

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7.

3,

9 and

12 are examples of __________ numbers.

Solution imaginary 8. In the complex number a  bi, a is the __________ part, and b is the __________ part.

Solution real, imaginary 9. If a  0 and b  0 in the complex number a  bi, the number is an __________ number.

Solution imaginary 10. If b  0 in the complex number a  bi, the number is a __________ number.

Solution real

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

11. The complex conjugate of 2  5i is __________.

Solution 2  5i 12. By definition, a  bi  __________.

Solution a 2  b2 13. The absolute value of a complex number is a __________ number.

Solution real 14. The product of two complex conjugates is a __________ number.

Solution real Practice Simplify the imaginary numbers. 15.

144 Solution

144 

1 144  12i

16.  225

Solution

 225   1 225  15i 17.  128

Solution

 128   1 64 2  8i 2 18.

108 Solution

108 

1 36 3  6i 3

19. 2 24

Solution

2 24  2 1 24  2i  2 6  4i 6

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240


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

20. 7 48

Solution

7 48  7 1 48  7i  4 3  28i 3 21.

25 4 Solution

1 25 1 4  10i 2  10  1  10

25 4  22. 3 8

1

Solution

3 8 1  3 1 8 1  i 2  3  2 2  6 2 23.

16 9 Solution

16 9 24.

1 16 1 9

4i 4  3i 3

6i 3  8i 4

6 1 64 Solution

6 1 64 25.

6i 1 64

50 9

Solution

50  9

26.  

1 

50 9

 i 

5 2 5 2  i 3 3

72 25

Solution

 

72   1  25

72 25

 i 

6 2 6 2   i 5 5

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241


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

27. 7 

3 8

Solution

7 

28. 5 

3  7 1  8

3 8

3

 7i 

8

2 2

 7i 

6 16

 

7 6 i 4

5 27

Solution

5 

5  5 1  27

5 27

 5i 

5 27

3 3

 5i 

15 81

5 15 i 9

Determine the real and imaginary part of each complex number. 29. 6 – 11i

Solution real part is 6; imaginary part is −11 30.

3  6i 5

Solution real part is 5 

31.

3 ; imaginary part is 6 5

2 i 3

Solution real part is

5; imaginary part is

2 3

32. 9  πi

Solution real part is 9; imaginary part is π 33. 

4 5

Solution real part is

4 ; imaginary part is 0 5

34. 6i

Solution real part is 0; imaginary part is 6

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242


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Find the values of x and y. 35. x  ( x  y )i  3  8i

Solution Equate real parts:

x  3 x  y  8

Equate imaginary parts:

3  y  8 y  5

36. x  5i  y  yi

Solution Equate imaginary parts:

5  y 5  y 5  y

Equate real parts:

x  5 37. 3x  2 yi  2  ( x  y )i

Solution Equate real parts: 3x  2

x 

Equate imaginary parts:

2 y  x  y 3 y  x

2 3

y   31 x y   31  23 y   92

2  x  y i  2  i 38.  x  3i  2  3i Solution Equate real parts: x  2

Equate imaginary parts:

x  y  1 2  y  1 y  3

Perform all operations. Give all answers in a + bi form. 39.

4 

100

Solution

4 

100 

1 4 

1 100  2i  10i  0  12i

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243


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

40.

9 

121

Solution

9 

121 

41. 2 72 

1 9 

1 121  3i  11i  0  8i

18

Solution

2 72  42.

18  2 1 72 

1 18  2i  6 2  3i 2  0  9i 2

27  3 75 Solution

27  3 75  43.

1 27  3 1 75  3i 3  3i  5 3  0  18i 3

6  4 2

Solution 6  4 6  1 4 6  2i    3  i 2 2 2 44.

8 

100 6

Solution

8 

100 6

45.

8 

1 100 8  10i 4 5   i 6 6 3 3

12  18 3 Solution

12  18 12  1 18 12  3i    4  i 2 3 3 3 46.

5 

200 20

Solution

5 

200 5   20

1 200 5  10i 2 1    20 20 4

2 i 2

47. 2  7i   3  i 

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244


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

2  7i   3  i   2  7i  3  i  5  6i

48.  7  2i   2  8i 

Solution

7  2i   2  8i   7  2i  2  8i  5  6i

49. 5  6i   7  4i 

Solution

5  6i   7  4i   5  6i  7  4i  2  10i

50.  11  2i    13  5i 

Solution

11  2i   13  5i   11  2i  13  5i  2  7i

51.

14i  2  2 

16

Solution

14i  2  2 

52. 5 

16   14i  2  2  4i   14i  2  2  4i  4  10i

64  23i  32

Solution

5  64  23i  32  5  8i  23i  32  5  8i  23i  32  37  15i 

53. 3 

 

4  2 

9

Solution

3  4  2  9  3  2i  2  3i  3  2i  2  3i  1  i 

54. 7 

 

25  8 

1

Solution

7  25  8  1  7  5i  8  i  7  5i  8  i  1  4i

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245


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

55. 4  7i   8  2i   5  4i 

Solution

4  7i   8  2i   5  4i   4  7i  8  2i  5  4i  7  i

56. 5  7i   4  2i   8  i 

Solution

5  7i   4  2i   8  i   5  7i  4  2i  8  i  9  4i

57. 3 

Solution

3 

 

36  5 

 

144

 

36  5 

 

144

  3  4i   4  6i   5  12i 

16  4 

16  4 

 3  4i  4  6i  5  12i  4  2i

58. 1 

Solution

1 

 

81  8 

 

121

 

81  8 

 

121

1  2 

1  2 

  1  i   2  9i   8  11i   1  i  2  9i  8  11i  5  19i

59. 53  5i 

Solution

53  5i   15  25i

60. 52  i 

Solution

52  i   10  5i

61. 7i 4  8i 

Solution

7i 4  8i   28i  56i 2  28i  56 1  28i  56  56  28i

62. 2i 3  7i 

Solution

2i 3  7i   6i  14i 2  6i  14 1  6i  14  14  6i

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246


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

63. 2  3i 3  5i 

Solution

2  3i 3  5i   6  19i  151  6  19i  15  9  19i

64. 5  7i 2  i 

Solution

5  7i 2  i   10  9i  7i 2  10  9i  71  10  9i  7  17  9i

65. 2  3i

2

Solution

2  3i 

2

66. 3  4i

 2  3i 2  3i   4  12i  9i 2  4  12i  9 1  4  12i  9  5  12i

2

Solution

3  4i 

67. 11 

2



36



36

  11  5i 2  6i   22  56i  30i  22  56i  301

25 2 

Solution

11 

 3  4i 3  4i   9  24i  16i 2  9  24i  16 1  9  24i  16  7  24i

25 2 

2

 22  56i  30  52  56i

68. 6 



49 6 

49

Solution

6  496  49  6  7i6  7i  36  49i  36  491  36  49  85  0i 2

69.

 16  32  9 Solution

 16  32 

9

  4i  32  3i   6  17i  12i  6  17i  121 2

 6  17i  12  6  17i

70. 12 



4 7 

25

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247


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

12  47 

25

  12  2i7  5i  84  74i  10i  84  74i  101 2

 84  74i  10  74  74i 71.

1 i Solution 1 1 i i i      0  i 2 1 i i i i

72.

3 i Solution 3 3 i 3i 3i      0  3i 2 i i i 1 i

73.

4 3i

Solution 4 4 4 i 4i 4i      0  i 2 3i 3i 3 i 3 3i 74.

10 7i Solution 10 10 i 10i 10i 10 i      0  2 7i 7i 7 i 7 7i

75.

1 2  i Solution

1  2  i 76.

12  i 

2  i 2  i 

2  i 22  i 2

2  i

4   1

2  i 2 1   i 5 5 5

2 3  i

Solution

2  3  i

23  i 

3  i 3  i 

23  i  2

3

 i

2

23  i  9   1

23  i  10

 3  i  5

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 

3 1  i 5 5

248


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

77.

2i 7  i Solution

2i  7  i 78.

2i 7  i 

7  i 7  i 

3i 2  5i 

3i  2  5i

2  i  3  i

 i

2

14i  2 1

49   1

14i  2 7i  1 1 7    i 50 25 25 25

2  5i 2  5i 

6i  15i 2 22  5i 

2

6i  15 4  25i

2

15 6 6i  15    i 29 29 29

2  i 3  i   6  5i  i 2  5  5i  5  5 i  1  1 i 10 10 10 2 2 9  i2 3  i 3  i 

3  i 1  i Solution

3  i  1  i 81.

7

2

2  i 3  i

Solution

80.

14i  2i 2

3i 2  5i

Solution

79.

3  i 1  i   3  4i  i 2  2  4i  2  4 i  1  2i 2 2 2 1  i2 1  i 1  i 

4  5i 2  3i

Solution

4  5i 2  3i   8  22i  15i 2  7  22i   7  22 i 4  5i  2  3i 13 13 13 4  9i 2 2  3i 2  3i 

82.

34  2i 2  4i Solution

34  2i  2  4i

34  2i 2  4i   68  140i  8i 2  60  140i  60  140 i  3  7i 20 20 20 4  16i 2 2  4i 2  4i 

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249


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

83.

5 

16

8 

4

Solution

5 

16

8 

4

5  4i  8  2i

5  4i 8  2i   40  22i  8i 2  48  22i   48  22 i 68 68 68 64  4i 2 8  2i 8  2i   

84.

3 

9

2 

1

Solution

85.

3 

9

2 

1

3  3i  2  i

3  3i 2  i   6  3i  3i 2  9  3i  9  3 i 5 5 5 4  i2 2  i 2  i 

2  i 3 3  i Solution

2  i 3  3  i

2  i 33  i   6  2i  3i 3  i 3  i 3  i 

9  i

2

3

2

 

86.

12 11  i 17 34

6 

3  3 3  2i 10

6  3 3 3  2  i 10 10

3  i 4  i 2 Solution 3  i 4  i

2

3  i  4  i 2 

4  i 24  i 2 

12  3i

2  4i  i 2 16  2i 2

2

12 

2  3 2  4i 18

12  2 4  3 2 i  18 18

Simplify each expression. 87. i 9

Solution

i 9  i 8i 

i  i  1 i  i 4

2

2

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250


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

88. i 26

Solution

i  i  1 i  11  1

i 26  i 24 i 2 

4

6

2

6 2

89. i 38

Solution

i  i  1 i  i  1

i 38  i 36 i 2 

4

9

2

9 2

2

90. i 99

Solution

i 99  i 96 i 3 

 i  i  1 i  i  i 4

24

3

24 3

3

91. i 87

Solution

i 87  i 84 i 3 

 i  i  1 i  i  i 4

21

3

21 3

3

92. i 44

Solution

 

i 44  i 4

11

 111  1

93. i 100

Solution

 

i 100  i 4

25

 125  1

94. i 201

Solution

  i1 ii

i 201  i 200 i  i 4

50

50

95. i 6

Solution

i 6 

1 i6

1  i2 i6  i2

i2 i8

i2  i 2  1 1

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251


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

96. i 0

Solution i0  1 97. i 10

Solution

1

i 10 

i 10

1  i2

i2

i 10  i 2

i 12

i2  i 2  1 1

98. i 31

Solution 1

i 31 

99.

i

31

1  i

i

31

 i

i i

32

i  i 1

1 i3

Solution 1 i 100.

3

1  i i

3

 i

i i

i  i  i 1

4

3 i5

Solution

3 i5 101.

3  i3 i5  i3

3i 3

i8

3i 3  3i 3  3i 1

4 i 10

Solution

4 i 10 102.

4  i 2 i 10  i 2

4i 2 i 12

4i 2  4 1  4 1

10 i 24

Solution 10 10   10 24 1 i

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252


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Write without absolute value symbols. 103. 3  4i

Solution

3  4i 

32  44 

9  16 

52  122 

25  144 

25  5

104. 5  12i

Solution

5  12i 

169  13

105. 2  3i

Solution

22  32 

4  9 

52   1

24  1 

2  3i 

13

106. 5  i

Solution

5  i  107. 7 

2

26

49

Solution

7  108. 2 

7  72  2

49  7  7i 

49  49 

98  7 2

4  16 

20  2 5

16

Solution

2 

109.

2  4 2

16  2  4i 

2

1 1  i 2 2 Solution

1 1  i  2 2

2

 1    2

2

 1     2

1 1   4 4

1  2

2 2

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253


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

110.

1 1  i 2 4 Solution

1 1  i  2 4 111.

2

 1    2

2

 1     4

1 1   4 16

5  16

5 4

6i

Solution

02   6

6i  0  6i 

2

0  36 

36  6

112. 5i

Solution

5i  0  5i  113.

0  25 

25  5

2 1  i Solution

2  1  i

114.

02  52 

2 1  i 

1  i 1  i 

2 1  i 

1  i

2

2 1  i 

2

 1  i 

12   1

2

2

 3     10 

2

3 3  i Solution

3  3  i

33  i 

3  i 3  i 

33  i  9  i2

33  i  10

9  3i   10

9    10 

2

81 9  100 100

90  100

9 3 10  10 10

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254

115.

3i 2  i


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3i  2  i

3i 2  i 

2  i 2  i 

3i 2  i  4  i2

3i 2  i  5

6i  3i 2 5

 

3 6  i 5 5

 3    5

9 36  25 25

45  25

2

116.

2

 6     5

45 3 5  5 5

5i i  2 Solution

5i 5i   i  2 2  i

5i  2  i 

5i  2  i 

2  i 2  i  4  i

2

5i  2  i  5

10i  5i 2 5

 1  2i

117.

1   2

1  4 

5

2

2

2

i  2 i  2 Solution

i  2  i  2

118.

i  2i  2  i 2  4i  4  3  4i  3  4 i  5 5 5 i2  4 i  2i  2

 3   5

4    5

9 16  25 25

25  25

1  1

2  i 2  i

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255


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

2  i  2  i

2  i 2  i   4  4i  i 2  3  4i  3  4 i  5 5 5 42  i 2 2  i 2  i 

 3   5

2

4    5

9 16  25 25

25  25

2

1  1

Factor each expression over the set of complex numbers. 119. x 2  4

Solution

    x  2i x  2i

x2  4  x2   4  x2  2i 2 120. 16a2  9

Solution

16a2  9  4a   9  4a  3i  2

2

2

 4a  3i 4a  3i 

121. 25p2  36q2

Solution 25 p2  36q2  5 p

2

 36q2

  5p  6qi   5p  6qi 5p  6qi  2

2

122. 100r 2  49s2

Solution 100r 2  49s2   10r 

2

 49s2

  10r   7si   10r  7si 10r  7si  2

2

123. 2 y 2  8z 2

Solution

2 y 2  8z2  2 y 2  4z2

  2 y  4z   2 y  2zi   2 y  2zi y  2zi 2

2

2

2

124. 12b2  75c2

Solution

12b2  75c2  3 4b2  25c2

  32b  25c   32b  5ci   32b  5ci2b  5ci 2

2

2

2

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256


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

125. 50m2  2n2

Solution

  25m  n   25m  ni   25m  ni5m  ni 

  44a   b   44a   bi    44a  bi 4a  bi 

50m2  2n2  2 25m2  n2

2

2

2

2

126. 64a4  4b2

Solution

64a4  4b2  4 16a4  b2

2

2

2

2

2

2

2

2

Fix It In exercises 127 and 128, identify the step the first error is made and fix it.

127. Multiply and write in standard form: 6 



9 5 

49

Solution Step 4 was incorrect. Step 1: 6  3i 5  7i  Step 2: 65  67 i    3i 5   3i 7i  Step 3: 30  42i  15i  21i 2 Step 4: 30  27i  21 Step 5: 9  27i 128. Divide and write in standard form:

11  10i 1  4i

Solution Step 5 was incorrect. Step 1:

Step 2:

11  10i 1  4i  1  4i 1  4i 11 1  114i    10i  1   10i 4i  1 1  14i    4i  1   4i 4i 

Step 3:

11  44i  10i  40 1  4i  4i  16

Step 4:

51  34i 17

Step 5: 3  2i

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257


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Applications In electronics, the formula V = IR is called Ohm’s Law. It gives the relationship in a circuit between the voltage V (in volts), the current I (in amperes), and the resistance R (in ohms). 129. Electronics Find V when I = 3 – 2i amperes and R = 3 ohms.

Solution

V  IR  3  2i 3  6i   9  18i  6i  12i 2  9  12i  12  21  12i

130. Electronics Find R when I = 2 – 3i amperes and V = 21 + i volts.

Solution

R 

V 21  i   I 2  3i

21  i 2  3i   42  63i  2i  3i 2  39  65i  3  5i 13 4  9i 2 2  3i 2  3i 

131. Electronics The impedance Z in an AC (alternating current) circuit is a measure of how much the circuit impedes (hinders) the flow of current through it. The impedance is related to the voltage V and the current I by the following formula. V = IZ If a circuit has a current of (0.5 + 2.0i) amps and an impedance of (0.4 – 3.0i) ohms, find the voltage.

Solution

V  IZ  0.5  2.0i 0.4  3.0i   0.2  1.5i  0.8i  6i 2  0.2  0.7i  6  6.2  0.7i

132. Fractals Complex numbers are fundamental in the creation of the intricate geometric shape shown below, called a fractal. The process of creating this image is based on the following sequence of steps, which begins by picking any complex number, which we will call z. 1.

Square z, and then add that result to z.

2. Square the result from step 1, and then add it to z. 3. Square the result from step 2, and then add it to z. If we begin with the complex number i, what is the result after performing steps 1, 2, and 3?

Solution 1.

i 2  i  1  i

2.

1  i   i  1  i 1  i   i  1  i  i  i2  i  i

3.

i   i  i 2  i  1  i

2

2

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258


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Discovery and Writing 133. Show that the addition of two complex numbers is commutative by adding the complex numbers a + bi and c + di in both orders and observing that the sums are equal.

Solution

a  bi   c  di   a  bi  c  di

c  di   a  bi   c  di  a  bi

 a  c  bi  di

 c  a  di  bi

 a  c  b  d  i

 c  a  d  b i

 a  c  b  d  i

134. Show that the multiplication of two complex numbers is commutative by multiplying the complex numbers a + bi and c + di in both orders and observing that the products are equal.

Solution

a  bi c  di   ac  adi  bci  bdi 2 c  di a  bi   ac  bci  adi  bdi 2  ac  ad  bc i  bd  ac  bc  ad  i  bd  ac  bd   ad  bc i  ac  bd   ad  bc i

135. Show that the addition of complex numbers is associative.

Solution

a  bi   c  di   e  fi   a  bi  c  di  e  fi    a  c  e  bi  di  fi

 a  c  e  b  d  f  i

a  bi   c  di   e  fi   a  bi  c  di  e  fi  a  c  e  bi  di  fi

 a  c  e  b  d  f  i 136. Explain how to determine whether two complex numbers are equal.

Solution Answers will vary. 137. Define the complex conjugate of a complex number.

Solution Answers will vary. 138. Explain how to divide two complex numbers.

Solution Answers will vary.

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259


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true.

300  10 3

139.

Solution

300 

False. 140.

3

1 100 3  10i 3

125  5i

Solution False. 141.

 i

3

125  5

  i

Solution True.

  i i     i i i i i2

142. 2  3i

3

 8  27i

Solution

False. 2  3i

3

 2  3i 2  3i 2  3i    5  12i 2  3i   46  9i

143. 4444i 4444  4444

Solution

 

True. 4444i 4444  4444 i 4 144.

10 7 

1111

 4444  11111  4444

70

Solution False.

10 7 

1 10 1 7  i 10  i 7  i 2 70   70

145. 5  6i 5  6i 2  i 2  i  is a real number.

Solution

True. 5  6i 5  6i  is a real number and 2  i 2  i  is a real number, so their product is too. 146. 81x 2  100 y 2 can be factored.

Solution True. 81x 2  100 y 2  9 x 

2

 10 y 2

  9x   10 yi   9x  10 yi 8x  10 yi  2

2

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260


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

EXERCISES 1.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Factor the trinomial. 24x2  2x  15

Solution

24x 2  2x  15  4x  36x  5

2. Factor the perfect square trinomial. x 2  18x  81

Solution

x2  18x  81   x  9 x  9   x  9

2

3. Solve 5x  7  2 2 for x.

Solution

5 x  7  2 2 5x  7  2 2 x 

7  2 2 5

4. If you take half of

4 and square it, what do you get? 5

Solution  1 4    2 5

2

2    5

2

4 25

5. Simplify. a.

b.

4 

( 4)2  4(1)(5) 2(1)

4 

( 4)2  4(1)( 5) 2(1)

Solution a.

b.

4 

( 4)2  4(1)( 5) 4   2(1)

16  20 4  6   5 2 2

4 

( 4)2  4(1)( 5) 4   2(1)

16  20 4  6   1 2 2

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261


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

6. Simplify.

18  5 27 9

Solution

18  5 27 18  5  3 3 18  15 3 6  5 3    9 9 9 9 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A quadratic equation is an equation that can be written in the form __________, where a  0.

Solution

ax2  bx  c  0 8. If a and b are real numbers and __________, then a  0 or b  0.

Solution ab  0 9. The equation x2  c has two roots. They are x  __________ and x = __________.

Solution

c,  c 10. The Quadratic Formula is __________ a  0.

Solution

x 

b 

b2  4ac 2a

11. If a, b, and c are real numbers and if b2  4ac  0, the two roots of the quadratic equation are repeated __________.

Solution rational numbers 12. If a, b, and c are real numbers and b2 – 4ac < 0, the two roots of the quadratic equation are __________.

Solution nonreal complex numbers

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262


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Practice Solve each equation by factoring. 13. x2  4x  0

Solution x2  4x  0

x  x  4  0 x  0 or

x  4  0

x  0

x  4

14. x2  15x Solution

x 2  15 x x 2  15 x  0

x  x  15  0 x  0 or

x  15  0

x  0

x  15

15. 3x2  21x  0 Solution

3 x 2  21x  0 3 x  x  7  0

3 x  0 or

x  7  0

x  0

x  7

16. 30x  6x2 Solution

30 x  6 x 2 0  6 x 2  30 x 0  6 x  x  5

6 x  0 or x  5  0 x  0

x  5

17. x2  144  0

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263


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

x 2  144  0

 x  12 x  12  0 x  12  0

x  12  0

or

x  12

x  12

18. 4x2  49  0 Solution

4 x 2  49  0

2x  72x  7  0 2x  7  0

or 2 x  7  0

2 x  7

2x  7

7 2

x 

x 

7 2

19. x2  x  6  0 Solution

x2  x  6  0

 x  2 x  3  0 x  2  0

or

x  3  0

x  2

x  3

20. x 2  8x  15  0 Solution

x 2  8 x  15  0

 x  5 x  3  0 x  5  0

or

x  3  0

x  5

x  3

21. 2x 2  x  10  0 Solution

2 x 2  x  10  0

2x  5 x  2  0 2x  5  0

or

x  2  0

2 x  5

x  2

 52

x  2

x 

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264


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

22. 3x2  4x  4  0 Solution

3x 2  4 x  4  0

3x  2 x  2  0 3 x  2  0 or

x  2  0

3x  2

x  2

2 3

x  2

x 

23. 5x2  13x  6  0 Solution

5 x 2  13 x  6  0

5x  3 x  2  0 5 x  3  0 or

x  2  0

5x  3

x  2

3 5

x  2

x 

24. 2x2  5x  12  0 Solution

2 x 2  5 x  12  0

2x  3 x  4  0

2 x  3  0 or x  4  0 2x  3

x  4

3 2

x  4

x 

25. 15x2  16x  15 Solution

15 x 2  16 x  15 15 x 2  16 x  15  0

3x  55x  3  0 3x  5  0

or 5 x  3  0

3 x  5

5x  3

x 

 53

x 

3 5

26. 6x2  25x  25

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265


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

6 x 2  25x  25 6 x 2  25x  25  0

3x  52x  5  0 3x  5  0 or 2x  5  0 3x  5 x 

2x  5

5 3

x 

5 2

27. 12x 2  9  24x Solution

12x 2  9  24 x 12x 2  24 x  9  0

3 4 x 2  8x  3  0

2x  12x  3  0 2x  1  0 or 2x  3  0 2x  1 x 

2x  3

1 2

x 

3 2

28. 24x2  6  24x Solution

24 x 2  6  24 x 24 x 2  24 x  6  0

6 4x2  4x  1  0

2x  12x  1  0 2x  1  0

or 2 x  1  0

2x  1

2x  1

x 

1 2

x 

1 2

Use the Square Root Property to solve each equation.

29. x2  9 Solution

x2  9 x  x  3

9 or

x   9 x  3

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266


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

1 64

30. x 2 

Solution

x 2  64 x  x 

64 or x   64 1 8

x   81

31. x2  169 Solution

x 2  169 x 

169 or

x  13i

x   169 x  13i

32. x 2  81 Solution

x 2  81 x 

81 or x   81

x  9i

x  9i

33. y 2  50  0 Solution

y 2  50  0 y 2  50 y 

50 or

y  5 2

y   50 y  5 2

34. x2  75  0 Solution

x 2  75  0 x 2  75 x 

75 or x   75

x  5 3

x  5 3

35. y 2  54  0

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267


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

y 2  54  0 y 2  54 y 

54

or

y   54

y  i 9 6

y  i 9 6

y  3i 6

y  3i 6

36. x2  125  0 Solution

x 2  125  0 x 2  125 x 

125

or x   125

x  i 25 5

x  i 25 5

x  5i 5

x  5i 5

37. 2x2  40 Solution

2x 2  40 x 2  20 x 

20 or x   20

x  2 5

x  2 5

38. 5x 2  400 Solution

5x 2  400 x 2  80 x 

80 or x   80

x  4 5

x  4 5

39. 2x2  90 Solution 2 x 2  90 x 2  45

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268


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 

45

or x   45

x  i 9 5

x  i 9 5

x  3i 5

x  3i 5

40. 5x2  200 Solution

5x 2  200 x 2  40 x 

40

x  i 4 x  2i

or x   40

x  i 4

10

x  2i

10

10

10

41. 4x 2  7 Solution

4x2  7 x2  x 

7 4

x 

7 2

7 4

or x  

7 4

x   27

42. 16x2  11 Solution

16 x 2  11 x2  x 

11 16

x 

11 4

or

11 16

x  

11 16

x   411

43. 9x2  7 Solution

9 x 2  7 x2   x  x  i x 

7 9

7 9 7 i 3

or

7 9

x    x  i x  

7 9

7 9 7 i 3

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269


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

44. 25x2  11 Solution

25 x 2  11 x2   x 

or

11

x  i x 

11 25

11 25

x   

11

x  i

25 11 i 5

x  

11 25

25 11 i 5

45. 2x2  13  0 Solution

2x 2  13  0 2x 2  13 x2  x 

13 2

x 

13

x 

26 2

2

or x  

13 2

x  

13

2 2

13 2

2

2 2

x   226

46. 3x2  11 Solution

3x 2  11 x2  x 

11 3

x 

11 3

x 

33 3

3 3

11 3

or x  

11 3

x  

11 3

x  

33 3

3 3

47. 2x2  15  0 Solution 2 x 2  15  0 2 x 2  15 x2  

15 2

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270


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 

15 2

or

15

x  i

2

2

2

x  

15 2

15

x  i

30 i 2

x 

x   

2

2

2

30 i 2

48. 5x2  11 Solution

5 x 2  11 x2   x 

11 5

11

x  i

5

x   

or 5

x  i

5

x 

55 i 5

 8  0

11 5

x  

11 2

11 5

5 5

55 i 5

2

49. x  1

Solution

 x  1  8  0 2  x  1  8 2

x  1 

8

x  1 

4

or

x  1   8 x  1   4

2

x  1  2 2

2

x  1  2 2

  98  0

50. y  2

2

Solution

 y  2  98  0 2  y  2  98 2

y  2 

98

y  2 

49 2

y  2  7 2 51.

or

y  2   98 y  2   49 2 y  2  7 2

 x  1  12  0 2

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271


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

 x  1  12  0 2  x  1  12 2

x  1 

12

or

x  1   12

x  1  i 4 3

x  1  i 4 3

x  1  2i 3

x  1  2i 3

  120  0

52. y  2

2

Solution

 y  2  120  0 2  y  2  120 2

y  2 

200

or

y  2   200

y  2  i 4 30

y  2  i 4 30

y  2  2i 30

y  2  2i 30

53. 2x  1

2

 27

Solution

2 x  1 2x  1 

2

 27

27

or 2 x  1   27

2x  1  3 3

2 x  1  3 3

2 x  1  3 3 x 

2 x  1  3 3

1  3 3 2

x 

1  3 3 2

  48  0

54. 5 y  2

2

Solution

5 y  2  48  0 2 5 y  2  48 2

5y  2 

48

or 5 y  2   48

5y  2  4 3

5 y  2  4 3

5 y  2  4 3 y 

2  4 3 5

5 y  2  4 3 y 

2  4 3 5

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272


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

55. 5x  1

2

 8

Solution

5x  1 5x  1 

8

 8

or 5 x  1   8

5 x  1  i x  

2

4

5 x  1  i

2

1 2i 2  5 5

x  

4

2

1 2i 2  5 5

  48  0

56. 7 y  2

2

Solution

7 y  2  48  0 2 7 y  2  48 2

7y  2 

48

or 7 y  2   48

7 y  2  i y  

2

57. 5 10x  1

7 y  2  i

16 3

2 4i 3  7 7

y  

16 3

2 4i 3  7 7

 25

Solution

5 10 x  1

 25

10x  1

 5

2 2

10 x  1 

or 10 x  1   5

5

10 x  1 

10 x  1 

5

x 

1  5 10

 14

58. 7 3x  4

2

x 

5

1  5 10

Solution

7 3 x  4

 14

3x  4

 2

2 2

3x  4 

or 3 x  4   2

2

3 x  4 

2

4  3

2

x 

3 x  4 

2

4  3

2

x 

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273


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

59. 3 8x  11

2

 9

Solution

8 x  11 

38 x  11

2

 9

8x  11

2

 3

3

or 8 x  11   3

8 x  11  i 3 11  i 3 8 i 3 11 x   8 8 x 

8 x  11  i 3 11  i 3 8 i 3 11 x   8 8 x 

  3  19

60. 4 6x  5

2

Solution

46 x  5

2

 3  19

46 x  5

 16

6x  5

 4

2 2

6x  5 

4

or

6 x  5  2i

6 x  5   4 6 x  5  2i

6 x  5  2i

6 x  5  2i

5  2i 6 5 1  i x  6 3 x 

5  2i 6 5 1  i x  6 3 x 

Complete the square to make each a perfect-square trinomial. 61. x2  6x

Solution

x 2  6 x   21 6

2

 x 2  6 x  32  x 2  6x  9

62. x2  8x

Solution

x 2  8x   21 8

2

 x 2  8x  42  x 2  8x  16

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274


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

63. x 2  4x

Solution

x 2  4 x   21  4

 x 2  4 x   2

2

2

 x2  4x  4 64. x2  12x

Solution

x 2  12 x   21  12

2

 x 2  12 x   6

2

 x 2  12 x  36 65. a2  5a

Solution 2

a

2

2

5  a  5a     2 25  a2  5a  4

2

9  t  9t    2 81  t 2  9t  4

 5a   21 5

2

66. t 2  9t

Solution

t

2

 9t   21 9

2

2

67. r 2  11r

Solution r 2  11r   21  11

2

 r 2  11r 

   r  11r 

121 4

 s2  7s 

   s  7s 

49 4

11 2

2

2

68. s2  7s

Solution s2  7 s   21  7

69. y 2 

2

7 2

2

2

3 y 4

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275


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

y2 

70. p2 

 1  3  3 y     4  2  4 

2

 y2 

3 3 y    4 8

 y2 

3 9 y  4 64

2

3 p 2

Solution 2

p

 1  3  3  p     2  2  4 

2

3 3  p    2 4

2

 p

 p2 

2

3 9 p  2 16

1 q 5

71. q2 

Solution

q

2

 1  1  1  q      5  2  5 

72. m2 

2

 q

2

 1  1  q    5  10 

2

 q2 

1 1 q  5 100

2 m 3

Solution 2

m

 1  2  2  m      3  2  3 

2 2

 m

 1  2  m    3 3

2

 m2 

2 1 m  3 9

73. x2  12x  8

Solution x 2  12 x  8 x 2  12 x  36  8  36

 x  6

2

x  6 

28

x  6  2 7 x  6  2 7

or

 28 x  6   28 x  6  2 7 x  6  2 7

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276


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

74. x2  6x  1

Solution x 2  6 x  1 x 2  6 x  9  1  9

 x  3 x  3 

2

or

8

x  3  2 2

 8 x  3   8 x  3  2 2

x  3  2 2

x  3  2 2

75. x2  10x  37  0

Solution x 2  10 x  37  0 x 2  10 x  37 x 2  10 x  25  37  25

 x  5 x  5 

12

2

 12

or

x  5   12

x  5  i 4 3

x  5  i 4 3

x  5  2i 3

x  5  2i 3

76. a2  16a  82  0

Solution

a2  16a  82  0 a2  16a  82 a2  16a  64  82  64

a  8 a  8 

18

2

 18

or a  8   18

a  8  i 9 2

a  8  i 9 2

a  8  3i 2

a  8  3i 2

77. x2  5  5x

Solution x 2  5  5 x x 2  5 x  5

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277


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 2  5x 

25 25  5  4 4 2

 5 x   2  x 

5  2

x 

5  2

5 4

5 4

or x 

5 2 5  x  2

x  5

5 5   2 4 5 5   2 2 5  x  2

5

78. x 2  1  4 x

Solution x 2  1  4 x x 2  4 x  1 x 2  4 x  4  1  4

 x  2 x  2 

2

 3

or x  2   3

3

x  2 

x  2 

3

3

79. y 2  11 y  49

Solution

y 2  11 y  49 y 2  11 y 

121 196 121    4 4 4 2

 11  y   2  y 

11  2

75 4

y  

11  i 2

y  

11 5 3  i 2 2

or 75 4

 

75 4

y 

11 75    2 4 y  

11  i 2

75

y  

11 5 3  i 2 2

4

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278


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

80. x2  5x  22

Solution

x 2  5x  22 x2  5x 

25 88 25    4 4 4 2

 5 x   2  x 

5  2

63 4

x 

5  i 2

x 

5 3 7  i 2 2

 

63 4

or x  63 4

5 63    2 4 x 

5  i 2

63

x 

5 3 7  i 2 2

4

81. 2x 2  20x  49

Solution

2 x 2  20 x  49 49 2 49 50 x 2  10 x  25    2 2 2 1  x  5  2 x 2  10 x  

x  5  x  5  x 

1 2 1

or

2

10  2

2 2

x 

10  2 2

x  5   x  5   x 

1 2 1 2

10 2   2 2 x 

10  2 2

82. 4x2  8x  7

Solution 4 x 2  8x  7 7 4 7 4 x 2  2x  1   4 4 2 11  x  1  4 x 2  2x 

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279


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  1  x  1  x 

2  2 x 

11 4

or

x  1  

11 4

11 2 11 2

11 2 2 11   x  2 2

2  11 2

x 

x  1  

2  11 2

83. 3x2  1  4x

Solution

3x 2  1  4 x 3x 2  4 x  1 4 1 x  3 3 4 4 1 4   x2  x  3 9 3 9 x2 

 2 x   3  x 

2  3

x 

2  3

2

7 9

7 9

or

7 3 2  x  3

x  x 

7

2   3

7 9

2 7   3 3 2  x  3

7

84. 3 x 2  4 x  5

Solution

3x 2  4 x  5 4 5 x  3 3 4 4 5 4 x2  x    3 9 3 9 x2 

 2 x   3  x 

2  3

x 

2  3

19 9

19 3 2  x  3

2

19 9

or x  x  19

2   3

19 9

2 19   3 3 2  x  3

19

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280


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

85. 2 x 2  3 x  1

Solution

2x 2  3x  1 2x 2  3x  1 3 1 x  2 2 3 9 1 9 2 x  x    2 16 2 16 x2 

 3 x   4   x 

3  4

x 

3  4

17 16

2

17 16

x 

or

17 4 3  17 x  4

x 

3   4

17 16

3 17   4 4 3  17 x  4

86. 2 x 2  5 x  14

Solution

2 x 2  5 x  14 5 x  7 2 5 25 112 25 x2  x    2 16 16 16 x2 

 5 x   2  x 

5  4

x 

5  4

137 16

137 4 5  137 x  4

2

137 16

or x  x 

5   4

137 16

5 137   4 4 5  137 x  4

Use the Quadratic Formula to solve each equation. 87. 9 x 2  18 x  14

Solution

9 x 2  18 x  14  9x 2  18 x  14  0  a  9, b  18, c  14

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281


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 

b 

 18  b2  4ac  2a

18 

180

2

63  i 5 18  6i 5  18 18

18

18  4914 18   29

324  504 18

  3i 5  1 3

5 i 3

88. 7 z 2  14 z  13

Solution

7 z 2  14z  13  7 z 2  14z  13  0  a  7, b  14, c  13 x 

b 

14  b2  4ac  2a

14 

168 14

14  4713 14   27 2

2 7  i 42 14  2i 42  14 14

196  364 14

  7  i 42  1  7

42 i 7

89. 2 x 2  14 x  30i

Solution

2 x 2  14 x  30  2 x 2  14 x  30  0  a  2, b  14, c  30 x 

b 

 14  b2  4ac  2a

14 

44 4

14  2i 4

11

14  4230 14   22 2

27  i 4

11

196  240 4

  7  i 11  7  2

2

11 i 2

90. 5 x 2  x  5

Solution

5x 2  x  5  5x 2  x  5  0  a  5, b  1, c  5 x  

b 

1  b2  4ac  2a

1 

99 10

1  3i 10

11

1  455 1  1  100  10 25 2

 

1 3 11  i 10 10

91. 3 x 2  5 x  1

Solution

3 x 2  5 x  1  3 x 2  5 x  1  0  a  3, b  5, c  1 x 

b 

5  b2  4ac  2a

5  431 5   23 2

25  12 5   6 6

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13

282


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

92. 2 x 2  5 x  11

Solution

2 x 2  5 x  11  2 x 2  5 x  11  0  a  2, b  5, c  11 x 

b 

 5  b2  4ac  2a

5  4211 5   22 2

25  88 5  113  4 4

93. x 2  1  7 x

Solution

x 2  1  7 x  x 2  7 x  1  0  a  1, b  7, c  1 x 

b 

 7   b2  4ac  2a

7  411 7  49  4 7  45   2 2 2 1 2

7  3 5 2

94. 13 x 2  1  10 x

Solution

13 x 2  1  10 x  13 x 2  10 x  1  0  a  13, b  10, c  1 x 

x 

b 

 10  b2  4ac  2a

10  4131 10  100  52 10  48   26 26 2 13 2

2 5  2 3 10  48 10  4 3   26 26 26

  5  2 3 13

95. 3 x 2  6 x  1

Solution

3 x 2  6 x  1  3 x 2  6 x  1  0  a  3, b  6, c  1 x 

x 

b 

6  b2  4ac  2a

6  431 6   23 2

2 3  6  24 6  2 6   6 6 6

6

36  12 6  24  6 6

  3  6 3

96. 2 x  x  3  1

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283


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

2x  x  3  1  2x 2  6 x  1  0  a  2, b  6, c  1 x  x 

b 

6  b2  4ac  2a

6  421 6  36  8 6  28   4 4 22 2

6  28 6  2 7 3    4 4 2

7

97. 7 x 2  2 x  2

Solution

7 x 2  2x  2  7 x 2  2x  2  0  a  7, b  2, c  2 x  x 

b 

 2  b2  4ac  2a

2 

60 14

2  472 2   27 2

4  56 2  60  14 14

2  2 15 1  15  14 7

 1 98. 5x  x    3 5  Solution  1 5x  x    3  5x 2  x  3  0  a  5, b  1, c  3 5 

x 

b 

 1  b2  4ac  2a

1  453 1  1  60 1  61   10 10 25 2

99. x 2  2 x  2  0

Solution

x 2  2 x  2  0  a  1, b  2, c  2 x 

b 

2  b2  4ac  2a

22  4 12 2 1

2 

4  8 2

2  4 2  2i  2 2  1  i

100. a2  4a  8  0

Solution

a2  4a  8  0  a  1, b  4, c  8 x 

b 

4  b2  4ac  2a

42  4 18 2 1

4 

16  32 4  16 4  4i   2 2 2  2  2i

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284


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

101. y 2  4 y  5  0

Solution

y 2  4 y  5  0  a  1, b  4, c  5 x 

b 

4  b2  4ac  2a

42  4 15 2 1

4 

16  20 4  4 4  2i   2 2 2  2  i

102. x 2  2 x  5  0

Solution

x 2  2x  5  0  a  1, b  2, c  5 x 

b 

2  b2  4ac  2a

22  4 15 2 1

2 

2  16 2  4i 4  20   2 2 2  1  2i

103. x 2  2 x  5

Solution

x 2  2 x  5  x 2  2x  5  0  a  1, b  2, c  5 x 

b 

 2  b2  4ac  2a

2  415 2   2 1 2

4  20 2  16 2  4i   2 2 2  1  2i

104. z 2  3z  8

Solution

z 2  3z  8  z 2  3z  8  0  a  1, b  3, c  8 x 

b 

 3  b2  4ac  2a

3  418 3   2 1 2

9  32 3  23  2 2

 105. x 2 

3  2

23 i 2

2 2 x   3 9

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285


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 2 2 2 2 x2  x    x2  x   0  9x 2  6x  2  0  a  9, b  6, c  2 3 9 3 9

x 

 6  b2  4ac  2a

b 

6  492 6   29 2

36  72 6  36  18 18 

106. x 2 

6  6i 1 1   i 18 3 3

5  x 4

Solution 5 5 x2   x  x2  x   0  4 x 2  4 x  5  0  a  4, b  4, c  5 4 4

x 

 4  b2  4ac  2a

b 

4  445 4   24 2

16  80 4  64  8 8 

4  8i 1   i 8 2

Solve each formula for the indicated variable. 107. h 

1 2 gt ; t 2

Solution

1 2 gt 2 2h  gt 2 h 

2h g

 

2h  g

 t2

2h  t g g g

 t

2hg  t g

108. x 2  y 2  r 2 ; x

Solution

x2  y 2  r 2 x2  r 2  y 2 x   r2  y 2

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286


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

109. h  64t  16t 2 ; t

Solution

h  64t  16t 2 16t 2  64t  h  0; a  16, b  64, c  h b2  4ac 2a

b 

t 

 64 

64 

2

4096  64h 32

6464  h

64 

64  416h 2 16

32 64  8 64  h 8    32

64  h 4

110. y  16x 2  4; x

Solution

y  16 x 2  4 y  4  16 x 2

 

111.

x2 a2

y  4  x2 16 y  4  x 16 y  4  x 4

y2 b2

 1; y

Solution

x2 a2

y2 b2 y2 b2 y2 b2

 1  1  

y2 

a

2

x2 a2  x2

a2 b2 a2  x 2

a

2

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287


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

b2 a2  x 2

y  

a b a

y  

112.

x2 a2

y2

2

2

 x2

a

 1; x

b2

Solution

x2 a2

y2

 1

b2 x2

y2

 1 

a2 x2

b

2

a

b2  y2

2

b2 a2 b2  y 2

x2 

b 2

x  

x2 a2

y2

a b2  y 2

x  

113.

2

2

b

a

b  y  2

2

b

 1; a

b2

Solution

x2

y2

 1 a2 b2  x2 y2  a2b2     a2b2  1 2   a2 b   2 2 2 2 b x  a y  a2b2 b2 x 2  a2b2  a2 y 2

b2 x 2  a2 b2  y 2 b2 x 2 2

b  

 y

2

2 2

b x 2

b

 y2

bx b2  y 2 b2  y 2

 a2  a  a

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288


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

114.

x2 a2

y2

 1; b

b2

Solution

x2

y2

 1 a2 b2  x2 y2  a2b2     a2b2  1 2  a2 b   2 2 b x  a2 y 2  a2b2 b2 x 2  a2b2  a2 y 2

b2 x 2  a2

ay 2

2

a2 y 2

b2 

x 2  a2 a2 y 2

b   b  

x 2  a2 x 2  a2

ay

x 2  a2

115. x 2  xy  y 2  0; x

Solution x 2  xy  y 2  0  a  1, b  y , c   y 2 x 

  

b2  4ac 2a

b   y  

 y   41 y 2  2 1

y 

y2  4y2 2

y 

5y2 2

2

y  y 2

5

116. x 2  3xy  y 2  0; y

Solution x 2  3 xy  y 2  0 

y 2  3 xy  x 2  0

a  1, b  3 x , c  x 2

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289


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 

 

b 

b2  4ac 2a

3x   41 x2  2 1

 3x  

2

3x 

9x 2  4 x 2 2

3x 

5x 2

3x  x 5  2 2 Use the discriminant to determine the number and type of roots. Do not solve the equation. 

117. x 2  6 x  9  0

Solution

x 2  6 x  9  0  a  1, b  6, c  9 b2  4ac  62  4 19  36  36  0 one repeated rational number 118. 3 x 2  2 x  21

Solution 3 x 2  2 x  21  3 x 2  2 x  21  0 a  3, b  2, c  21 b2  4ac  2

 4 3 21

2

 4  252  248 two different nonreal complex numbers

119. 3 x 2  2 x  5  0

Solution

3 x 2  2 x  5  0  a  3, b  2, c  5 b2  4ac   2

2

 435

 4  60  56 two different nonreal complex numbers 120. 9 x 2  42 x  49  0

Solution 9 x 2  42 x  49  0 a  9, b  42, c  49 b2  4ac  42

2

 4949

 1764  1764  0 one repeated rational number

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290


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

121. 10 x 2  29 x  21

Solution 10 x 2  29 x  21  10 x 2  29 x  21  0 a  10, b  29, c  21 b2  4ac  29

2

 4 10 21

 841  840  1681 two different rational numbers

122. 10 x 2  x  21

Solution 10 x 2  x  21  10 x 2  x  21  0 a  10, b  1, c  21 b2  4ac   1

 4  10 21

2

 1  840  841 two different rational numbers

123. x 2  5 x  2  0

Solution

x 2  5 x  2  0  a  1, b  5, c  2 b2  4ac   5

2

 4 12  25  8  17

two different irrational numbers 124. 8 x 2  2 x  13

Solution

8 x 2  2 x  13  8 x 2  2 x  13  0 a  8, b  2, c  13 b2  4ac   2

2

 4 8 13

 4  416  412 two different nonreal complex numbers 125. Find two values of k such that x 2  kx  3k  5  0 will have two roots that are equal.

Solution x 2  kx  3k  5  0 a  1, b  k , c  3k  5 Set the discriminant equal to 0:

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291


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

b2  4ac  0 k 2  4 13k  5  0 k 2  43k  5  0 k 2  12k  20  0

k  2k  10  0 k  2 or k  10

126. For what value(s) of b will the solutions of x 2  2bx  b2  0 be equal?

Solution x 2  2bx  b2  0 a  1, b  2b, c  b2 Set the discriminant equal to 0:

b2  4ac  0

2b  41b2   0 2

4b2  4b2  0 0  0 True for all values of b Change each rational equation to quadratic form and solve it by the most efficient method. 12 127. x  1  x Solution

12 x  12  x  x  1  x   x x  1 

x 2  x  12 x 2  x  12  0

 x  4 x  3  0 x  4  0 x  4 128. x  2 

or

x  3  0 x  0

15 x

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292


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

15 x  15  x  x  2  x   x x  2 

x 2  2 x  15 x 2  2 x  15  0

 x  3 x  5  0 x  3  0

or

x  5  0

x  3

x  5

3  10 x

129. 8x 

Solution

3  10 x  3 x  8 x    x  10 x  8x 

8 x 2  3  10 x 8 x 2  10 x  3  0

4 x  12x  3  0 4x  1  0

or 2x  3  0

4 x  1

2x  3

x  130. 15x 

 41

x 

3 2

4  4 x

Solution

4  4 x  4 x  15 x    x 4 x  15 x 

15 x 2  4  4 x 15 x 2  4 x  4  0

5x  23x  2  0 5x  2  0

or 3x  2  0

5x  2

3x  2

x 

 52

x 

2 3

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293


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

131.

5 4   6 x x2

Solution

5 4   6 x x2 5  4   6 x2    x2  2 x x  5x  4  6x 2 6x 2  5x  4  0

3x  42x  1  0

132.

3x  4  0

or 2 x  1  0

3 x  4

2x  1

x   43

x 

6 x

2

1 2

1  12 x

Solution 6 1   12 x x2  6 1    x 2 12 x2  2 x x 

 

6  x  12 x 2 0  12 x 2  x  6

0  3 x  24 x  3 3x  2  0

or 4 x  3  0

3 x  2

4x  3

x 

 23

x 

3 4

 13  10 133. x  30    x x   Solution  13  10 x  30    x x   10 30 x  13  x  10  x 30 x  13  x    x

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294


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

30 x 2  13x  10 30 x 2  13x  10 30x 2  13x  10  0

5x  26x  5  0

5x  2  0

or 6x  5  0

5x  2

6x  5

x   52

x 

5 6

 17  10 134. x  20    x x  Solution  17  10 x  20    x x   10 20 x  17  x  10  x 20 x  17  x    x

20 x 2  17 x  10 20x 2  17 x  10  0

5x  24x  5  0

5x  2  0

or 4 x  5  0

5x  2

4x  5

x  135.

 52

x 

5 4

1 3   2 x x  2 Solution

1 3   2 x x  2 1 3  x  x  2    x  x  22 x  2 x 1 x  2  3x  2 x  x  2 x  2  3x  2x 2  4 x 0  2x 2  2

0  2 x  1 x  1 x  1  0

or x  1  0

x  1

x  1

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295


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

136.

1 1 5   x  1 x  4 4 Solution

1 1 5   x  1 x  4 4  1 1  5 4 x  1 x  4    4 x  1 x  4 1 4 4 x  x    4 x  4  4 x  1  5 x  1 x  4

4 x  16  4 x  4  5x 2  25x  20 0  5x 2  33x  40 0  5x  8 x  5 5x  8  0 or x  5  0 5x  8

x  5

8 5

x  5

x  137.

1 5   1 x  1 2x  4 Solution 1 1   1 x  1 2x  4    x  12x  4 x 1 1  2x 5 4    x  12x  4 1   12 x  4  5 x  1   x  12 x  4 2x  4  5x  5  2x 2  2x  4 0  2x 2  9x  5

0  2 x  1 x  5

2x  1  0

or x  5  0

2x  1

x  5

 21

x  5

x 

138.

x 2x  1 x  2 Solution

10 x  2

x 2 x  1 x  2 x 2 x  1

 x  2 x  2

10 x  2

  x  2

10 x  2

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296


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 2x  1  10 2x 2  x  10 2x 2  x  10  0

2x  5 x  2  0 2x  5  0

or x  2  0

2x  5

x  2

x   52

x  2

Since x  2 does not check, the only solution is x   52 . 139. x  1 

x  2 3  x  1 x  1

Solution 3 x  2  x  1 x  1    x  1 x 1 1  xx  21    x  1 x 3 1   x  1 

 x  1 x  1  x  2  3 x2  1  x  2  3 x2  x  2  0

 x  2 x  1  0 x  2  0

or

x  1  0

x  2

x  1

Since x  1 does not check, the only solution is x  2. 140.

1 1 1   4  y 4 y  2 Solution

1 1 1   4  y 4 y  2  1  1 1  44  y  y  2   44  y  y  2   4 4 2 y y      4 y  2 1  4  y  y  2 1  44  y  1 4 y  8   y 2  2 y  8  16  4 y y 2  6 y  16  0

 y  8 y  2  0 y  8  0 y  8

or

y  2  0 y  2

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297


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

141.

4  a a  2  2a 3

Solution 4  a a  2  2a 3 4  a a  2 6a   6a   2a   3 

12  3a  2a2  4a 0  2a2  7a  12 a  2, b  7, c  12 a 

142.

b 

 7  b2  4ac  2a

7  4212 7  145  4 22 2

a  2a  4  aa  3 10

5

Solution

a  2a  4  aa  3

10 5  a  2a  4  aa  3    10   10  10 5     a  2a  4  2aa  3 a2  2a  8  2a2  6a 0  a2  8a  8 a  1, b  8, c  8 a 

143. x 

b 

  8  b2  4ac  2a

8  418 8  32 8  4 2    4  2 2 2 2 2 1 2

36  0 x

Solution

36  0 x  36  x x    x  0 x   x 

x 2  36  0 x 2  36 x   36 x  6i

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298


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

144. x 

5  2 x

Solution

5  2 x  5 x x    x  2 x  x 

x 2  5  2x x 2  2 x  5 x 2  2 x  1  5  1

 x  1

2

 4

x  1   4 x  1  2i Fix It In exercises 145 and 146, identify the step the first error is made and fix it. 145. Solve by completing the square: x 2  6 x  34  0

Solution Step 4 was incorrect. Step 1: x 2  6 x  34 Step 2: x 2  6 x  9  34  9 Step 3:

 x  3

2

 25

Step 4: x  3  5i Step 5: x  3  5i 146. Solve x2  6x  2 by using the quadratic formula. To do so, identify a, b, and c. Then substitute into the formula and simplify.

Solution Step 3 was incorrect. Step 1: a  1; b  6; c  2 Step 2: x 

 6 

6  412 2 1 2

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299


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Step 3: x  Step 4: x 

6 

24 2

6  2 6 2

Step 5: x  3 

6

Discovery and Writing 147. Explain why the Zero-Factor Theorem is true.

Solution Answers may vary. 148. Explain how to complete the square on x2 – 17x.

Solution Answers may vary. 149. If r1 and r2 are the roots of ax 2  bx  c  0, show that r1  r2   ab .

Solution If r1 and r2 are the roots of ax 2  bx  c  0, then their values are

b2  4ac b  and r2  2a

b 

r1 

r1  r2 

b2  4ac b   2a

b 

b2  4ac . 2a b2  4ac b 2b    2a 2a a

150. If r1 and r2 are the roots of ax 2  bx  c  0, show that r1r2 

c . a

Solution

b2  4ac b  and r2  2a

b 

r1  r1r2 

b 

b2  4ac b   2a

b2  4ac . 2a

b2  4ac 2a

b  b  4ac b  b  4ac   2

2

4a2

b   b2  4ac  2

4a2

2

b2  b2  4ac 4a2

  b  b  4ac  4ac  c 2

2

4a2

4a2

a

In Exercises 151 and 152, a stone is thrown straight upward, higher than the top of a tree. The stone is even with the top of the tree at time t1 on the way up and at time t2 on the way down. If the height of the tree is h feet, both t1 and t2 are solutions of h = v0t – 16t2. 151. Show that the tree is 16t1t2 feet tall.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Rewrite the equation as 16t 2  v 0t  h  0 and solve for t using the quadratic formula.

a  16, b  v0 , c  h t 

b 

b2  4ac 2a

 v0  

v0   416h v0   2 16 2

v02  64h 32

Since t1 and t2 are the solutions to the equation, we have

t1 

v0 

v02  64h

32

16t1t2  16 

v0 

and t2 

v02  64h

32

v0 

v02  64h

32 v0 

v02  64h

32

. Calculate 16t1t2 :  16 

v02  v02  64h

1024 16.64h 1024h    h 1024 1024

Thus, h  16t1t2 . 152. Show that v0 is 16(t1 + t2) feet per second.

Solution Proceed as in #135 to calculate t1 and t2. Then  v  v 2  64 h v  v 2  64 h  2v 0 16 t 1  t2   16 0 320  0 320   16 32   Thus, v 0  16t1  t2 .

 

32v 0 32

 v0 .

Critical Thinking In Exercises 153–156, match each quadratic equation on the left with the easiest strategy to use to solve it on the right. 153. 6 x 2  76  0

a. Factoring

154. 6 x 2  35 x  6  0

b. Square Root Property

155. x 2  6 x  6

c. Completing the Square

156. 6 x 2  6 x  1

d. Quadratic Formula

Solution 153. b 154. a 155. c 156. d

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Determine if the statement is true or false. If the statement is false, then correct it and make it true. 157. 1492 x 2  1984 x – 1776  0 has real number solutions.

Solution 1492 x 2  1984 x – 1776  0 a  1492, b  1984, c  1776

b2  4ac   1984

2

 4 1492 1776

 3,936,256  10,599,168  14,535,424 The solutions are real numbers. True.

158. 2004 x 2  10 x  1994  0 has real number solutions.

Solution 2004 x 2  10 x  1994  0 a  2004, b  10, c  1994

b2  4ac   10

2

 42004 1994

 100  15,983,904  15,983,804 The solutions are real numbers. False.

EXERCISES 1.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

If the length of a rectangle is 15 feet and its width is

12 feet, what is the area of the 5

rectangle?

Solution 12  36ft 2 15  5 2. Write and algebraic expression that represents the area of a rectangle if its width is x and its length is (5x – 2).

Solution

x 5x  2  5x 2  2x

3. The shorter leg of a right triangle measures 10 yards and the longer leg measures 20 yards. Find the hypotenuse of the right triangle.

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302


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 102  202  x 2 100  400  x 2 500  x 2

x 

500 or x   500 (however, x can’t be negative)

x 

100  5

x  10 5 yards

4. Distance equals rate times time, d = rt. a) What does r equal? b) What does t equal?

Solution a.

r 

d t

b.

t 

d t

5. Solve for t: –16t2 + 96t = 0

Solution 16t 2  96t  0 16t t  6  0

16t  0

or t  6  0

t  0

t  6

6. What would be the most efficient method to use to solve an application problem in which the equation –4.9t2 + 28.6t + 67.3 = 0 occurs?

Solution The quadratic formula Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The formula for the area of a rectangle is __________.

Solution A = lw 8. The __________ Theorem states that the sum of the squares of the lengths of a right triangle equals the square of the length of the hypotenuse.

Solution d = rt, Pythagorean

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303


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Practice Solve each problem. 9. Geometric problem A rectangle is 4 feet longer than it is wide. If its area is 32 square feet, find its dimensions.

Solution Let w  the width of the rectangle. Then w  4  the length.

Width  Length  Area w w  4  32 w 2  4w  32 w 2  4w  32  0

w  8w  4  0 w  8  0

or

w  4  0 w  4

w  8

Since the width cannot be negative, the only reasonable solution is w = 4. The dimensions are 4 feet by 8 feet. 10. Geometric problem A rectangle is five times as long as it is wide. If the area is 125 square feet, find its perimeter.

Solution Let w = the width of the rectangle. Then 5w = the length.

Width  Length  Area w  5w  125 5w 2  125 w 2  25

w 

25

or

w   25

w  5 w  5 Since the width cannot be negative, the only reasonable solution is w = 5. The dimensions are 5 feet by 25 feet, and the perimeter is 60 feet. 11. Jumbotron The length of a rectangular jumbotron is 88 feet more than its width. If the jumbotron has an area of 11,520 square feet, find its dimensions.

Solution Let w = the width of the screen. Then w + 88 = the length.

Width  Length  Area w w  88  11520 w 2  88w  11520

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304


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

w 2  88w  11520  0  a  1, b  88, c  11520 b 

w 

 88  b2  4ac  2a

88  4111520 88  53824  2 2 1 2

w = 72 or w = –160; Since the screen cannot have a negative width, the solution is w = 72 and the dimensions of the screen are 160 ft by 72 ft. 12. IMAX screen A large movie screen is in the Panasonic IMAX theater at Darling Harbor, Sydney, Australia. The rectangular screen has an area of 11,349 square feet. Find the dimensions of the screen if it is 20 feet longer than it is wide.

Solution Let w = the width of the screen. Then w + 20 = the length.

Width  Length  Area w w  20  11349 w 2  20w  11349 w 2  20w  11349  0  a  1, b  20, c  11349 b 

w 

20  b2  4ac  2a

20  4111349 20  45796  2 2 1 2

w = 97 or w = –117; Since the screen cannot have a negative width, the solution is w = 97 and the dimensions of the screen are 117 ft by 97 ft. 13. Geometric problem The side of a square is 4 centimeters shorter than the side of a second square. If the sum of their areas is 106 square centimeters, find the length of one side of the larger square.

Solution Let s = the side of the second square. Then s – 4 = the side of the first square.

Area of first

Area of second

 106

s  4  s2  106 2

s2  8s  16  s2  106 2s2  8s  90  0

2 s2  4s  45  0 2 s  5 s  9  0 s  5  0

or

s  9  0

s  5 s  9 Since the side cannot be negative, the only reasonable solution is s = 9. The larger square has a side of length 9 cm.

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305


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

14. Geometric problem If two opposite sides of a square are increased by 10 meters and the other sides are decreased by 8 meters, the area of the rectangle that is formed is 63 square meters. Find the area of the original square.

Solution Let s = the side of the original square. Then the new rectangle has dimensions of s + 10 and s – 8.

Length  Width  Area

s  10s  8  63 s2  2s  80  63 s2  2s  143  0

s  13s  11  0 s  13  0

or

s  11  0

s  13 s  11 Since the side cannot be negative, the only reasonable solution is s = 11. The original area was 112 = 121 m2 15. Geometric problem Find the dimensions of a rectangle whose area is 180 cm2 and whose perimeter is 54 cm.

Solution

P  2l  2w, so l 

P  2w 54  2w .  2 2

Length  Width  Area 54  2w  w  180 2 54  2w w  360 54w  2w 2  360 0  2w 2  54w  360

0  2 w 2  27w  180 0  2w  12w  15

w  12  0

or

w  15  0

w  12 w  15 The dimensions are 12 cm by 15 cm. 16. Flags In 1912, an order by President Taft fixed the width and length of the U.S. flag in the ratio of 1 to 1.9. If 100 square feet of cloth are to be used to make a U.S. flag, estimate its dimensions to the nearest 41 foot.

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306


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let the dimensions be x and 1.9x.

Width  Length  Area x  1.9 x   100 1.9 x 2  100 100 1.9 100 x   1.9 x  7.25 1.9 x  1.97.25  13.75 x2 

Since the dimensions cannot be negative, the only reasonable solution 7 41 ft by 13 43 ft. 17. Metal fabrication A piece of tin, 12 inches on a side, is to have four equal squares cut from its corners, as in the illustration. If the edges are then to be folded up to make a box with a floor area of 64 square inches, find the depth of the box.

Solution The floor area of the box is a square with a side of length 12 – 2x.

Floor area  64

12  2x 

2

 64

144  48 x  4 x

2

 64

4 x 2  48x  80  64

4 x 2  12 x  12  0 4 x  2 x  10  0 x  2  0 or x  10  0 x  2

x  10

The solution x = 10 does not make sense in the problem, so the depth is 2 inches. 18. Making gutters A piece of sheet metal, 18 inches wide, is bent to form the gutter shown in the illustration. If the cross-sectional area is 36 square inches, find the depth of the gutter.

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307


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

Cross-sectional area  36

x  18  2x   36 18x  2x 2  36 0  2x 2  18x  36

0  2 x 2  9x  18

0  2 x  3 x  6 x  3  0 or x  6  0 x  3

x  6

Both solutions are valid, so the depth of the gutter is either 3 inches or 6 inches. 19. Parking lot A rectangular parking lot at PetSmart is 480 feet by 550 feet. Determine the diagonal of the parking lot.

Solution Using the Pythagorean Theorem:

height 2  width2  diagonal2 4802  5502  diagonal2 532,900  x 2

x 

532,900

or

x   532,900

x  730 x  730 Since lengths are positive, the answer is x  730 feet. 20. Photograph The diagonal of a square photograph of Katy Perry measures 8 2 inches. Find the length of one of its sides.

Solution Using the Pythagorean Theorem: side2  side2  diagonal2

x2  x2  8 2

2

2 x 2  128 x  8

x 2  64 or x  8

Since lengths are positive, the answer is x  8 inches

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

21. Baseball A baseball diamond is a square, 90 feet on a side. A shortstop for the Los Angeles Dodgers fields a grounder at a point halfway between second base and third base. How far will he have to throw the ball to make an out at first base?

Solution

A right triangle is formed in which the longer leg is 90 feet and the shorter leg is 45 feet (half of 90 feet). The hypotenuse will represent the distance that the ball will be thrown.

902  452  x 2 10,125  x 2 x 

10,125 or x   10,125

x  45 5

or x  45 5

Since lengths are positive, the answer is x  45 5 feet 22. Baseball A baseball diamond is a square, 90 feet on a side. A shortstop for the Chicago Cubs fields a grounder at a point one third of the way between second base and third base. How far will he have to throw the ball to make an out at first base?

Solution

A right triangle is formed in which the longer leg is 90 feet and the shorter leg is 30 feet (one-third of 90 feet). The hypotenuse will represent the distance that the ball will be thrown.

902  302  x 2 9,000  x 2 x 

9000 or x   9000

x  30 10

or x  30 10

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309


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Since lengths are positive, the answer is x  30 10 feet 23. Geometric problem The base of a triangle is onethird as long as its height. If the area of the triangle is 24 square meters, how long is its base?

Solution Let h = the height of the triangle. Then 31 h = the base of the triangle. 1 2

 Base  Height  Area 1 2

 31 h  h  24 1 2 h 6 2

h

h 

 24  144

144 or h   144

h  12

h  12

Since the height cannot be negative, the only reasonable solution is h  12. The base has a length of 4 meters. 24. Geometric problem The base of a triangle is onehalf as long as its height. If the area of the triangle is 100 square yards, find its height.

Solution Let h = the height of the triangle. Then 21 h = the base of the triangle. 1 2

 Base  Height  Area 1 2

 21 h  h  100 1 2 h 4 2

 100  400

h

h 

400 or h   400

h  20

h  20

Since the height cannot be negative, the only reasonable solution is h  20. The base has a length of 20 yards. 25. Right triangle If one leg of a right triangle is 14 meters shorter than the other leg, and the hypotenuse is 26 meters, find the length of the two legs.

Solution Let the legs have lengths x and x – 14. x 2   x  14

2

 262

x 2  x 2  28 x  196  676 2 x 2  28 x  480  0

0

2 x 2  14 x  240

2 x  24 x  10  0

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  24  0

or x  10  0

x  24

x  10

Since lengths are positive, the answer is x = 24, and the legs have length 10 meters and 24 meters. 26. Right triangle If one leg of a right triangle is five times the other leg, and the hypotenuse is 10 26 centimeters, find the length of the two legs.

Solution Let the legs have lengths x and 5x. x 2  5 x 

2

10 26 

2

x 2  25 x 2  2600 26 x 2  2600 x 2  100

x 

100 or x   100

x  10 x  10 Since lengths are positive, the answer is x = 10, and the legs have length 10 cm and 50 cm. 27. Manufacturing A manufacturer of television sets for a news studio received an order for sets with a 46-inch screen (measured along the diagonal). If the televisions are 17 21 inches wider than they are high, find the dimensions of the screen to the nearest tenth of an inch.

Solution Let x = the height of the screen. Then x + 17.5 = the width. Use the Pythagorean Theorem:

height2  width2  diagonal2 x 2   x  17.5

2

 462

x 2  x 2  35 x  306.25  2116 2x 2  35x  1809.75  0 a  2, b  35, c  1809.75 x  

b  35 

b2  4ac 2a

352  42 1809.75 22

35  125.312 15703  4 4 The only positive solution is 22.6. The dimensions are 22.6 inches by 40.1 inches. 

35 

28. Finding dimensions An oriental rug is 2 feet longer than it is wide. If the diagonal of the rug is 12 feet, to the nearest tenth of a foot, find its dimensions.

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311


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the width of the rug. Then x + 2 = the length. Use the Pythagorean Theorem:

width2  height2  diagonal 2 x 2   x  2

2

 122

x 2  x 2  4 x  4  144 2x 2  4 x  140  0 a  2, b  4, c  140 x  

b2  4ac 2a

b 

42  42 140

4 

22

4  33.705 4 4 The only positive solution is 7.4. The dimensions are 7.4 feet by 9.4 feet. 29. Cycling rates A cyclist rides from DeKalb to Rockford, a distance of 40 miles. His return trip takes 2 hours longer, because his speed decreases by 10 mph. How fast does he ride each way? 

4 

1136

Solution Let r = the cyclist’s rate from DeKalb to Rockford. Then his return rate is r – 10.

Return time  First time  2 40 40   2 r  10 r  40  40 r r  10  r r  10  2 r  10  r 

40r  40r  10  2r r  10 40r  40r  400  2r 2  20r 0  2r 2  20r  400

0  2r  20r  10 r  20  0

or

r  10  0

r  20

First trip Return trip

r  10 Rate

Time

Dist.

r

40 r

40

r  10

40 r  10

40

Since r = –10 does not make sense, the solution is r = 20. The cyclist rides 20 mph going and 10 mph returning. 30. Travel times Jake drives a moped from one town to another, a distance of 120 kilometers. He drives 10 kilometers per hour faster on the return trip, cutting 1 hour off the time. How fast does he drive each way?

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312


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let r = the farmer’s first rate. Then his return rate is r + 10.

Return time  First time  1 120 120   1 r  10 r  120  120 r r  10  r r  10  1 r  10  r 

120r  120r  10  r r  10 120r  120r  1200  r 2  10r

r 2  10r  1200  0

r  30r  40  0

r  30  0

or

r  40  0

r  30

First trip Return trip

r  40 Rate

Time

Dist.

r

120 r

120

r  10

120 r  10

120

Since r = –40 does not make sense, the solution is r = 30. The farmer drives 30 kph going and 40 kph returning. 31. Uniform motion problem If the speed were increased by 10 mph, a 420-mile trip would take 1 hour less time. How long will the trip take at the slower speed?

Solution Let r = the slower rate. Then the faster rate is r + 10.

Faster time  Slower time  1 420 420   1 r  10 r

r r  10

 420  420  r r  10  1 r  10  r 

420r  420r  10  r r  10 420r  420r  4200  r 2  10r

r 2  10r  4200  0

r  60r  70  0

r  60  0

or

r  70  0

r  60

Slower trip Faster trip

r  70 Rate

Time

Dist.

r

420 r

420

r  10

420 r  10

420

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Since r = –70 does not make sense, the solution is r = 60. The slower speed results in a trip of length 7 hours. 32. Uniform motion problem By increasing her usual speed by 25 kilometers per hour, a bus driver decreases the time on a 25-kilometer trip by 10 minutes. Find the usual speed.

Solution Let r = the driver’s slower rate. Then her faster rate is r + 25. 10 Faster time  Slower time  60 25 25 1   r  25 r 6  25 25 1  6r r  25   6r r  25 r  25 6  r

150r  150r  25  r r  25

150r  150r  3750  r 2  25r r 2  25r  3750  0

r  50r  75  0

r  50  0

or

r  75  0

r  50

r  75 Rate

Time

Dist.

r

25 r

25

r  25

25 r  25

25

Slower trip Faster trip

Since r = –75 does not make sense, the solution is r = 50. The driver’s usual speed is 50 kph. 33. Falling coins An object falls 16t2 feet in t seconds. If a penny is dropped from the top of the Sears Tower in Chicago, from a height of 1454 feet, how long will it take for the penny to hit the ground? Round to one decimal place.

Solution Set s  1454:

s  16t 2 1454  16t 2 1454  t2 16 1454 16 t  9.5 t 

or

1454 16 t  9.5 t  

t = –9.5 does not make sense, so it takes it about 9.5 seconds to hit the ground.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

34. Stunt act According to the Guinness Book of World Records, stuntman Dan Koko fell a distance of 312 feet into an airbag after jumping from the Vegas World Hotel and Casino. If Dan fell 16t2 feet in t seconds, to the nearest tenth of a second, how long did he fall?

Solution Set d  312:

d  16t 2 312  16t 2 312  t2 16 312 312 t   or 16 16 t  4.4 t  4.4 t = –4.4 does not make sense, so the fall lasted about 4.4 seconds. t 

35. Tybee Island lighthouse Kylie accidentally drops her car keys from the top of the Tybee Island lighthouse, 44.2 meters high. If the keys fall 4.9t2 meters per second, where t represents time in seconds, how long will it take her keys to hit the ground? Round to the nearest second.

Solution Set h  0 :

h  4.9t 2  44.2 0  4.9t 2  44.2 4.9t 2  44.2  0 4.9t 2  44.2 t2  t 

44.2 4.9

44.2 4.9

or t  

44.2 4.9

t  3 or t  3 t  3 does not make sense, so it takes about 3 seconds for the keys to hit the ground. 36. Brooklyn Bridge Brad accidentally drops his water bottle from the Brooklyn Bridge, 84.28 meters high. If the bottle falls 4.9t2 meters per second, where t represents time in seconds, how long will it take the water bottle to land in New York City’s East River? Round to one decimal place.

Solution Set h  0 :

h  4.9t 2  84.28 0  4.9t 2  84.28

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

4.9t 2  84.28  0 4.9t 2  84.28 t2 

84.28 4.9

84.28 84.28 or t   4.9 4.9 t  4.1 or t  4.1 t = 4.1 does not make sense, so it takes about 4.1 seconds for the keys to hit the ground. t 

37. Cliff jumping Jeff jumped off of a cliff into the ocean while vacationing in Maui, Hawaii. His height h above the water in feet after t seconds is represented by the equation h = –16t2 + 5t + 30. After how many seconds did he hit the water? Round to the nearest tenth.

Solution Set h  0 :

h  16t 2  5t  30 0  16t 2  5t  30 t 

5 

25  4 1630 2 16

5  1945 32 t  1.5 or t  1.2 t = 1.2 doesn’t make sense, so it takes about 1.5 seconds for Jeff to hit the water. t 

38. Platform diving An athlete dives from a 32 feet platform with an initial velocity of 7 feet per second. The formula h = –16t2 + 7t + 32 can be used to determine the height of the diver in feet after t seconds. When did the athlete reach the pool water? Round to two decimal places.

Solution Set h  0 :

h  16t 2  7t  32 0  16t 2  7t  32 t 

7 

49  4( 16)(32) 2( 16)

7 

2097 32 t  1.65 or t  1.21 t  1.2 doesn’t make sense, so it takes about 1.65 seconds for the diver to hit the water. t 

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

39. Ballistics The height of a projectile fired upward with an initial velocity of 400 feet per second is given by the formula h = –16t2 + 400t, where h is the height in feet and t is the time in seconds. Find the time required for the projectile to return to earth.

Solution Set h  0: h  16t 2  400t 0  16t 2  400t

16t 2  400t  0 16t t  25  0

16t  0 or t  25  0 t  0

t  25

t  0 represents when the projectile was fired, so it returns to earth after 25 seconds. 40. Ballistics The height of an object tossed upward with an initial velocity of 104 feet per second is given by the formula h = –16t2 + 104t, where h is the height in feet and t is the time in seconds. Find the time required for the object to return to its point of departure.

Solution Set h  0:

h  16t 2  104t 0  16t 2  104t 16t 2  104t  0 8t 2t  13  0

8t  0 or 2t  13  0 t  0

t 

13 2

 6.5

t  0 represents when the projectile was fired, so it returns after 6.5 seconds. 41. Ballistics The height of an object thrown upward with an initial velocity of 32 feet per second is given by the formula h = –16t2 + 32t, where t is the time in seconds. How long will it take the object to reach a height of 16 feet?

Solution Set h  16:

h  16t 2  32t 16  16t 2  32t 16t 2  32t  16  0 16t  1t  1  0

t  1  0 or t  1  0 t  1

t  1

It takes the object 1 second to reach a height of 16 feet.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

42. Cruise ship anchor A cruise ship drops anchor in a harbor. The formula h = –16t2 + 50 can be used to determine the height h in feet of the anchor at time t in seconds after it is released. When is the anchor 5 feet above the water? Round to the nearest tenth.

Solution Set h  5 :

h  16t 2  50 5  16t 2  50 0  16t 2  45 16t 2  45 t 2  2.8125 t  1.7 or t  1.7 t  1.7 doesn’t make sense, so it takes about 1.7 seconds for the anchor to be 5 feet above water. 43. Water balloon toss A water balloon is tossed from the window of college student’s dormitory. The equation h = –4.9t2 + 5.2t + 10.1 can be used to determine the height h in meters of the water balloon at time t in seconds after it is released. When will the water balloon hit the ground? Round to the nearest tenth.

Solution Set h  0 :

h  4.9t 2  5.2t  10.1 0  4.9t 2  5.2t  10.1

t 

5.2 

5.2  44.910.1 2 4.9 2

5.2  225 9.8 t  2.1 or t  1 t  1 doesn’t make sense, so it takes about 2.1 seconds for the water balloon to hit the ground. t 

44. Beachball toss A beachball is tossed from the top of a swimming pool sliding board. The formula h = –4.9t2 + 8.1t + 5.275 can be used to determine the height h in meters of the beachball at time t in seconds after it is released. When will the beachball hit the pool water? Round to the nearest tenth.

Solution Set h  0 :

h  4.9t 2  8.1t  5.275 0  4.9t 2  8.1t  5.275 t  t 

8.1 

(8.1)2  4( 4.9)(5.275) 4( 4.9)

169 8.1  9.8

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

t  2.2 or t  0.5 t  0.5 doesn’t make sense, so it takes about 2.2 seconds for the beachball to hit the pool water. 45. Setting fares A bus company has 3000 passengers daily, paying a 25¢ fare. For each nickel increase in fare, the company projects that it will lose 80 passengers. What fare increase will produce $994 in daily revenue?

Solution Let x = the number of nickel increases. The new fare = 25 + 5x (in cents), while the number of passengers = 3000 – 80x.

Number of Passengers

 Fare  Revenue

3000  80x 25  5x   99400 75000  13000x  400x 2  99400 400x 2  13000 x  24400  0

200 2x 2  65x  122  0 2002x  61 x  2  0 2 x  61  0

x  2  0

or

x  2 x  30.5 Since you cannot have half of a nickel increase, x  30.5 does not make sense. Thus, there should be 2 nickel increases, for a fare increase of 10 cents. 46. Jazz concerts A jazz group on tour has been drawing average crowds of 500 persons. It is projected that for every $1 increase in the $12 ticket price, the average attendance will decrease by 50. At what ticket price will nightly receipts be $5600?

Solution Let x = the number of dollar increases. The new price = 12 + x, while the number attending = 500 – 50x. Number of attending

 Price  Revenue

500  50 x 12  x   5600 6000  100 x  50 x 2  5600

50x 2  100x  400  0

50 x 2  2x  8  0 50 x  4 x  2  0

x  4  0 x  4

or

x  2  0 x  2

Since you cannot have a negative number of increases, x  4 does not make sense. Thus, there should be an increase of 2 dollars, for a ticket price of $14.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

47. Concert receipts Tickets for the annual symphony orchestra pops concert cost $15, and the average attendance at the concerts has been 1200 persons. Management projects that for each 50¢ decrease in ticket price, 40 more patrons will attend. How many people attended the concert if the receipts were $17,280?

Solution Let x = the number of $0.50 decreases. The new price = 15 – 0.5x, while the number attending = 1200 + 40x.

Number of attending

 Price  Revenue

1200  40x15  0.5x  17280 18000  20 x 2  17280 20 x 2  720 x 2  36 x 

36 or x   36

x  6

x  6

x = –6 does not make sense. Thus, there should be six 50-cent decreases, for a ticket price of $12 and an attendance of 1440 people. 48. Projecting demand The Vilas County News earns a profit of $20 per year for each of its 3000 subscribers. Management projects that the profit per subscriber would increase by 1¢ for each additional subscriber over the current 3000. How many subscribers are needed to bring a total profit of $120,000?

Solution Let x = the number of subscribers over 3000. The new profit = 20 + 0.01x, while the number subscribing = 3000 + x.

Number subscribing

 Profit  Total profit

3000  x 20  0.01x   120000 60000  50 x  0.01x 2  120000 0.01x 2  50 x  60000  0 x 2  5000 x  6000000  0

 x  6000 x  1000  0

x  6000  0

or

x  1000  0

x  6000 x  1000 x = –6000 does not make sense. The increase should be 1000, for a total of 4000. 49. Investment problems Morgan and Kyung each have a bank CD. Morgan’s is $1000 larger than Kyung’s, but the interest rate is 1% less. Last year Morgan received interest of $280, and Kyung received $240. Find the rate of interest for each CD.

Solution Let x = Chloe’s principal.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Morgan’s rate

Chloe’s rate

 0.01

280 240   0.01 x  100 x x  x  1000

 240  280  x  x  1000  0.01 x  1000 x  

280x  240 x  1000  0.01x  x  1000

0.01x x

2

2

 50x  240,000  0

 5000x  24,000,000  0

 x  3000 x  8000  0

x  3000  0

or x  8000  0

x  3000

x  8000 

I

P

r

Chloe

240

x

240 x

Morgan

280

x + 1000

280 x  1000

x = –8000 does not make sense. The principal amounts were $3000 and $4000. The interest rates were 8% for Chloe and 7% for Morgan.

50. Investment problem Safa and Laura have both invested some money. Safa invested $3000 more than Laura and at a 2% higher interest rate. If Safa received $800 annual interest and Laura received $400, how much did Safa invest?

Solution Let x = Laura’s principal.

Scott’s rate

Laura’s rate

 0.02

800 400   0.02 x  3000 x x  x  3000

 400  800  x  x  3000  0.02 x  3000 x  

800 x  400 x  3000  0.02 x  x  3000 800 x  400 x  1,200,000  0.02 x 2  60 x

0.02x 2  340 x  1,200,000  0 x 2  17,000 x  60,000,000  0

 x  5000 x  12,000  0

x  5000  0 x  5000

or

x  12,000  0 x  12,000 

Laura invested either $5000 or $12,000, so Scott invested either $8000 or $15,000.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

I

P

r

Laura

400

x

400 x

Scott

800

x + 3000

800 x  3000

51. Buying microwave ovens Some mathematics professors would like to purchase a $150 microwave oven for the department workroom. If four of the professors don’t contribute, everyone’s share will increase by $10. How many professors are in the department?

Solution Let x = the total number of professors.

New share with lower number

Original

 10

share

150 150   10 x  4 x  150  150 x  x  4  x  x  4  10 x  4  x 

150 x  150 x  4  10 x  x  4 150 x  150 x  600  10 x 2  40 x 0  10 x 2  40 x  600 0  x 2  4 x  60

0   x  10 x  6 x  10  0

or

x  6  0

x  10 x  6 x = –6 does not make sense, so there are 10 professors in the department. 52. Digital cameras A merchant could sell one model of digital cameras at list price for $180. If he had three more cameras, he could sell each one for $10 less and still receive $180. Find the list price of each camera.

Solution Let x = the actual number of cameras.

Actual price per camera

 10 

New price per camera

180 180  10  x x  3  180  180  10   x  x  3 x  x  3  3 x x  

180 x  3  10 x  x  3  180 x 10 x 2  30 x  540  0 x 2  3 x  54  0

 x  6 x  9  0

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  6  0

or

x  9  0

x  6 x  9 x = –9 does not make sense, so there are 6 cameras, with each costing $30. 53. Filling storage tanks Two pipes are used to fill a water storage tank. The first pipe can fill the tank in 4 hours, and the two pipes together can fill the tank in 2 hours less time than the second pipe alone. How long would it take for the second pipe to fill the tank?

Solution Let x = time for the second pipe to fill tank.

First in

1 hour

Second in 1 hour

Total in 1 hour

1 1 1   4 x x  2 1 1 1 4 x  x  2    4 x  x  2  x x  2 4

x  x  2  4 x  2  4 x

x 2  2x  4 x  8  4 x x 2  2x  8  0

 x  4 x  2  0

x  4  0 or x  2  0 x  4

x  2

Since x = –2 does not make sense, the solution is x = 4. It takes the second pipe 4 hours to fill the tank alone. 54. Filling swimming pools A hose can fill a swimming pool in 6 hours. Another hose needs 3 more hours to fill the pool than the two hoses combined. How long would it take the second hose to fill the pool?

Solution Let x = time for the both hoses to fill pool.

First in 1 hour

Second in 1 hour

Total in 1 hour

1 1 1   x  3 x 6 1 1  1 6 x  x  3    6 x  x  3  x  3 x 6 x  x  3  6 x  6 x  3 x 2  3 x  6 x  6 x  18 x 2  3 x  18  0

 x  3 x  6  0 x  3  0 x  3

or

x  6  0 x  6

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Since x = –6 does not make sense, the solution is x = 3. It takes the second hose 6 hours to fill the pool alone. 55. Mowing lawns Kristy can mow a lawn in 1 hour less time than her brother Steven. Together they can finish the job in 5 hours. How long would it take Kristy if she worked alone?

Solution Let x = time for the Steven to mow lawn.

Steven in 1 hour

Kristy in

1 hour

Total in 1 hour

1 1 1   x x  1 5 1  1 1 5 x  x  1    5 x  x  1  1 5 x x    5 x  1  5 x  x  x  1 5x  5  5x  x 2  x 0  x 2  11x  5 a  1, b  11, c  5 x   

b 

b2  4ac 2a

 11  11 

11  415 2 1 2

101

 10.5 or 0.5 2 x = 0.5 does not make sense, so Kristy could mow the lawn in about 9.5 hours alone. 56. Cleaning the garage Working together, Sarah and Heidi can clean the garage in 2 hours. If they work alone, it takes Heidi 3 hours longer than it takes Sarah. How long would it take Heidi to clean the garage alone?

Solution Let x = time for the Sarah to clean it.

Sarah in 1 hour

Heidi in 1 hour

Total in 1 hour

1 1 1   x x  3 2 1 1  1 2 x  x  3    2 x  x  3  x  3 2 x 2 x  3  2 x  x  x  3 2x  6  2x  x 2  3x 0  x2  x  6

0   x  3 x  2

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  3  0

or

x  2  0

x  2 x  3 x = –2 does not make sense. It would take Heidi 6 hours to clean the garage alone. 57. Planting windscreens A farmer intends to construct a windscreen by planting trees in a quarter-mile row. His daughter points out that 44 fewer trees will be needed if they are planted 1 foot farther apart. If her dad takes her advice, how many trees will be needed? A row starts and ends with a tree. (Hint: 1 mile = 5280 feet.)

Solution The number of trees is the length of the row divided by the space between the trees, plus 1. Let x = the original spacing.

Original number

 44 

1320  1  44  x 1320  44  x  1320  x  x  1  44   x  

New number 1320  1 x  1 1320 x  1 1320 x  x  1 x  1

1320 x  1  44 x  x  1  1320 x 44 x 2  44 x  1320  0 x 2  x  30  0

 x  5 x  6  0 x  5  0

or

x  6  0

x  5 x  6 x = –6 does not make sense. The original spacing was 5 feet, resulting in 265 trees, so the new spacing will require 221 trees. 58. Angle between spokes If a wagon wheel had 10 more spokes, the angle between spokes would decrease by 6°. How many spokes does the wheel have?

Solution Let x = the actual number of spokes.

Actual angles between spokes

 6 

New angle between spokes

360 360  6  x x  10  360  360  6  x  x  10 x  x  10 x x  10  

360 x  10  6 x  x  10  360 x 6 x 2  60 x  3600  0 x 2  10 x  600  0

 x  20 x  30  0

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325


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  20  0

or

x  30  0

x  20 x  30 x = –30 does not make sense, so there are 20 spokes. 59. Architecture A golden rectangle is one of the most visually appealing of all geometric forms. The front of the Parthenon, built in Athens in the 5th century B.C. and shown in the illustration, is a golden rectangle. In a golden rectangle, the length l and the height h of the rectangle must satisfy the following equation. If a rectangular billboard is to have a height of 15 feet, how long should it be if it is to form a golden rectangle? Round to the nearest tenth of a foot. l h  h l  h

Solution Let h = 15: l h  h l  h 15 l  15 l  15

l 15  15l  15  l  15 15 l l  15  152

15l  15 

l2  15l  225  0 a  1, b  15, c  225

l  

b 

b2  4ac 2a

 15 

2

15  33.541 2 2 The only positive solution is l  24.3 ft. 

15 

15  41225 2 1

1125

AB 60. Golden ratio Rectangle ABCD, shown here, will be a golden rectangle if AD 

BC BE

where AE = AD. Let AE = 1 and find the ratio of AB to AD.

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326


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution If AE = 1, then AD = BC = 1 and BE = x – 1. AB BC  AD BE x 1  1 x  1 x 1 x  1   x  1  1 x  1 x2  x  1

x2  x  1  0 a  1, b  1, c  1 x  

b2  4ac 2a

b   1 

2

1  2.236 2 2 The only positive solution is x = 1.618. 1.618 , or 1.618 to 1. The ratio is about 1 

1 

1  411 2 1

5

61. Automobile engines As the piston shown moves upward, it pushes a cylinder of a gasoline/air mixture that is ignited by the spark plug. The formula that gives the volume of a cylinder is V = πr2h, where r is the radius and h the height. Find the radius of the piston (to the nearest hundredth of an inch) if it displaces 47.75 cubic inches of gasoline/air mixture as it moves from its lowest to its highest point.

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327


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

V   r 2h 47.75   r 2 5.25 47.75  r2 5.25 47.75   r 5.25 1.70  r The radius is about 1.70 in. 62. History One of the important cities of the ancient world was Babylon. Greek historians wrote that the city was square-shaped. Its area numerically exceeded its perimeter by about 124. Find its dimensions in miles. (Round to the nearest tenth.)

Solution Let x = the length of a side of the city.

Area  Perimeter  124 x 2  4 x  124 x 2  4 x  124  0 a  1, b  4, c  124 x   

b 

 4  4 

b2  4ac 2a

4  41124 2 1 2

512

 13.3 or 9.3 2 The dimensions were 13.3 mi by 13.3 mi. Discovery and Writing 63. Summarize the general strategy used to solve application problems in this section.

Solution Answers may vary. 64. Describe why it is important to check your solutions to an application problem.

Solution Answers may vary. 65. Which of the preceding application problems did you find the hardest? Why?

Solution Answers may vary.

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328


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

66. Is it possible for a rectangle to have a width that is 3 units shorter than its diagonal and a length that is 4 units longer than its diagonal?

Solution Let x = the length of the diagonal.

Using the Pythagorean Theorem:

 x  4   x  3 2

2

 x2

x 2  8 x  16  x 2  6 x  9  x 2 x 2  2 x  25  0

a  1, b  2, c  25 x  

b 

2 

b2  4ac 2a

22  4 125

2 

21

96 2

This does not equal a real number, so it is not possible.

EXERCISES 1.6 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Factor completely: 4 x 4  36 x 2

Solution

4x4  36x2  4x2 x2  9  4x2  x  3 x  3 2. Factor completely: 2 y 3  3 y 2  35 y

Solution

2 y 3  3 y 2  35 y  y 2 y 2  3 y  35  y  y  52 y  7 3. Factor by grouping: z 3  5z 2  2z  10

Solution

z3  5z2  2z  10  z2  z  5  2 z  5  z2  2  z  5

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329


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

3

4. Simplify:  32 5

Solution

32 5   5 32  3

3

  2

3

 8

1

5. If u  x 3 what is u2?

Solution 1

If u  x 3 , then u2 

x   x 13

2

23

6. Perform the operation and simplify: ( 3x  1  6)2

Solution

 3x  1  6   3x  1  6 3x  1  6  3x  1  12 3x  1  36 2

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. Equal powers of equal real numbers are __________.

Solution equal 8. If a and b are real numbers and a = b then a2 = __________.

Solution

b2 9. False solutions that don’t satisfy the equation are called __________ solutions.

Solution extraneous 10. Radical equations contain radicals with variables in their__________.

Solution radicands Practice Use factoring to solve each equation. 11. x 3  9 x 2  20 x  0

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330


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

x 3  9 x 2  20 x  0

0

x x 2  9 x  20

x  x  5 x  4  0 x  0 or

x  5  0

x  0

or

x  4  0

x  5

x  4

12. x 3  4 x 2  21x  0

Solution

x 3  4 x 2  21x  0

x x 2  4 x  21  0 x  x  7 x  3  0 x  0 or

x  7  0

x  0

or

x  3  0

x  7

x  3

13. 6a3  5a2  4a  0

Solution

6a3  5a2  4a  0

a 6a2  5a  4  0 a2a  13a  4  0 a  0 or 2a  1  0 a  0

2a  1

a  0

 21

a 

or 3a  4  0 3a  4 4 3

a 

14. 8b3  10b2  3b  0

Solution 8b3  10b2  3b  0

0

b 8b2  10b  3

b4b  32b  1  0

b  0 or 4b  3  0 or 2b  1  0 b  0

4b  3

b  0

3 4

b 

2b  1 b 

1 2

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331


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

15. y 4  26 y 2  25  0

Solution y 4  26 y 2  25  0

 y  25 y  1  0 2

2

y 2  25  0

or

y2  1  0

y 2  25

y2  1

y  5

y  1

16. y 4  13 y 2  36  0

Solution y 4  13 y 2  36  0

 y  4 y  9  0 2

2

y2  4  0

or

y2  9  0

y2  4

y2  9

y  2

y  3

17. 2 y 4  46 y 2  180

Solution

2 y 4  46 y 2  180

 0 2 y  18 y  5  0

2 y 4  23 y 2  90 2

2

y 2  18  0

or

y 2  18 y  

y2  5  0 y2  5

18

y   5

y  3 2

y   5

18. 2 x 4  102 x 2  196

Solution 2 x 4  102 x 2  196

0 2 x  49 x  2  0 2 x 4  51x 2  98 2

x 2  49  0

2

or

x2  2  0

x 2  49

x2  2

x  7

x   2

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332


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

19. x 4  8 x 2  9

Solution

x 4  8x 2  9 x 4  8x 2  9  0

 x  9 x  1  0 2

2

x2  9  0

x2  1  0

or

x2  9

x 2  1

x   9

x   1

x  3

x  i

20. x 4  12 x 2  64

Solution

x 4  12 x 2  64 x 4  12 x 2  64  0

 x  16 x  4  0 2

2

x 2  16  0

or

x2  4  0

x 2  16

x 2  4

x   16

x  

x  4

x  2i

4

21. 4 y 4  7 y 2  36  0

Solution

4 y 4  7 y 2  36  0

4 y  9 y  4  0 2

2

4y2  9  0 y2 

or

y2  4  0 y 2  4

9 4

y  

9 4

y   32

y   4 y  2i

22. 9 y 4  56 y 2  225  0

Solution

9 y 4  56 y 2  225  0

9 y  25 y  9  0 2

2

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333


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

9 y 2  25  0 y2 

or

y2  9  0 y 2  9

25 9

y  

25 9

y  

9

y  3i

y   53

Solve each equation by factoring or by making an appropriate substitution. 23. 2 x 3  3 x 2  4 x  6

Solution

2x 3  3x 2  4 x  6 2x 3  3x 2  4 x  6  0 x 2 2 x  3  22 x  3  0

2x  3 x 2  2  0

2x  3  0

or x 2  2  0

2 x  3

x2  2

x 

3 2

x   2

24. 3 x 3  x 2  12 x  4

Solution 3 x 3  x 2  12 x  4 3 x 3  x 2  12 x  4  0 x 2 3 x  1  43 x  1  0

3x  1 x 2  4  0

3 x 3  x 2  12x  4 3x 3  x 2  12x  4  0 x 2 3x  1  43x  1  0

3x  1 x 2  4  0

2 3 x  1  0 or x  4  0

or

x2  4

1 or 3

x  2

3x  1 x 

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334


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

25. x 3  x 2  9 x  9  0

Solution

x 3  x 2  9x  9  0 x 2  x  1  9 x  1  0

 x  1 x 2  9  0 or

x2  9  0

x  1 or

x 2  9

x  1  0

x   9 x   3i 26. x 3  5 x 2  8 x  40

Solution

x 3  5 x 2  8 x  40 x 3  5 x 2  8 x  40  0 x 2  x  5  8 x  5  0

 x  5 x 2  8  0

x  5  0

or x 2  8  0

x  5

x 2  8 x   2i 2

27. x 4  37 x 2  36  0

Solution

x 4  37 x 2  36  0

 x  36 x  1  0 2

2

x 2  36  0

or x 2  1  0

x 2  36

x2  1

x  6

x  1

28. x 4  50 x 2  49  0

Solution x 4  50 x 2  49  0

 x  49 x  1  0 2

2

x 2  49  0

or

x2  1  0

x 2  49

x2  1

x  7

x  1

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335


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

29. 2m2 3  3m1 3  2  0

Solution 2m2 3  3m1 3  2  0

2m



13

0

2m1 3  1  0 m1 3 

 1 m1 3  2

or m1 3  2  0 m 1 3  2

1 2

m     3

13

1 2

m   2

3

3

m  8

1 8

m 

3

13

Both answers check.

30. 6t 2 5  11t 1 5  3  0

Solution

2t

6t 2 5  11t 1 5  3  0 15



2t 1 5  3  0

or 3t 1 5  1  0

t 1 5   32

t 1 5   31

 3 3t 1 5  1  0

t    

5

t    

t   243 32

1 t   243

15

5

3 2

15

5

1 3

5

Both answers check.

31. x  13 x 1 2  12  0

Solution

x

x  13 x 1 2  12  0 12



x 1 2  12  0

or

x1 2  1  0

x 1 2  12

 12 x 1 2  1  0

x1 2  1

 x   12

 x   1

x  144

x  1

12

2

2

12

2

2

Both answers check. 12

32. p  p

 20  0

Solution

p

p  p1 2  20  0 12



 5 p1 2  4

0

p1 2  5  0

or

p 1 2  5

p1 2  4  0 p1 2  4

 p   5

 p   4 

p  25

p  16

12

2

2

12

2

2

p  25 does not check and is extraneous. 12

33. 6p  p

 1

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336


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

2 p

12

6 p  p1 2  1

2 p1 2  1  0

6 p  p1 2  1  0

p1 2   21



or 3p1 2  1  0 p1 2 

 p    

 1 3 p1 2  1  0

2

12

1 2

1 4

p 

p    

2

12

1 4

p 

1 3

2

1 3

2

1 9

p 

does not check and is extraneous.

34. 3r  r 1 2  2

Solution

3r

12

3r  r 1 2  2

3r 1 2  2  0

or r 1 2  1  0

3r  r 1 2  2  0

r 1 2   23

r1 2  1



r     

 2 r1 2  1  0

2

12

2 3

4 9

r   1 2

12

2

r  1

4 9

r  r 

2

does not check and is extraneous.

35. 2t 1 3  3t 1 6  2  0

Solution 2t 1 3  3t 1 6  2  0

2t

16



0

 1 t1 6  2

2t 1 6  1  0 t1 6 

or t 1 6  2  0 t 1 6  2

1 2

t     16

6

t 

1 2

6

t   2 6

16

6

t  61

1 64

r  64 does not check and is extraneous.

36. z 3  7 z 3 2  8  0

Solution

z3  7z3 2  8  0



z3 2  1 z3 2  8  0

z3 2  1  0

or z 3 2  8  0

z 3 2  1

z3 2  8

z   1 32

2

2

z   8 32

2

z3  1

z 3  64

z  1

z  4

2

z  1 does not check and is extraneous.

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337


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

37. x 2  10 x 1  16  0

Solution x 2  10 x 1  16  0

x



1

0

x 1  8  0

or

x 1  8

 8 x 1  2

x  1

1

x 1  2

 8

x 

x 1  2  0

x 

1

1

1 8

1

 2

x 

1

1 2

Both answers check.

38. 2 y 2  9 y 1  5  0

Solution 2 y 2  9 y 1  5  0

2 y



1

0

2 y 1  1  0

 1 y 1  5

y 1 

y  1

1

or

y 1   5

1 2

 1 2

y  2

y 1  5  0

1

y  1

1

  5

1

x   51

Both answers check.

39. z 3 2  z 1 2  0

Solution

z3 2  z 1 2  0

z 1 2 z2 2  1  0 z 1 2  z  1  0 z1 2  0

or z  1  0

z   0

z  1

12

2

2

z  1

z  0

Both answers check. 40. r 5 2  r 3 2  0

Solution r5 2  r3 2  0

r3 2 r2 2  1  0 r 3 2 r  1  0

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338


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

r3 2  0

or r  1  0

r   0

r  1

32

2

2

r  1

r3  0

r  1

r  0

Both answers check.

Find all real solutions of each equation. 41.

x  2  3  2 Solution

x  2  3  2 x  2  5

 x  2  5 2

2

x  2  25 x  27 The solution checks. 42.

a  3  5  0 Solution

a  3  5  0 a  3  5

 a  3  5 2

2

a  3  25 a  28 The solution checks. 43. 3 x  1 

6

Solution

3 x  1 

6

3 x  1   6  2

2

9 x  1  6 9x  9  6

9 x  3 x   31 The solution checks.

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339


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

44.

x  3  2 x Solution

x  3  2

x

 x  3   2 x  2

2

x  3  4x 3  3x 1  x The solution checks. 45.

5a  2 

a  6

Solution

5a  2 

a  6

 5a  2   a  6  2

2

5a  2  a  6 4a  8 a  2 The solution checks. 46.

16x  4 

x  4

Solution

16 x  4 

x  4

 16x  4    x  4  2

2

16 x  4  x  4 15 x  0 x  0 The solution checks. 47. 2 x 2  3 

16 x  3

Solution

2 x2  3 

16 x  3

2 x  3   16x  3 2

2

2

4 x 2  3  16 x  3 4x

2

 12  16 x  3

4 x 2  16 x  15  0

2x  32x  5  0

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340


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

2x  3  0

or 2 x  5  0

 32

x   52

x 

Both solutions check.

7 x  11

x2  1 

48.

6

Solution

7 x  11

x2  1 

6 2

 7 x  11   x  1     6   7 x  11 x2  1  6 2 6 x  6  7 x  11 2

2

6x 2  7 x  5  0

2x  13x  5  0

2x  1  0 or 3x  5  0

x 

1 2

x   53

Both solutions check.

x 2  21  x  3

49.

Solution

x 2  21  x  3

 x  21   x  3 2

2

2

x 2  21  x 2  6 x  9 21  6 x  9 12  6 x 2  x The solution checks.

5  x 2   x  1

50.

Solution

5  x 2    x  1 5  x2

   x  1 2

2

5  x 2   x  1

2

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341


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

5  x 2  x 2  2x  1 0  2x 2  2x  4

0  2 x  2 x  1 x  2  0

x  1  0

or

x  2

x  1

x  1 does not check and is extraneous.

51.

x  37  x  5 Solution

x  37  x  5

 x  37    x  5 2

2

x  37  x 2  10 x  25 0  x 2  11x  12

0   x  12 x  1 x  12  0

x  1  0

or

x  12

x  1

x  1 does not check and is extraneous.

52.

10  x  x  4  0 Solution

10  x  x  4  0 10  x  x  4

 10  x    x  4 2

2

10  x  x 2  8 x  16 0  x2  7 x  6

0   x  6 x  1 x  6  0

or

x  1  0

x  6

x  1

x  6 does not check and is extraneous. 53.

3z  1  z  1 Solution 3z  1  z  1

 3z  1  z  1 2

2

3z  1  z 2  2z  1

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342


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

0  z 2  5z

0  z  z  5 z  0 or

z  5  0

z  0

z  5

z  0 does not check and is extraneous. 54.

y  2  4  y

Solution

y  2  4  y

 y  2   4  y  2

2

y  2  16  8 y  y 2 0  y 2  9 y  14

0   y  2 y  7

y  2  0 or

y  7  0

y  0

y  7

y  7 does not check and is extraneous. 55. x 

7 x  12  0

Solution

x 

7 x  12  0 x 

7 x  12

x2 

 7 x  12

2

x 2  7 x  12 x 2  7 x  12  0

 x  4 x  3  0 x  4  0 or

x  3  0

x  4

x  3

Both solution checks. 56. x 

4x  4  0

Solution x 

4x  4  0 x 

4x  4

x2 

 4x  4

2

x2  4x  4

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343


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x2  4x  4  0

 x  2 x  2  0 x  2  0 or

x  2  0

x  2

x  2

The solution checks.

6x  6  3 5

57. x  4  Solution

x  4  x  1 

6x  6  3 5 6x  6 5 2

   x  1   6x 5 6    x  6 6 x 2  2x  1  5 2 5 x  10 x  5  6 x  6 2

5x 2  4 x  1  0

5x  1 x  1  0 5 x  1  0 or x  1  0 x 

x  1

1 5

Both solutions check.

8x  43  1  x 3

58.

Solution 8 x  43  1  x 3 8 x  43  x  1 3    

8 x  43    3 

2

  x  1

2

8 x  43  x2  2x  1 3 8 x  43  3 x 2  6 x  3 0  3 x 2  2 x  40

0  3 x  10 x  4

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344


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

3x  10  0

or x  4  0

x  4

 10 3

x 

does not check and is extraneous. x   10 3 59.

x2  1  2 2 x  2 Solution

x2  1  2 2 x  2    

2

2 x2  1    2 2 x  2  2 x  1  8 x  2 x 2  1  8 x  16

x 2  8 x  15  0

 x  3 x  5  0 x  3  0 or

x  5  0

x  3

x  5

Both solutions check. 60.

x2  1 3x  5

2

Solution

x2  1 3x  5    

2

2

2 x2  1    2 3 x  5  x2  1  2 3x  5 x 2  1  6 x  10

 

x2  6x  9  0

 x  3 x  3  0 x  3  0 or

x  3  0

x  3

x  3

The solutions check.

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345


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

61.

2p  1  1 

p

Solution

2p  1  1 

p

 2p  1  1   p  2

2

2p  1  2 2p  1  1  p p  2  2 2p  1

 p  2

2

 2 2p  1

p2  4 p  4  42p  1

2

p2  4 p  4  8p  4 p2  4 p  0

p p  4  0 p  0 or p  4  0 p  0

p  4

Both solutions check. 62.

r 

r  2  2

Solution r 

r  2  2 r  2 

r  2

 r   2 

r  2

2

2

r  4  4 r  2  r  2 4 r  2  6

4 r  2  6 2

2

16r  2  36 16r  32  36 16r  4 r 

4 16

1 4

The solutions check. 63.

x  3 

2x  8  1

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346


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

x  3 

2x  8  1

 x  3   2x  8  1 2

2

x  3  2x  8  2 2x  8  1 2 2x  8  x  6

2 2x  8   x  6 2

2

42 x  8  x 2  12 x  36 8 x  32  x 2  12x  36 0  x2  4x  4

0   x  2 x  2 x  2  0

or

x  2  0

x  2

x  2

The solution checks. 64.

x  2  1 

2x  5

Solution

x  2  1 

2x  5

 x  2  1   2x  5 2

2

x  2  2 x  2  1  2x  5 2 x  2  x  2

2 x  2   x  2 2

2

4 x  2  x 2  4 x  4 4x  8  x2  4x  4 0  x2  4

0   x  2 x  2 x  2  0

or

x  2  0

x  2

x  2

Both solutions check. 65.

y  8 

y  4  2

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347


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

y  8 

y  4  2 y  8 

y  4  2

 y  8   y  4  2 2

y  8  y  4  4 4

2

y  4  4

y  4  8

4 y  4   8 2

2

16 y  4  64

16 y  64  64 16 y  128 y  8 The solution does not c he c k .  No so lu t ion .

66.

z  5  2 

z  3

Solution

z  5  2 

z  3

 z  5  2   z  3 2

2

z  5  22 z  5  4  z  3

12  4 z  5

122  4 z  5 144  16 z  5

2

144  16z  80 64  16z 4  z

The solution checks. 67.

2b  3 

b  1 

b  2

Solution 2b  3 

 2b  3  2b  3  2

b  1  b  1

b  2

   b  2 2

2

2b  3b  1  b  1  b  2

3b  4  2 2b2  5b  3  b  2 2b  6  2 2b2  5b  3

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348


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

2b  6

2

 2 2b2  5b  3

2

4b2  24b  36  4 2b2  5b  3

4b2  24b  36  8b2  20b  12 0  4b2  4b  24

0  4b  3b  2 b  3  0 or b  2  0 b  3

b  2

b   2 does not check, so it is an extraneous solution.

a  1 

68.

3a 

5a  1

Solution

a  1 

3a 

 a1

3a

5a  1

   5a  1 2

2

a  1  2 3aa  1  3a  5a  1 4a  1  2 3a2  3a  5a  1 2 3a2  3a  a

2 3a  3a   a 2

2

4 3a2  3a

2

a

2

12a2  12a  a2 11a2  12a  0

a 11a  12  0 a  0 or 11a  12  0 a  0

a   12 11

does not check, so it is an extraneous solution. a   12 11

b 

69.

b  8  2

Solution

b    

b  b 

b  8  2 b  8  

2

 22

b  8  4 b  8  4 

b

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349


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

 b  8   4  b  2

2

b  8  16  8 b  b 8 b  8 b  1

 b  1 2

2

b  1  The solution checks.

x  19 

70.

x  2 

3

Solution

x  19    

x  2  x  2  

x  19  x  19 

2

3

 3

2

x  2  3 x  19  3 

x  2

 x  19  3 

x  2

2

2

x  19  9  6 x  2  x  2 12  6 x  2 2 

x  2

22 

 x  2

2

4  x  2 6  x  The solutions checks. 71.

3

7x  1  4

Solution 3

7x  1  4

 7 x  1  4 3

3

3

7 x  1  64 7 x  63 x  9 The solution checks.

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350


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

72.

3

11a  40  5

Solution 3

11a  40  5

 11a  40   5 3

3

3

11a  40  125 11a  165 a  15 The solution checks. 73.

3

x3  7  x  1

Solution 3

x3  7  x  1

 x  7    x  1 3

3

3

3

x 3  7  x 3  3x 2  3x  1 0  3x 2  3x  6

0  3 x  2 x  1 x  2  0

or x  1  0

x  2

x  1

Both solutions check. 74.

3

x3  7  1  x

Solution 3

x3  7  1  x 3

x3  7  x  1

 x  7    x  1 3

3

3

3

x 3  7  x 3  3x 2  3x  1

3x 2  3x  6  0

3 x  2 x  1  0 x  2  0 or x  2

x  1  0 x  1

Both solutions check.

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351


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

75.

3

8x 3  61  2x  1

Solution 3

8x 3  61  2 x  1

 8x  61  2x  1 3

3

3

3

8x 3  61  8x 3  12 x 2  6 x  1 0  12 x 2  6 x  60

0  62 x  5 x  2 2x  5  0

or x  2  0

x   52

x  2

Both solutions check. 76.

3

8x 3  37  2x  1

Solution 3

3

8x 3  37  2 x  1

8x 3  37

  2x  1 3

3

8x 3  37  8x 3  12 x 2  6 x  1 12 x 2  6 x  36  0

62x  3 x  2  0 2x  3  0

or x  2  0

 32

x  2

x 

Both solutions check. 77. 4 30t  25  5

Solution 4

30t  25  5

 30t  25  5 4

4

4

30t  25  626 30t  600 t  20 The solution checks.

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352


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

78. 4 3z  1  2

Solution 4

3z  1  2

 3z  1   2 4

4

4

3z  1  16 3z  15 z  5 The solution checks. 79.

5

5

14

2 x  11 

5

2x  11 

Solution 5

14

 2x  11   14  5

5

5

5

2 x  11  14 2 x  25 x 

25 2

The solution checks. 80.

5

x 2  24  1

Solution 5

5

x 2  24  1

x 2  24

  1 5

5

x 2  24  1 x 2  25 x  5 Both solutions check. Fix It In exercises 81 and 82, identify the step the first error is made and fix it. 81. Solve the radical equation:

2x  1  x  2

Solution Step 3 was incorrect. Step 1:

2x  1  x  2

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353


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Step 2:

 2 x  1    x  2 2

2

Step 3: 2 x  1  x 2  4 x  4 Step 4: 0  x 2  6 x  5 Step 5: 0   x  5 x  1 Step 6: x  5,  1, however, x   1 is an extraneous solution, so x  5 82. Solve for x by making a substitution: x 2 3  7 x 1 3  8

Solution Step 5 was incorrect. Step 1: x 2 3  7 x 1 3  8  0 Step 2: Let u = x 1 3 Step 3: u2  7u  8  0 Step 4: u  8u  1  0 Step 5: u  8 or u = 1 Step 6: x 1 3  8 or x 1 3  1 Step 7: x  512 or x  1

Applications 83. Height of a bridge The distance d (in feet) that an object will fall in t seconds is given by the following formula. To find the height of a bridge above a river, a man drops a stone into the water. (See the illustration.) If it takes the stone 5 seconds to hit the water, how high is the bridge?

t 

d 16

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354


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution t 

d 16

5 

d 16 2

 d  5     16    d 25  16 400  d  The bridge is 400 feet high. 2

84. Horizon distance The higher a lookout tower, the farther an observer can see. (See the illustration.) The distance d (called the horizon distance, measured in miles) is related to the height h of the observer (measured in feet) by the following formula.

d 

1.5h

How tall must a tower be for the observer to see 30 miles?

Solution

d 

1.5h

30 

1.5h

302 

 1.5h

2

900  1.5h 600  h The tower must be 600 feet tall. 85. Carpentry During construction, carpenters often brace walls, as shown in the illustration. The appropriate length of the brace is given by the following formula.

l 

f 2  h2

If a carpenter nails a 10-foot brace to the wall 6 feet above the floor, how far from the base of the wall should he nail the brace to the floor?

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355


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

l 

f 2  h2

10 

f 2  62

102 

f 2  36

2

100  f 2  36 64  f 2 8  f He should nail the brace to the floor 8 feet from the wall. 86. Windmills The power generated by a windmill is related to the velocity of the wind by the following formula where P is the power (in watts) and v is the velocity of the wind (in mph).

 

3

P 0.02

To the nearest 10 watts, find the power generated when the velocity of the wind is 31 mph.

Solution

v 

3

P 0.02

31 

3

P 0.02

 P   31   3  0.02    P 29791  0.02 297910.02  p

3

3

600  P The power generated is about 600 watts.

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356


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

87. Diamonds The effective rate of interest r earned by an investment is given by the following formula where P is the initial investment that grows to value A after n years.

r 

n

A  1 P

If a diamond buyer got $4000 for a 1.03-carat diamond that he had purchased 4 years earlier and earned an annual rate of return of 6.5% on the investment, what did he originally pay for the diamond?

Solution

r 

n

A  1 P

0.065 

4

4000  1 P

1.065 

4

4000 P 4

   1.065   4 4000 P   4000 1.286466  P 1.286466P  4000 4

P  3109 The original price was about $3109. 88. Theater productions The ropes, pulleys, and sandbags shown in the illustration are part of a mechanical system used to raise and lower scenery for a stage play. For the scenery to be in the proper position, the following formula must apply: w2 

w 12  w32

If w2 = 12.5 lb and w3 = 7.5 lb, find w1.

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357


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

w2 

w12  w32

12.5 

w12  7.5

12.5

2

2

 w  56.25 2 1

2

156.25  w12  56.25 100  w12  100  w1 w1  10 lb Discovery and Writing 89. Explain the Power Property of Real Numbers.

Solution Answers may vary. 90. Describe what it means for an equation to be quadratic in form.

Solution Answers may vary. 91. Identify two methods that can be used to solve the equation x 4  6 x 2  7  0. Compare and contrast the two methods.

Solution Answers may vary. 92. Outline a strategy that can be used to solve radical equations.

Solution Answers may vary. 93. Explain why squaring both sides of an equation might introduce extraneous roots.

Solution Answers may vary. 94. Can cubing both sides of an equation introduce extraneous roots? Explain.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 95. Factoring can be used to solve x 4  6 x 3  5  0.

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358


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution False. The equation is not quadratic in form. x 4  6 x 2  5  0 can be solved by factoring. 96. The first step used to solve the equation 4 x 4  2x 2 is to divide both sides of the equation by 2 x 2 .

Solution False. The first step is to rewrite the equation so that zero is on one side of the equation. 2

1

1

97. To solve the equation 5 y 3  4 y 3  1  0, we can make the substitution u  y 3 .

Solution True. 1

1

98. The equation x 8  7 x 4  12  0 is quadratic in form.

Solution True.

x  2 then x  16.

99. If

Solution False.

x  2 x  4 x  16 x  256

100. To solve the radical equation

z 

z  2  2, we square each term individually.

Solution False. Isolate one of the radicals first, then square both sides of the equation. 101. To solve the radical equation both sides.

x  1 

2x  3  1, the first step is to square

Solution False. Isolate one of the radicals first, then square both sides of the equation. 102. IF

999

x 2  1, then x  i .

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359


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

True.

999

999

x2

x 2  1

999

  1

999

x 2  1  x  i

EXERCISES 1.7 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Express the graph shown on the number line using interval notation.

Solution

, 3

2. Express the graph shown on the number line using interval notation.

Solution

5, 2

3. Solve the linear equation: 2 x  5  5  3 x  3

Solution

2 x  5  5  3 x  3

2 x  10  5  3 x  9 2 x  10  4  3 x  x  10  4 x  6 x  6 4. Solve the quadratic equation: 3 x 2  7 x  6  0

Solution 3x 2  7 x  6  0

3x  2 x  3  0 3x  2  0

or

x  3  0

3x  2

or

x  3

x 

2 3

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360


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

5. Is –3 a solution of 4 x  1 x  2  0?

Solution

   

Is  4 3 1  3  2  0 ?

13  1  0 13  0 True Yes, –3 is a solution. 6. Is –3 a solution of

3  0? x  3

Solution 3 Is  0? 3  3 No, since

3 is undefined. 0

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If x > y, then x lies to the __________ of y on a number line.

Solution right 8. a < b, __________, a > b.

Solution a=b 9. If a < b and b < c, then __________.

Solution a<c 10. If a < b then a + c < __________.

Solution b+c 11. If a < b then a – c < __________.

Solution b–c 12. If a < b and c > 0, then ac __________ bc.

Solution < © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

361


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

13. If a < b and c < 0, then ac __________ bc.

Solution > 14. If a < b and c < 0, then

a b __________ . c c

Solution > 15. 3 x  5  12 and ax  c  0a  0 are examples of __________ inequalities.

Solution linear 16. ax2  bx  c  0 a  0 and 3x2  6x  0 are examples of __________ inequalities.

Solution quadratic 17. If two inequalities have the same solution set, they are called __________ inequalities.

Solution equivalent 18. An inequality that contains a fraction with a polynomial numerator and denominator is called a __________ inequality.

Solution rational Practice Solve each linear inequality and graph its solution set on a number line. Write the solution

set in interval notation. 19. 3 x  2  5

Solution 3x  2  5 3x  3 x  1 

, 1

20. 2 x  4  6

Solution 2 x  4  6 2x  2 x  1 

1, 

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362


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

21. 3 x  2  5

Solution 3x  2  5 3x  3

x  1  1,  

22. 2 x  4  6

Solution 2 x  4  6 2 x  2 x  1 

,  1

23. 5 x  3  2

Solution 5 x  3  2 5 x  5 x  1 

, 1

24. 4 x  3  4

Solution 4 x  3  4 4 x  1 x   41 

  ,  1 4

25. 5 x  3  2

Solution 5 x  3  2 5 x   5

x  1  1,  

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363


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

26. 4 x  3  4

Solution 4 x  3  4 4 x  1 x   41 

,   1 4

27. 2 x  3  2 x  3

Solution

2 x  3  2 x  3 2 x  6  2 x  6 4 x  12

x  3   , 3

28. 3 x  2  2 x  5

Solution

3 x  2  2 x  5 3 x  6  2 x  10

x  4   , 4

29.

3 x  4  2 5

Solution 3 x  4  2 5 3  5 x  4   52 5 

3 x  20  10 3 x  10 x   10  3

  ,  10 3

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364


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

30.

1 x  3  5 4 Solution 1 x  3  5 4 1  4 x  3  45 4  x  12  20

x  32  32,  

31.

x  3 2x  4  4 3

Solution x  3 2x  4  4 3 x  3 2x  4  12  12  4 3 3 x  3  42 x  4 3 x  9  8 x  16 5 x  25

x  5  5,  

32.

x  2 x  1  5 2

Solution x  2 x  1  5 2 x  2 x  1  10  10  5 2 2 x  2  5 x  1 2x  4  5x  5 3 x  9

x  3   , 3

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365


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

33.

6 x  4 5

3 x  2 4

Solution

6 x  4

3 x  2

5 4 6 x  24 3x  6 20   20  5 4 4 6 x  24  53 x  6 24 x  96  15 x  30 9 x  126

x  14  14,  

34.

3 x  3 2

2 x  7  3

Solution

3 x  3

2 x  7

2 3 3x  9 2 x  14  6  6  2 3 33 x  9  22 x  14 9 x  27  4 x  28 5x  1 x 

35.

1 5

,  1 5

5 4 a  3  a  a  3  1  9 3

Solution 5 4 a  3  a  a  3  1  9 3 5  4  9  a  3  a  9  a  3  1 9  3 

5a  3  9a  12a  3  9 5a  15  9a  12a  36  9

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366


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

16a  60 60 16 15 a  4  a 

36.

,  15 4

2 3 y  y    y  5 3 2

Solution 2 3 y  y    y  5 3 2 2   3  6 y  y   6   y  5 3   2  4 y  6 y  9 y  5 2 y  9 y  45 7 y  45 y 

37.

45 7

,  45 7

2 3 3 2 1 a  a  a    3 4 5 3 3

Solution

2 3 3 2 1 a  a  a    3 4 5 3 3 3  2 3  2 1 60 a  a   60   a     4  3 3 3 5   2 40a  45a  36 a    20 3  5a  36a  24  20 41a  44  a   44 41

  ,  44 41

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367


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solve each compound inequality and graph its solution set on a number line. Write the solution set in interval notation. 38.

1 2 1 1 b  b   b  1  b 4 3 2 2

Solution 1 2 1 1 b  b   b  1  b 4 3 2 2 1 1  2 1 12 b  b    12  b  1  b 4 3 2 2     3b  8b  6  6b  1  12b 11b  6  6b  6  12b 7b  12 b   12  7

,   12 7

39. 4  2 x  8  10

Solution 4  2 x  8  10 12 

2x

 18

12 

x

9 

6, 9

40. 3  2 x  2  6

Solution 3  2x  2  6 1 

2x

 4

1 2

x

 2   21 , 2

41. 9 

x  4  2 2

Solution x  4  2 2 18  x  4  2 9 

22 

x

8 

x

 8

 22  8, 22

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368


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  2  6 6

42. 5 

Solution x  2 5   6 6 30  x  2  36 32 

x

 38  32, 38

4  x  5 3

43. 0 

Solution 4  x 0   5 3 0  4  x  15 4 

x

 11

4 

x

 11

11 

x

4  11, 4

5  x  10 2

44. 0 

Solution 5  x  10 2 0  5  x  20 0 

5 

x

 25

x

 25  5, 25

5

45. 2 

1  x  10 2

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369


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 1  x  10 2 4  1  x  20 2 

5 

x

 21

x

 21  5, 21

5

46. 2 

1  x  10 2

Solution 1  x 2   10 2 4  1  x  20 5 

x

 19

x

 19

19 

x

5

5  19, 5

47. 3 x  2 x   x

Solution 3 x  2 x   x

3 x  2 x and 2 x   x x  0

x  0

x  0

x  0

x  0   , 0

48. 3 x  2 x   x

Solution 3 x  2 x   x

3 x  2 x and 2 x   x x  0

x  0

x  0

x  0

x  0  0, 

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370


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

49. x  2 x  3 x

Solution x  2x  3x

x  2 x and 2 x  3 x x  0

x  0

x  0

x  0

x  0  0, 

50. x  2 x  3 x

Solution x  2x  3x

x  2 x and 2 x  3 x x  0

x  0

x  0

x  0

x  0   , 0

51. 2 x  1  3 x  2  12

Solution 2 x  1  3 x  2  12

2 x  1  3 x  2 and 3 x  2  12  x  3

3 x  14

x  3

x 

14 3

x  3 x 

14 3

x  3 and x 

14 3

Solution set: 3, 14 3

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371


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

52. 2  x  3 x  5  18

Solution 2  x  3 x  5  18

2  x  3 x  5 and 3 x  5  18 4 x  3

3 x  13

x   43

x 

13 3

x   43

x 

13 3

x   43 and

x 

13 3

Solution set:  43 , 13 3

53. 2  x  3 x  2  5 x  2

Solution 2  x  3x  2  5x  2

2  x  3 x  2 and 3 x  2  5 x  2 2 x  4

2 x  4

x  2

x  2

x  2 x  2

x  2

and x  2 Solution set: 2,  54. x  2 x  3  4 x  7

Solution x  2x  3  4 x  7

x  2 x  3 and 2 x  3  4 x  7 x  3

2 x  10

x  3

x  5

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372


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  3 x  5 x  3 and

x  5

Solution set:  ,  3 55. 3  x  7 x  2  5 x  10

Solution 3  x  7 x  2  5 x  10

3  x  7 x  2 and 7 x  2  5 x  10 6 x  5 x  x 

2 x  8 x  4

5 6

5 6

x  4

x 

5 6

and x  4

Solution set: 4, 65

56. 2  x  3 x  1  10 x

Solution 2  x  3 x  1  10 x

2  x  3 x  1 and 3 x  1  10 x 4 x  1 x  x 

1 4

x 

1 7

1 4

7 x  1 x 

1 7

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373


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 

1 4

and

x 

1 7

Solution set:

 ,  1 4

57. x  x  1  2 x  3

Solution

x  x  1  2x  3 x  x  1

and x  1  2 x  3

0  1

x  2

true for all real

x  2

numbers x 0  1 x  2

0  1

and x  2 Solution set:   2,   58.  x  2 x  1  3 x  1

Solution  x  2 x  1  3 x  1  x  2 x  1 and 2 x  1  3 x  1 x  1

x  0

x  1

x  0 x  1 and

x  0

Solution set: 1,  

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374


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

59. x 2  7 x  12  0

Solution

x 2  7 x  12  0

 x  3 x  4  0 factors  0: x  3, x  4

intervals:  , 4,  4, 3,  3,  interval

,  4 4,  3 3,  

test number

value of x

2

 7 x  12

–5

+2

–3.5

–0.25

0

+12

Solution set:  4,  3

60. x 2  13 x  12  0

Solution

x 2  13 x  12  0

 x  12 x  1  0 factors  0: x  12, x  1

intervals:  , 1,  1, 12,  12,  interval

, 1 1, 12 12,  

value of

test number

x 2  13 x  12

0

+12

2

–10

13

+12

Solution set: 1, 12

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375


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

61. x 2  5 x  6  0

Solution

x 2  5x  6  0

 x  3 x  2  0

factors  0: x  3, x  2

intervals:  , 2, 2, 3, 3,  interval

value of

test number

x 2  5x  6

0

+6

2.5

–0.25

4

+2

, 2 2, 3 3, 

Solution set:  , 2  3,  

62. 6 x 2  5 x  6  0

Solution

6x 2  5x  6  0

2x  33x  2  0 factors  0: x   32 , x 

2 3

intervals: ,  32 ,  32 , 23 ,

2 3

interval



test number

,    ,   , 

  , 

value of 6x

2

 5x  6

3 2

–2

+8

3 2 2 3

0

–6

1

+5

2 3

  , 

Solution: ,  32 

2 3

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376


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

63. x 2  5 x  6  0

Solution

x 2  5x  6  0

 x  3 x  2  0 factors  0: x  3, x  2

intervals:  ,  3,  3,  2,  2,  interval

, 3 3, 2 2,  

test number

value of x

2

 5x  6

–4

+2

–2.5

–0.25

0

+6

Solution set:  3,  2

64. x 2  9 x  20  0

Solution

x 2  9 x  20  0

 x  4 x  5  0 factors  0: x  4, x  5

intervals:  , 5,  5, 4,  4,  interval

, 5 5,  4 4, 

test number

value of x

2

 9 x  20

–6

+2

–4.5

–0.25

0

+20

Solution:  , 5   4,  

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377


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

65. 2 x 2  5 x  3  0

Solution

2x 2  5x  3  0

2x  1 x  3  0 1 factors  0: x   , x  3 2  1  1  intervals:  ,  ,   , 3, 3,  2  2   value of

interval

test number

 1  ,   2 

−1

4

 1    , 3  2 

0

–34

3, 

4

9

 1 Solution:  ,    3,  2 

2x

2

 5x  3

66. 3 x 2  5 x  2  0

Solution

3x 2  5x  2  0

3x  1 x  2  0 1 , x  2 3   1  1 intervals:  , 2,  2, ,  ,   3 3     value of test interval 2 number 3x  5x  2 factors  0: x 

, 2

−3

10

 1  2,  3 

0

–2

1   ,  3 

1

6

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378


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

 1 Solution:  2,  3 

67. 6 x 2  5 x  1  0

Solution

6x2  5x  1  0

2x  13x  1  0 factors  0: x   21 , x   31

 



intervals: ,  21 ,  21 ,  31 ,  31 ,  test number

interval

,    ,    ,  1 2

value of 6x

2

 5x  1

1 2

−1

+2

1 3

–0.4

–0.04

0

+1

1 3

 

Solution: ,  21   31 , 

68. x 2  9 x  20  0

Solution

x 2  9x  20  0

 x  5 x  4  0 factors  0: x  5, x  4

intervals:  , 5,  5, 4,  4,  interval

, 5 5, 4 4, 

value of

test number

x 2  9 x  20

−6

+2

–4.5

–0.25

0

+20

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379


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution:  5,  4

69. 6 x 2  5 x  1

Solution

6 x 2  5 x  1 6x 2  5x  1  0

2x  13x  1  0 factors  0: x 

intervals: ,

,  ,   , 

1 3

  , ,  ,  

1 , 3

1 3

1 2

1 2

value of

test number

interval

1 3

1 ,x 2

6x

2

 5x  1

1 3

0

+1

1 2

0.4

–0.04

1

+2

1 2

Solution:

,  1 3

1 2

70. 2 x 2  3  x

Solution

2x 2  3  x 2x 2  x  3  0

2x  3 x  1  0

factors  0: x   32 , x  1

 

intervals: ,  32 ,  32 , 1 ,  1, 

value of

test number

2x 2  x  3

–2

+3

3 2

0

–3

1, 

2

+7

interval

,    , 1 3 2

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380


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution: ,  32   1, 

71. 4x 2  4x  1  0

Solution

4x2  4x  1  0

2x  12x  1  0 1 2   1  1 intervals:  , ,  ,   2 2     factors  0: x 

value of

test number

interval

, 12  12, 

4x

2

 4x  1

0

1

1

1

 1  1 Solution:  ,    ,   2  2 

72. 9 x 2  24 x  16

Solution

9 x 2  24 x  16 9 x 2  24 x  16  0

3x  43x  4  0 factors  0: x   43 , x   43

 

intervals: ,  43 ,  43 , 

interval

,     ,  4 3

4 3

test number

value of 9x

2

 24 x  16

–2

+4

0

+16

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381


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

 

Solution: ,  43   43 , 

73. 9 x 2  24 x  16

Solution

9x 2  24 x  16 9x 2  24 x  16  0

3x  43x  4  0 factors  0: x 

4 3

4 3

test number

,   ,  4 3

4 3

Solution: x 

  , 

intervals: , 43 ,

interval

4 , x 3

value of 9x

2

 24 x  16

0

+16

2

+4

4 , or  43 , 43  3

74. 25x2  20x  4

Solution

25x 2  20 x  4  0

5x  25x  2  0

2 5  2  2  intervals:  ,  ,   ,   5    5  factors  0: x  

value of

interval

test number

25 x 2  20 x  4

 2  ,   5 

−1

49

 2    ,   5 

0

4

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382


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

 2 2 Solution:   ,    5 5

75. x 2  2x  1

Solution

x 2  2x  1  0

 x  1 x  1  0

factors  0: x  1

intervals:  , 1, 1,  

interval

value of

test number

x 2  2x  1

0

1

2

1

, 1 1,  Solution:  , 

76.  x 2  6x  9

Solution

0  x2  6x  9

0   x  3 x  3 factors  0: x  3

intervals:  , 3, 3,  interval

test number

, 3 3, 

value of x

2

 6x  9

0

9

4

1

Solution:  , 

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383


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

77. x2  3  0

Solution x2  3  0 x2  3  0 x2  3 x   3



intervals: ,  3 ,  3, test number

interval

  3, 

3,

value of x2  3

–2 +1 ,  3 0 –3  3, 3 2 +1  3,  Solution:  ,  3   3, 

78. x2  7  0

Solution x2  7  0 x2  7  0 x2  7 x2   7



intervals: ,  7 ,  7, interval

test number

–3 ,  7  0  7, 7 3  7,  Solution:   7, 7 

  7, 

7,

value of x2  7 +2 –7 +2

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384


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

79. x2  11  0

Solution x 2  11  0 x 2  11  0 x 2  11 x  

11



intervals: ,  11 ,  11, test number

interval

–4 ,  11 0  11, 11 4  11,  Solution:   11, 11

  11, 

11 ,

value of x 2  11 +5 –11 +5

80. x2  20  0

Solution x 2  20  0 x 2  20  0 x 2  20 x   20  2 5





interval

test number

value of

, 2 5

–5

+5

intervals: , 2 5 , 2 5, 2 5 , 2 5, 

x2  20

0 –20 2 5, 2 5 5 +5  2 5,  Solution:  , 2 5   2 5, 

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385


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solve the rational inequality and graph its solution set on a number line. Write the solution set in interval notation. x  3  0 81. x  2 Solution x  3  0 x  2 factors  0: x  3, x  2

intervals:  ,  3,  3, 2, 2,  

interval

, 3 3, 2 2,  Solution:  3, 2

82.

test number

sign of xx  23

−4

+

0

3

+

x  3  0 x  2 Solution x  3  0 x  2 factors  0: x  3, x  2

intervals:  ,  3,  3, 2, 2,  interval

test number

−4 ,  3 0 3, 2 3 2,  Solution:  ,  3  2,  

83.

sign of xx  23 + – +

x2  x  0 x  1

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386


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

x2  x x2  1 x  x  1

 x  1 x  1

 0  0

factors  0: x  0, x  1, x  1

intervals:  ,  1,  1, 0, 0, 1,  1,  

interval

,  1 1, 0 0, 1 1, 

test number

sign of x 2  x

−2

+

 21

+

1 2

2

+

2

x 1

Solution:  ,  1   1, 0   1,  

84.

x2  4 x2  9

 0

Solution

x2  4 x2  9

 0 

 x  2 x  2  0  x  3 x  3

factors  0: x  2, x  3

intervals:  , 3,  3, 2,  2, 2, 2, 3, 3,  interval

test number

−4 , 3 –2.5 3, 2 0 2, 2 2.5 2, 3 4 3,  Solution:  3,  2  2, 3

2

sign of x 2  4 x 9

+ – + – +

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387


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

85.

x 2  5x  6 x2  x  6

 0

Solution

x 2  5x  6 x

2

 x  6

 0 

 x  3 x  2  0  x  3 x  2

factors  0: x  3, x  2

intervals:  , 3,  3, 2,  2, 2, 2,  interval

, 3 3, 2 2, 2 2, 

test number

sign of x 2 5 x  6

−4

+

–2.5

+

0

3

+

2

x  x 6

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:  ,  3   3,  2  2,  

86.

x 2  10x  25 x 2  x  12

 0

Solution

x 2  10 x  25 x

2

 x  12

 0 

 x  5 x  5  0  x  3 x  4

factors  0: x  3, x  4, x  5

intervals:  , 5,  5, 3,  3, 4, 4,  interval

, 5 5, 3 3, 4 4, 

test number

sign of x 2 10 x  25

−6

+

–4

+

0

5

+

2

x  x  12

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0.

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388


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution:  5,  5   3, 4

87.

6x2  x  1 x2  4x  4

 0

Solution

6x2  x  1 x2  4x  4

 0 

2x  13x  1  0  x  2 x  2

1 ,x 2

factors  0: x 

  31 , x  2



  , 

intervals:  ,  2, 2,  31 ,  31 , 21 , test number

sign of 62x  x  1

−3

+

1 3

–1

+

1 2

0

1

+

interval

, 2

2,    ,   ,  1 3

1 2

2

x  4x  4

  , 

Solution:  ,  2  2,  31 

88.

6x 2  3x  3 x 2  2x  8 6x 2  3x  3 x

 2x  8

1 2

 0

Solution 2

1 2

 0 

32 x  1 x  1

 x  2 x  4

 0

factors  0: x   21 , x  1, x  2, x  4



intervals:  ,  2, 2,  21 ,  21 , 1 ,  1, 4, 4,  interval

, 2

2,    , 1 1 2

1 2

1, 4 4, 

test number

sign of 6 x2  3 x  3

−3

+

–1

0

+

2

5

+

2

x  2x  8

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389


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution: 2,  21   1, 4

89.

3  2 x Solution 3  2 x 3  2  0 x 3  2x  0 x factors  0: x 

3 ,x 2

  , 

intervals:  , 0, 0, 32 ,

3 2

interval

test number

sign of 3 x2 x

, 0

−1

1

+

2

0,   ,  3 2

3 2

Solution: 0, 32

90.

 0

3  2 x Solution 3  2 x 3  2  0 x 3  2x  0 x factors  0: x 

3 ,x 2

 0

  , 

intervals:  , 0, 0, 32 ,

3 2

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390


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

interval

test number

sign of 3 x2 x

, 0

−1

1

+

2

0,   ,  3 2

3 2

Solution:  , 0 

91.

 ,  3 2

20  10 x Solution

20  10 x 20  10  0 x 20  10 x  0 x factors  0: x  0, x  2 intervals:  , 0, 0, 2, 2,   interval

, 0 0, 2 2, 

test number

sign of

20  10x x

−1

1

+

3

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:  , 0  2,  

92.

21  7 x

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391


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

21  7 x 21  7  0 x 21  7 x  0 x factors  0: x  0, x  3 intervals:  , 0, 0, 3, 3,   interval

, 0 0, 3 3, 

test number

sign of

21  7x x

–1

1

+

4

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution: 0, 3

93.

4  2 x  4 Solution 4  2 x  4 4  2  0 x  4 4  2x  8  0 x  4 12  2 x  0 x  4 factors  0: x  4, x  6 intervals:  , 4, 4, 6, 6,   interval

, 4 4, 6 6, 

test number

sign of

12  2 x x  4

0

5

+

7

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392


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Do not include endpoints which make the numerator or denominator equal to 0. Solution: 4, 6

94.

15  5 x  2 Solution 15  5 x  2 15  5  0 x  2 15  5 x  10  0 x  2 25  5 x  0 x  2 factors  0: x  2, x  5

intervals:  , 5,  5,  2,  2,   Interval

Test number

Sign of 25  5 x x  2

−6

−3

+

0

, 5 5, 2 2, 

Do not include endpoints which make the numerator or denominator equal to 0. Solution:  5,  2

95.

3  5 x  2 Solution

3  5 x  2 3  5  0 x  2 5 x  2 3   0 x  2 x  2

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393


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

3  5 x  10  0 x  2 13  5 x  0 x  2 factors  0: x 

13 , x 5

intervals:  , 2, 2,

 2

  , 

13 , 5

13 5

interval

test number

sign of 13x52x

, 2

0

11 5

+

3

2,   ,  13 5

13 5

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. ,  Solution:  , 2   13 5

96.

3  4 x  2 Solution

3  4 x  2 3  4  0 x  2 4 x  2 3   0 x  2 x  2 3  4x  8  0 x  2 4 x  5  0 x  2 factors  0: x  2, x   45

 

intervals:  ,  2, 2,  45 ,  45 , 

interval

test number

sign of 4x x 25

, 2

–3

 47

+

0

2,    ,  5 4

5 4

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394


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:  ,  2   45 , 

97.

2x  3 x  1 Solution 2x  3 x  1 2x  3  0 x  1 2x  3x  3  0 x  1 x  3  0 x  1 factors  0: x  1, x  3

intervals:  , 1,  1, 3, 3,   interval

, 1 1, 3 3, 

test number

sign of

x  3 x  1

0

2

+

4

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:  , 1  3  

98.

2 x  3  4 x  3

Solution 2 x  3  4 x  3 2 x  3  4  0 x  3 2 x  3  4 x  12  0 x  3

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395


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

6 x  9  0 x  3

factors  0: x  3, x  

3 2

 3  3  intervals: , 3 , 3 ,  3,  ,   ,   2  2  



6 x  9 x  3

interval

test number

,  3

−4

 3  3,   2 

−2

+

 3    ,   2 

0

sign of

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0.

 3 Solution:  3,   2  99.

6 x

2

 1

 1

Solution 6 x

2

 1

 1

6

 1  0  1 x2  1 6   0 x2  1 x2  1 7  x2  0 x2  1 7  x2  0  x  1 x  1 x

2

factors  0: x   7, x  1





 

intervals: ,  7 ,  7, 1 , 1, 1 , 1,

  7, 

7,

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396


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

test number

interval

,  7  7, 1 1, 1

1, 7   7, 

6

x2  1 –

−2

+

0

2

+

3

Solution: ,  7  1, 1 

100.

7  x2

−3

sign of

 7, 

 1

x2  1

Solution 6 x

2

 1

 1

6

 1  0  1 x2  1 6   0 2 x  1 x2  1 7  x2  0 x2  1 7  x2  0  x  1 x  1 x

2

factors  0: x   7, x  1





test number

sign of

 

intervals: ,  7 ,  7, 1 , 1, 1 , 1, interval

,  7  7, 1 1, 1

1, 7  7, 

  7, 

7,

7  x2 x2  1

−3

−2

+

0

2

+

3

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397


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

 

Solution:  7, 1  1,

7

Fix It In exercises 101 and 102, identify the step the first error is made and fix it. 101. Solve the linear inequality: 3 x  4 x  1  5 x  1. Write the solution set using interval notation.

Solution Step 4 was not correct. Step 1: 3 x  4 x  4  5 x  5 Step 2: x  4  5 x  5 Step 3: 4 x  9 Step 4: x  

9 4

 9  Step 5:   ,    4  102. Solve the rational inequality:

6  3. Write the solution set using interval notation. x

Solution Step 5 was not correct. Step 1:

6  3  0 x

Step 2:

6  3x  0 x

Step 3: We establish three intervals:  , 0, 0, 2, 2,   Step 4: The numbers in the interval 0, 2 satisfy the inequality. Step 5: The solution set is 0, 2

Applications Solve each problem. 103. Golfing lessons Macy decides to take golfing lessons. If her new set of golf clubs cost $250 and private lessons are $60 per hour lesson, what is the maximum number of lessons she can take if the total spent for lessons and purchasing clubs is at most $970?

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398


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the number of lessons.

Total cost  970 250  60x  970 60 x  720 x  12 She can take at most 12 lessons. 104. Surfing lessons Dylan and Dusty plan to take weekly surfing lessons together. If the 2-hour lessons are $40 per person and they plan to spend $200 each on new surfboards, what is the maximum number of lessons the two can take if the total amount spent for lessons and surfboards is at most $960?

Solution Let x = the number of lessons.

Total cost  960 400  80 x  960 80 x  560 x  7 She can take at most 7 lessons. 105. Long distance A long-distance telephone call costs 40¢ for the first three minutes and 10¢ for each additional minute. At most how many minutes can a person talk and not exceed $2?

Solution Let x = the number of minutes after 3 minutes. The total cost = 40 + 10x cents.

Total cost  200 40  10 x  200 10 x  160 x  16 A person can talk for up to 16 minutes after the initial 3 minutes, for a total of up to 19 minutes for less than $2. 106. Buying a computer A student who can afford to spend up to $2000 sees the ad shown in the illustration. If she buys a touch-screen laptop, how many games can she buy?

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399


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the number of games. Then the total cost  1695.95  19.95 x.

Total cost  2000 1695.95  19.95 x  2000 19.95 x  304.05 x  15.2 She can buy up to 15 games. 107. Musical items Andy can spend up to $275 on a guitar and some music books. If he can buy a guitar for $150 and music books for $9.75, what is the greatest number of music books that he can buy?

Solution Let x = the number of books. Then the total cost = 150 + 9.75x.

Total cost  275 150  9.75 x  275 9.75 x  125 x  12.8 He can buy up to 12 books. 108. Buying a digital camera Audrey wants to spend less than $600 for a digital camera and some batteries. If the camera of her choice costs $425 and batteries cost $7.50 each, how many batteries can she buy?

Solution Let x = the number of DVDs. Then the total cost = 425 + 7.50x.

Total cost  600 425  7.50 x  600 7.50 x  175 x  23.3 She can buy up to 23 DVDs. 109. Buying a refrigerator Madeline, who has $1200 to spend, wants to buy a refrigerator. Refer to the following table and write an inequality that shows how much she can pay p for the refrigerator.

State sales tax

6.5%

City sales tax

0.25%

Solution Let p = the price of the refrigerator. Then the total cost = p + 0.065p + 0.0025p. Total cost  1200 p  0.065 p  0.0025p  1200 1.0675p  1200 p  1124.122 p  $1124.12

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400


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

110. Renting a rototiller The cost of renting a rototiller is $17.50 for the first hour and $8.95 for each additional hour. How long, to the nearest hour, can a person have the rototiller if the cost must be less than $75?

Solution Let x = the number of hours after the first. Then the total cost = 17.50 + 8.95x.

Total cost  75 17.50  8.95 x  75 8.95 x  57.50 x  6.4 A person could have the rototiller for up to 6 hours after the first hour, for a total of up to 7 hours. 111. Profit Profit occurs when revenue exceeds cost. If the revenue R in dollars from producing and selling x Hugo Boss polo shirts is R = 26x dollars and the cost C is C = 6x + 3660 dollars, what production level produces a profit?

Solution R  C

26 x  6 x  3660 26 x  3660 x  183 112. Profit The revenue R in dollars of producing and selling x Yankee candles is R = 19x and the cost C is C = 3x + 2800. At what production level will revenue exceed cost and the company obtain a profit?

Solution R  C

19x  3x  2800 16 x  2800 x  175 113. Real estate taxes A city council has proposed the following two methods of taxing real estate:

Method 1

$2200 + 4% of assessed value

Method 2

$1200 + 6% of assessed value

For what range of assessments a would the first method benefit the taxpayer?

Solution Let a = the assessed value. Find when Method 1 < Method 2: 2000  0.04a  1200  0.06a 1000  0.02a 50000  a The first method will benefit the taxpayer when a > $50,000.

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401


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

114. Medical plans A college provides its employees with a choice of the two medical plans shown in the following table. For what size hospital bills is Plan 2 better for the employee than Plan 1? (Hint: The cost to the employee includes both the deductible payment and the employee’s coinsurance payment.)

Plan 1

Plan 2

Employee pays $100

Employee pays $200

Plan pays 70% of the rest

Plan pays 80% of the rest

Solution Let b = the hospital bill. Find when Cost of Plan 1 > Cost of Plan 2: 100  0.30( b  100)  200  0.20( b  200) 100  0.30b  30  200  0.20b  40 0.1b  90 b  900 Plan 2 is better for bills over $900.

115. Medical plans To save costs, the college in Exercise 96 raised the employee deductible, as shown in the following table. For what size hospital bills is Plan 2 better for the employee than Plan 1? (Hint: The cost to the employee includes both the deductible payment and the employee’s coinsurance payment.)

Plan 1

Plan 2

Employee pays $200

Employee pays $400

Plan pays 70% of the rest

Plan pays 80% of the rest

Solution Let b = the hospital bill. Find when Cost of Plan 1 > Cost of Plan 2:

200  0.30b  200  400  0.20b  400 200  0.03b  60  400  0.02b  80 0.1b  180

b  1800 Plan 2 is better for bills over $1,800. 116. Geometry The perimeter of a rectangle is to be between 180 inches and 200 inches. Find the range of values for its length l when its width is 40 inches.

Solution Let P = the perimeter. Then the length is equal to 180

P

P  2w P  80 , or . 2 2

200

180  80  P  80  200  80 100

 P  80 

120

100 2

P  80 2

120 2

50

length

60

The length is between 50 and 60 inches.

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402


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

117. Geometry The perimeter of an equilateral triangle is to be between 50 centimeters and 60 centimeters. Find the range of lengths of one side s.

Solution Let P = the perimeter. Then the length of one side is equal to

50

P

 60

50 3 16 23

P 3

P . 3

60 3

 lenght  20

The length of a side is between 16 23 and 20 cm. 118. Geometry The perimeter of a square is to be from 25 meters to 60 meters. Find the range of values for its area A.

Solution Let P = the perimeter. Then the length of one side is equal to P A  s2    4

2

25

P

60

25 4

P 4

60 4

152

 

625 16

 Area  225

25 4

2

 P 4

2

P , so the area is equal to 4

The area is between 625 m2 and 225 m2 . 16 119. Projectile height If a Nerf sports bash ball is projected from ground level with an initial velocity of 160 feet per second, its height s in feet t seconds after being projected is given by the equation s  16t 2  160t. When will the height of the bash ball exceed 144 feet?

Solution

16t 2  160t  144 16t 2  160t  144  0

16 t 2  10t  9  0 t

2

 10t  9  0

t  1t  9  0

factors  0: t  1, t  9

intervals:  , 1,  1, 9, 9, 

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403


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

interval

value of

test number

t  10t  9

0

+9

2

–7

10

+9

, 1 1, 9 9, 

2

Solution:  1, 9. It will exceed 144 ft between 1 and 9 seconds. 120. Projectile height If a jumbo hypercharged pop sky ball is projected from ground level with an initial velocity of 192 feet per second, its height s in feet t seconds after being projected is given by the equation s  16t 2  240t. When will the height of the pop sky ball exceed 576 feet?

Solution

16t 2  240t  576 16t 2  240t  576  0

16 t 2  15t  36  0 t

2

 15t  36  0

t  3t  12  0 factors  0: t  3, t  12

intervals:  , 3, 3, 12,  12,  interval

value of

test number

t  15t  36

0

+36

4

–8

13

+10

, 3 3, 12 12, 

2

Solution: 3, 12. It will exceed 576 ft between 3 and 12 seconds.

Discovery and Writing 121. The techniques used for solving linear equations and linear inequalities are similar, yet different. Explain.

Solution Answers may vary. 122. When graphing the solution set of an inequality, what does a bracket indicate on the number line? What does a parenthesis indicate on the number line?

Solution Answers may vary.

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404


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

123. What is a quadratic inequality? Give two examples.

Solution Answers may vary. 124. What is a rational inequality? Give two examples.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 125. The solution set of the inequality x 2  100  0 is  , 10 .

Solution False. x 2  100  0

 x  10 x  10  0 factors  0: x  10, x  10

intervals:  ,  10,   10, 10,  10,  interval

value of

test number

x2  100

–11

+21

0

–100

11

+21

, 10 10, 10 10,  Solution: 10, 10

126. The solution set of x2  0 is all real numbers.

Solution False. The solution set of x2  0 is all real numbers except 0. 1  2, the first step is multiply both sides by x  10 x  10 to clear the inequality of fractions.

127. To solve the rational inequality

Solution False. The first step is top subtract 2 from both sides of the equation. 128. The solution set of the inequality

100  10 is [0,10]. x

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405


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

100  10 x

False.

100  10  0 x 100  10 x  0 x factors  0: x  0, x  10

intervals:  , 0, 0, 10,  10,  interval

test number

sign of 100 x 10 x

–1

1

+

11

, 0 0, 10 10,  Solution: 0, 10

EXERCISES 1.8 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Identify two real numbers whose absolute value is 12.

Solution 12,  12 2. Tell whether the statement is true or false.  7   6

Solution

 7   6 7  6 True 3. Solve each equation. a.

3x  1  8 4

b.

3x  1  8 4

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406


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution a.

3x  1  8 4 3x  1  32 3x  33 x  11

b.

3x  1  8 4 3 x  1  32 3 x  31 3x 

31 3

4. Solve each equation. a.

6x  2  2x  5

b.

6 x  2   2 x  5

Solution a. 6x  2  2x  5

4 x  2  5 4 x  7 x  

7 4

b. 6 x  2   2 x  5

6 x  2  2 x  5 8x  2  5 8x  3 x 

3 8

5. Solve and write the solution set in interval notation. 9  2 x  5  9

Solution 9  2x  5  9

4  2x  14 2  x  7 2, 7

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407


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

6. Solve and write the solution set in interval notation. 2 x  5  9 or 2 x  5  9

Solution 2x  5  9 or 2x  5  9

2x  4 or 2x  14 x  2 or x  7

, 2  7, 

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If x  0, then x  __________.

Solution x 8. If x  0, then x  __________.

Solution

x

9.

x  k is equivalent to __________.

Solution x  k or x  k 10. a  b is equivalent to a = b or __________.

Solution a  b 11.

x  k is equivalent to __________.

Solution k  x  k 12.

x  k is equivalent to __________.

Solution x  k or x  k 13.

x  k is equivalent to __________.

Solution x  k or x  k

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408


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

a 2  __________.

14.

Solution a

Practice Write each expression without absolute value symbols. 15. 6

Solution 7  7

16.  15

Solution 9  9

17. 0

Solution 0  0

18. 3  5

Solution 3  5  2  2

19. 5   3

Solution 5  3  5  3  2

20.  3  5

Solution 3  5  3  5  8

21.   2

Solution

  2     2    2

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409


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

22.   4

Solution

  4     4  4  

23. x  5 and x  5

Solution x  5 

x  5  x  5

24. x  5 and x  5

Solution x  5 

x  5

   x  5  5  x

25. x 3

Solution

x

3

x3 if x  0   3  x if x  0

26. 2x

Solution

2 x if x  0 2x   2 x if x  0 Solve each equation with one absolute value. 27. x  2  2

Solution

x  2  2 x  2  2 or x  2  2 x  0

x  4

28. 2 x  5  3

Solution

2x  5  3 2x  5  3

or 2x  5  3

2x  2

2x  8

x  1

x  4

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410


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

29. 3 x  1  7  2

Solution

3x  1  7  2 3x  1  5 3x  1  5 or 3 x  1  5 3x  6

3x  4

x  2

x   43

30. 7 x  5  5  8

Solution

7x  5  5  8 7x  5  3 7 x  5  3 or 7 x  5  3 7x  8 x  31.

7x  2

8 7

x 

2 7

3x  4  5 2 Solution

3x  4  5 2 3x  4 3x  4  5 or  5 2 2 3x  4  10 3x  4  10 3x  14

3x  6

14 3

x  2

x  32.

10 x  1 9  2 2 Solution 10 x  1 9  2 2

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411


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

10 x  1 9  2 2 10 x  1  9

or

10 x  1 9   2 2 10 x  1  9

10 x  8 x  33.

8 10

10 x  10 

x  1

4 5

2x  4  7  9 5 Solution

2x  4  6  8 5 2x  4  2 5 2x  4 2x  4  2 or  2 5 5 2x  4  10 2x  4  10

34.

2x  14

2x  6

x  7

x  3

3x  11  15  14 7 Solution

3x  11  15  14 7 3x  11  1 7 3x  11  1 7 3x  11  7

3x  11  1 7 3x  11  7

3x  4

3x  18

 43

x  6

x  35.

or

x  3  5 4 Solution

x  3  2 4 An absolute value can never equal a negative number. no solution

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412


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

36.

x  5  11  10 2 Solution

x  5  3  2 2 x  5  1 2 An absolute value can never equal a negative number. no solution 37.

x  5  0 3 Solution

x  5  0 3 x  5  0 or 3 x  5  0 x  5 38.

x  5  0 3 x  5  0 x  5

x  7  0 9 Solution

x  7  0 9 x  7  0 9 x  7  0

or

x  7  0 9 x  7  0

x  7 39.

x  7

4x  2  3 x Solution 4x  2 x

 3

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413


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

4x  2 4x  2  3 or  3 x x 4 x  2  3x 4 x  2  3x

40.

x  2

7x  2

x  2

x 

2 x  3

2 7

 6

3x Solution

2 x  3 3x

 6

2x  6 2x  6  6  6 or 3x 3x 2x  6  18x 2x  6  18x 16x  6 x  41.

20 x  6

 83

x 

3 10

x  x

Solution x  x

True for all x  0. 42. x  x  4

Solution

x  x  2 x  x  2 x   x  2 or x    x  2 2x  2

x  x  2

x  1

0  2  not true

Solve each equation with two absolute values. 43. x  3 

x

Solution x  3  x

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414


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  3  x or x  3   x 0  3

2 x  3

not true

x   32

44. x  5  5  x

Solution

x  5  5  x x  5  5  x or x  5  5  x  2x  0

x  5  5  x

x  0

0  10 not true

45. x  3  2 x  3

Solution

x  3  2x  3 x  3  2x  3 or x  3  2x  3 x  6

x  3  2x  3

x  6

3x  0 x  0

46. x  2  3 x  8

Solution

x  2  3x  8 x  2  3x  8 or x  2  3x  8 2x  10

x  2  3x  8

x  5

4 x  6 x   32

47. x  2  x  2

Solution

x  2  x  2 x  2  x  2 or x  2   x  2 0  4 not ture

x  2  x  2 2x  0 x  0

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415


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

48. 2 x  3  3 x  5

Solution

2x  3  3x  5 2x  3  3x  5 or 2x  3  3x  5  x  2

2x  3  3x  5

x  2

5x  8 8 5

x  49.

x  3  2x  3 2 Solution

x  3  2x  3 2 x  3  2 x  3 or 2 x  3  4x  6 3 x  9 x  3

x  3  2 x  3 2 x  3  2 x  3 2 x  3  4 x  6 5x  3 x 

50.

3 5

x  2  6  x 3 Solution

x  2  6  x 3 x  2  6  x or 2 x  2  6  x 3 x  2  18  3x

51.

x  2  6  x  2 x  2  6  x 3 x  2  18  3x

4 x  20

2x  16

x  5

x  8

3x  1 2x  3  2 3

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416


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3x  1 2x  3  2 3 3x  1 2x  3  2 3  3x  1 2x  3  6   3   2

or

33 x  1  22 x  3

33 x  1  22 x  3

9x  3  4 x  6

9x  3  4 x  6

5x  9

13 x  3

9 5

3 x   13

x  52.

3x  1 2x  3   2 3  3x  1 2x  3  6    3   2

5x  2  3

x  1 4

Solution

5x  2  3

x  1 4

5x  2 x  1 5x  2 x  1  or   3 4 3 4  5x  2  5x  2 x  1 x  1 12  12     3 4 3 4    

45x  2  3 x  1

45x  2  3 x  1

20 x  8  3x  3

20 x  8  3x  3

17 x  11

23x  5

11  17

5 x   23

x 

Solve each absolute value inequality. Express the solution set in interval notation, and graph it. 53. x  3  6

Solution

x  3  6 6  x  3  6 3 

x

3, 9

 9

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417


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

54. x  2  4

Solution

x  2  4 x  2  4 or x  2  4 x  6

x  2

, 2  6, 

55. x  3  6

Solution

x  3  6 x  3  6 or x  3  6 x  3

x  9

, 9  3, 

56. x  2  4

Solution

x  2  4 4  x  2  4 6 

 2

x 6, 2

57. 2 x  4  10

Solution

2 x  4  10 2 x  4  10 or 2 x  4  10 2x  6

2 x  14

x  3

x  7

, 7  3, 

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418


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

58. 5 x  2  7

Solution

5x  2  7 7  5 x  2  7 5 

5x

 9

1 

x

1, 

9 5

9 5

59. 3 x  5  1  9

Solution 3x  5  1  9 3x  5  8

8  3x  5  8 13 

3x

 3

 13 3

x

 1

 , 1 13 3

60. 2 x  7  3  2

Solution

2x  7  3  2 2x  7  5 2x  7  5

or 2 x  7  5

2 x  12

2x  2

x  6

x  1

, 1  6, 

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419


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

61. x  3  0

Solution x  3  0 x  3  0

x  3  0

or

x  3

x  3

, 3  3, 

62. x  3  0

Solution

x  3  0 0

 x  3 0

3

 3

x 3, 3

63.

5x  2  1 3 Solution 5x  2 3

1 

 1

5x  2 3

 1

3  5x  2  3 5 

5x

 1

1 

x

1, 

1 5

1 5

64.

3x  2  2 4

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420


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 3x  2 4 3x  2 4

 2 3x  2 4

 2 or

 2

3x  2  8

3x  2  8

3x  6

3x  10

x  2

x   10 3

,    2,  10 3

65. 3

3x  1  5 2

Solution

3 3 x2 1  5 3x  1 2

6 

3x  1 2 3x  1 2

5 3 3x  1 2

5 3

 

6  3 x2 1  6  53

 6  53

9x  3  10

or

9x  3  10

9x  13

9x  7

13 9

x   97

x 

.     ,  7 9

66. 2

  53

13 9

8x  2  1 5

Solution

2 8 x5 2  1 8x  2 5

1 2

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421


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

1 1 8x  2   5 2 2  1 8x  2 1 10    10   10  5 2  2 16 x  4  5 5 

9

16 x

1

9  16

x

1 16

 9 , 1   16 16 

67.

x  1

 3

2 Solution

x  1

 3 2 x  1  6 6  x  1  6 5 

68.

x

5, 7

2x  3

 7

 1

3 Solution

2x  3

 1 3 2x  3  3

2x  3  3 or 2x  3  3 2x  6

2x  0

x  3

x  0

. 0  3, 

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422


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solve each compound inequality with absolute value. Express the solution set in interval notation, and graph it. 69. 0  2 x  1  3

Solution

0  2x  1  3 0  2x  1

2x  1  3

and

1 2x  1  0

 2

2x  1  3

2x  1  0

or 2 x  1  0

2 x  1

2 x  1

4 

2x

 2

x   21

x   21

2 

x

 1

3  2 x  1  3

(1)

(2) (1)

(2)

(1) and (2)

2,     , 1 1 2

1 2

70. 0  2 x  3  1

Solution

0  2x  3  1 0  2x  3

and

2x  3  1

1 2x  3  0

 2 2 x  3  1

2 x  3  0 or 2 x  3  0

1  2 x  3  1

2x  3 x 

3 2

2x  3

2 

2x

 4

3 2

1 

x

 2

x 

(1)

(2) (1)

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423


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

(2) (1) and (2)

1,    , 2 3 2

3 2

71. 8  3 x  1  3

Solution

8  3x  1  3 3x  1  3

and

1 3x  1  3

2 3 x  1

3 x  1  3 or 3 x  1  3 3x  4

3 x  2

4 3

 32

x 

x 

8  3x  1  8

8  3 x  1  8

(1)

7 

3x

 9

 73

x

 3

(2)

(1)

(2) (1) and (2)

 ,     , 3 7 3

2 3

4 3

72. 8  4 x  1  5

Solution

8  4x  1  5 4x  1  5

and

1 4 x  1  5 4 x  1  5 or 4 x  1  5

8  4x  1

 2 4 x  1

 8

8  4 x  1  8

4x  6

4 x  4

7 

4x

 9

3 2

x  1

 47 

x

x 

(1)

9 4

(2)

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424


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

(1)

(2)

(1) and (2)

73. 2 

 ,  1   ,  3 2

7 4

9 4

x  5  4 3

Solution

2  2 

x  5  4 3

x  5 3

and

1 x 3 5  2 x  5  2 3 x  5  6

or

x  11

x  5  2 3 x  5  6 x  1

(1)

x  5  4 3 x  5 3

 2

 4

x  5  4 3 12  x  5  12 4  7 

x

 17

(2) (1) (2)

(1) and (2)

74. 3 

7,  1   11, 17

x  3  5 2

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425


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

x  3  5 2

3  3 

x  3 2

x  3  5 2

and

1 x 2 3  3 x  3  3 or 2 x  3  6

x  3  3 2 x  3  6

x  9

x  3 2

 2

 5

x  3  5 2 10  x  3  10 5 

x  3

7 

(1)

 13

x

(2) (1) (2)

(1) and (2) 

75. 10 

7, 3   9, 13

x  2  4 2

Solution

10 

x  2  4 2

x  2  4 2

and

1 x 2 2  4 x  2  4 2 x  2  8 x  10

(1)

or

x  2  4 2 x  2  8 x  6

10 

x  2 2

x  2 2

 2

 10

x  2  10 2 20  x  2  20

10  18 

x

 22

(2)

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426


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

(1) (2) (1) and (2)

76. 5 

18, 6   10, 22

x  2  1 3

Solution

5 

x  2  1 3

x  2  1 3

and

1 x 3 2  1 x  2  1 or 3 x  2  3 x  1

x  2  1 3 x  2  3 x  5

(1)

x  2 3

5 

x  2 3

 2

 5

x  2  5 3 15  x  2  15

5 

17 

x

 13

(2) (1) (2)

(1) and (2)

77. 2 

17, 5   1, 13

x  1  3 3

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427


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

x  1  3 3

2  2 

x  1 3

x  1  3 3

and

1 x 3 1  2 x  1  2 or 3 x  1  6

x  1  2 3 x  1  6

x  5

x  1  3 3

 2

x  7

x  1  3 3 9  x  1  9

3 

10 

(1)

 8

x

(2) (1)

(2) (1) and (2)

78. 8 

10, 7   5, 8

3x  1  2 2

Solution

8 

3x  1  2 2

3x  1  2 2

and

1 3x 2 1  2 3x  1 3x  1  2 or  2 2 2 3x  1  4 3 x  1  4

8 

3x  1 2

 2

3x  1  8 2

3x  1  8 2 16  3x  1  16 8 

3x  3

3 x  5

17 

3x

 15

x  1

x   53

 17  3

x

 5

(1)

(2)

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428


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

(1)

(2)

(1) and (2)

 ,     1, 5 17 3

5 3

Solve each inequality and express the solution using interval notation. 79. x  1 

x

Solution

x  1  x

 x  1  x2 2  x  1  x2 2

x 2  2x  1  x 2 2x  1 x   21 Solution:  21 ,  80. x  1 

x  2

Solution

x  1  x  2

 x  1   x  2 2 2  x  1   x  2 2

2

x 2  2x  1  x 2  4 x  4 2x  3 x   32 Solution:  32 , 

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429


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

81. 2 x  1  2 x  1

Solution

2 x  1  2x  1

2 x  1  2x  1 2 2 2x  1  2x  1 2

2

4x2  4x  1  4x2  4x  1 8x  0

x  0

Solution:  , 0

82. 3 x  2  3 x  1

Solution

3x  2  3x  1

3x  2  3x  1 2 2 3x  2  3x  1 2

2

9x 2  12x  4  9x 2  6x  1 18x  3 1 6

x 

Solution: , 61  83. x  1 

x

Solution

x  1  x

 x  1  x2 2  x  1  x2 2

x 2  2x  1  x 2 2x  1

x   21

Solution: ,  21

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430


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

84. x  2 

x  1

Solution

x  2  x  1

 x  2   x  1 2 2  x  2   x  1 2

2

x 2  4 x  4  x 2  2x  1 2x  3 x   32

Solution: ,  32

85. 2 x  1  2 x  1

Solution

2 x  1  2x  1

2 x  1  2 x  1 2 2 2 x  1  2 x  1 2

2

4x2  4x  1  4x2  4x  1 8x  0 x  0

Solution: 0,  

86. 3 x  2  3 x  1

Solution

3x  2  3x  1

3x  2  3x  1 2 2 3x  2  3x  1 2

2

9x 2  12x  4  9x 2  6x  1 18x  3 x  Solution:

 , 

1 6

1 6

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431


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Fix It In exercises 87 and 88, identify the step the first error is made and fix it. 87. Solve the absolute value equation:

4x  5 3

 4  8.

Solution Step 3 was incorrect. Step 1:

4x  5 3

 4

Step 2: 4 x  5  12 Step 3: 4 x  5  12 or 4 x  5  12 Step 4: 4 x  17 or 4 x  7 Step 5: x 

17 7 or x   4 4

88. Solve the absolute value inequality: 4 x  5  7  5. Write the solution set using interval notation.

Solution Step 2 was incorrect. Step 1: 4 x  5  2 Step 2: 2  4 x  5  2 Step 3: 3  4 x  7 Step 4:

3 7  x  4 4

3 7 Step 5:  ,  4 4 Applications 89. Finding temperature ranges The temperatures on a summer day satisfy the inequality t  78  8, where t is the temperature in degrees Fahrenheit. Express this range

without using absolute value symbols.

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432


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

t  78  8 8  t  78  8 70 

t

 86

90. Finding operating temperatures A tablet has an operating temperature of t  40  80, where t is the temperature in degrees Fahrenheit. Express this range

without using absolute value symbols.

Solution

t  40  80 80  t  40  80 40 

t

 120

91. Range of camber angles The specifications for a certain car state that the camber angle c of its wheels should be 0.6  0.5. Express this range with an inequality containing an absolute value.

Solution

0.6  0.5  1.1 0.6  0.5  1.1 0.1 

c

 1.1

0.6  0.5 

c

 0.6  0.5

 0.5  c  0.6  0.5 c  0.6  0.5 92. Tolerance of a sheet of steel A sheet of steel is to be 0.25 inch thick, with a tolerance of 0.015 inch. Express this specification with an inequality containing an absolute value.

Solution

0.25 0.015  0.265 0.25 0.015  0.235 0.235 

x

 0.265

0.25  0.015 

x

 0.25  0.015

 0.015  x  0.25  0.015 x  0.25  0.015 in. 93. Humidity level A Steinway piano should be placed in an environment where the relative humidity h is between 38% and 72%. Express this range with an inequality containing an absolute value.

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433


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

38  72 110   55 2 2 38  55  17 72  55  17 38 

h

 72

55  17 

h

 55  17

 17  h  55  17 h  55  17 94. Light bulbs A light bulb is expected to last h hours, where h  1500  200. Express this range without using absolute value symbols.

Solution

h  1500  200 200  h  1500  200 1300 

h

 1700

95. Error analysis In a lab, students measured the percent of copper p in a sample of copper sulfate. The students know that copper sulfate is actually 25.46% copper by mass. They are to compare their results to the actual value and find the amount of experimental error. a. Which measurements shown in the illustration satisfy the absolute value inequality p  25.46  1.00? b. What can be said about the amount of error for each of the trials listed in part a?

Solution

p  25.46  1.00 1.00  p  25.46  1.00 24.46 

p

 26.46

a. 24.76% and 26.45% are within the range.

b. The error is less than 1%.

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434


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

96. Error analysis See Exercise 95. a. Which measurements satisfy the absolute value inequality p  25.46  1.00? b. What can be said about the amount of error for each of the trials listed in part a?

Solution

p  25.46  1.00 p  25.46  1.00

or p  25.46  1.00

p  26.46

p  24.46

a. 22.91% and 26.49% are within the range.

b. The error is More than 1%.

97. Physical therapist income The yearly income range in dollars of a physical therapist can be modeled by the inequality

x  93,500 2

 13,250. What is the income range? Write

the answer using interval notation. This is according to ZipRecruiter.

Solution

x  93,500  13,250 2 x  93,500  13,250 2 26,500  x  93,500  26,500 13,250 

67,000  x  120,000 67,000, 120,000 98. Plumber income The yearly income range in dollars of a plumber can be modeled by the inequality

x  51,500 4

 4,375. What is the income range? Write the answer using

interval notation. This is according to ZipRecruiter.

Solution x  51,500  4375 4 x  51,500  4375 4 17,500  x  51,500  17,500 4375 

34,000  x  69,000 34,000, 69,000

Discovery and Writing 99. Explain how to find the absolute value of a number.

Solution Answers may vary.

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435


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

100. Explain why the equation x  9  0 has no solution.

Solution Answers may vary. 101. If k > 0, explain the differences between the solution sets of x

 k and x

 k.

Solution Answers may vary. 102. If k < 0, explain why the solution set of x  k has no solution.

Solution Answers may vary. 103. If k < 0, explain why the solution set of x  k is all real numbers.

Solution Answers may vary. 104. Explain how to solve an inequality with two absolute values.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 105. Absolute value equations always have two solutions.

Solution False. Absolute value equations can have zero, one, or two solutions. 106.  x  x

Solution False.  x

 x only when x  0.

107. The solution set of x  5 is 5,  .

Solution False.

x  5 x  5 or x  5

, 5  5, 

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436


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

108. a  b  a  b

Solution

False. a  b  a  b . 109. x  555  554 has no solution.

Solution

True. x  555  554

x  1 This inequality is never true. 110. The solution set of x  555  554 is all real numbers.

Solution

True. x  555  554

x  1 This inequality is always true.

CHAPTER REVIEW SOLUTIONS Exercises Find the restrictions on x, if any. 1.

4 x  9  11

Solution 3x  7  4 no restrictions on x 2.

x 

1  2 x

Solution

1  2 x restrictions: x  0 x 

3.

4x

 x  1

 8

Solution

1  4 x  1 restrictions: x  1

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437


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

4.

1 2  x  2 x  3 Solution

1 2  x  2 x  3 restrictions: x  2, x  3 Solve each equation and classify it as an identity, a conditional equation, or a contradiction. 5.

39 x  4  28

Solution

39x  4  28 27 x  12  28 27 x  16 x 

16 27

conditional equation 6.

3 a  7a  11 2 Solution

3 a  7a  11 2 3 2  a  2  7a  11 2 3a  14a  154 11a  154 a   154  14 11 conditional equation 7.

83 x  5  4  x  3  12

Solution

83x  5  4 x  3  12 24 x  40  4 x  12  12 20 x  52  12 20 x  64 x 

64 20

16 5

 conditional equation

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438


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

8.

x  3 x  3   2 x  4 x  2 Solution

x  3 x  3   2 x  4 x  2    x  4 x  2 xx  43  xx  32    x  4 x  2  2  

 x  2 x  3   x  4 x  3   x2  6x  8  2 x 2  5x  6  x 2  7 x  12  2x 2  12 x  16 2x 2  12 x  18  2x 2  12 x  16

18  16  no solution, contradiction 9.

3 1  x  1 2 Solution

3 1  x  1 2 3 1  2 x  1  2 x  1  x  1 2 6  x  1 7  x conditional equation 10.

8x2  72x  8x 9  x Solution

8x 2  72x  8x 9  x 2 x  9  x   8x 9  x  8x9  72 x 8x 2  72x  72x  8x 2 all real numbers except –9, indentity 11.

3x 5   3 x  1 x  3

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439


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3x 5   3 x  1 x  3    x  1 x  3 x 3x 1  x 5 3    x  1 x  3  3   3x  x  3  5 x  1 

 x  2x  3  3 2

3 x 2  9x  5 x  5  3x 2  6 x  9 4 x  5  6x  9 2 x  14 x  7  conditional equation 12. x 

1 2x 2  2x  3 2x  3

Solution

1 2x 2  2x  3 2x  3 2   2x  3 x  2x 1 3   2x  3  2x2x 3   x 

2x  3 x  1  2x2 2x 2  3x  1  2x 2 3x  1 x 

13.

4 x2  13x  48

1 3

 conditional equation

1 x2  x  6

2 x2  18x  32

Solution

4 2

1 2

2 2

 13x  48 x  x  6 x  18x  32 4 1 2    x  16 x  3  x  3 x  2  x  16 x  2 x

4 x  2   x  16  2 x  3

{multiply by common denominator}

4 x  8  x  16  2x  6 3x  8  2x  6 x  2  conditional equation

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440


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

14.

a  1 2a  1 2  a   a  3 3  a a  3 Solution

2a  1 2  a a  1   3  a a  3 a  3 1  2a 2  a a  1   a  3 a  3 a  3 a  1a  3  1  2aa  3  2  aa  3 {multiply by common denominator} a2  3a  a  3  a  3  2a2  6a  2a  6  a2  3a a2  9a  6  a2  a  6 9a  6  a  6 0  8a 0  a  conditional equation Solve each formula for the indicated variable. 15. C 

5 F  32; F 9

Solution

5 F  32 9 9 9 5 C   F  32 5 5 9 9 C  F  32 5 C 

9 C  32  F 5 16. Pn  1 

si ;f f

Solution

Pn  l  Pn  l 

si f

f Pn  l  f  f Pn  l  si

f Pn  l Pn  l

si f

si f

si Pn  l

f 

si Pn  l

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441


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

17.

1 1 1   ; f1 f f1 f2 Solution

1 1 1   f f1 f2 f f1 f2 

1 1 1  f f1 f2    f f2   f1

f1 f2  f f2  f f1 f1 f2  f f1  f f2

f1 f2  f   f f2

f1 f2  f  f2  f

f1 

18. S 

f f2 f2  f f f2 f2  f

a  lr ;l 1  r

Solution

a  lr 1  r a  lr S1  r   1  r  1  r S  1  r   a  lr S 

S  Sr  a  lr lr  a  S  Sr lr a  S  Sr  r r a  S  Sr l  r 19. Test scores Gabrielle took four tests in an English class. On each successive test, her score improved by 4 points. If her mean score was 66%, what did she score on the first test?

Solution Let x = the score on the first exam. Then his scores on the following tests were x  4, x  8 and x  12.

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442


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Sum of scores  66 4 x  x  4  x  8  x  12  66 4 4 x  24  66 4 4 x  24  264 4 x  240 x  60 His score on the first test was 60%. 20. Fencing a garden A homeowner has 100 feet of fencing to enclose a rectangular garden. If the garden is to be 5 feet longer than it is wide, find its dimensions.

Solution Let w = the width. Then w + 5 = the length.

Perimeter  100

2w  2w  5  100

2w  2w  10  100 4w  10  100 4w  90 w  22.5 The dimensions are 22.5 ft by 27.5 ft. 21. Travel Two shoppers leave a shopping center by car traveling in opposite directions. If one car averages 45 mph and the other 50 mph, how long will it take for the cars to be 285 miles apart?

Solution Let t = the time the cars travel.

Distance 1st car travels

Distance 2nd car travels

 Total distance

45t  50t  285 95t  285 t  3  They will be 285 miles apart after 3 hours. 22. Travel Two taxis leave an airport and travel in the same direction. If the average speed of one taxi is 40 mph and the average speed of the other taxi is 46 mph, how long will it take before the cars are 3 miles apart?

Solution Let t = the time the cars travel.

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443


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Distance 1st car travels

Distance 2nd

 Distance between them

car travels

46t  40t  3 6t  3 t  0.5  They will be 3 miles apart after 0.5 hours. 23. Preparing a solution A liter of fluid is 50% alcohol. How much water must be added to dilute it to a 20% solution?

Solution Let x = the liters of water added.

Liters of alcohol at start

Liters of

alcohol added

Liters of alcohol at end

0.50 1  0  0.20 1  x  0.50  0.20  0.20 x 0.30  0.20 x 1.5  x  1.5 liters of water should be added. 24. Washing windows Scott can wash 37 windows in 3 hours, and Bill can wash 27 windows in 2 hours. How long will it take the two of them to wash 100 windows?

Solution Let x = hours for both working together.

Number Scott washes in 1 hour

Number of hours

Number Bill washes in 1 hour

Number of hours

 100 windows

37 27 x  x  100 3 2  37 27  6 x  x   6 100 2   3 74 x  81x  600 155 x  600 x 

600  3.9 155

They can wash 100 windows together in about 3.9 hours 25. Filling a tank A tank can be filled in 9 hours by one pipe and in 12 hours by another. How long will it take both pipes to fill the empty tank?

Solution Let x = hours for both pipes to fill the tank.

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444


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

1st pipe in 1 hour

2nd pipe

Total in

in 1 hour

1 hour

1 1 1   9 12 x 1  1 1 36 x     36 x   9 12   x 4 x  3x  36 7 x  36 x 

36 1  5 7 7

The tank can be filled in 5

1 hours. 7

26. Producing brass How many ounces of pure zinc must be alloyed with 20 ounces of brass that is 30% zinc and 70% copper to produce brass that is 40% zinc?

Solution Let x = the ounces of pure zinc added.

Ounces of

zinc at start

Ounces of zinc added

Ounces of zinc at end

0.3020  x  0.4020  x  6  x  8  0.40 x 0.60x  2 6 x  20 x  3

20 1  3 6 3

1 ounces of zinc should be added. 3

27. Lending money A bank lends $10,000, part of it at 11% annual interest and the rest at 14%. If the annual income is $1265, how much was loaned at each rate?

Solution Let x = the amount invested at 11%. Then 10,000 – x = the amount invested at 14%.

Interest at 11%

Interest at 14%

Total interest

0.11x  0.14 10,000  x   1,265 0.11x  1,400  0.14 x  1,265 0.03x 

 135

x  4,500 $4,500 was invested at 11% and $5,500 was invested at 14%.

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445


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

28. Producing oriental rugs An oriental rug manufacturer can use one loom with a setup cost of $750 that can weave a rug for $115. Another loom, with a setup cost of $950, can produce a rug for $95. How many rugs are produced if the costs are the same on each loom?

Solution Let x = # of rugs for equal costs.

Cost of 1st loom  Cost of 2nd loom 750  115x  950  95x 20x  200 x  10 The costs are the same on either loom for 10 rugs. Perform all operations and express all answers in a + bi form. 29. 3 300

Solution

3 300  3 1 100 3  30i 3 30. 

45 4

Solution

 31.

45   1  4

9 5 4

 

3 5 i 2

2  3i   4  2i  Solution

2  3i   4  2i   2  3i  4  2i  2  i

32. 3 

  16  2

36 

Solution

3  36   16  2  3  6i  4i  2  3  6i  4i  2  5  2i

33. 2  3i   4  2i 

Solution

2  3i   4  2i   2  3i  4  2i  2  5i

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446


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

34. 5  11i 5  11i 

Solution

5  11i 5  11i   25  55i  55i  121i 2  25  1211  146  146  0i

35. 8  3i

2

Solution

8  3i 

 8  3i 8  3i   64  24i  24i  9i 2  64  48i  9 1  55  48i



2

36. 3 

9 2 

25

Solution

3  92  25  3  3i2  5i  6  9i  15i  6  9i  15  21  9i 2

37.

3 i Solution 3 3i 3i 3i     0  3i 2 i ii 1 i

38. 

5 6i

Solution 

39.

5 5  i 5i 5i 5        0  i 2 6i 6i  i  6 6 6i

3 1  i Solution

3  1  i 40.

3 1  i 

1  i 1  i 

3 1  i  2

1  i

2

3  3i 3 3   i 2 2 2

2i 2  i Solution

2i  2  i

2i 2  i 

2  i 2  i 

4i  2i 2 2

2

 i

2

2  4i 2 4    i 5 5 5

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447


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

41.

3  i 3  i Solution

3  i 3  i   9  6i  i 2  8  6i  8  6 i  4  3 i 10 10 10 5 5 32  i 2 3  i 3  i 

3  i  3  i 42.

3  2i 1  i Solution

3  2i 1  i   3  5i  2i 2  1  5i  1  5 i 2 2 2 12  i 2 1  i 1  i 

3  2i  1  i

43. Simplify: i53.

Solution

  i  1 i 0i

i 53  i 52i  i 4

13

13

44. Simplify: i 103 .

Solution

i 103  i 100 i 3 

i  i  1 i  0  i 4

25

3

25 3

2

45.  3

i

Solution

2 i

3

 

2 i 3

i  i

 

2i i

 

4

2i  0  2i 1

46. 3  i

Solution

3  i 

47.

32   1

2

9  1

10  0i

1  i 1  i Solution

1  i 1  i

1  i 1  i   1  2i  i 2  2i  0  i  2 12  i 2 1  i 1  i 

02  12  1

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448


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

48. Factor 64r2 + 9s2 over the set of complex numbers.

Solution

64r 2  9s2  64r 2  9s 2

  8r   3si   8r  3si 8r  3si  2

2

Solve each equation by factoring. 49. 2x2  x  6  0

Solution

2x 2  x  6  0

2x  3 x  2  0 2x  3  0

or x  2  0

2x  3

x  2

x   32

x  2

50. 12x2  13x  4

Solution

12x 2  13x  4 12x 2  13x  4  0

4x  13x  4  0

4 x  1  0 or 3x  4  0 4x  1

3x  4

1 4

x   43

x 

51. 5x2  8x  0

Solution

5x 2  8x  0

x 5 x  8  0

x  0 or 5 x  8  0 x  0

5x  8

x  0

x 

8 5

52. 27x2  30x  8

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449


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

27 x 2  30 x  8 27 x 2  30x  8  0

9x  43x  2  0 9x  4  0 or 3x  2  0 9x  4

3x  2

4 9

x 

x 

2 3

Solve each equation by using the Square Root Property. 53. 3x2  24

Solution

2x 2  16 x2  8 x2   8 x  2 2 54. 12x2  60

Solution

12x 2  60 x 2  5 x 2   5 x  i 5

55. 4z  5

2

 32

Solution

4z  5  32 2 4z  5   32 2

4z  5  4 2 4z  5  4 2 z 

5  4 2 4

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450


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

56. 5x  7

2

 45

Solution

5x  7  45 2 5x  7   45 2

5x  7  3i 5 5x  7  3i 5 x 

7  3i 5 7 3 5   i 5 5 5

Solve each equation by completing the square. 57. x2  8x  15  0

Solution

x2  8x  15  0 x2  8x  15 x2  8x  16  15  16

 x  4 x  4 

2

 1

1 or x  4   1

x  4  1

x  4  1

x  5

x  3

58. 3x2  18x  24

Solution

3x2  18x  24 3x2  18x 24  3 3 2 x  6x  8 x 2  6x  9  8  9

 x  3 x  3 

2

 1

1 or x  3   1

x  3  1 x  2

x  3  1 x  4

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451


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

59. 5x2  x  1  0

Solution

5x 2  x  1  0 5x 2  x  1 1 1 x  5 5 1 1 1 1 x2  x    5 100 5 100 x2 

2

 1 x   10   x 

1  10

x 

1  10

21 100

21 100

or x 

21 10 1  21 x  10

x 

1 21   10 100 1 21   10 10 1  21 x  10

60. 5x2  x  0

Solution

5x2  x  0 1 x  0 5 1 1 1 x2  x   0  5 100 100 x2 

 1  x   10  

2

1 100

1 1 1 1    or x  10 100 10 100 1 1 1 1 x   x    10 10 10 10 2 1 0 x   x   0 10 5 10 x 

61. Solve: 3x2  2x  1  0.

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452


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3x 2  2x  1  0 3x 2  2x  1 2 1 x   3 3 2 1 1 1 x2  x     3 9 3 9 x2 

2

 1 x   3  x 

1  3

x 

1  3

2 9

2 i 3 1 2 x  i  3 3

 

2 9

or x  x 

1 2    3 9 1 2 i   3 3 1 2 x  i  3 3

Use the Quadratic Formula to solve each equation. 62. x2  5x  14  0

Solution

x 2  5x  14  0  a  1, b  5, c  14 x 

b 

5  b2  4ac  2a

5  4114 5   2 1 2

25  56 5  81  2 2 

x 

5  9 2

4 5  9 5  9 14   2 or x    7 2 2 2 2

63. 3x2  25x  18

Solution

3x2  25x  18  3x2  25x  18  0  a  3, b  25, c  18 x 

b 

 25  b2  4ac  2a

25  4318 25   23 2

625  216 6

25 

841 6

25  29 6

25  29 54 25  29 4 2 x    9 or x     6 6 6 6 3

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453


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

64. 5x2  1  x

Solution

5x 2  1  x  5x 2  x  1  0  a  5, b  1, c  1 x 

b 

 1  b2  4ac  2a

1  451 1  1  20 1  21   10 10 25 2

65. 5  a2  2a

Solution

5  a2  2a  a2  2a  5  0  a  1, b  2, c  5 a 

b 

2  b2  4ac  2a

2  415 2  4  20 2  24   2 2 2 1 2

2  2 6  1  2

6

66. Solve: 3x2  4  2x.

Solution

3x 2  4  2x  3x 2  2x  4  0  a  3, b  2, c  4 x   

b 

b2  4ac 2a

 2 

2  434 2   23 2

2 2 11 1   i  6 6 3

4  48 2  44  6 6

11 i 3

67. Calculate the discriminant associated with the equation 6x2  5x  1  0.

Solution

6 x 2  5x  1  0 a  6, b  5, c  1 b2  4ac  5

2

 46 1  25  24  1

68. Determine the number and nature of the roots of the equation in Exercise 67.

3x2  18x  24 Solution two different rational numbers

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454


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

69. Find the value of k that will make the roots of kx2  4x  12  0 equal.

Solution kx 2  4 x  12  0 a  k , b  4, c  12 Set the discriminant equal to 0:

b2  4ac  0 42  4k  12  0 16  48k  0 48k  16 k 

1 3

70. Find the values of k that will make the roots of 4 y 2  k  2 y  1  k equal.

Solution

4 y 2  k  2 y  1  k

4 y 2  k  2 y  1  k  0

a  4, b  k  2, c  1  k Set the discriminant equal to 0: b2  4ac  0

k  2  441  k   0 2

k 2  4k  4  16  16k  0 k 2  12k  20  0

k  10k  2  0

71.

k  10  0

or k  2  0

k  10

k  2

3x 2x   x  3 2 x  1 Solution 3x 2x   x  3 x  1 2  3x 2x   2 x  1   2 x  1 x  3 x  1  2

 x  1  3x  22 x   2 x 2  4 x  3 3x 2  3x  4 x  2x 2  8x  6

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455


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x2  x  6  0

 x  3 x  2  0 x  3  0

or

x  2  0

x  3 72. Solve:

x  2

4 4   5. a  4 a  1

Solution

4 4   5 a  4 a  1 a  4a  1 a 4 4  a4 1  a  4a  15

4a  1  4a  4  5 a2  5a  4

4a  4  4a  16  5a2  25a  20 0  5a2  33a  40 0  5a  8a  5 5a  8  0 or a  5  0 5a  8

a  5

8 5

a  5

a 

73. Fencing a field A farmer wishes to enclose a rectangular garden with 300 yards of fencing. A river runs along one side of the garden, so no fencing is needed there. Find the dimensions of the rectangle if the area is 10,450 square yards.

Solution Let x = one side of the garden.

Area  10450

x 300  2 x   10450 2 x 2  300 x  10450 0  2 x 2  300 x  10450

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456


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

0  2 x 2  150 x  5225 0  2 x  95 x  55 x  95  0

or x  55  0

x  95

x  55

The dimensions are 95 yards by 110 yards or 55 yards by 190 yards. 74. Flying rates A jet plane, flying 120 mph faster than a propeller-driven plane, travels 3520 miles in 3 hours less time than the propeller plane requires to fly the same distance. How fast does each plane fly?

Solution Let r = the rate of the propeller-driven plane. Then the rate of the jet plane is r + 120.

Jet time  Propeller time  3 3520 3520   3 r  120 r  3520  3520 r r  120  r r  120  3 r  120  r 

3520r  3520r  120  3r r  120 3520r  3520r  422400  3r 2  360r

3r 2  360r  422,400  0 3r  320r  440  0 r  320  0

or r  440  0

r  320

r  440

Rate

Time

Dist.

Propeller

r

3520 r

3520

Jet

r + 120

3520 r  120

3520

Since r  440 does not make sense, the solution is r  320. The prop. plane's rate is 320 mph, while the jet plane's rate is 440 mph. 75. Flight of a ball A ball thrown into the air reaches a height h (in feet) according to the formula h  16t 2  64t, where t is the time elapsed since the ball was thrown. Find the shortest time it will take the ball to reach a height of 48 feet.

Solution Set h  48:

h  16t 2  64t 48  16t 2  64t 16t 2  64t  48  0

16t  1t  3  0

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457


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

t  1  0 or t  3  0 t  1

t  3

The shortest time required for the ball to reach a height of 48 feet is 1 second. 76. Width of a walk A bricklayer built a walk of uniform width around a rectangular pool. If the area of the walk is 117 square feet and the dimensions of the pool are 16 feet by 20 feet, how wide is the walk?

Solution Let x = the width of the walk. Then the total dimensions are 16  2x by 20  2x.

Total area

Area of

pool

Area

of walk

16  2x20  2x   1620  117 320  72x  4 x 2  320  117 4 x 2  72x  117  0

2x  392x  3  0 2x  39  0

or 2x  3  0

x   39 2

x 

3 2

Since x   39 does not make sense, the only solution is x  2

3 . The walk is 1 21 2

feet wide.

Solve each equation. 77. x3  4x2  12x  0

Solution

x 3  4 x 2  12 x  0

x x 2  4 x  12  0 x  x  6 x  2  0

x  0 or x  6  0 x  0

or x  2  0

x  6

x  2

78. 3x3  4x2  4x  0

Solution

3x 3  4 x 2  4 x  0

x 3x 2  4 x  4  0 x 3x  2 x  2  0 x  0 or 3x  2  0 or x  2  0 x  0

x 

2 3

x  2

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458


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

79. x4  2x2  1  0

Solution

x 4  2x 2  1  0

 x  1 x  1  0 2

2

x2  1  0

or x 2  1  0

x2  0

x2  1

x  1

x  1

80. x4  36  35x2

Solution

x 4  36  35x 2 x 4  35x 2  36  0

 x  36 x  1  0 2

2

x 2  36  0

or x 2  1  0

x 2  36

x2  1

x  6i

x  1

12

81. a  a

6  0

Solution

a

12

a  a1 2  6  0

a1 2  2  0

or a1 2  3  0

 2 a1 2  3  0

a1 2  2

a1 2  3



a   2 12

2

a   3

2

12

a  4

2

2

a  9

a  4 does not check and is extraneous. 23

82. x

 x1 3  6  0

Solution

x2 3  x1 3  6  0



x1 3  2 x1 3  3  0

x1 3  2  0

or x 1 3  3  0

x1 3  2

x 1 3  3

 x   2

 x   3

x  8

x  27

13

3

3

13

3

3

Both answers check.

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459


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

83. 6 y 2  13 y 1  5  0

Solution

6 y 2  13 y 1  5  0



3 y 1  1 2 y 1  5  0

3 y 1  1  0

or 2 y 1  5  0

y 1  31

y 1   52

y  1

1

 1 3

1

y  1

y  3

1

 

  52

y  

1

2 5

Both answers check. 84.

5x  11  5  3 Solution

5x  11  5  3 5x  11  2

 5x  11  2 2

2

5x  11  4 x  3 The solution checks. 85.

x  1  x  7 Solution

x  1  x  7 x  1  7  x

 x  1   7  x  2

2

x  1  49  14 x  x 2 0  x 2  15x  50

0   x  5 x  10 x  5  0 or x  10  0 x  5

x  10

x = 10 does not check and is extraneous

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460


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

86.

a  9 

a  3

Solution

a  9 

a  3

a  9  3 

a

 a  9   3  a  2

2

a  9  9  6 a  a 0  6 a

02  6 a

2

0  36a  a  0 The solution checks. 87.

5  x 

5  x  4

Solution

5  x 

5  x  4 5  x  4 

5  x

 5  x   4  5  x  2

2

5  x  16  8 5  x  5  x 8 5  x  16  2 x

8 5  x   16  2x 2

2

645  x  256  64x  4x2 320  64x  4x2  64x  256 0  4x2  64

0  4 x  4 x  4 x  4  0

or x  4  0

x  4

x  4

Both solutions check. 88.

y  5 

y  1

Solution y  5 

y  1

y  5  1 

y

 y  5   1 

y

2

2

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461


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

y  5  1  2

y  y

y  4

2

2 y   4 2

2

4 y  16 y  4 The solution does not check.  No solution. 89.

3

4x  9  3  2

Solution 3

4x  9  3  2 3

4 x  9  1

 4x  9  1 3

3

3

4 x  9  1 4x  8 x  2 The solution checks. 90.

4

x  2  3  5

Solution 4

x  2  3  5 4

x  2  2

 x  2  2 4

4

4

x  2  16 x  18 The solution checks.

Solve each inequality; graph the solution set and write the answer in interval notation. 91. 2 x  9  5

Solution 2x  9  5

2x  14

x  7   , 7

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462


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

92. 5 x  3  2

Solution 5x  3  2

5 x  1 x   51   51 , 

93.

5 x  1 2

 x

Solution

5 x  1

 x 2 5 x  1  2 x 5x  5  2x 3x  5 x 

94.

5 3

 , 53

1 2 1 1 x  x  x    x  1 4 3 2 2 Solution 1 2 1 1  x  x  x  x  1 4 3 2 2 1  1  2 1 x  1 12 x  x  x   12  3 2 4  2  3x  8x  12x  6  6 x  1

 x  6  6x  6 12  7 x 

 12 12   x   ,   7 7 

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463


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

3  x  4 2

95. 0 

Solution 3  x 0   4 2 0  3  x  8

3 

 5  3, 5

x

96. 2  a  3a  2  5a  2

Solution 2  a  3a  2  5a  2

2  a  3a  2 and 3a  2  5a  2 4  2a

4  2a

a  2

a  2

a  2 a  2 a  2 and

a  2

Solution set: 2,  97.  x  2 x  4  0

Solution

 x  2 x  4  0 factors  0: x  2, x  4

intervals: , 2,  2, 4, 4,  interval

, 2 2, 4 4, 

value of

test number

 x  2 x  4

–3

+7

0

–8

5

+7

Solution:  ,  2  4,  

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464


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

98.  x  1 x  4  0

Solution

 x  1 x  4  0 factors  0: x  1, x  4

intervals:  , 4,  4, 1,  1,  interval

, 4 4, 1 1, 

value of

test number

 x  1 x  4

–5

+6

0

–4

2

+8

Solution:  4, 1

99. x2  2x  3  0

Solution

x 2  2x  3  0

 x  3 x  1  0 factors  0: x  3, x  1

intervals:  ,  1,  1, 3, 3,  interval

,  1 1, 3 3, 

value of

test number

x 2  2x  3

–2

+5

0

–3

4

+5

Solution: 1, 3

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465


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

100. 2x2  x  3

Solution 2x 2  x  3  0

2 x  3 x  1  0 factors  0: x   32 , x  1

 

intervals: ,  32 ,  32 , 1 ,  1,  value of

test number

2x2  x  3

–2

+3

3 2

0

–3

1, 

2

+7

interval

,    , 1 3 2

Solution: ,  32   1,  

101.

x  2  0 x  3 Solution x  2  0 x  3 factors  0: x  2, x  3

intervals:  , 2,   2, 3, 3,  interval

, 2   2, 3 3, 

test number

sign of

–3

+

0

4

+

x  2 x  3

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:  ,  2  3,  

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466


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

102.

x  1  0 x  4 Solution x  1  0 x  4 factors  0: x  1, x  4

intervals:  ,  4,   4, 1,  1,  interval

, 4  4, 1 1, 

test number

sign of

–5

+

0

2

+

x  1 x  4

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:  4, 1

103.

x2  x  2  0 x  3 Solution

x2  x  2  0 x  3  x  2 x  1  0 x  3 factors  0: x  2, x  1, x  3

intervals:  , 2,  2, 1,  1, 3, 3,  interval

, 2  2, 1 1, 3 3, 

test number

sign of

x2  x  2 x  3

–3

0

+

2

4

+

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0.

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467


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution: 2, 1  3,  

104.

5  2 x Solution 5  2 x 5  2  0 x 5  2x  0 x factors  0: x 

5 ,x 2

 0

  , 

intervals:  , 0, 0, 52 ,

5 2

interval

test number

value of

, 0

–1

–7

1

+3

3

 31

0,   ,  5 2

5 2

Solution:  , 0 

5  2x x

 ,  5 2

Solve each equation or inequality. 105. x  1  6

Solution

x  1  6 x  1  6 or x  1  6 x  5 106.

x  7

3x  11  1  0 7

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468


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3x  11  1  0 7 3x  11  1 7 3x  11  1 7 3x  11  7

107.

or

3x  11  1 7 3x  11  7

3x  4

3x  18

x   43

x  6

2a  6  6  0 3a Solution

2a  6  6  0 3a 2a  6  6 3a 2a  6 or  6 3a 2a  6  18a 16a  6 a  a 

6  16  83

2a  6  6 3a 2a  6  18a 20a  6 a  a 

6 20 3 10

108. 2 x  1  2 x  1

Solution

2x  1  2x  1 2 x  1  2 x  1 or 2 x  1   2 x  1 0  2

2 x  1  2 x  1

never true

4x  0 x  0

109. 3 x  11  16  5

Solution

3 x  11  16  5 3 x  11  11 An absolute value can never equal a negative number. no solution

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469


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

110.

x  3  3

Solution

x  3  3 3  x  3  3 6 

111.

 0

x

6, 0

3x  7  1

Solution

3x  7  1 3x  7  1

or 3 x  7  1

3x  8

3x  6

8 3

x  2

x 

, 2  83 , 

112.

x  2  5  6 3 Solution x 2 3

 5  6 x 2 3

1 

 1

x 2 3

 1

3  x  2  3 5 

x

5, 1

 1

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470


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

113.

x  3  8 4 Solution x 3 4 x 3 4

 8

 8

or

x 3 4

 8

x  3  32

x  3  32

x  35

x  29

, 29  35, 

114. 1  2 x  3  4

Solution

1  2x  3  4 1  2x  3

and

1 2x  3  1 2x  3  1

2x  3  4

 2

or 2x  3  1

2x  3  4

4  2x  3  4

2 x  2

2 x  4

7 

2x

 1

x  1

x  2

 72 

x

1 2

(2)

(1) (1)

(2) (1) and (2)

 , 2  1,  7 2

1 2

115. 0  3 x  4  7

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471


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

0  3x  4  7 0  3x  4

1 3x  4  0

 2

3x  4  0 or 3 x  4  0 3x  4 x 

4 3

3x  4  7

and

3x  4  7

7  3x  4  7

3x  4

3 

3x

 11

4 3

1 

x

x 

(1)

11 3

(2) (1) (2)

(1) and (2) 

1,    ,  4 3

4 3

11 3

CHAPTER TEST SOLUTIONS Find all restrictions on x. 1.

x

x  x  1 Solution x

x  x  1

 2

 2

restrictions: x  0, x  1 2.

4  3  7 3x  2 Solution

4  3  7 3x  2 restrictions: x  23

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472


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solve each equation. 3.

7 2a  5  7  6a  8

Solution

72a  5  7  6a  8 14a  35  7  6a  48 8a  20 20 5  8 2

a 

4.

1 1 3   a  2 5a 2a Solution

1 1 3   5a 2a a  2  1 1  3  10aa  2   10aa  2  5a  2a a  2 10a 1  2a  2  15a  2 10a  2a  4  15a  30 34  7a 34  a 7 5. Solve for x: z 

x  

.

Solution

z 

x  

az  a 

x  

za  x  

za    x 6. Solve for a:

1 1 1   . a b c

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473


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

1 1 1   a b c 1 1 1 abc   abc    a c b bc  ac  ab

bc  ac  b

ac  b bc  c  b c  b bc  a c  b 7. Test scores A student’s average on three tests is 75. If the final is to count as two one-hour tests, what grade must the student make to bring the average up to 80?

Solution Let x = the score on the final exam. Note: This score is counted twice. Sum of scores  80 5 75  75  75  x  x  80 5 2 x  225  80 5 2 x  225  400 2 x  175 x  87.5 The student needs to score 87.5. 8. Investment A woman invested part of $20,000 at 6% interest and the rest at 7%. If her annual interest is $1260, how much did she invest at 6%?

Solution Let x = the amount invested at 6%. Then 20,000 – x = the amount invested at 7%. Interest at 6%

Interest at 7%

Total Interest

0.06 x  0.07 20,000  x   1,260 0.06 x  1,400  0.07 x  1,260 0.01x  140 x  14,000 $14,000 was invested at 6%.

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474


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Simplify the imaginary numbers. 9.

3 96 Solution

3 96  3 1 16 6  12i 6 10.

18 5

Solution

18  5

18  5  5  5

90  25

1 9 10 25

3 10 i 5

Perform each operation and write all answers in a + bi form. 11.

4  5i   3  7i  Solution

4  5i   3  7i   4  5i  3  7i  7  12i

12.

4  5i 3  7i  Solution

4  5i 3  7i   12  43i  35i 2  12  43i  35  23  43i

13.

2 2  i Solution

2  2  i 14.

2 2  i 

2  i 2  i 

4  2i 22  i 2

4  2i 4 2   i 5 5 5

1  i 1  i Solution

1  i  1  i

1  i 1  i   1  2i  i 2  2i  0  i 2 12  i 2 1  i 1  i 

Simplify each expression. 15. i 13

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475


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

 

3

i 13  i 12i  i 4 i  13 i  i 16. 7i 4

Solution

1i 4  1  1  1 Solve each equation. 17. 4x2  8x  3  0

Solution

4 x 2  8x  3  0

2x  32x  1  0 2x  3  0 or 2x  1  0 2x  3 3 2

x 

2x  1 x 

1 2

18. 2b2  12  5b

Solution

2b2  12  5b 2b2  5b  12  0

2b  3b  4  0 2b  3  0 or b  4  0 2b  3

b  4

3 2

b  4

b 

19. 5x2  135

Solution

5x 2  135 x 2  27 x   27 x  3i 3

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476


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

20. Use completing the square to solve x2  14x  23.

Solution

x 2  14x  23 x 2  14x  49  23  49

 x  7

2

x  7 

 72

72

or x  7   72

x  7  6 2

x  7  6 2

x  7  6 2

x  7  6 2

21. Use the Quadratic Formula to solve 3x2  5x  9  0.

Solution

3x 2  5x  9  0  a  3, b  5, c  9 x 

22.

b 

5  b2  4ac  2a

3 x

2

 5x  14

5  439 5   23 2

25  108 5  133  6 6

4 x

2

 5x  6

Solution

3 x

2

 5x  14 3

 x  7 x  2

4 x

2

 5x  6 4

 x  2 x  3  x  7 x  2 x  3 x  7 3 x  2   x  7 x  2 x  3 x  24 x  3       3 x  3  4 x  7 3x  9  4 x  28 37  x

23. Find k such that x 2  k  1 x  k  4  0 will have two equal roots.

Solution

x 2  k  1 x  k  4  0

a  1, b  k  1, c  k  4 Set the discriminant equal to 0:

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477


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

b2  4ac  0

k  1  41k  4  0 2

k 2  2k  1  4k  16  0 k 2  2k  15  0

k  5k  3  0 k  5  0 or k  3  0 k  5

k  3

24. Height of a projectile The height h (in feet) of a projectile shot up into the air, at time t (in seconds), is given by the formula h  16t2  128t. Find the time t required for the projectile to return to its starting point.

Solution

Set h  0: h  16t 2  128t 0  16t 2  128t 0  16t 2 t  8 16t  0 or t  8  0 t  0

t  8

The projectile will return after 8 seconds. Find each absolute value. 25. 5  12i

Solution

5  12i 

26.

52   12

2

25  144 

169  13

1 3  i Solution

1 3  i

13  i 

3  i 3  i 

3  i 32  i 2

3  i  10

3 1 i    10 10

3    10 

2

 1     10 

9 1  100 100

10  100

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2

10 10

478


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solve each equation. 27. z4  13z2  36  0

Solution

z 4  13z 2  36  0

z  4z  9  0 4

z2  4  0

or z 2  9  0

z2  4

z2  9

z  2

z  3

2

28. 2 p2 5  p1 5  1  0

Solution

2 p

2p2 5  p1 5  1  0 15



 1 p1 5  1  0

2p1 5  1  0

or p1 5  1  0

p1 5   21

p1 5  1

 p     15

5

1 2

5

 p   1 15

1 p   32

5

5

p  1

Both answers check.

x  5  12

29.

Solution

x  5  12

x  5

  12 2

2

x  5  144 x  139 The answer checks. 30.

2z  3  1 

z  1

Solution 2z  3  1 

z  1

 2z  3    1 

z  1

2

2

2z  3  1  2 z  1  z  1 2 z  1  z  1

2 z  1  z  1 2

2

4 z  1  z 2  2z  1 4 z  4  z 2  2z  1

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479


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

0  z2  2z  3

0   z  1 z  3 z  1  0

or z  3  0

z  1

z  3

The answer z  3 is extraneous.

Solve each inequality; graph the solution set and write the answer using interval notation. 31. 5 x  3  7

Solution 5x  3  7

5x  10

x  2   , 2

32.

x  3 2x  4  4 3 Solution

x  3 2x  4  4 3 x 3 12  4  12  2 x3 4

3 x  3  42x  4 3x  9  8x  16 5x  25

x  5   , 5

33. 5  2 x  1  7

Solution 5  2x  1  7

6 

2x

3 

x

 8

 4  3, 4

34. 1  x  3 x  3  4 x  2

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480


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

1  x  3x  3  4 x  2 1  x  3x  3 and 3x  3  4x  2 2x  4

x  1

x  2

x  1

x  2 x  1 x  2 and

x  1

Solution: 2, 

35. x2  7x  8  0

Solution

x2  7x  8  0

 x  1 x  8  0

factors  0: x  1, x  8

intervals:  ,  1,  1, 8, 8,  interval

, 1 1, 8 8, 

value of

test number

x2  7 x  8

–2

+10

0

–8

9

+10

Solution: , 1  8, 

36.

x  2  0 x  1 Solution

x  2  0 x  1 factors  0: x  2, x  1

intervals:  , 2,  2, 1,  1, 

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481


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

interval

test number

sign of xx  21

–3

+

0

2

+

, 2 2, 1 1, 

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0.

Solution:   2, 1

Solve each equation. 37.

3x  2  4 2 Solution

3x  2  4 2 3x  2 2

3x  2 2

 4 or

 4

3x  2  8

3x  2  8

3x  6

3x  10

x  2

x   10 3

38. x  3  x  3

Solution

x  3  x  3 x  3  x  3 or

x  3   x  3

0  6

x  3  x  3

not true

2x  0 x  0

Solve each inequality; graph the solution set and write the answer using interval notation. 39. 2 x  5  2

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482


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

2x  5  2 2 x  5  2 or 2 x  5  2 2x  7

2x  3

7 2

x 

x 

,    ,  3 2

40.

3 2

7 2

2x  3  5 3 Solution

2x  3  5 3 2x  3 3

5 

 5

15  2 x  3  15 18 

2x

 12

9 

x

 6

9, 6

CUMULATIVE REVIEW EXERCISES

Consider the set 5,  3,  2, 0, 1, 1.

2, 2, 25 , 5, 6, 11 .

Which numbers are even integers?

Solution even integers: –2, 0, 2, 6 2. Which numbers are prime numbers?

Solution prime numbers: 2, 5, 11 Write each inequality as an interval and graph it. 3.

4  x  7 Solution

4  x  7  4, 7 

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483


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

4.

x  2 or x  0

Solution

x  2 or x  0   , 0  2, 

Determine which property of the real numbers justifies each expression. 5.

a  b  c  c  a  b Solution Commutative Property of Addition

6. If x < 3 and 3 < y, then x < y.

Solution Transitive Property Simplify each expression. Assume that all variables represent positive numbers. Give all answers with positive exponents. 7.

81a  4

12

Solution

12

 

12

81a4

8.

81 a4

 

12

 

12

2    9a2   

 9a2

Solution

 

81 a4

9.

2   81 a2   

12

a b  3 2

 81a2

2

Solution

a b  3 2

2

 4x4  10.   2   12x y 

  b 

 a3

2

2

2

 a6b4

2

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484


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

 4x4    2   12x y 

2

 4x 0 y 2  11.  2   x y 

2

2

 12x 2 y     4 x 4   

 3y     2 x 

9y2

x4

2

Solution

 4x0 y 2   2   x y 

2

2

 x2 y     4 x0 y 2   

2

 x2     4 y   

x4 16 y 2

2

 4 x 5 y 2  12.  2 3   6x y  Solution

2

 4 x 5 y 2   2 3   6x y  13.

2

 2y5     3x 3   

4 y 10 9x 6

a b ab  12

2

12

2

Solution

a b ab   ab a b  a b 12

14.

2

12

a b c 12 12

2

2

2

3 3

2

Solution

a b c  abc 12 12

2

2

Rationalize each denominator and simplify. 15.

3 3 Solution

3 3 16.

3 3 3 3

3 3  3

3

2 3

4x

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485


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 2 3

17.

4x

3

2 2x 2 3

4x

3

2x 2

3

2 2x 2 3

8x 3

3

3

2 2x 2  2x

2x 2 x

3 y 

3

Solution

3 y 

3 y 

3

3

 y  3 y  3

  3 y  3 y  3 y   3 

3 y 

3

2

2

2

3x

18.

x  1 Solution

3x x  1

 x  1  3x  x  1  3x  x  1 x  1  x  1 x  1  x   1 3x

2

2

Simplify each expression and combine like terms.

75  3 5

19.

Solution

75  3 5  18 

20.

25 3  3 5  5 3  3 5

8  2 2

Solution

18  21.

8  2 2 

 2  3

9 2 

4 2  2 2  3 2  2 2  2 2  3 2

2

Solution

 2  3   2  3 2  3  2

22. 3 



5 3 

5

4  2 6 

9  5  2 6

Solution

3  53  5  9  25  9  5  4

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486


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Perform the operations and simplify when necessary.

23. 3x2  2x  5  3 x2  2x  1

Solution

3x  2x  5  3 x  2x  1  3x  2x  5  3x  6x  3  8x  8 2

2

2

  10x  5x  x  x  9x  4x

24. 5x 2 2x2  x  x x2  x3

Solution

2

5x 2 2 x 2  x  x x 2  x 3

4

3

3

4

4

3

25. 3 x  52 x  7

Solution

3x  52x  7  6x2  21x  10x  35  6x2  11x  35



26. z  2 z2  z  2

Solution

 z  2 z2  z  2  z3  z2  2z  2z2  2z  4  z3  z2  4

27. 3 x  2 6 x 3  x 2  x  2

Solution

2x 2  x  1 3x  2 6x 3  x 2  x  2 6x3  4x2  3x 2   3x

2

x

 2x 3x  2 3x  2 0

28. x 2  2 3 x 4  7 x 2  x  2

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487


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3x 2  1 

x x2  2

x 2  2 3x 4  0 x 3  7 x 2  x  2 3x 4

 6x2 x2  x  2 x2

 2  x

Factor each polynomial. 29. 3t2  6t

Solution

3t 2  6t  3t t  2

30. 3x2  10x  8

Solution

3x 2  10x  8  3x  2 x  4

31. x8  2x 4  1

Solution

x 8  2x 4  1 

 x  1 x  1   x  1 x  1 x  1 x  1   x  1  x  1 x  1 x  1 x  1   x  1  x  1  x  1 4

4

2

2 2

2

2

2

2 2

2

2

32. x 6  1

Solution

x6  1 

 x   1   x  1 x  1   x  1 x  x  1 x  1 x  x  1 3

2

2

3

3

2

2

Perform the operations and simplify. 33.

x2  4 x 2  5x  6

x 2  2x  15 x 2  3x  10

Solution

x2  4 x2  5x  6

x 2  2 x  15 x 2  3 x  10

 x  2 x  2   x  5 x  3  x  5  x  2 x  3  x  5 x  2 x  5

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488


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

34.

6x 3  x 2  x 3x 2  x  x  2 x2  4x  4 Solution

x 6x2  x  1 6x 3  x 2  x 3x 2  x x2  4x  4    x  2 x  2 3x 2  x x2  4x  4 x 2x  13x  1  x  2 x  2    2x  1 x  2 x  2 x 3x  1 35.

2 5x  x  3 x  3 Solution

2 5x   x  3 x  3

2 x  3

 x  3 x  3

5x  x  3

 x  3 x  3

 

36.

2x  6

 x  3 x  3

5x 2  15x

 x  3 x  3

5x 2  17 x  6

 x  3 x  3

 x  2 x  3  1  2 x  3 x  4  Solution

 x  2 x  3 x  2 x  3 x2  4  x  2  x2  x  7      1     2   2 x  3 x  4 x  3 x  4 x  3   x  2 x  2  x 2  4    

 x2  x  7

 x  3 x  2

1 1  a b 37. 1 ab Solution 1 1  ab a1  b1 a b  1 1 ab ab ab

38.

 

  ba  ba 1

x 1  y 1 x  y Solution

1  1 xy x1  y1 x 1  y 1 y  x 1 x y      x  y x  y xy xy  x  y  xy  x  y 

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489


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solve each equation. 39.

3x x  x  5 x  5 Solution

3x x  x  5 x  5 3x  x  5  x  x  5 3x 2  15x  x 2  5 x 2x 2  20 x  0

2 x  x  10  0

2x  0 or x  10  0 x  0

x  10

40. 82 x  3  35 x  2  4

Solution

82 x  3  35 x  2  4 16 x  24  15 x  6  4 x  34

Solve each formula for the indicated variable. 41.

1 1 1   ;R R R1 R2 Solution

1 1 1   R R1 R2 RR1R2 

 1 1 1   RR1R2    R R R  1 2

R1R2  RR2  RR1

R1R2  R R2  R1 

RR2  R1  R1R2  R2  R1 R2  R1 R1R2  R R2  R1

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490


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

42. S 

a  lr ;r 1  r

Solution

a  lr 1  r a  lr S1  r   1  r  1  r S  1  r   a  lr S 

S  Sr  a  lr S  a  Sr  lr

S  a  r S  l S  a  r S  l 43. Gardening A gardener wishes to enclose her rectangular raspberry patch with 40 feet of fencing. The raspberry bushes are planted along the garage, so no fencing is needed on that side. Find the dimensions if the total area is to be 192 square feet.

Solution

Area  192

x 40  2 x   192 40 x  2 x 2  192

0  2 x 2  40 x  192 0  2 x  8 x  12 x  8  0 or x  8

x  12  0 x  12

If x  8, then the dimensions are 8 feet by 24 feet. If x  12, then the dimensions are 12 feet by 16 feet. 44. Financial planning A college student invested part of a $25,000 inheritance at 7% interest and the rest at 6%. If his annual interest is $1670, how much did he invest at 6%?

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491


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution Let x = the amount invested at 6%. Then 25,000 – x = the amount invested at 7%.

Interest at 6%

Interest at 7%

Total

interest

0.06 x  0.07 25,000  x   1,670 0.06 x  1,750  0.07 x  1,670 0.01x  80 x  8,000  $8,000 was invested at 6%. Perform the operations. If the result is not real, express the answer in a + bi form. 45.

2  i 2  i Solution

2  i  2  i

46.

2  i 2  i   4  4i  i 2  3  4i  3  4 i 5 5 5 4  i2 2  i 2  i 

i 3  i 

1  i 1  i  Solution

3i  i 1  2i  i    1  i 1  i  1  i 1  i 1  i 1  i  1  i 1  i  i 3  i 

i 3  i  1  i  1  i 

2

2

2

2

1  3i 2i   2i  6i 2  6  2i  3  1 i 4

1  2i 2  i 4

4

2

2

47. 3  4i

Solution

32  42 

3  4i  48.

5 i7

9  16 

25  5

 5i

Solution

5 i

7

 5i 

5i 7

i i

 5i 

5i i

8

 5i 

5i

i  4

2

 5i 

5i 12

 5i  5i  5i  10i  0  10i

Solve each equation. 49. 15x2  16x  7  0

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492


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

15x 2  16x  7  0

5x  73x  1  0 5x  7  0 or 3x  1  0 x 

50. 7 x  4

2

7 5

x   31

 8

Solution

7 x  4  8 2 7 x  4   8 7 x  4  2i 2 2

7 x  4  2i 2 x 

51.

4 2 2  i 7 7

x  3 6   1 x  1 x Solution

x  3 6   1 x  1 x x  3 6 x  x  1    x  x  1 1 x x  1 x  x  3  6 x  1  x 2  x

x 2  3x  6x  6  x 2  x 2 x  6 x  3 52. x4  36  13x2

Solution

x 4  36  13x 2 x 4  13x 2  36  0

 x  4 x  9  0 2

2

x2  4  0

or x 2  9  0

x2  4

x2  9

x  2

x  3

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493


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

y  2 

53.

11  y  5

Solution

y  2 

11  y  5

y  2  5   11  y

 y  2  5   11  y  2

y  2  10

2

y  2  25  11  y 10

y  2  2 y  16

10 y  2  2 y  16 2

2

100 y  2  4 y 2  64 y  256

100 y  200  4 y 2  64 y  256 0  4 y 2  36 y  56 0  4 y  2 y  7 y  2  0 or

y  7  0

y  2

y  7

Both solutions check. 23

54. z

 13z1 3  36  0

Solution

z2 3  13z 1 3  36  0

z z



13

 4 z1 3  9  0

13

 4  0

or z 1 3  9  0

z1 3  4

z1 3  9

z   4

z   9

13

3

13

3

z  64

3

3

z  729

Both solutions check. Solve each inequality; graph the solution set and write the answer using interval notation. 55. 5 x  7  4

Solution 5x  7  4

5x  11 x 

11 5

 , 115 

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494


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

56. x2  8x  15  0

Solution

x2  8x  15  0

 x  3 x  5  0 factors  0: x  3, x  5

intervals: , 3, 3, 5, 5,  interval

, 3 3, 5 5, 

value of

test number

x  8x  15

0

+15

4

1

6

+3

2

Solution:  , 3  5,  

57.

x2  4x  3  0 x  2 Solution

x2  4x  3  0 x  2  x  3 x  1  0 x  2 factors  0: x  3, x  1, x  2

intervals:  , 3,  3,  1, 1, 2, 2,  interval

, 3 3,  1 1, 2 2, 

test number

sign of

x2  4x  3 x  2

–4

–2

+

0

3

+

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:

3,  1  2, 

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495


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

58.

9  x x Solution

9  x x 9  x  0 x 9  x2  0 x 3  x3  x   0 x factors  0: x  3, x  3, x  0

intervals:  , 3,  3, 0, 0, 3, 3,  interval

test number

, 3 3, 0 0, 3 3,  Solution:

sign of

9  x2 x

–4

+

–1

1

+

4

, 3  0, 3

59. 2 x  3  5

Solution

2x  3  5 2 x  3  5 or 2 x  3  5 2x  8

2 x  2

x  4

x  1

,  1  4, 

60.

3x  5  2 2

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496


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

3x  5  2 2 3x  5  2 2 4  3x  5  4 2  1

3x

 9

1 3

x

 3

 , 3 1 3

GROUP ACTIVITY SOLUTIONS Health and Body Mass Index (BMI) Real-World Example of a Quadratic Equation The American Heart Association recommends that we know our “numbers.” In addition to cholesterol, blood pressure, and glucose numbers, our Body Mass Index (BMI) is an important number to know. It is a measure of the level of body fat. A high BMI is related to a greater risk of developing heart disease, osteoarthritis, diabetes, stroke, and certain cancers.

Group Activity BMI is calculated according the following formula.

BMI 

703w h2

w is weight in pounds; h is height in inches

BMI  18.5  underweight

 

18.5  BMI  24.9  normal 25.0  BMI  29.9  overweight

BMI  30  obese

a. The approximate weight and BMI of seven people are shown in the table. Determine the approximate height of each in inches. Round to the nearest inch.

Person A B C D E F G

BMI 17.9 30.8 21.0 26.0 28.8 22.1 21.9

Weight in Pounds 118 240 130 166 224 150 120

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497


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

b. Based on the height calculated in part a for Person A, a healthy BMI would be 20. How much weight should Person A strive to gain to reach that BMI number? c. Based on the height calculated in part a for Person E, a healthy BMI would be 24. How much weight should Person E strive to lose to reach that BMI number?

Solution a. We are given BMI 

703w h2

Solve the equation for height.

BMI  h2  703w h2 

703w BMI

h  

703w , however, negative heights will not make sense. BMI

So, h 

703w BMI

Person

BMI

w

A

17.9

118

68 inches

B

30.8

240

74 inches

C

21

130

66 inches

D

26

166

67 inches

E

28.8

224

74 inches

F

22.1

150

69 inches

G

21.9

120

62 inches

h 

703w BMI

b. Solve the BMI equation for weight: w 

BMI  h2 703

For person A, h  68. If BMI  20 then:

w 

20  682  132 pounds. 703

Person A currently weighs 118 pounds and would need to gain 14 pounds to reach a BMI of 20. c. For person E, h  74. If BMI  27 then:

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498


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

w 

24  742  187 pounds. 703

Person E currently weighs 224 pounds and would need to lose 37 pounds to reach a BMI of 24.

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499


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution and Answer Guide GUSTAFSON/HUGHES, COLLEGE ALGEBRA 2023, 9780357723654; CHAPTER 2: FUNCTIONS AND G RAPHS

TABLE OF CONTENTS End of Section Exercise Solutions ................................................................................................. 500 Exercises 2.1 .............................................................................................................................................. 500 Exercises 2.2 ............................................................................................................................................. 534 Exercises 2.3 ............................................................................................................................................. 573 Exercises 2.4 ............................................................................................................................................. 600 Exercises 2.5 ............................................................................................................................................. 638 Exercises 2.6 ............................................................................................................................................. 678 Chapter Review Solutions ...............................................................................................................693 Chapter Test Solutions ....................................................................................................................727 Group Activity Solutions..................................................................................................................739

END OF SECTION EXERCISE SOLUTIONS EXERCISES 2.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Given two sets: A = {(Backpack, $36), (Backpack, $40)} B = {(Amazon Fire TV Stick, $25), (Laptop Cooling Pad, $25)} In which set is each input paired with exactly one output? Solution B

2. How many y values correspond to an x-value of 16? a)

y 

b)

y x

x

2

Solution a. Since

16  4, then 16 only corresponds to 1 y-value.

b. Since (4)2  16 and (  4)2  16, then 16 corresponds to 2 y-values.

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500


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

3. Evaluate 2x2 – 3x + 4 at x = –2. Solution 2(–2)2 – 3(–2) + 4 = 2(4) + 6 + 4 = 18 4. Given 5x2. Substitute x + h in for x and simplify. Solution

5  x  h  5  x  h  x  h  5 x 2  2 xh  h2  5 x 2  10 xh  5h2 2

5. Identify the values of x that make the denominator of

x 4 equal 0. x 2  6x  8

Solution

x 2  6 x  8   x  4  x  2 So, 4 and 2 will make the denominator 0.

6. Given 2 x  7. Identify the values of x for which 2 x  7  0. Write the answer using interval notation.

Solution 2x  7  0 2 x  7 7 x 2  7   ,    2  Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A correspondence that assigns exactly one value of y to any number x is called a __________.

Solution function 8. A set of ordered pairs is called a __________.

Solution relation 9. The set of input numbers x in a function is called the __________ of the function.

Solution domain

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501


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

10. The set of all output values y in a function is called the __________ of the function.

Solution range 11. The statement “y is a function of x” can be written as the equation __________.

Solution

y  f x

12. In the function of Exercise 11, __________ is called the independent variable.

Solution x 13. In the function of Exercise 5, y is called the __________ variable.

Solution dependent 14. The expression

f  x  h  f  x  h

, h  0, is called the __________.

Solution difference quotient Practice A relation is given. (a) State the domain and range. (b) Determine if the relation is a function. 15. {(2, 3), (3, 4), (4, 5), (5, 6)}

Solution

D  2, 3, 4, 5 ; R  3, 4, 5, 6

Each element of the domain is paired with only one element of the range. Function. 16. {(5, 4), (6, 4), (7, 4), (8, 4)}

Solution

D  5, 6, 7, 8 ; R  4

Each element of the domain is paired with only one element of the range. Function. 17. {(1, 3), (1, 4), (2, 5), (–5, 2)}

Solution

D  1, 2,  5 ; R  3, 4, 5, 2

1 is both paired with 3 and 4. Not a function. 18. {(–1, 2), (2, –1), (0, 1), (0, 3)}

Solution

D  1, 2, 0 ; R  2,  1, 1, 3

0 is both paired with 1 and 3. Not a function.

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502


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

In Exercises 19–22, a relation is given. (a Write the ordered pairs of the relation. (b the domain and the range. (c Determine whether the relation is a function. 19. University

State

Mascot

Solution {(LSU, Tigers), (Georgia, Bulldogs), (MSU, Bulldogs), (Auburn, Tigers)} D = {LSU, Georgia, MSU, Auburn}; R = {Tigers, Bulldogs} Each element of the domain is paired with only one element of the range. Function. 20.

City

State

Solution {(Jackson, Louisiana), (Jackson, Mississippi), (Jackson, Tennessee), (Alexandria, Virginia)} D= {Jackson, Alexandria}; R = {Louisiana, Mississippi, Tennessee, Virginia} Jackson is paired with Louisiana, Mississippi, and Tennessee. Not a function. 21.

Golf Score 76 76 78 80

Date September 9 October 12 May 10 June 1

Solution {(76, September 9), (76, October 12), (78, May 10), (80, June 1)} D = {79, 78, 80}; R = {September 9, October 12, May 10, June 1} 76 is paired with September 9 and October 12. Not a function. 22.

Occupation

Median Salary

Architect

$73,090

Dentist

$149,310

Microbiologist

$66,260

Actuary

$93,680

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503


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution {(Architect, $73,090), (Dentist, $149,310), (Microbiologist, $66,260), (Actuary, $93,680)} D = {Architect, Dentist, Microbiologist, Actuary}; R = {$73,090, $149,310, $66,260, $93,680} Each element of the domain is paired with only one element of the range. Function.

Assume that all variables represent real numbers. Determine whether each equation determines y to be a function of x. 23. y  x

Solution y  x Each value of x is paired with only one value of y. function 24. y  2x  0

Solution y  2x  0

y  2x Each value of x is paired with only one value of y. function 25. y 2  x

Solution

y2  x y x At least one value of x is paired with more than one value of y. not a function 26. y 2  4 x  1

Solution

y 2  4x  1 y 2  4x  1 y 

4x  1

At least one value of x is paired with more than one value of y. not a function 27. y  x 2

Solution

y  x2 Each value of x is paired with only one value of y. function

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504


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

28. y  1  5x 3

Solution

y  1  5x 3 Each value of x is paired with only one value of y. function 29. y  x

Solution y x y x

At least one value of x is paired with more than one value of y. not a function 30. 2 y  x  4

Solution 2 y  x4 x4 2 x 4 y  2

y 

At least one value of x is paired with more than one value of y. not a function 31. x  2  y

Solution

x 2  y y  x 2 Each value of x is paired with only one value of y. function 32. y  x  3

Solution

y x 3 y  x 3 Each value of x is paired with only one value of y. function

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505


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

33. x  y

Solution x  y y  x y  x

At least one value of x is paired with more than one value of y. not a function 34. y  x  2

Solution

y  x 2

y    x  2

At least one value of x is paired with more than one value of y. not a function 35. y  7

Solution y  7; Each value of x is paired with only one value of y. function 36. x  7

Solution x  7; At least one value of x is paired with more than one value of y. not a function 37. y  7 

x

Solution y 7  y 

x x 7

Each value of x is paired with only one value of y. function 38. y  3 x  8

Solution y3x 8 y  3 x 8

Each value of x is paired with only one value of y. function

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506


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

39. x 2  y 2  25

Solution

x 2  y 2  25 y 2   x 2  25 y    x 2  25 At least one value of x is paired with more than one value of y. not a function 40.  x  1  y 2  16 2

Solution

 x  1  y  16 2

2

y 2  16   x  1 y 

2

16   x  1

2

At least one value of x is paired with more than one value of y. not a function

Evaluate the functions at the given values of the independent variable. 41. f  x   7 x  8 a.

f  2

b.

f  3 

Solution

f  x   7x  8 f 2  7 2  8  14  8 6

f  3   7  3   8  21  8  29

42. f  x   5 x  2 a.

f 4

b.

f  5 

Solution

f  x   5 x  2 f  4   5  4   2  20  2  18

f  5   5  5   2  25  2  27

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507


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

43. f  x   x 2  2 x  9 a.

f  3

b.

f  2 

Solution

f  x   x 2  2x  9 f  3   32  2  3   9  969 6

f  2    2   2  2   9 2

 449  9

44. f  x   2 x 2  x  1 a. b.

 1 f  2

f  4 

Solution

f  x   2x 2  x  1 2

 1  1 1 f    2    1 2 2 2      1 1  2    1 4 2 1 1   1 2 2 1

f  4   2  4    4   1 2

 2  16   4  1  32  5  37

45. f  x   2  x  3   1 2

a. b.

f (0)

f  1

Solution f  x   2  x  3   1 2

f  0   2  3  1 2

 2  9  1  18  1  17

f  1  2  1  3   1 2

 2  2   1 2

 2  4   1  8  1  7

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508


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

46. f  x   a. b.

2 1 x  4  1  2

f (5)

f  2 

Solution 2 1 f  x    x  4  1 2 2 1 5  4  1  2 1 2   1  1 2 1  1 2 1  2

f 5 

2 1 2  4   1  2 2 1   6   1 2 1   36   1 2  18  1  17

f  2  

47. f  x   x 3  2 x 2  x  1 a. b.

f (1)

f  1

Solution

f  x   x 3  2x 2  x  1

f  1  13  2  1  1  1 2

 1 2 3

f  1   1  2  1   1  1 3

2

 1  2  1  1 3

48. f  x    x 3  x 2  5 x  1 a. b.

f (2)

f  3 

Solution

f  x    x 3  x 2  5x  1 f  2   23  22  5  2   1  8  4  10  1  3

f  3     3    3   5  3   1 3

2

 27  9  15  1 2

49. f  x   2  x  1  2 3

a. b.

f (3)

f  2 

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509


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution f  x   2  x  1  2 3

f  3   2  3  1  2 3

f  2   2  2  1  2 3

 2  2   2

 2  3  2

3

3

 2  27   2

 16  2  14

 54  2  56

3 1  x  5  4 5 f (0)

 

50. f x  a. b.

f  6 

Solution

f  x 

3 1 x  5  4  5

3 1 0  5  4  5 1   125   4 5  25  4

f 0  

 21

3 1  6  5   4 5 1   1  4 5 1   4 5 21  5

f  6  

51. f  x   2 x  3  2 a. b.

f (4)

f  5 

Solution

f x  2 x  3  2

f 4  2 4  3  2

 2  1  2 4

f  5   2 5  3  2

 2 8  2  18

52. f  x   3 x  2  4 a. b.

f (7)

f  4 

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510


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

f  x   3 x  2  4

f  7   3 7  2  4

 3  9   4  23

f  4   3 4  2  4

 3  2   4  2

 

53. f x  2 x  4  5 a. b.

f (12)

f  3 

Solution

f  x   2 x  4  5 f  12  2 12  4  5  2 16  5

 2  1  5

 13

 2  5  7

 2  4   5

 

54. f x  a. b.

f  3  2 3  4  5

x 8 5

f (19)

f  4 

Solution

f  x  x  8  5 f  19  

19  8  5

f  4  

4  8  5

 27  5

 4 5

 3 3 5

7

 

3 55. f x  5 x  6

a. b.

f (16)

f  54 

Solution

f  x   53 x  6

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511


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

f  16   5 3 16  6

f  54   5 3 54  6

 53 8  2  6

 5 3 27  2  6

 5  23 2  6

 5  3 3 2  6

 10 3 2  6

 15 3 2  6

 

3 56. f x  x  1  3

a. b.

f (7)

f  28

Solution

f  x  3 x  1  3 f 7   3 7  1  3

57. f  x  

f  28   3 28  1  3

 38 3

 3 27  3

 23

 3  3

 1

 6

3 x

a.

 1 f  5

b.

 3 f    2

Solution 3 f x  x

 1 3 f  5 1   5  3 5  15

58. f  x  

 3 3 f    2 3    2  2  3   3  2

x x 4

a.

f 6

b.

f  4 

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512


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution f x 

x x 4

4 4  4 4  8 1  2

6 64 6  2 3

f 6 

a.

2x x 9 f 4

b.

f  2 

 

59. f x 

f  4  

2

Solution

f  x 

2x x 9 2

f 4 

2 4

4 9 8  7

 

60. f x  

2

f  2   

2  2 

 2  9 2

4 5

x x  25 2

a.

f 6

b.

f  4 

Solution

f  x  

x x  25 2

6 62  25 6  11

f 6  

f  4   

4

 4   25 2

4 16  25 4  9 

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513


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Evaluate the functions at the given values of the independent variable. 61. f  x   6 x  7 a.

f x 

b.

f  3x 

c.

f  x  3

Solution

f  x   6x  7 f x   6 x   7  6 x  7

f  x  3  6  x  3  7

f 3x   6 3x   7

 6 x  18  7  6 x  11

 18 x  7

62. f  x   4 x  3 a.

f x 

b.

f  4 x 

c.

f  x  3

Solution

f  x   4 x  3 f   x   4   x   3

f   4 x   4  4 x   3

 4x  3

 16 x  3

f  x  3   4  x  3   3  4 x  12  3  4 x  15

63. f  x   x 2  2 x  3 a.

f x 

b.

f x2

c.

f  x  5

 

Solution

f  x   x 2  2x  3

f x   x   2 x   3 2

 x 2  2x  3

     2 x   3

f x2  x2

2

2

 x 4  2x 2  3

f  x  5   x  5  2  x  5  3 2

 x 2  10 x  25  2 x  10  3  x 2  12 x  32

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514


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

64. f  x   x 2  2 x  1 a.

f x 

b.

f x2

c.

f  x  5

 

Solution

f  x   x 2  2x  1

     2x   1

f x   x   2 x   1 2

f x2  x2

 x 2  2x  1

2

f  x  5   x  5  2  x  5  1 2

2

 x 2  10 x  25  2 x  10  1

 x 4  2x 2  1

 x 2  12 x  26 65. f  x   3 x 2  2 x  5 a.

f x 

b.

f 2x 3

c.

f  x  2

 

Solution

f  x   3 x 2  2 x  5 f   x   3   x   2   x   5 2

   2 2x   5  3  4 x   4 x  5

f 2 x 3  3 2 x 3

 3 x 2  2 x  5

5

2

3

3

 12 x 5  4 x 3  5

f  x  2   3  x  2   2  x  2   5 2

 3 x 2  4 x  4  2 x  4  5  3 x  12 x  12  2 x  9 2

 3 x 2  10 x  3

66. f  x   3 x 2  5 x  12 a.

f x 

b.

f 3x 2

c.

f  x  2

 

Solution

f  x   3 x 2  5 x  12

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515


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

f   x   3   x   5   x   12 2

 3 x 2  5 x  12

   5 3x   12  3  9 x   15 x  12 2

f 3x 2  3 3x 2

4

2

2

 27 x 4  15 x 2  12

f  x  2   3  x  2   5  x  2   12 2

 3 x 2  4 x  4  5 x  10  12  3 x  12 x  12  5 x  22 2

 3 x 2  17 x  34

67. f  x   x 3  3 x 2  x  4 a.

f x 

b.

f x2

c.

f  x  1

 

Solution

f  x   x 3  3x 2  x  4 f x   x   3 x   x   4 3

2

  x 3  3x 2  x  4

     3 x   x   4

f x2  x2

3

2

2

2

 x6  3x 4  x 2  4

f  x  1   x  1  3  x  1   x  1  4 3

2

 x 3  3x 2  3x  1  3x 2  6x  3  x  1  4  x 3  2x  5 68. f  x   2 x 3  2 x 2  5 x  1 a.

f x 

b.

f x3

c.

f  x  1

 

Solution

f  x   2 x 3  2 x 2  5 x  1 f   x   2   x   2   x   5   x   1 3

2

 2x  2x  5x  1 3

2

 

   2x   5x   1

f x 3  2 x 3

3

3

2

3

 2 x 9  2 x 6  5 x 3  1

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516


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

f  x  1  2  x  1  2  x  1  5  x  1  1 3

2

 2 x 3  3 x 2  3 x  1  2 x 2  4 x  2  5 x  5  1  2 x  6 x  6 x  2  2 x 2  x  4 3

2

 2 x 3  4 x 2  7 x  6

69. f  x   x 4  3 x 2  4 a.

f x 

b.

f 2x 

c.

f x2

 

Solution

f  x   x 4  3x 2  4 f x   x   3 x   4 4

2

 x4  3x2  4

 

f x2  x2

f 2x    2x   3 2x   4 4

2

 16 x 4  12 x 2  4

  3 x   4 4

2

2

 x 8  3x 4  4 70. f  x   2 x 4  2 x 2  5 a.

f x 

b.

f  3 x 

c.

f x4

 

Solution

f  x   2 x 4  2 x 2  5

f   x   2   x   2   x   5

f   3 x   2  3 x   2  3 x   5

 2 x 4  2 x 2  5

 162 x 4  18 x 2  5

4

 

2

4

2

   2 x   5

f x 4  2 x 4

4

4

2

 2 x 16  2 x 8  5

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517


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

71. f  x  

1 2

x 2

a.

f x 

b.

f 4x 

c.

f 100x 2

Solution 1 f x  x 2 2 f x  

1 x  2 2

1 4x  2 2 1   2 x  2 2  x 2

f 4 x  

1 100 x 2  2 2 1   10 x  2 2  5x  2

f 100 x 2 

 

72. f x   x  3 a.

f x 

b.

f 9x 2  3

c.

 1  f  x2  3 16  

Solution

f  x   x  3 f  x    x  3

f 9x 2  3   9x 2  3  3  3 x

 1  1 2 f  x2  3   x 33 16  16  1  x 4

 

3 73. f x  2 x  5

a.

f x 

b.

f  8 x 

c.

f x3

 

Solution

f  x   23 x  5

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518


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

f   x   23  x  5

f  8 x   2 3  8 x  5

 2 3 x  5

 4 3 x  5

 

3

f x3  2 x3  5  2x  5

 

3 74. f x  3 x  2

a.

f x 

b.

f  27 x 

c.

f 64 x 6

Solution

f  x   33 x  2 f   x   3 3  x  2

f  27 x   3 3 27 x  2

3

3

3 x 2

a.

3x x2  9 f x 

b.

f x2

c.

f  x  2

 9 x  2

f 64 x 6  3 3 64 x 6  2  12 x 2  2

 

75. f x 

 

Solution

f  x 

3x x 9 2

f x  

3 x 

x   9

2

3 x x2  9

 

f x2  

3x 2

x   9 2

2

3x 2 x4  9

f  x  2   

 

76. f x  

3  x  2

 x  2  9 2

3x  6 x  4x  4  9 3x  6 2

x2  4x  5

5x x2  9

a.

f x 

b.

f x3

c.

f  x  4

 

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519


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

5x x 9

f  x  

2

f x    

5 x 

x   9 2

 

f x3  

5x x2  9



5x 3

x   9 3

2

5x 3 x6  9

f  x  4    

5  x  4

 x  4  9 2

5 x  20 x  8 x  16  9 5 x  20 2

x 2  8x  7

Find the domain of each function. 77. f  x   3 x  5

Solution

f  x   3 x  5  domain   ,  

78. f  x   5 x  2

Solution

f  x   5 x  2  domain   ,  

79. f  x   x 2  x  1

Solution

f  x   x 2  x  1  domain   ,  

80. f  x   x 3  3 x  2

Solution

f  x   x 3  3 x  2  domain   ,  

 

81. f x 

x 2

Solution

f  x 

x 2  x 20

domain  2,  

 

82. f x  2x  3

Solution

f  x   2x  3  2x  3  0

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520


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

 3  domain    ,   2  

 

83. f x  4  x

Solution

f  x  4  x  4  x  0 domain    , 4 

 

84. f x  3 2  x

Solution

f  x  3 2  x  2  x  0 domain    , 2

 

85. f x 

x2  1

Solution

f  x   x2  1  x2  1  0 domain    ,  1   1,  

 

86. f x 

x 2  2x  3

Solution

f  x   x 2  2x  3  x 2  2x  3  0 domain    ,  1  3,  

 

3 87. f x  x  1

Solution

f  x   3 x  1  domain   ,  

 

3 88. f x  5  x

Solution

f  x   3 5  x  domain   ,  

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521


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

89. f  x  

3 x1

Solution f x 

3  x  1 x1

domain    ,  1    1,  

 

90. f x 

7 x3

Solution

f  x 

7  x  3 x3

domain    ,  3    3,  

 

91. f x 

x x 3

Solution

f  x 

x x3 x 3

domain    , 3    3,   x2 x1 Solution

92. f  x   f x 

x2  x1 x1

domain    , 1   1,  

 

93. f x 

x x 4 2

Solution f x 

x x  x  4  x  2  x  2 2

x  2, x  2

domain    ,  2     2, 2    2,  

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522


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

 

94. f x 

2x x 9 2

Solution f x 

2x 2x  x  9  x  3  x  3  2

x  3, x  3

domain    ,  3     3, 3    3,  

 

95. f x 

1 x  4x  5 2

Solution f x 

1 1  x  4 x  5  x  1 x  5  2

x  1, x  5

domain    ,  1    1, 5    5,  

 

96. f x 

x 2 x  5x  14 2

Solution f x 

x 2 x 2  5 x  14

x 2

 x  7  x  2

x  7, x  2

domain   ,  2   2, 7    7,  

 

97. f x 

x1 3x  2x  1 2

Solution

f x  x

x1 x1  3 x  2 x  1  3 x  1 x  1 2

1 , x  1 3

  1  1 domain   ,  1   1,    ,   3 3    

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523


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

 

98. f x 

x 2x  16x  30 2

Solution f x 

x 2 x  16 x  30 2

x

2  x  5  x  3 

x  3, x  5

domain   , 3    3, 5    5,  

99. f  x  

2x x 4

Solution f x  x 4 0

2x x 4

x4

domain   4,  

100. f  x   

3 6x

Solution

f x  

3 6 x

6 x 0  x  6 x 6

domain   , 6  101. f  x   x  3

Solution

f  x   x  3  domain    ,  

102. f  x   2 x  1

Solution

f  x   2 x  1  domain    ,  

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524


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Evaluate the difference quotient for each function f (x). 103. f  x   3 x  1

Solution

f  x  h  f  x  h

3  x  h  1  3 x  1 3 x  3h  1  3 x  1           h h 3x  3h  1  3x  1 3h   3 h h

104. f  x   5 x  1

Solution

f  x  h  f  x  h

5  x  h  1  5 x  1 5 x  5h  1  5 x  1           h h 5 x  5h  1  5 x  1 5h   5 h h

105. f  x   7 x  8

Solution

f  x  h  f  x  h

 7  x  h  8   7 x  8  7 x  7h  8   7 x  8           h h 7 x  7h  8  7 x  8 7h    7 h h

106. f  x   8 x  1

Solution

f  x  h  f  x  h

 8  x  h  1   8 x  1  8 x  8h  1   8 x  1           h h 8 x  8h  1  8 x  1 8h    8 h h

107. f  x   x 2  1

Solution

f  x  h  f  x  h

 x  h 2  1   x 2  1 2 2 2      x  2 xh  h  1   x  1     h h x 2  2 xh  h2  1  x 2  1  h 2 h  2 x  h 2 xh  h    2x  h h h

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525


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

108. f  x   x 2  3

Solution

f  x  h  f  x  h

 x  h 2  3   x 2  3 2 2 2      x  2 xh  h  3   x  3    h h x 2  2 xh  h2  3  x 2  3  h 2 h  2 x  h 2 xh  h    2x  h h h

109. f  x   4 x 2  6

Solution

f  x  h  f  x  h

4 x  h 2  6   4 x 2  6  2 2 2     4 x  8 xh  4h  6  4 x  6      h h 4 x 2  8 xh  4h2  6  4 x 2  6  h 2 h 8x  4h  8x  4h 8 xh  4h   h h

110. f  x   5 x 2  3

Solution

f  x  h  f  x  h

5 x  h 2  3  5 x 2  3 2 2 2     5 x  10 xh  5h  3  5 x  3      h h 5 x 2  10 xh  5h2  3  5 x 2  3  h 2 h  10x  5h  10x  5h 10 xh  5h   h h

111. f  x   x 2  3 x  7

Solution f  x  h  f  x  h

 x  h 2  3 x  h  7   x 2  3x  7           h  x 2  2 xh  h2  3 x  3h  7    x 2  3 x  7       h x 2  2 xh  h2  3 x  3h  7  x 2  3 x  7  h 2 xh  h2  3h h  2 x  h  3     2x  h  3 h h

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526


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

112. f  x   x 2  5 x  1

Solution f  x  h  f  x  h

 x  h 2  5 x  h  1   x 2  5 x  1          h  x 2  2 xh  h2  5 x  5h  1   x 2  5 x  1      h x 2  2 xh  h2  5 x  5h  1  x 2  5 x  1  h 2 h 2 x  h  5  2 xh  h  5h    2x  h  5 h h

113. f  x   2 x 2  4 x  2

Solution f  x  h  f  x  h

2 x  h 2  4 x  h  2   2 x 2  4 x  2            h 2 x 2  4 xh  2h2  4 x  4h  2  2 x 2  4 x  2      h 2 x 2  4 xh  2h2  4 x  4h  2  2 x 2  4 x  2  h 4 xh  2h2  4h h  4 x  2h  4     4 x  2h  4 h h

114. f  x   3 x 2  2 x  3

Solution f  x  h  f  x  h

3 x  h 2  2 x  h  3  3 x 2  2 x  3            h 3 x 2  6 xh  3h2  2 x  2h  3  3 x 2  2 x  3      h

3 x 2  6 xh  3h2  2 x  2h  3  3 x 2  2 x  3 h 2 6 xh  3h  2h h  6 x  3h  2     6 x  3h  2 h h 

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527


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

115. f  x    x 2  x  3

Solution f  x  h  f  x  h

  x  h 2  x  h  3    x 2  x  3           h   x 2  2 xh  h2  x  h  3    x 2  x  3      h  x 2  2 xh  h2  x  h  3  x 2  x  3  h 2 h  2x  h  1  2x  h  1 2 xh  h  h   h h

116. f  x   3 x 2  5 x  1

Solution f  x  h  f  x  h

 3 x  h 2  5 x  h  1   3 x 2  5 x  1           h  3 x 2  6 xh  3h2  5 x  5h  1   3 x 2  5 x  1      h 3 x 2  6 xh  3h2  5 x  5h  1  3 x 2  5 x  1  h 6 xh  3h2  5h h  6 x  3h  5     6 x  3h  5 h h

117. f  x   x 3

Solution f  x  h  f  x  h

 x  h  x   x  3x h  3xh  h    x   3

3

3

2

2

3

3

h h 3 x 2 h  3 xh2  h3  h h 3 x 2  3 xh  h2   3 x 2  3 xh  h2 h

118. f  x    x 3

Solution f  x  h  f  x  h

3

  x  h   x 3

   x  3x h  3xh  h   x 3

2

2

3

3

h h 3 x 2 h  3 xh2  h3  h h 3 x 2  3 xh  h2   3 x 2  3 xh  h2 h

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528


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

1 x

119. f  x  

Solution

f  x  h  f  x  h

1 1  1  1   x  x  h    xh x  xh x   h h  x  x  h 

 

120. f x 

x   x  h xh  x  h 

h

xh  x  h



1

x  x  4

x

Solution

f  x  h  f  x  h

xh x h

Fix It In exercises 121 and 122, identify the step the first error is made and fix it. 121. Given f (x) = 2x2 – 3x + 4. Find f (x – 5).

Solution Step 2 was incorrect. Step 1: 2  x  5   3  x  5   4 2

Step 2: 2 x 2  10x  25  3x  15  4 Step 3: 2 x 2  20 x  50  3 x  19 Step 4: 2 x 2  23 x  69 122. Given f(x) = 3x2 – 2x + 1. Find the difference quotient

f  x  h  f  x  h

.

Solution Step 5 was incorrect.

2

Step 1:

Step 2: Step 3:

3  x  h  2  x  h  1  3 3x 2  2 x  1 h

3 x 2  2 xh  h2  2 x  2h  1  3 x 2  2 x  1 h 3 x 2  6 xh  3h2  2h  3 x 2 h

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529


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Step 4:

6 xh  3h2  2h h

Step 5: 6 x  3h  2

Applications 123. Target heart rate The target heart rate f(x), in beats per minute, at which a person should train to get an effective workout is a function of the person’s age x in years. If f(x) = –0.6x + 132, find the target heart rate for a 25-year-old college student.

Solution

f  x   0.6 x  132

f  25  0.6  25  132  117 124. Temperature conversion

The Fahrenheit temperature reading F is a function of the

Celsius reading C. The function can be written as F C   95 C  32.

Find the Fahrenheit temperature for the Celsius temperatures: C = 0; C = –40; C = 10.

Solution 9 F C   C  32 5 9 F  0    0   32 5  32

9  40   32 5  40

F  40  

9  10   32 5  50

F  10  

125. Free-falling objects The velocity v of a falling object is a function of the time t it has been falling. If v as a function of t can be expressed as v(t) = –32t + 15, where v is in feet per second and t is in seconds, when will the velocity be 0?

Solution

v  t   32t  15 v t   0

32t  15  0 t

15 seconds 32

126. Cliff divers The height s, in feet, of a cliff diver is a function of the time t in seconds he has been falling. If s as a function of t can be expressed as s(t) = –16t2 + 10t + 300, what is the height of the diver at 3 seconds?

Solution

s  t   16t 2  10t  300

s  3   16  3   10  3   300  186 ft 2

127. Go green The typical family in the USA uses about 300 gallons of water a day. If the number of gallons g used expressed in terms of days d is g(d) = 300d, find the number of gallons used in one year.

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530


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

g  d   300d

g  365   300  365  109,500 gallons 128. Volume of a basketball The volume V of a sphere can be expressed in terms of its radius r according to the function V  r   43  r 3 . Find the volume of a men’s NCAA basketball if

the diameter of the ball is 29.5 centimeters. Round to the nearest cubic centimeter.

Solution 4 3 r 3 3 4  29.5  3 V  29.5       13, 442 cm 3  2  V r  

129. Formulas The area A of a rectangle is determined by the length and width. If the length of a rectangle is x inches and the width is 5 inches more than the length, express the area as a function of the length.

Solution Let x = the length. Then x + 5 = the width. A  x   x  x  5  x 2  5x 130. Formulas The volume V of a rectangular box is determined by the length, the width, and the height. For a particular set of boxes, the height is 4 feet, the length is given as x feet, and the width is 3x feet. Express the volume as a function of x.

Solution

V  x   x  3 x  4   12 x 2

131. Cost of t-shirts A chapter of Phi Theta Kappa, an honors society for two-year college students, is purchasing t-shirts for each of its members. A local company has agreed to make the shirts for $8 each plus a graphic arts fee of $75. a. Write a function that describes the cost C for the shirts in terms of x, the number of t-shirts ordered. b. Find the total cost of 85 t-shirts.

Solution a.

C  x   8 x  75

b.

C  85   8  85   75  $755

132. Service projects The Circle “K” Club is planning a service project for children at a local children’s home. They plan to rent a “Dora the Explorer Moonwalk” for the event. The cost of the moonwalk will include a $60 delivery fee and $45 for each hour it is used. Write a function that describes the cost C for renting the moonwalk in terms of x, the number of hours used.

Solution

C  x   45 x  60

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531


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

133. Cell phone plans A grandmother agrees to purchase a cell phone for emergency use only. A cellular provider now offers such a plan for $9.99 per month and $0.07 for each minute x the phone is used. a. Write a function that describes the monthly cost C in terms of the time in minutes x the phone is used. b. If the grandmother uses her phone for 20 minutes during the first month, what was her bill?

Solution a.

C  x   0.07 x  9.99

b.

C  20   0.07  20   9.99  $11.39

134. Concessions A concessionaire at a football game pays a vendor $40 per game for selling hot dogs at $2.50 each. a. Write a function that describes the income I the vendor earns for the concessionaire during the game if the vendor sells x hot dogs. b. Find the income if the vendor sells 175 hot dogs.

Solution a.

I  x   2.5 x  40

b.

I  175   2.5  175   40  $397.50

Discovery and Writing 135. Using words, state three real-life correspondences that represent relations.

Solution Answers may vary. 136. Using words, state three real-life correspondences that represent functions.

Solution Answers may vary. 137. Explain why some equations represent y as a function of x, and some do not.

Solution Answers may vary. 138. Explain why all functions are relations, but not all relations are functions.

Solution Answers may vary. 139. Describe what is meant by the domain of a function.

Solution Answers may vary.

 

140. Are the domains of the functions f x 

x  4 and g  x   4  x the same or different?

Solution They are different. 10 is in the domain of f(x), but not in the domain of g(x).

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532


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

 

141. Describe how you would find the domain of f x 

x 2  16.

Solution Answers may vary. 142. Describe how you would find the domain of f  x  

x3 . x2  5x  6

Solution Answers may vary.

 

143. Are the functions f  x   x  3 and g x 

x2  9 the same? Explain why or why not. x 3

Solution They are different. The domain of f(x) is the set of all real numbers, but 3 is not in the domain of g(x). 144. Explain why the difference quotient for f(x) = 5 is 0.

Solution

f  x  h  f  x  h

55 0   0. h h

Critical Thinking In Exercises 145–152, match the function with its range. Some answers can be repeated. 145. f  x   5 x

a.

 , 0  0,  

146. f  x   x 2  1

b.

 , 0

147. f  x    x 2

c.

  1,  

148. f  x   x 3

d.

 1,  

149. f  x   x

e.

 ,  

f.

0,  

 

150. f x 

x 1

 

3 151. f x  x

152. f  x  

1 x

Solution 145. e 146. d 147. b

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533


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

148. e 149. f 150. c 151. e 152. a

EXERCISES 2.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Solve the equation 3x – 5y = 10 for y.

Solution 3 x  5 y  10

5 y  3 x  10 y 

3 x 2 5

2. Solve the equation Ax + By = C for y.

Solution Ax  By  C

By  C  Ax C A  x B B A C y  x B B y 

3. If y = −2x + 7, find the y-values that correspond to the x-values of −2, −1, 0, 1, and 2.

Solution y  2 x  7

4. If y 

x

2x  7

–2

11

–1

9

0

7

1

5

2

3

2 x  5, find the y-values that correspond to the x-values of –6, –3, 0, 3, and 6. 3

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534


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

y

2 x  5, 3 x

2 x 5 3

–6

–9

–3

–7

0

–5

3

–3

6

–1

5. Given 6x – 7y = –42. a. If x = 0, what is y? b. If y = 0, what is x?

Solution a. 6  0   7 y  42 y 6

b.

6 x  7  0    42 x  7

288.

6. Simplify

Solution 288 

144

2  12 2.

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The coordinate axes divide the plane into four __________.

Solution quadrants 8. The coordinate axes intersect at the __________.

Solution origin 9. The positive direction on the x-axis is __________.

Solution to the right

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535


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

10. The positive direction on the y-axis is __________.

Solution upward 11. The x-coordinate is the __________ coordinate in an ordered pair.

Solution first 12. The y-coordinate is the __________ coordinate in an ordered pair.

Solution second 13. A __________ equation is an equation whose graph is a line.

Solution linear 14. The point where a line intersects the __________ is called the y-intercept.

Solution y-axis 15. The point where a line intersects the x-axis is called the __________.

Solution x-intercept 16. The graph of the equation x = a will be a __________ line.

Solution vertical 17. The graph of the equation y = b will be a __________ line.

Solution horizontal 18. Complete the Distance Formula: d = __________.

Solution

x  x    y  y  2

2

1

2

2

1

19. If a point divides a segment into two equal segments, the point is called the __________ of the segment.

Solution midpoint 20. The midpoint of the segment joining P(x1, y1) and Q(x2, y2) is M = __________.

Solution  x  x2 y 1  y 2  , M   1  2   2

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536


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Practice Refer to the illustration and determine the coordinates of each point.

21. A

Solution A(2, 3) 22. B

Solution B(–3, 5) 23. C

Solution C(–2, –3) 24. D

Solution D(4, –5) 25. E

Solution E(0, 0) 26. F

Solution F(–4, 0) 27. G

Solution G(–5, –5) 28. H

Solution H(2, –2)

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537


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Graph each point. Indicate the quadrant in which the point lies, or the axis on which it lies. 29-36 Solution

29. (2, 5)

Solution QI 30. (–3, 4)

Solution QII 31. (–4, –5)

Solution QIII 32. (6, 2)

Solution QI 33. (5, 2)

Solution QI 34. (3, –4)

Solution QIV 35. (4, 0)

Solution + x-axis 36. (0, 2)

Solution + y-axis

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538


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Graph the line represented by the equation by plotting points. 37. y = 2x + 7

Solution y  2x  7

y  2x  7 X

y

0

7

–2

3

38. y = –4x – 3

Solution y  3  4 x

y  4 x  3 X

y

0

–3

–1

1

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539


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

39. y + 5x = 5

Solution y  5x  5

y  5 x  5 X

y

0

5

1

0

40. y – 3x = 6

Solution y  3x  6

y  3x  6

41. y 

X

y

0

6

–2

0

1 x1 3

Solution

y

1 x1 3

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540


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

x

y

0

1

3

2

42. y  

1 x2 4

Solution 1 y   x2 4 x

y

0

2

4

1

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541


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

43. 6x – 3y = 10

Solution 6 x  3 y  10

3 y  6 x  10 y  2x  x

10 3

y

0

2

10 3 2 3

44. 4x + 8y = 1

Solution 4x  8 y  1  0

8 y  4 x  1 y 

1 1 x 2 8

X

y

0

1 8

4

15 8

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542


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

45. 2 x  5 y  10  0

Solution 2 x  5 y  10  0

2 x  5 y  10 5 y  2 x  10 y  x

y

0

–2

5

0

2 x 2 5

46. 3 x  2 y  6  0

Solution 3x  2 y  6  0

3 x  2 y  6 2 y  3 x  6 y 

x

y

0

–3

–2

0

3 x 3 2

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543


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

47. 3x = 6y – 1

Solution 3x  6 y  1

6 y  3 x  1 y 

1 1 x 2 6

x

y

0

1 6

–2

5 6

48. 2x + 1 = 4y

Solution 2x  1  4 y

4 y  2 x  1 y 

1 1 x 2 4

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544


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

X

y

0

1 4

–2

3 4

49. 2(x + y + 1) = x + 2

Solution

2  x  y  1  x  2 2x  2 y  2  x  2 2 y  x y 

1 x 2

X

y

0

0

–2

1

50. 5(x + 2) = 3y – x

Solution

5  x  2  3 y  x 5 x  10  3 y  x 3 y  6 x  10 10 y  2x  3

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545


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

x

y

0

10 3

–2

2 3

Find the x- and y-intercepts and use them to graph each equation. 51. x + y = 5

Solution x y 5

x y 5

x 0  5 x 5

0 y 5 y 5

5, 0

0, 5

52. x – y = 3

Solution xy 3

xy 3

x 0  3

0 y  3

x3

y  3

 3, 0

y  3

0,  3

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546


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

53. 2x – y = 4

Solution 2x  y  4

2x  y  4

2x  0  4

2 0  y  4

2x  4

y  4

x2

y  4

 2, 0

0,  4 

54. 3x + y = 9

Solution 3x  y  9

3x  y  9

3x  0  9

3 0  y  9

3x  9

y 9

x3

0, 9

 3, 0

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547


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

55. 3x + 2y = 6

Solution 3x  2 y  6

3x  2 y  6

3x  2 0  6

3 0   2 y  6

3x  6

2y  6

x2

y 3

 2, 0

0, 3

56. 2x – 3y = 6

Solution 2x  3 y  6

2x  3 y  6

2x  3 0  6

2 0  3 y  6

2x  6

3 y  6

x3

y  2

 3, 0

0,  2

57. 4x – 5y = 20

Solution 4 x  5 y  20

4 x  5 y  20

4 x  5  0   20

4  0   5 y  20

4 x  20

 5 y  20

x 5

y  4

5, 0

0,  4

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548


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

58. 3x – 5y = 15

Solution 3 x  5 y  15

3 x  5 y  15

3 x  5  0   15

3  0   5 y  15

3 x  15

 5 y  15

x 5

y  3

5, 0

0,  3

Graph each equation. 59. 2 x  3 y  5

Solution 2x  3 y  5

3 y  2 x  5 y 

2 5 x 3 3

x

Y

1

1

4

–1

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549


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

60. 4 x  3 y  10

Solution 4 x  3 y  10

3 y  4 x  10 y

4 10 x 3 3

x

y

1

–2

4

2

61. 2 x  3 y  9  0

Solution 2x  3 y  9  0

3 y  2x  9 y

2 x 3 3

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550


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

x

y

3

–2

–3

–5

62. 4 x  7 y  12  0

Solution 4 x  7 y  12  0

7 y  4 x  12 y  x

y

3

0

–4

4

4 12 x 7 7

63. y = 3

Solution y=3

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551


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

64. x = –4

Solution x = –4

65. x 

12 0 5

Solution

x

12 0 5 12 x 5

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552


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

66. y 

10 0 3

Solution

y

10 0 3 10 y  3

67. 3x + 5 = –1

Solution 3 x  5  1

3 x  6  x  2

68. 7y – 1 = 6

Solution 7y  1  6

7y  7  y  1

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553


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

69. 3(y + 2) = y

Solution

3  y  2  y 3y  6  y 2 y  6  y  3

70. 4 + 3y = 3(x + y)

Solution

4  3y  3x  y  4  3 y  3x  3 y 4  3x  x 

4 3

71. 3(y + 2x) = 6x + y

Solution

3 y  2 x   6 x  y 3 y  6x  6x  y 2y  0  y  0

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554


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

72. 5(y – x) = x + 5y

Solution

5 y  x  x  5y 5 y  5x  x  5 y 0  6x  x  0

Use a graphing calculator to graph each equation and then find the x-coordinate of the x-intercept to the nearest hundredth. 73. y = 3.7x – 4.5

Solution y  3.7 x  4.5

x-int: x  1.22

74. y 

3 5 x 5 4

Solution

y

3 5 x 5 4

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555


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

x-int: x  2.08

75. 1.5x – 3y = 7

Solution 1.5 x  3 y  7

3 y  1.5 x  7 7 y  0.5 x  3

x-int: x  4.67

76. 0.3x + y = 7.5

Solution 0.3x  y  7.5

y  0.3x  7.5

x-int: x  25.00

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556


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Find the distance between P and O (0, 0). 77. P(4, –3)

Solution

x  x    y  y    4  0    3  0   4   3  2

d

2

1

2

2

2

2

2

2

1

16  9  25  5

78. P(–5, 12)

Solution

x  x    y  y    5  0    12  0    5    12 

d

2

2

1

2

2

1

2

2

2

2

 25  144 

169  13

79. P(5, 0)

Solution

x  x    y  y   5  0  0  0  5  0

d

2

2

1

2

2

2

1

2

2

2

 25  0  25  5

80. P(6, –8)

Solution

x  x    y  y    6  0    8  0    6    8 

d

2

2

1

2

2

2

 36  64 

2

1

2

2

100  10

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557


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

81. P(–3, 2)

Solution

x  x    y  y    3  0    2  0    3    2  2

d

2

1

2

2

1

2

2

2

2

 94 

13

82. P(1, 1)

Solution

x  x    y  y    1  0   1  0   1   1 2

d

2

1

2

2

2

2

2

1

2

1 1  2

83. P  3,  6

Solution

x  x    y  y    3  0    6  0    3    6  2

d

2

1

2

2

1

2

2

2

2

 9  36  45  3 5

84. P  6,  2

Solution

x  x    y  y    6  0    2  0    6    2 

d 

2

2

1

2

2

2

2

1

2

2

 36  4  40  4  10  2 10

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558


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

85. P

 3, 1

Solution

d

x  x    y  y  2

2

1

2

 3  0   1  0   3    1 2

2

2

1

2

2

 31  4  2 86. P

 7, 2 

Solution

d

x  x    y  y  2

2

1

2

 7  0   2  0   7    2 2

2

2

1

2

2

 72  9 3 Find the distance between P and Q. 87. P(3, 7); Q(6, 3)

Solution

x  x    y  y    3  6   7  3   3    4 

d

2

2

1

2

2

2

1

2

2

2

 9  16  25  5

88. P(4, 9); Q(9, 21)

Solution

x  x    y  y    4  9    9  21   5    12 

d

2

2

1

2

2

2

 25  144 

2

1

2

2

169  13

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

559


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

89. P(4, –6); Q(–1, 6)

Solution

x  x    y  y   4   1    6  6   5    12 

d

2

2

2

1

2

1

2

2

2

2

 25  144  169  13 90. P(0, 5); Q(6, –3)

Solution

x  x    y  y   0  6  5   3     6    8 

d

2

2

1

2

2

1

2

2

2

2

 36  64 

100  10

91. P(–2, –15); Q(–6, –21)

Solution

x  x    y  y    2   6     15   21 

d

2

2

2

1

2

1

2

 4   6

16  36  52  2 13

2

2

2

92. P(–7, 11); Q(–11, 7)

Solution

x  x    y  y    7   11    11  7   4  4

d

2

2

2

1

2

1

2

2

2

2

16  16  32  4 2

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560


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

93. P(3, –3); Q(–5, 5)

Solution

x  x    y  y   3   5     3  5   8    8 

d

2

2

2

1

2

1

2

2

2

2

 64  64 

128  8 2

94. P(6, –3); Q(–3, 2)

Solution

x  x    y  y   6   3     3  2   9    5 

d

2

2

2

1

2

1

2

2

2

2

 81  25 

106

3  95. P  1,  3  ; Q  ,  6  2  Solution d

x  x    y  y  2

2

1

2

2

2

1

2  3   1    3   6  2 

2

2  1       3  2

1 9  4

37  4

37 2

 1  96. P  3,  1 ; Q   ,  2  2   Solution d

x  x    y  y  2

2

1

2

2

2

1

2   1    3       1   2   2  

2

2  5       1  2

25 1 4

29  4

29 2

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561


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

97. P  ,  2  ; Q  , 5

Solution d

x  x    y  y 

      2  5

 0    7 

2

2

1

2

2

2

2

1

2

2

 0  49  49  7

98. P

 5, 0 ; Q 0, 2

Solution

d

x  x    y  y 

 5  0  0  2 2    

 5    2

2

2

1

2

2

1

2

2

2

 54  9  3 Find the midpoint of the line segment PQ. 99. P(2, 4); Q(6, 8)

Solution  x  x2 y 1  y 2  2  6 4  8  8 12  , M  1   M  2 , 2   M  2 , 2   M  4, 6  2 2       100. P(3, –6); Q(–1, –6)

Solution

 3   1 6   6    x  x2 y 1  y 2   2 12    M , M  1 , ,   M    M  1,  6    2  2 2 2 2   2  

101. P(2, –5); Q(–2, 7)

Solution

 2   2   5  7   x  x2 y 1  y 2  0 2   M  ,   M  0, 1 M  1 , ,   M   2  2 2  2 2  2  

102. P(0, 3); Q(–10, –13)

Solution

 0   10  3   13    x  x2 y 1  y 2   10 10    M M  1 , , ,   M   M  5,  5     2 2 2 2 2 2      

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562


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

103. P(–8, 5); Q(6, –4)

Solution

 8  6 5   4    x  x2 y 1  y 2   2 1   1   M M  1 , , ,   M  1,    M    2 2 2 2 2 2 2        

104. P(3, –2); Q(2, –3)

Solution

 3  2 2   3    x  x2 y 1  y 2   5 5  5 5   M , M  1 , ,   M    M ,     2  2 2 2 2 2  2  2  

1 2   17 18  105. P  ,   ; Q  ,   5 3 5  3 Solution  1 17 2 18   20      6   x 1  x2 y 1  y 2  3 3 5 5   M  , 5   M  3, 2  M  , ,   M  2  2  2  2 2   2        

106. P  5.6, 1.7  ; Q  4.4, 8.3 

Solution

 5.6   4.4  1.7  8.3   x  x2 y 1  y 2   10 10    M M  1 , , ,   M  5, 5    M    2 2 2 2 2 2     

107. P  0, 0  ; Q

 5, 5 

Solution

0  5 0  5   5 5  x  x2 y 1  y 2  , , , M  1   M    M    2 2  2  2   2  2   108. P

 3, 0 ; Q 0,  5 

Solution

 

 0  5   3  5  3  x  x2 y 1  y 2  5  3 0   M   M  M  1 , , , ,   M    2  2 2  2 2  2   2  2     

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563


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

One endpoint P and the midpoint M of line segment PQ are given. Find the coordinates of the other endpoint, Q. 109. P(1, 4); M(3, 5)

Solution

Let Q have coordinates  x, y  :

 x  x2 y 1  y 2  , M  1    3, 5  2   2 x 1  x2 y 1  y2 3 5 2 2 1 x 4 y 3 5 2 2 1 x  6 4  y  10 x 5 y 6 Q  5, 6  110. P(2, –7); M(–5, 6)

Solution

Let Q have coordinates  x, y  :

 x  x2 y 1  y 2  , M  1    5, 6  2   2 x 1  x2 y 1  y2  5 6 2 2 2 x 7  y  5 6 2 2 2  x  10 7  y  12 x  12 y  19 Q  12, 19  111. P(5, –5); M(5, 5)

Solution

Let Q have coordinates  x, y  :

 x  x2 y 1  y 2  , M  1    5, 5 2   2 x 1  x2 y 1  y2 5 5 2 2 5 x 5  y 5 5 2 2 5  x  10 5  y  10 x 5 y  15 Q  5, 15 

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564


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

112. P(–7, 3); M(0, 0)

Solution

Let Q have coordinates  x, y  :

 x  x2 y 1  y 2  , M  1    0, 0  2   2 x 1  x2 y 1  y2 0 0 2 2 7  x 3 y 0 0 2 2 7  x  0 3 y 0 x7 y  3 Q  7,  3  113. Show that a triangle with vertices at (13, –2), (9, –8), and (5, –2) is isosceles.

Solution Let the points be identified as A(13, –2), B(9, –8), and C(5, –2). AB 

 x  x    y  y    13  9   2   8   16  36  52  2 13

BC 

 x  x    y  y   9  5   8   2   16  36  52  2 13

2

2

1

2

2

2

2

1

2

2

2

2

1

2

2

1

Since AB and BC have the same length, the triangle is isosceles. 114. Show that a triangle with vertices at (–1, 2), (3, 1), and (4, 5) is isosceles.

Solution Let the points be identified as A(–1, 2), B(3, 1), and C(4, 5).

 x  x    y  y    1  3   2  1  16  1  17 BC   x  x    y  y    3  4    1  5   1  16  17 AB 

2

2

1

2

1

2

1

2

1

2

2

2

2

2

2

2

Since AB and BC have the same length, the triangle is isosceles. 115. In the illustration, points M and N are the midpoints of AC and BC, respectively. Find the length of MN.

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565


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution  2  6 4  10   8 14   4  6 6  10   10 16  M , ,  ,    4, 7  ; N    ,    5, 8  2 2 2 2 2   2 2      2 MN 

 x  x    y  y    4  5   7  8  1  1  2 2

2

1

2

2

2

2

1

116. In the illustration, points M and N are the midpoints of AC and BC, respectively. Show

 

that d MN  21 d AB  .

Solution

0  b 0  c b c a  b 0  c a  b c M , , ,     , ; N    2 2 2 2 2   2 2      2 AB 

 x  x    y  y   0  a   0  0  a  a

MN 

x  x    y  y 

2

2

1

2

2

2

2

1

2

2

2

2

1

1

2

2

2

b a  b c c         2  2 2 2

a2 a 1  0   AB 4 2 2

117. In the illustration, point M is the midpoint of the hypotenuse of right triangle AOB. Show that the area of rectangle OLMN is one-half of the area of triangle AOB.

Solution

0  a b  0 a b a   b , M    ,  ; L   , 0  ; N   0,  2  2 2  2 2   2 1 1 1 1  base  height  OAOB    a  b   ab 2 2 2 2 a b 1 1 Area of OLMN  lenght  width  OL ON     ab   Area of AOB  2 2 4 2 Area of AOB 

118. Rectangle ABCD in the illustration is twice as long as it is wide, and its sides are parallel to the coordinate axes. If the perimeter is 42, find the coordinates of point C.

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566


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution Let x = the width (from A to D). Then the length (from A to B) = 2x. Perimeter  42 x  2 x  x  2 x  42 6 x  42 x7 Thus, the distance from A to D is 7 and the distance from A to B is 2(7) = 14. Thus, the x-coordinate of C is –3 + 14, or 11. The y-coordinate of C is –2 + 7, or 5. Point C then has coordinates (11, 5).

Fix It In exercises 119 and 120, identify the step the first error is made and fix it. 119. Given 4 x  5 y  9. Determine the x- and y-intercepts and graph the line.

Solution Step 2 was incorrect.

 9  Step 1: The x-intercept is   , 0  .  4   9 Step 2: The y-intercept is  0,  .  5 Step 3:

120. Find the distance between the two the points, P  4,  5  and Q  2, 1 . To do so, label the points, substitute into distance formula, and simplify.

Solution Step 5 was incorrect. Step 1: P  4, 5   P  x1 , y 1  ; Q  2, 1  Q  x2 , y 2 

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567


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Step 2:

 2  4    1  5

Step 3:

 2   6

Step 4:

40

2

2

2

2

Step 5: 2 10

Applications 121. House appreciation A lake house purchased for $325,000 is expected to appreciate according to the formula f(x) = 17,500x + 325,000, where f (x) is the value of the lake house after x years. Find the value of the house 5 years later.

Solution y  17500 x  325000

y  17500  5   325000 y  87500  325000

y  412500 The value will be $412,500.

122. Car depreciation A car purchased for $24,000 is expected to depreciate according to the formula f (x) = –1920x + 24,000, where f (x) is the value of the car after x years. When will the car be worthless?

Solution set y  0 : y  1920 x  24, 000 0  1920 x  24, 000 1920 x  24, 000 x  12.5 The car will be worthless after 12.5 years. 123. Demand equations The number of photo scanners that consumers buy depends on price. The higher the price, the fewer photo scanners people will buy. The equation that relates price to the number of photo scanners sold at that price is called a demand equation. If the demand for a photo scanner is p   101 q  170, where p is the price and q is the number of photo scanners sold at that price, how many photo scanners will be sold at a price of $150?

Solution

1 q  170 10 1 150   q  170 10 1 q  20 10 q  200 p

200 scanners will be sold.

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568


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

124. Supply equations The number of television sets that manufacturers produce depends on price. The higher the price, the more TVs manufacturers will produce. The equation that relates price to the number of TVs produced at that price is called a supply equation. If the supply equation for a 25-inch TV is p  101 q  130, where p is the price and q is the number of TVs produced for sale at that price, how many TVs will be produced if the price is $150?

Solution

1 q  130 10 1 q  130 150  10 1 20  q 10 200  q p

200 TVs will be produced. 125. Meshing gears The rotational speed V of a large gear (with N teeth) is related to the speed v of the smaller gear (with n teeth) by the equation V  nvN . If the larger gear in the illustration is making 60 revolutions per minute, how fast is the smaller gear spinning?

Solution nv V  N 12v 60  20 1200  12v 100  v The smaller gear is spinning at 100 rpm. 126. Social services The number f (x) of incidents requiring social work intervention appears to be related to x, the money spent on crisis intervention, by the equation f(x) = 430 – 0.005x. What expenditure would reduce the number of incidents to 350?

Solution

f  x   430  0.005 x

350  430  0.005d 0.005d  80 d  16, 000 $16,000 would reduce the number to 350.

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569


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

127. Football Suppose Dak Prescott, quarterback for the Dallas Cowboys, throws a football to his wide receiver. If Tony’s location on the football field is represented by Q(30, 25), 30 yards from the end zone and 25 yards from the sideline, and his wide receiver location is represented by R(0, 10), on the end zone line and 10 yards from the sideline, find the actual distance the football was thrown.

Solution d

x  x    y  y 

0  30    10  25

 30    15

2

2

1

2

2

1

2

2

2

2

 900  225 

1125  15 5 yd

128. Baseball If home plate on a baseball field is represented by the origin and second base is represented by the P(90, 90), find the actual distance between home plate and second base. The units are in feet.

Solution d 

x  x    y  y 

90  0   90  0 

90   90 

2

2

1

2

2

2

2

1

2

2

 8100  8100 

16200  90 2 ft

129. Football Use the information stated in Exercise 127. If Dak Prescott’s pass is intercepted at the midpoint between the wide receiver and Tony, find the point of interception.

Solution  x  x2 y 1  y 2   30  0 25  10   30 35  , , , M  1   M    M   M  15, 17.5 2 2 2 2    2 2    130. Baseball Use the information stated in Exercise 128 to identify the midpoint between home plate and second base.

Solution  x  x2 y 1  y 2   90  0 90  0   90 90  , , , M  1   M    M   M  45, 45 2 2 2 2    2 2    131. Navigation See the illustration. An ocean liner is located 23 miles east and 72 miles north of Pigeon Cove Lighthouse, and its home port is 47 miles west and 84 miles south of the lighthouse. How far is the ship from port?

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

570


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

d 2  702  1562 d 2  4900  24, 336 d 2  29, 236 d 

29, 236

d  171 miles

132. Engineering Two holes are to be drilled at locations specified by the engineering drawing shown in the illustration. Find the distance between the centers of the holes.

Solution

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571


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

d 2  102  32 d 2  100  9 d 2  109 d 

109

d  10.4 mm

Discovery and Writing 133. Explain how to determine the quadrant in which the point P(a, b) lies.

Solution Answers may vary. 134. Explain how to graph a line using the intercept method.

Solution Answers may vary. 135. In Figure 2-17, show that d(PM) + d(MQ) = d(PQ).

Solution Answers may vary. 136. Use the result of Exercise 135 to explain why point M is the midpoint of segment PQ.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 137. The domain of y = 8 is  ,   .

Solution True. 138. The range of y = 8 is {8}.

Solution True. 139. All linear equations are functions.

Solution False. Vertical lines have equations that are not functions. 140. Three points are always required to graph a line.

Solution False. Only two points are required to graph a line.

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572


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

141. A line can have at most one y-intercept.

Solution False. The vertical line x = 0 has infinitely many y-intercepts. 142. A line must have at least one x-intercept.

Solution False. Most horizontal lines have no x-intercept. 143. The distance between the origin and P(x, y) is

x2  y 2 .

Solution True. 144. The midpoint between P(x, y) and Q(–x, –y) is the origin.

Solution True.

EXERCISES 2.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Evaluate:

5   11 2   10 

Solution 6 1  12 2 2. Evaluate:

3   3 6   1

Solution 0 0 7 3. Evaluate:

6   1

3   3

Solution

7  undefined 0

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573


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

4. If  x1 , y 1    5,  6  and  x2 , y 2    2,  9  , find

y2  y 1 x2  x 1

.

Solution

9  6 3  1 25 3 5. Fill in the blanks. a. _____________ lines do not intersect. b. _____________ lines intersect to form right angles.

Solution a. parallel b. perpendicular 6. The numbers 

7 5 and are negative reciprocals. Find their product. 5 7

Solution

7 5   1 5 7

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The slope of a nonvertical line is defined to be the change in y __________ by the change in x.

Solution divided 8. The change in __________ is often called the rise.

Solution y 9. The change in x is often called the __________.

Solution run 10. When computing the slope from the coordinates of two points, always subtract the y-values and the x-values in the __________.

Solution same order 11. The symbol Δy means __________ y.

Solution the change in

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574


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

12. The slope of a __________ line is 0.

Solution horizontal 13. The slope of a __________ line is undefined.

Solution vertical 14. If the slopes of two lines are equal, the lines are __________.

Solution parallel 15. If the product of the slopes of two lines is –1, the lines are __________.

Solution perpendicular 16. The average rate of change of a function f on the interval  x 1 , x2  is represented by the formula ___________.

Solution

f  x2   f  x 1  x2  x 1

Practice Find the slope of the line passing through each pair of points, if possible. 17. P(2, 2); Q(–1, –1)

Solution

m

y2  y 1 x2  x 1

2   1 2   1

3 1 3

18. P(3, –1); Q(5, 3)

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575


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution m

y2  y1 x2  x 1

3   1 53

4 2 2

19. P(–6, 3); Q(6, –2)

Solution y  y1 5 2  3 5 m 2    12 x2  x 1 6   6  12 20. P(2, 5); Q(3, 10)

Solution y  y 1 10  5 5 m 2   5 32 1 x2  x 1 21. P(3, –2); Q(–1, 5)

Solution m

y2  y 1 x2  x 1

5   2  1  3

7 7  4 4

22. P(3, 7); Q(6, 16)

Solution y  y 1 16  7 9 m 2   3 63 3 x2  x 1

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576


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

23. P(8, –7); Q(4, 1)

Solution m

y2  y 1 x2  x 1

1   7  48

8  2 4

24. P(5, 17); Q(17, 17)

Solution y  y 1 17  17 0 m 2   0 17  5 12 x2  x 1 25. P(–7, –14); Q(2, –14)

Solution

m

y2  y 1 x2  x1

14   14  2   7 

0 0 9

26. P(–4, 3); Q(–4, –3)

Solution y  y1 3  3 6 m 2    und. 0 x2  x 1 4    4  27. P(–5, 3); Q(–5, –2)

Solution y  y1 2  3 5 m 2    und. 0 x2  x 1 5   5 

  7, 2

28. P 2, 7 ; Q

Solution m

y2  y 1 x2  x 1

2 7 7 2

 1

3 2 5 7 29. P  ,  ; Q  ,  2 3 2 3 Solution 7 2 5 5  5 m  3 3  3  3  5 3 2 1 3 x2  x 1  2 2 2 y2  y1

 2 1 3 5 30. P   ,  ; Q  ,   5 3 5 3   

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577


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution 5 1 6   3 3 m   3  2 x2  x 1 5 3  2    3 5  5 

y2  y1

31. P(a + b, c); Q(b + c, a) assume c ≠ a

Solution y  y1 ac  m 2 x2  x 1  b c    a  b 

ac  1 ca

32. P(b, 0); Q(a + b, a) assume a ≠ 0

Solution y  y1 a0 a m 2   1 x2  x 1 a  b   b a Find two points on the line and use slope formula to find the slope of the line. 33. y = 3x + 2

Solution y = 3x + 2 x

y

0

2

1

5

m

y2  y1 x2  x 1

52 10 3  3 1

34. f(x) = 5x – 8

Solution y  5x  8 x

y

0

–8

1

–3

m

y2  y 1 x2  x 1

 

 3   8  10 5 5 1

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578


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

35. f(x) = 4x – 6

Solution y  4x  6 x

y

0

–6

1

–2

m

y2  y 1

x2  x 1

 

36. f x  

 2   6 

10 4  4 1

1 x 5 3

Solution

y

1 x 5 3

x

y

0

5

3

4

m

y2  y 1

45 30 1 1   3 3

x2  x 1

37. 5x – 10y = 3

Solution 5 x  10 y  3 x 0

y

1 5

1 m

3 10

y2  y1 x2  x 1 1  3     5  10 

10 5 1  10  1 2

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579


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

38. 8y + 2x = 5

Solution 8 y  2x  5 x

y

0

5 8

1

3 8 y2  y1

m

x2  x 1 3 5  8 8  10 2  1  8  1 4

39. 3(y + 2) = 2x – 3

Solution

3  y  2  2 x  3 3 y  2 x  9 x

y

0

–3

3

–1 y2  y1

m 

x2  x 1

1   3 

2  3

30

40. 4(x – 2) = 3y + 2

Solution

4  x  2  3 y  2 4 x  3 y  10 x

y

10 3

0

1

–2

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580


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y2  y1

m

x2  x 1

 10  2      3   10 4 4  3  1 3

41. 3(y + x) = 3(x – 1)

Solution

3  y  x   3  x  1 3 y  3 y  1 x

y

0

1

1

1 y2  y 1

m

x2  x 1

1   1

10 0  0 1

42. 2x + 5 = 2(y + x)

Solution

2x  5  2  y  x  5  2y 5  y 2 x

y

0

5 2

1

5 2

m

y2  y1

x2  x 1 5 5   2 2 10 0  0 1

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581


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Find the slope of the line, if possible. 43. y = 7

Solution horizontal  m  0

44. 2y = 5

Solution 2y  5 5 y  2 horizontal  m  0

45. f  x  

1 4

Solution 1 1 f x   y  4 4 horizontal  m  0 46. f  x   

Solution

f x    y  

horizontal  m  0

47. x  

1 2

Solution 1 x 2 vertical  m is undefined. 48. x – 7 = 0

Solution x 7 0 x7 vertical  m is undefined. Determine whether the slope of the line is positive, negative, 0, or undefined. 49.

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582


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution The slope is negative. 50.

Solution The slope is zero. 51.

Solution The slope is positive. 52.

Solution The slope is positive. 53.

Solution The slope is undefined. 54.

Solution The slope is negative.

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583


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Find the average rate of change of the function from x1 to x2. 55. Find the average rate of change of f  x   2 x  5 from x1  3 to x2  5.

Solution

f  x2   f  5   5 f  x1   f  3  1 f  x2   f  x 1  x2  x 1

51 4  2 53 2

56. Find the average rate of change of f  x   3 x  2 from x1  2 to x2  1.

Solution

f  x2   f  1  5

f  x 1   f  2   8 f  x2   f  x 1  x2  x 1

58

3  3 1

1   2 

57. Find the average rate of change of f  x   x 2 from x1  4 to x2  3.

Solution

f  x 2   f  3   9

f  x 1   f  4   16 f  x2   f  x 1  x2  x 1

9  16

 3   4 

7  7 1

58. Find the average rate of change of f  x   2 x 2 from x1  2 to x2  4.

Solution

f  x2   f  4   32 f  x1   f  2  8 f  x2   f  x 1  x2  x 1

32  8 24   12 42 2

59. Find the average rate of change of f  x   x 3  1 from x1  1 to x2  2.

Solution

f  x2   f  2   7

f  x 1   f   1   2 f  x2   f  x 1  x2  x 1

7   2  2   1

9 3 3

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584


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

60. Find the average rate of change of f  x   x 3  4 from x1  3 to x2  0.

Solution

f  x2   f  0   4

f  x 1   f  3   23 f  x2   f  x 1  x2  x 1

4   23  0   3 

27 9 3

 

61. Find the average rate of change of f x 

x from x1  16 to x2  25.

Solution

f  x2   f  25   5 f  x 1   f  16   4 f  x2   f  x 1  x2  x 1

54 1  25  16 9

 

62. The average rate of change of f x  2 x from x 1 

1 9 to x2  . 4 4

Solution 9 f  x2   f    3 4  1 f  x1   f    1 4 f  x2   f  x 1  x2  x 1

31 2  1 9 1 2  4 4

 

3 63. Find the average rate of change of f x  3 x from x1  1 to x2  8.

Solution

f  x2   f  8   6 f  x 1   f  1  3 f  x2   f  x 1  x2  x 1

63 3  81 7

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585


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

64. Find the average rate of change of f  x  

13 x from x1  8 to x2  1. 2

Solution 1 2 f  x 1   f   8   1 f  x2   f  1  

f  x2   f  x 1  x2  x 1

1 3 1  3 1 3 2   2    7 2 7 14 1   8  

Determine whether the lines with the given slopes are parallel, perpendicular, or neither. 65. m1  3; m2  

1 3

Solution

 1 m1m2  3     1  3 perpendicular 66. m1 

2 3 ; m2  3 2

Solution

2 3   1  1 3 2

m1  m2 ; m1m2  neither

67. m1  8; m2  2 2

Solution

m1  8  2 2  m2 parallel 68. m1  1; m2  1

Solution

m1m2  1  1  1

perpendicular 69. m1   2; m2 

2 2

Solution  2 m1m2   2    1  2    perpendicular

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586


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

70. m1  2 7; m2  28

Solution

m2  28  2 7  m1 parallel 71. m1  0.125; m2  8

Solution

m1m2  0.125  8  1

perpendicular 72. m1  0.125; m2 

1 8

Solution m1  0.125  parallel

1  m2 8

73. m1  ab1 ; m2  a 1b  a  b, b  0 

Solution

m1m2  ab1 a 1b  a0 b0  1 perpendicular 1

a b 74. m1    ; m2    a  0, b  0, a  b  b a  

Solution 1

a b b m1        m2 a a b b b m1m2      1  neither a a Determine whether the line through the given points and the line through R(–3, 5) and S(2, 7) are parallel, perpendicular, or neither. (For Exercises 75-80 use the slope of line through R and S calculated below:) mRS 

y2  y1 x2  x 1

75

2   3 

2 5

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587


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

75. P(2, 4); Q(7, 6)

Solution

mPQ 

y2  y 1 x2  x 1

64 2   mRS  parallel 72 5

76. P(–3, 8); Q(–13, 4)

Solution y  y1 48 2 4 mPQ  2     mRS  parallel x2  x 1 13   3  10 5 77. P(–4, 6); Q(–2, 1)

Solution y  y1 16 5 5 mPQ  2      perpendicular 2 2 x2  x 1 2   4  78. P(0, –9); Q(4, 1)

Solution mPQ 

y2  y 1 x2  x 1

1   9  40

79. P(a, a); Q(3a, 6a)

10 5   neither 4 2

(a ≠ 0)

Solution

mPQ 

y2  y 1 x2  x 1

6a  a 5a 5    neither 3a  a 2a 2

80. P(b, b); Q(–b, 6b)

(b ≠ 0)

Solution

mPQ 

y2  y 1 x2  x 1

6b  b 5b 5     perpendicular 2 b  b 2b

Lines PQ and RS are either parallel or perpendicular. Find x or y. 81. Parallel: P(–3, 7); Q(2, 9); R(10, –4); S(x, –6)

Solution

mPQ 

y2  y 1 x2  x 1

97

2   3

y  y 1 6   4  2 2 ; mRS  2   x2  x 1 x  10 x  10 5

2 1 2   ; x  10  5  x  5 5 1 5

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588


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

82. Parallel: P(2, –3); Q(5, 7); R(3, –1); S(6, y)

Solution mPQ 

y2  y1 x2  x 1

7   3 

10 ; 3

52

mRS 

y2  y1 x2  x 1

y   1 63

y1 3

10  y  1  y  9 83. Perpendicular: P(2, –7); Q(1, 0); R(–9, 5); S(–2, y)

Solution mPQ  7 

y2  y 1 x2  x 1

0   7 

y  y1 y 5 y 5 7  7; mRS  2   x2  x 1 1 7 2   9 

12

7 1 ; Perp. slope  ; y  5  1  y  6 1 7

84. Perpendicular: P(1, –2); Q(3, 4); R(x, 6); S(6, 5)

Solution y2  y 1

4   2 

y  y1 5  6 6 1  3; mRS  2   x2  x 1 x2  x 1 31 2 6 x 6x 3 1 1 3  ; Perp. slope    ; 6  x  3  x  3 1 3 3 mPQ 

Find the slopes of lines PQ and PR, and determine whether points P, Q, and R lie on the same line. 85. P(–2, 8); Q(–6, 9); R(2, 5)

Solution y  y1 98 1 1    mPQ  2 x2  x 1 4 6   2  4

mPR 

y2  y 1 x2  x 1

58

2   2

3 3    not on same line 4 4

86. P(1, –1); Q(3, –2); R(–3, 0)

Solution

mPQ  mPR 

y2  y 1 x2  x 1 y2  y 1 x2  x 1

 

2   1 31

0   1 3  1

1 1  2 2

1 1    not on same line 4 4

87. P(–a, a); Q(0, 0); R(a, –a)

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589


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution y  y1 a 0a    1 mPQ  2 x2  x1 0   a  a

mPR 

y2  y 1 x2  x 1

a  a

a   a 

2a  1  on same line 2a

88. P(a, a + b); Q(a + b, b); R(a – b, a)

Solution

mPQ 

y2  y1

b  a  b 

a a  b b

a  b   a a   a  b  b y y m     1  not on same line x x  a  b   a b x2  x 1 2

1

2

1

PR

Determine which, if any, of the three lines PQ, PR, and QR are perpendicular. 89. P(5, 4); Q(2, –5); R(8, –3)

Solution mPQ  mPR  mQR 

y2  y 1 x2  x 1 y2  y1 x2  x 1 y2  y1 x2  x 1

5  4 9  3 25 3

7 3  4 7   85 3 3

3   5  82

2 1   None are perpendicular. 6 3

90. P(8, –2); Q(4, 6); R(6, 7)

Solution mPQ  mPR  mQR 

y2  y 1 x2  x 1 y2  y1 x2  x 1 y2  y 1 x2  x 1

  

6   2  48

7   2  68

8  2 4

9 9  2 2

7 6 1   PQ and QR are perpendicular. 64 2

91. P(1, 3); Q(1, 9); R(7, 3)

Solution

mPQ  mPR  mQR 

y2  y1 x2  x 1 y2  y 1 x2  x 1 y2  y1 x2  x 1

93 6   undefined  vertical 1 1 0

33 0   0  horizontal 71 6

3  9 6   1  PQ and PR are perpendicular. 71 6

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590


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

92. P(2, –3); Q(–3, 2); R(3, 8)

Solution mPQ  mPR  mQR 

y2  y 1 x2  x 1 y2  y 1 x2  x 1 y2  y 1 x2  x 1

  

2   3 

5  1 5

11  11 1

6  1  PQ and QR are perpendicular. 6

3  2

8   3  32 82

3   3 

93. P(0, 0); Q(a, b); R(–b, a)

Solution

mPQ  mPR  mQR 

y2  y 1 x2  x 1 y2  y 1 x2  x 1 y2  y 1 x2  x 1

b0 b  a0 a

a0 a a   b b  0 b

ab ab   PQ and PR are perpendicular. b  a b  a

94. P(a, b); Q(–b, a); R(a – b, a + b)

Solution y  y1 ab ab ab mPQ  2    x2  x 1 ab b  a   a  b 

a  b   b  a   a x x  a  b   a b b a  b   a  b  PR and QR are perpendicular. y y   x x  a  b    b  a

mPR  mQR

y2  y1 2

1

2

1

2

1

95. Right triangles Show that the points A(–1, –1), B(–3, 4), and C(4, 1) are the vertices of a right triangle.

Solution

mAB  mAC 

y2  y1 x2  x 1 y2  y 1 x2  x 1

 

96. Right triangles right triangle.

4   1

3   1 1   1

4   1

5 5  2 2

2  AB and AC are perpendicular.  right triangle 5

Show that the points D(0, 1), E(–1, 3), and F(3, 5) are the vertices of a

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591


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

mDE  mEF 

y2  y 1 x2  x 1 y2  y1 x2  x 1

97. Squares square.

 

mBC  mCD  mDA 

53

2 1   DE and EF are perpendicular.  right triangle 4 2

3   1

Show that the points A(1, –1), B(3, 0), C(2, 2), and D(0, 1) are the vertices of a

Solution mAB 

31 2   2 1  0 1

y2  y 1 x2  x 1 y2  y1 x2  x 1 y2  y 1 x2  x 1 y2  y1 x2  x 1

0   1

1 ; d  A, B   2

31

 1  3   1  0  5 2

2

20 2   2; d  B, C   2  3 1

 3  2  0  2  5

12 1 1   ; d  C, D   0  2 2 2

 2  0   2  1  5

1   1 01

2  2; d  D, A  1

2

2

2

2

 1  0   1  1  5 2

2

Adjacent sides are perpendicular and congruent, so the figure is a square. 98. Squares Show that the points E(–1, –1), F(3, 0), G(2, 4), and H(–2, 3) are the vertices of a square.

Solution

mEF  mFG  mGH  mHE 

y2  y1 x2  x 1 y2  y1 x2  x 1 y2  y1 x2  x 1 y2  y 1 x2  x 1

0   1 3   1

1 ; d E, F   4

 1  3   1  0  17 2

2

40 4   4; d  F , G   2  3 1

 3  2  0  4   17

34 1 1   ; d G, H   2  2 4 4

 2   2    4  3  17

3   1

2   1

2

4  4; d  H, E   1

2

2

2

 1   2    1  3  17 2

2

Adjacent sides are perpendicular and congruent, so the figure is a square. 99. Parallelograms Show that the points A(–2, –2), B(3, 3), C(2, 6), and D(–3, 1) are the vertices of a parallelogram. (Show that both pairs of opposite sides are parallel.)

Solution

mAB  mBC 

y2  y 1 x2  x 1 y2  y 1 x2  x 1

 

3   2 3   2

5 1 5

63 3   3 2  3 1

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592


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

mCD  mDA 

y2  y 1 x2  x 1 y2  y 1 x2  x 1

 

5 16  1 3  2 5 1   2 

3   2 

3  3 1

Opposite sides are parallel, so the figure is a parallelogram. 100. Trapezoids Show that points E(1, –2), F(5, 1), G(3, 4), and H(–3, 4) are the vertices of a trapezoid. (Show that only one pair of opposite sides is parallel.)

Solution mEF 

y2  y 1 x2  x 1 y2  y 1

1   2  51

3 4

41 3 3 mFG     3  5 2 2 x2  x 1 y  y1 44 0 mGH  2   0 x2  x 1  3  3 6 mHE 

y2  y1 x2  x 1

4   2  3  1

6 3  2 4

Exactly one pair of sides is parallel, so the figure is a trapezoid. 101. Geometry In the illustration, points M and N are midpoints of CB and BA, respectively. Show that MN is parallel to AC.

Solution 5  7 9  5  12 14   1 7 3  5 8 8 M , ,   M ,   M  6, 7  ; N    N  ,   N  4, 4  2 2 2 2 2 2       2 2 y  y 1 4  7 3 3 y  y1 9  3 6 3 mMN  2    ; mAC  2     MN  AC x2  x 1 x2  x 1 4  6 2 2 51 4 2 102. Geometry

In the illustration, d(AB) = d(AC). Show that AD is perpendicular to BC.

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593


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution d  AB  

0  a   0  0  a  0  a  a d  AC    0  b    0  c   b  c . From the given information, a  b  c . mAD 

y2  y 1 x2  x 1

mAD mBC 

2

2

2

2

2

2

2

2

c0

a  b  0

2

2

2

y  y1 0  c c c ; mBC  2   ab x2  x 1 a  b a  b

c c c2   2  a  b a  b a  b2

c2

 b c  b 2

2

2

 2

c2 c2   1 b2  c 2  b2 c2

Thus, AD is perpendicular to BC.

Fix It In exercises 103 and 104, identify the step the first error is made and fix it. 103. Write slope formula. Then use the formula to find the slope of the line passing through the points P  3,  8 and Q  5, 6  .

Solution Step 2 was incorrect. Step 1: m 

Step 2: m 

Step 3: m 

y2  y1 x2  x 1

6   8

5   3 14 2

Step 4: m  7 104. Write the average rate of change formula and use it to calculate the average rate of change of f  x   x 3  2 from x 1  1 to x2  2.

Solution Step 3 was incorrect. Step 1:

Step 2:

f  x2   f  x 1  x2  x 1

f  2  f  1 2   1

23  2   1  2 3

Step 3:

3

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594


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Step 4:

82 1 2 3

Step 5:

9 3

Step 6: 6

Applications 105. Rate of growth When a college started an aviation program, the administration agreed to predict enrollments using a straight-line method. If the enrollment during the first year was 14, and the enrollment during the fifth year was 42, find the average rate of growth per year (the slope of the line). See the illustration.

Solution y  y 1 42  14 28 m 2   7 51 4 x2  x 1 The rate of growth was 7 students per year. 106. Rate of growth A small business predicts sales according to a straight-line method. If sales were $50,000 in the first year and $110,000 in the third year, find the average rate of growth in dollars per year (the slope of the line).

Solution y  y 1 110, 000  50, 000 m 2  x2  x 1 31 60, 000   30, 000 2 The sales increased $30,000 per year. 107. Rate of decrease The price of computers has been dropping steadily for the past ten years. If a desktop PC cost $6700 ten years ago, and the same computing power cost $2200 three years ago, find the average rate of decrease per year. (Assume a straightline model.)

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595


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution y  y 1 6700  2200 m 2  x2  x 1 10  3 4500   642.86 7 The cost decreased about $642.86 per year. 108. Hospital costs The table shows the changing mean daily cost for a hospital room. For the ten-year period, find the rate of change per year of the portion of the room cost that is absorbed by the hospital.

Year

Total Cost to the Hospital

Amount Passed on to Patient

2012

$459

$212

2017

$670

$295

2022

$812

$307

Solution The cost absorbed by the hospital was $247 in 2000, $375 in 2005 and $505 in 2010. m

y2  y1 x2  x 1

505  247 258   25.8 2010  2000 10

The cost absorbed by the hospital increased by $25.80 per year. 109. Charting temperature changes The following Fahrenheit temperature readings were recorded over a four-hour period.

Time

12:00

1:00

2:00

3:00

4:00

Temperature

47

53

59

65

71

Let t represent the time (in hours), with 12:00 corresponding to t = 0. Let T represent the temperature. Plot the points (t, T), and draw the line through those points. Explain the meaning of

T . t

6. Solution

T  the hourly rate of change of temperature. t

Let t  x and T  y .

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596


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

110. Tracking the Dow The Dow Jones Industrial Averages at the close of trade on three consecutive days were as follows:

Day

Monday

Tuesday

Wednesday

Close

12,981

12,964

12,947

Let d represent the day, with d = 0 corresponding to Monday, and let D represent the Dow Jones average. Plot the points (d, D), and draw the graph. Explain the meaning of

D . d

Solution

D  the daily rate of change of the Dow Jones average. d

111. Speed of an airplane A pilot files a flight plan indicating her intention to fly at a constant speed of 590 mph. Write an equation that expresses the distance traveled in terms of the flying time. Then graph the equation and interpret the slope of the line. (Hint: d = rt.)

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597


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution D  590t; The slope is the speed of the plane.

112. Growth of savings A student deposits $25 each month in a Holiday Club account at her bank. The account pays no interest. Write an equation that expresses the amount A in her account in terms of the number of deposits n. Then graph the line, and interpret the slope of the line.

Solution A  25n; The slope is the monthly increase of the value of the account.

Discovery and Writing 113. Explain what the slope of a line is.

Solution Answers may vary.

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598


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

114. How do you determine the slope of a line?

Solution Answers may vary. 115. Explain why the slope of a vertical line is undefined.

Solution Answers may vary. 116. Explain how to determine whether two lines are parallel, perpendicular, or neither.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 117. Slope formula is m 

x2  x 1 y2  y 1

.

Solution

False. m 

y2  y 1 x2  x1

.

118. The slope of a linear function is never undefined.

Solution True.

  152  152  119. The slope of the line passing through  5,  is 0.  and   5, 99 99     Solution

True.  y  0. 

 152   152  , 5  and  ,  5  is undefined. 120. The slope of the line passing through   99   99  Solution

True.  x  0.  121. The slope of the line parallel to f  x    is undefined.

Solution False. The line will be horizontal, so the slope is 0. 122. The slope of the line perpendicular to f  x    is 0.

Solution False. The line will be vertical, so the slope is undefined.

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599


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

123. If the price of a movie ticket increased from $8.95 in 2018 to $12.25 in 2022, then the average rate of change in ticket price during this time period was approximately $0.55/year.

Solution

m

y2  y1 x2  x 1

10.25  6.95 3.30   0.55. True. 2014  2008 6

124. If the cost of college tuition in 2018 was $8015 and in 2022 was $11,139, then the average rate of change in tuition during this time period was $781/year.

Solution

m

y2  y 1 x2  x 1

9139  6015  781. True. 2014  2010

EXERCISES 2.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Solve the equation 7 x  14 y  28 for y and identify the coefficient of x and the constant term?

Solution 7 x  14 y  28

14 y  7 x  28 1 x 2 2 1 Coefficient: Constant: –2 2 y

2. What axis do the points  0, 4  ,  0, 2  ,  0,  2  , and  0,  4  all lie?

Solution y-axis

1  3. Determine the slope of the line passing through  1,  3  and  ,  2  . 2  Solution m

2    3 

1   1 2 1 2 2 m  1  3 3 3 2

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600


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

4. Clear the equation

y  y1 x  x1

 m of fractions and write the equation you obtain.

Solution

y2  y 1 x2  x 1

m y2  y 1

x  x  x  x  mx  x  y  y  m x  x  2

1

2

1

2

1

2

1

2

1

5. Substitute  x1 , y 1    2,  4  and m   3 into y  y 1  m  x  x1  and write the equation you get.

Solution

y   4   3  x  2  y  4  3  x  2 

6. Simplify the expression 

2  x  3  1. 5

Solution

2 2 6 2 1 x  3  1   x   1   x   5 5 5 5 5

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The equation y = mx + b is called the ____________ form of the equation of a line.

Solution slope-intercept 8. In the equation y = mx + b, ____________ is the slope of the graph of the line.

Solution m 9. In the equation y = mx + b, (0, b) is the ____________.

Solution y-intercept 10. The formula for the point-slope form of a line is ____________.

Solution

y  y 1  m  x  x1 

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601


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

11. The standard form of an equation of a line is ____________.

Solution Ax  By  C 12. In statistics a ____________ line is a linear equation that best fits given data.

Solution regression Practice Use slope-intercept form to write an equation of the line with the given properties. 13. m  0; b  9

Solution y  9 14. m  0; b 

4 5

Solution

y 

4 5

15. m  3; b  2

Solution y  mx  b

y  3x  2 16. m  7; b  8

Solution y  7 x  8 17. m  5; b  

1 5

Solution y  mx  b

y  5 x 

1 5

1 2 18. m   ; b  3 3 Solution y  mx  b

y 

1 2 x 3 3

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602


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

19. m  0.3; b  2.7

Solution y  0.3x  2.7 20. m  2; b  2

Solution y  mx  b

y  2x  2 21. m  5; y-intercept  0,  8 

Solution y  5x  8 22. m   4; y-intercept  0, 11

Solution y  4 x  11 23. m  

1 ; y-intercept  0,  12 10

Solution

y 

24. m 

1 x  12 10

 1 6 ; y-intercept  0,   2 7  

Solution

y

6 1 x 7 2

25. P  3,  4  ; m  2

Solution y  mx  b

4  2  3   b

4  6  b 10  b y  mx  b y  2 x  10

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603


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Use slope-intercept form to write an equation of a line passing through the given point and having the given slope. See Example 1. 26. P  3, 5  ; m  3

Solution y  3 x  4 27. P  5, 1 ; m  1

Solution y  x 6

28. P 3,  7 ; m  

2 3

Solution

2 y    x  9 3  1 4 29. P  2,   ; m  2 5  Solution y  mx  b

1 4   2  b 2 5 1 8   b 2 5 1 8   b 2 5 21  b 10 y  mx  b 

y 

4 21 x 5 10

 1  3 30. P  , 2  ; m   3 4   Solution y  mx  b

3 1  b 4 3 1 2 b 4 2

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604


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

9 b 4 y  mx  b 3 9 y  x 4 4 Find the slope and the y-intercept of the lines determined by the given equations. 31. y  13x 

5 6

Solution

5 6  5 m  13,  0,   6 y  13 x 

2 4 x 9 3

32. y 

Solution

2 4 x 9 3 2  4 m  ,  0,   9  3 y 

33. 3 x  2 y  8

Solution 3x  2 y  8

2 y  3x  8 y m

3 x 4 2

3 ,  0,  4  2

34. 2x  4 y  12

Solution 2 x  4 y  12

4 y  2 x  12 y  m

1 x3 2

1 ,  0, 3  2

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605


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

35. 2  x  3 y   5

Solution

2  x  3 y   5 2 x  6 y  5 6 y  2 x  5 1 5 y  x 3 6

1  5 m   ,  0,   3  6 36. 5  2 x  3 y   4

Solution

5  2x  3 y   4 10 x  15 y  4 15 y  10 x  4 2 4 y  x 3 15

m

37. x 

2  4 ,  0,   3  15  2y  4 7

Solution 2y  4 x 7 7x  2 y  4  2 y  7 x  4 7 y  x2 2 m

7 ,  0, 2  2

38. 3 x  4  

2  y  3

Solution 3x  4  

5

2  y  3

5 15 x  20  2  y  3 

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606


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

15 x  20  2 y  6 2 y  15 x  14

y  m

15 x 7 2

15 ,  0,  7  2

Find the slope and y-intercept and then use them to draw the graph of the line. 39. y  3 x  2

Solution

y  3 x  2  m  3,  0, 2 

40. y  4 x  4

Solution

y  4 x  4  m  4,  0,  4 

41. x  y  1

Solution xy 1

y  x  1  m  1,  0,  1

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607


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

42. x  y  2

Solution xy 2

y   x  2  m  1,  0, 2

43. 3 x  5 y  10

Solution 3 x  5 y  10

5 y  3 x  10 y 

3 3 x  2  m   ,  0, 2 5 5

44. 5 x  3 y  6

Solution 5x  3 y  6

3 y  5 x  6 y 

5 5 x  2  m  ,  0,  2 3 3

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608


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

45. x 

3 y 3 2

Solution 3 y 3 2 2x  3 y  6 x

3 y  2 x  6 2 2 y  x  2  m  ,  0, 2  3 3

46. x  

4 y 2 5

Solution

4 y 2 5 5x  4 y  10 x

4 y  5x  10 y 

5 5 5  5 x   m   ,  0,  4 2 4  2

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609


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

47. 3  y  4   2  x  3 

Solution

3  y  4   2  x  3  3 y  12  2 x  6 3 y  2 x  18 y 

2 2 x  6  m   ,  0, 6  3 3

48. 4  2 x  3   3  3 y  8 

Solution

4  2 x  3   3  3 y  8  8 x  12  9 y  24 9 y  8 x  36 y 

8 8 x  4: m   ,  0,  4  9 9

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610


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Determine whether the graphs of each pair of equations are parallel, perpendicular, or neither. 49. y  3 x  4, y  3 x  7

Solution y  3x  4

y  3x  7 m3

m3

The lines are parallel. 50. y  4 x  13, y 

1 x  13 4

Solution y  4 x  13 m4

1 x  13 4 1 m 4 y 

The lines are neither. 51. x  y  2, y  x  5

Solution xy 2

y  x 5

y  x  2 m1 m  1 The lines are perpendicular. 52. x  y  2, y  x  3

Solution x  y 2 y  x3 m1  y  x  2 y  x 2 m1 The lines are parallel. 53. y  3 x  7, 2 y  6 x  9

Solution y  3x  7 m3

2 y  6x  9 9 y  3x  2 m2

The lines are parallel. 54. 2 x  3 y  9, 3 x  2 y  5

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611


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution 2x  3 y  9 3 y  2 x  9 2 y   x3 3

m

3x  2 y  5  2 y  3 x  5 y 

2 3

m

3 5 x 2 2

3 2

The lines are perpendicular. 55. 3 x  6 y  1, y 

1 x 2

Solution 1 x 2 1 m 2

3x  6 y  1 6 y  3 x  1 y 

m

y 

1 1 x 2 6

1 2

The lines are neither. 56. x  3 y  4, y  3 x  7

Solution x  3y  4 3 y   x  4 1 4 y  x 3 3

m

y  3 x  7 m  3

1 3

The lines are perpendicular. 57. y  3, x  4

Solution y 3

x4

horizontal

vertical

The lines are perpendicular. 58. y  3, y  7

Solution y  3

y  7

horizontal

horizontal

The lines are parallel.

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612


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

59. x 

y 2 , 3  y  3  x  0 3

Solution

y 2 3 3x  y  2

3  y  3  x  0

x

3y  9  x  0 3 y  x  9

 y  3x  2 y  3x  2

y 

m3

1 x3 3

1 3 The lines are perpendicular. m

60. 2 y  8, 3  2  x   3  y  2 

Solution 2y  8

3  2  x   3  y  2

y 4

6  3x  3 y  6

horizontal

 3 y  3 x y x

m 1 neither Use the given properties to write an equation for the line in standard form. 61. m  2 passing through P  2, 4 

Solution

y  y 1  m  x  x1  y  4  2  x  2

y  4  2x  4 2 x  y  0 2x  y  0

62. m  3 passing through P  3, 5 

Solution

y  y 1  m  x  x1  y  5  3  x  3  y  5  3 x  9

3 x  y  14

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613


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

 3 1 63. m  2 passing through P   ,   2 2 Solution

y  y 1  m  x  x1 

 1 3  2 x   2 2   1 y   2x  3 2 2 y  1  4x  6 y

4 x  2 y  7 4 x  2 y  7

1  64. m  6 passing through P  ,  2  4  Solution

y  y 1  m  x  x1 

 1 y  2  6  x   4  3 y  2  6 x  2 2 y  4  12 x  3 12 x  2 y  1

65. m 

2 passing through P  1, 1 5

Solution

y  y 1  m  x  x1 

2  x  1 5 5  y  1  2  x  1 y 1

5 y  5  2x  2 2 x  5 y  7 2 x  5 y  7

66. m  

1 passing through P  2,  3 5

Solution

y  y 1  m  x  x1  y 3 

1  x  2 5

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614


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

5  y  3    x  2 5 y  15   x  2 x  5 y  17

67. m  0 passing through P  6,  3 

Solution

y  y 1  m  x  x1  y  3  0  x  6 y 30 y  3

68. m  0 passing through P  7, 5 

Solution

y  y 1  m  x  x1  y  5  0  x  7 y 5  0 y 5

69. m is undefined passing through P  6,  3 

Solution m is und  vertical x  constant x  6 70. m is undefined passing through P  6,  1

Solution m is und  vertical x  constant x 6 71. m   passing through P  , 0 

Solution

y  y 1  m  x  x1  y  0   x    y  x 2

 x  y   2

x  y  2

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615


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

72. m   passing through P  0,  

Solution

y  y 1  m  x  x1  y      x  0

y   x  x  y  

 x  y   Write an equation in standard form for each line shown. 73.

Solution From the graph, m 

2 and the line passes through  2, 5 . 3

y  y 1  m  x  x1  2  x  2 3 2 3  y  5  3   x  2 3 3 y  15  2  x  2  y 5 

3 y  15  2 x  4 2 x  3 y  11 2 x  3 y  11

74.

Solution From the graph, m  

2 and the line passes through  3, 2  . 3

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616


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y  y 1  m  x  x1  2  x  3 3  2 3  y  2  3      x  3  3 y 2  

3 y  6  2  x  3  3 y  6  2 x  6

2x  3 y  0

Write an equation of a line in slope-intercept form that passes through the two given points. 75. P(0, 0), Q(4, 4)

Solution

m

y2  y 1 x2  x 1

40 4  1 40 4

y  y 1  m  x  x1  y  0  1  x  0 y x 76. P(–5, –5), Q(0, 0)

Solution

m

y2  y1 x2  x 1

0   5  0   5 

5 1 5

y  y 1  m  x  x1  y  0  1  x  0 y x 77. P(3, 4), Q(0, –3)

Solution

m

y2  y 1 x2  x 1

3  4 7 7   03 3 3

y  y 1  m  x  x1  7  x  0 3 7 y  x 3 3

y 3

78. P(4, 0), Q(6, –8)

Solution

m

y2  y 1 x2  x 1

8  0 8   4 64 2

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617


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y  y 1  m  x  x1  y  0  4  x  4  y  4 x  16 Write an equation in slope-intercept form of each line shown. 79.

Solution From the graph, m   y  y 1  m  x  x1 

9 and the line passes through  2, 4  . 5

9  x  2 5 9 18 y 4   x  5 5 9 18 y  x 4 5 5 9 2 y  x 5 5 y 4  

80.

Solution From the graph, m  y  y 1  m  x  x1 

8 and the line passes through  2, 3 . 5

8  x  2 5 8 16 y 3  x  5 5 8 16 y  x 3 5 5 8 1 y  x 5 5 y 3 

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618


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Write an equation of the line in slope-intercept form that passes through the given point and is parallel to the given line. 81. P(0, 0), y = 4x – 7

Solution y  4x  7

y  y 1  m  x  x1 

m4

y  0  4  x  0

Use m  4.

y  4x

82. P(0, 0), x = –3y – 12

Solution x  3 y  12 3 y   x  12

y  y 1  m  x  x1  1  x  0 3 1 y  x 3

y 0  

1 x 4 3 1 m 3 1 Use m   . 3 y 

83. P(2, 5), 4x – y = 7

Solution 4x  y  7

y  y 1  m  x  x1 

 y  4 x  7

y  5  4  x  2

y  4x  7

y  5  4x  8

m4

y  4x  3

Use m  4.

84. P(–6, 3), y + 3x = –12

Solution y  3 x  12 y  3 x  12 m  3 Use m  3.

y  y 1  m  x  x1  y  3  3  x  6  y  3  3 x  18 y  3 x  15

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619


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

85. P  4,  2  , x 

5 y 2 4

Solution 5 x  y 2 4 4x  5 y  8 5 y  4 x  8 4 8 y  x 5 5 4 m 5 4 Use m  . 5 86. P  1,  5  , x  

y  y 1  m  x  x1  4  x  4 5 4 16 y 2 x 5 5 4 26 y  x 5 5 y 2

3 y 5 4

Solution 3 x   y 5 4 4 x  3 y  20 3 y  4 x  20 4 20 y  x 3 3 4 m 3 4 Use m   . 3

y  y 1  m  x  x1  4  x  1 3 4 4 y 5   x  5 3 4 11 y  x 3 3 y 5  

Write an equation of the line in slope-intercept form that passes through the given point and is perpendicular to the given line. 87. P(0, 0), y = 4x – 7

Solution y  4x  7

y  y 1  m  x  x1 

m4

y 0  

Use m  

1 . 4

1  x  0 4 1 y  x 4

88. P(0, 0), x = –3y – 12

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620


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

y  y 1  m  x  x1 

x  3 y  12

y  0  3  x  0

3 y   x  12 1 x 4 3 1 m 3 Use m  3. y 

y  3x

89. P(2, 5), 4x – y = 7

Solution

y  y 1  m  x  x1 

4x  y  7  y  4 x  7 y  4x  7 m4 Use m  

1  x  2 4 1 1 y 5   x  4 2 1 11 y  x 4 2

y 5  

1 . 4

90. P(–6, 3), y + 3x = –12

Solution y  3 x  12

y  3 x  12 m  3 1 Use m  . 3

91. P  4,  2  , x 

Solution 5 x  y 2 4 4x  5 y  8 5 y  4 x  8 4 8 y  x 5 5 4 m 5 5 Use m   . 4

y  y 1  m  x  x1  1  x  6 3 1 y 3 x 2 3 1 y  x 5 3 y 3

5 y 2 4

y  y 1  m  x  x1  5  x  4 4 5 y 2   x 5 4 5 y   x3 4 y 2 

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621


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

92. P  1,  5  , x  

3 y 5 4

Solution 3 x   y 5 4 4 x  3 y  20 3 y  4 x  20 4 20 y  x 3 3 4 m 3 3 Use m  . 4

y  y 1  m  x  x1  3  x  1 4 3 3 y 5  x  4 4 3 23 y  x 4 4 y 5 

93. Find an equation of the line perpendicular to the line y = 3 and passing through the midpoint of the segment joining (2, 4) and (–6, 10).

Solution Since y = 3 is the equation of a horizontal line, any perpendicular line will be vertical. Find the midpoint: x

2   6  2

 2; y 

4  10 7 2

The vertical line through  2, 7  is x  2. 94. Find an equation of the line parallel to the line y = –8 and passing through the midpoint of the segment joining (–4, 2) and (–2, 8).

Solution Since y = –8 is the equation of a horizontal line, any parallel line will be horizontal. Find the midpoint: x

 4   2  2

 3; y 

28 5 2

The horizontal line through  3, 5  is y  5. 95. Find an equation of the line parallel to the line x = 3 and passing through the midpoint of the segment joining (2, –4) and (8, 12).

Solution Since x = 3 is the equation of a vertical line, any parallel line will be vertical. Find the midpoint: x

28 4  12  5; y  4 2 2

The vertical line through  5, 4  is x  5.

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622


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

96. Find an equation of the line perpendicular to the line x = 3 and passing through the midpoint of the segment joining (–2, 2) and (4, –8).

Solution Since x = 3 is the equation of a vertical line, any perpendicular line will be horizontal. Find the midpoint: x

2   8  2  4  1; y   3 2 2

The horizontal line through  1,  3  : y  3.

Fix It In exercises 97 and 98, identify the step the first error is made and fix it. 97. Find an equation of the line in standard form that passes through  1, 2 and having a slope of  41 .

Solution Step 4 was incorrect. Step 1: y  2  

1  x  1 4

Step 2: y  2  

1 1 x 4 4

Step 3: y  

1 7 x 4 4

Step 4: x  4 y  7 98. Find an equation of the line in standard form that passes through the point  2, 41  and perpendicular to y  43 x  5

Solution Step 3 was incorrect. Step 1: y 

1 3    x  2 4 4

Step 2: y 

1 3 3  x 4 4 2

Step 3: y  

3 5 x 4 4

Step 4: 3x  4 y  5

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623


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Applications In Exercises 99–109, assume straight-line depreciation or straight-line appreciation. 99. Depreciation A Subaru Outback was purchased for $24,300. Its salvage value at the end of 7-year is expected to be $1900. Find a depreciation equation.

Solution Let x = the number of years the truck has been owned and let y = the value of the

truck. Then two points on the line are given:  0, 24300  and  7, 1900  .

m

24300  1900 22400   3200 07 7 y  y 1  m  x  x1 

y  24300  3200  x  0

y  24300  3200 x y  3200 x  24300 100. Depreciation A small business purchases the laptop computer shown. It will be depreciated over a 4-year period, when its salvage value will be $300. Find a depreciation equation.

Solution Let x = the number of years the laptop has been owned and let y = the value of the laptop. Then two points on the line are given:  0, 2700  and  4, 300  .

2700  300 2400   600 04 4 y  y 1  m  x  x1 

m

y  2700  600  x  0 

y  2700  600 x y  600 x  2700 101. Appreciation A condominium in San Diego was purchased for $475,000. The owners expect the condominium to double in value in 10 years. Find an appreciation equation.

Solution Let x = the number of years the building has been owned and let y = the value of the building. Then two points on the line are given:  0, 475000  and  10, 950000  .

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624


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

m

950000  475000 475000  10  0 10  47500 y  y 1  m  x  x1 

y  475000  47500  x  0  y  475000  47500 x y  47500 x  475000 102. Appreciation A house purchased for $112,000 is expected to double in value in 12 years. Find an appreciation equation.

Solution Let x = the number of years the house has been owned and let y = the value of the house. Then two points on the line are given:  0, 112000  and  12, 224000  .

m

224000  112000 112000 28000   12  0 12 3 y  y 1  m  x  x1 

28000  x  0 3 28000 y  112000  x 3 28000 y  x  112000 3 y  112000 

103. Depreciation

Find a depreciation equation for the TV in the following want ad.

Solution Let x = the number of years the TV has been owned and let y = the value of the TV. Then two points on the line are given:  0, 1900  and  3, 1190  .

m

1900  1190 710 710   03 3 3

y  y 1  m  x  x1  710  x  0 3 710 y  1900   x 3 710 y  x  1900 3 y  1900  

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625


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

104. Depreciation A Bose Wave Radio cost $555 when new and is expected to be worth $80 after 5 years. What will it be worth after 3 years?

Solution Let x = the number of years the radio has been owned and let y = the value of the radio. Then two points on the line are given:  0, 555  and  5, 80  .

555  80 475   95 5 05 y  y 1  m  x  x1 

m

y  555  95  x  0 

y  555  95 x y  95 x  555 Let x  3 and find the value of y : y  95 x  555

 95  3   555  270

It will be worth $270.

105. Salvage value A copier cost $1050 when new and will be depreciated at the rate of $120 per year. If the useful life of the copier is 8 years, find its salvage value.

Solution Let x = the number of years the copier has been owned and let y = the value of the copier. Then one point on the line is given: (0, 1050). Since the copier depreciates by $120 per year, m  120.

y  y 1  m  x  x1 

y  1050  120  x  0  y  1050  120 x y  120 x  1050 Let x  8 and find the value of y : y  120 x  1050  120  8   1050  90

The salvage value will be $90.

106. Rate of depreciation A jet ski that cost $13,800 when new will have no salvage value after 6 years. Find its annual rate of depreciation.

Solution Let x = the number of years the jet ski has been owned and let y = its value. Then two two points on the line are given:  0, 13800  and  6, 0  .

m

13800  0 13800   2300 6 06

The jet ski depreciates at a rate or $2300 per year.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

626


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

107. Value of an antique An antique table is expected to appreciate $40 each year. If the table will be worth $450 in 2 years, what will it be worth in 13 years?

Solution Let x = the number of years the table has been owned and let y = the value of the

table. Then one point on the line is given:  2, 450 . Since the table appreciates by $40

per year, m  40.

y  y 1  m  x  x1 

y  450  40  x  2 

y  450  40 x  80 y  40 x  370 Let x  13 and find the value of y : y  40 x  370

 40  13   370  890

The value will be $890.

108. Value of an antique An antique clock is expected to be worth $350 after 2 years and $530 after 5 years. What will the clock be worth after 7 years?

Solution Let x = the number of years the clock has been owned and let y = the value of the clock. Then two points on the line are given:  2, 350  and  5, 530  .

m

530  350 180   60 52 3

y  y 1  m  x  x1 

y  350  60  x  2 

y  350  60 x  120 y  60 x  230 Let x  7 and find the value of y : y  60 x  230

 60  7   230  650

It will be worth $650.

109. Purchase price of real estate A cottage that was purchased 3 years ago is now appraised at $47,700. If the property has been appreciating $3500 per year, find its original purchase price.

Solution Let x = the number of years the cottage has been owned and let y = the value of the

cottage. Then one point on the line is given:  3, 47700  . Since the cottage appreciates by $3500 per year, m  3500.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

627


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y  y 1  m  x  x1 

y  47700  3500  x  3 y  47700  3500 x  10500 y  3500 x  37200 Let x  0 and find the value of y : y  3500 x  37200  3500  0   37200  37200

The purchase price was $37,200.

110. Computer repair A computer repair company charges a fixed amount, plus an hourly rate, for a service call. Use the information in the illustration to find the hourly rate

Solution Let x = the number of hours of service needed and let y = the total charge. Then two

points on the line are given:  2, 70  and  4, 105  m

105  70 35   17.50 42 2

y  y 1  m  x  x1 

y  70  17.50  x  2  y  70  17.50 x  35 y  17.50 x  35 The hourly charge is $17.50.

111. Automobile repair An auto repair shop charges an hourly rate, plus the cost of parts. If the cost of labor for a 1 21 -hour radiator repair is $69, find the cost of labor for a 5-hour transmission overhaul.

Solution Let x = the hours of labor and let y = the labor charge. Then m = the hourly charge. y  mx

y  46 x

46  m

The charge will be $230.

69  m  1.5 

y  46  5   230

112. Printer charges A printer charges a fixed setup cost, plus $1 for every 100 copies. If 700 copies cost $52, how much will it cost to print 1000 copies?

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628


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution Let x = the number of hundreds of copies and let y = the total charge. Then m = the charge per copy and b = the fixed charge.

y  mx  b

y  x  45

y  1x  b

y  10  45  55

52  1  7   b

The charge will be $55.

45  b 113. Predicting fires A local fire department recognizes that city growth and the number of reported fires are related by a linear equation. City records show that 300 fires were reported in a year when the local population was 57,000 persons, and 325 fires were reported in a year when the population was 59,000 persons. How many fires can be expected in the year when the population reaches 100,000 persons?

Solution Let x = the number of fires and let y = the population. Then two points on the line are given:  300, 57000 and  325, 59000  .

m

59000  57000 2000   80 325  300 25 y  y 1  m  x  x1 

y  57000  80  x  300 

y  57000  80 x  24000 y  80 x  33000

Let y  100000 and find the value of x: y  80 x  33000 100000  80 x  33000 67000  80 x 837.5  x  There will be about 838 fires when the population is 100,000.

114. Estimating the cost of rain gutter A neighbor tells you that an installer of rain gutter charges $60, plus a dollar amount per foot. If the neighbor paid $435 for the installation of 250 feet of gutter, how much will it cost you to have 300 feet installed?

Solution Let x = the number of feet of gutter and let y = the total charge. Then m = the charge per foot. One point on the line is given:  250, 435  y  mx  b

435  m  250   60 375  250m 1.5  m Let x  300 and find the value of y :

y  1.5 x  60

 1.5  300   60  510

It will cost $510.

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629


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

115. Converting temperatures Water freezes at 32 F, or 0 C. Water boils at 212 F, or 100 C. Find a formula for converting a temperature from degrees Fahrenheit to degrees Celsius.

Solution Let F replace x and C replace y. Then two points on the line are given:

 32, 0 and  212, 100 .

100  0 100 5   212  32 180 9 C  C1  m  F  F1  m

5  F  32 9 5 C   F  32  9

C 0 

116. Converting units A speed of 1 mile per hour is equal to 88 feet per minute, and of course, 0 miles per hour is 0 feet per minute. Find an equation for converting a speed x, in miles per hour, to the corresponding speed y, in feet per minute.

Solution

Two points on the line are given:  1, 88  and  0, 0  .

m

88  0 88   88 10 1

y  y 1  m  x  x1  y  0  88  x  0  y  88 x

117. Smoking The percent y of 18- to 25-year-old smokers in the United States has been declining at a constant rate since 1974. If about 47% of this group smoked in 1974 and about 29% smoked in 1994, find a linear equation that models this decline. If this trend continues, estimate what percent will smoke in 2024.

Solution Let y = the percent who smoke and let x = the # of years since 1974. Two points are given:  0, 47  and  20, 29  .

m

29  47 18 9   20  0 20 10

y  y 1  m  x  x1 

9  x  0 10 9 y  x  47 10

y  47  

Let x  50: 9 y    50   47  45  47  2 10 2% will smoke in 2024.

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630


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

118. Forensic science Scientists believe there is a linear relationship between the height h (in centimeters) of a male and the length f (in centimeters) of his femur bone. Use the data in the table to find a linear equation that expresses the height h in terms of f. Round all constants to the nearest thousandth. How tall would you expect a man to be if his femur measures 50 cm? Round to the nearest centimeter.

Person

Length of Femur ( f )

Height (h)

A

62.5 cm

200 cm

B

40.2 cm

150 cm

Solution Let f replace x and h replace y. Then two points on the line are given:

62.5, 200 and  40.2, 150 .

150  200 50   2.242 40.2  62.5 22.3 h  h1  m  f  f1 

m

h  200  2.242  f  62.5 

h  200  2.242f  140.125 h  2.242f  59.875 Let f  50:

h  2.242  50   59.875  172

He would be about 172 cm tall.

119. Predicting stock prices The value of the stock of ABC Corporation has been increasing by the same fixed dollar amount each year. The pattern is expected to continue. Let 2015 be the base year corresponding to x = 0 with x = 1, 2, 3,  corresponding to later years. ABC stock was selling at $37 21 in 2015 and at $45 in 2017. If y represents the price of ABC stock, find the equation y = mx + b that relates x and y, and predict the price in the year 2025.

Solution

Two points on the line are given:  0, 37.5  and  2, 45  . 45  37.5 7.5   3.75 20 2 y  y 1  m  x  x1 

m

y  37.5  3.75  x  0 

y  3.75 x  37.5 Let x  10 and find the value of y : y  3.75 x  37.5  3.75  10   37.5

 37.5  37.5  75 The price will be $75 in the year 2020.

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631


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

120. Estimating inventory Inventory of unsold goods showed a surplus of 375 units in January and 264 in April. Assume that the relationship between inventory and time is given by the equation of a line, and estimate the expected inventory in March. Because March lies between January and April, this estimation is called interpolation.

Solution Let January be represented by x = 0, and later months by x = 1, 2, 3, … Let y represent the inventory. Then two points on the line are given:  0, 375  and  3, 264  . 375  264 111   37 03 3 y  y 1  m  x  x1 

m

y  375  37  x  0 

y  37 x  375

Let x  2 and find the value of y : y  37 x  375

 37  2   375  301

The March inventory will be about 301.

121. Oil depletion When a Petroland oil well was first brought on line, it produced 1900 barrels of crude oil per day. In each later year, owners expect its daily production to drop by 70 barrels. Find the daily production after 3 21 years.

Solution The equation describing the production is y  70 x  1900, where x represents the number of years and y is the level of production. Let x  3 21  72 . y  70 x  1900 7  70    1900  1655 2 The production will be 1655 barrels per day.

122. Waste management The corrosive waste in industrial sewage limits the useful life of the piping in a waste processing plant to 12 years. The piping system was originally worth $137,000, and it will cost the company $33,000 to remove it at the end of its 12-year useful life. Find a depreciation equation.

Solution Let x = the number of years the piping has been owned and let y = the value of the piping. Then two points on the line are given:  0, 137000  and  12,  33000  . m

33000  137000 42500  12  0 3 y  y 1  m  x  x1  42500  x  0 3 42500 y  x  137000 3

y  137000  

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632


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

123. Crickets The table shows the approximate chirping rate at various temperatures for one type of cricket.

Temperature (F)

Chirps per Minute

50

20

60

80

70

115

80

150

100

250

a. Construct a scattergram below.

b. Assume a linear relationship and write a regression equation. c. Estimate the chirping rate at a temperature of 90F.

Solution a.

b. Use  50, 20  and  100, 250  for the regression line.

m

250  20 230 23   100  50 50 5

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633


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y  y 1  m  x  x1  23  x  50  5 23 y  20  x  230 5 23 y  x  210 5 y  20 

c.

23 90  210  204 5 The rate will be about 204 chirps per minute. y 

124. Fishing The table shows the lengths and weights of seven muskies captured by the Department of Natural Resources in Catfish Lake in Eagle River, Wisconsin.

Musky

Length (in.)

Weight (lb)

1

26

5

2

27

8

3

29

9

4

33

12

5

35

14

6

36

14

7

38

19

a. Construct a scattergram for the data. b. Assume a linear relationship and write a regression equation. c. Estimate the weight of a musky that is 32 inches long.

Solution a.

b. Use  26, 5  and  38, 19  for the regression line.

m

19  5 14 7   38  26 12 6

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634


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y  y 1  m  x  x1  7  x  26  6 7 91 y 5  x  6 3 7 76 y  x 6 3 y 5 

c.

7 76 32    12  6 3 The weight will be about 12 pounds. y 

125. Use the linear regression feature on a graphing calculator to determine an equation of the line that best fits the data given in Exercise 123. Round to the hundredths.

Solution y  4.44 x  196.62 126. Use the linear regression feature on a graphing calculator to determine an equation of the line that best fits the data given in Exercise 124. Round to the hundredths.

Solution y  0.96 x  19.22 Discovery and Writing 127. Explain how to find an equation of a line passing through two given points.

Solution Answers may vary. 128. Explain how to find an equation of a line that passes through a given point and is parallel to a given line.

Solution Answers may vary. 129. Describe how to find an equation of a line that passes through a given point and is perpendicular to a given line.

Solution Answers may vary. 130. In straight-line depreciation, explain why the slope of the line is called the rate of depreciation.

Solution Answers may vary. 131. Prove that an equation of a line with x-intercept of (a, 0) and y-intercept of (0, b) can be written in the form

x y  1 a b

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

635


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

m

b0 b  0a a

y  y 1  m  x  x1  b  x  0 a b y b   x a ay  ab  bx bx  ay  ab bx  ay ab  ab ab x y  1 a b y b  

132. Find the x- and y-intercepts of the line bx + ay = ab.

Solution x-intercept bx  ay  ab

y -intercept bx  ay  ab

bx  a  0   ab

b  0   ay  ab

bx  ab

ay  ab

x a

y b

a, 0

0, b

Investigate the properties of slope and the y-intercept by experimenting with the following problems. 133. Graph y = mx + 2 for several positive values of m. What do you notice?

Solution Answers may vary. 134. Graph y = mx + 2 for several negative values of m. What do you notice?

Solution Answers may vary. 135. Graph y = 2x + b for several increasing positive values of b. What do you notice?

Solution Answers may vary. 136. Graph y = 2x + b for several decreasing negative values of b. What do you notice?

Solution Answers may vary.

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636


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 137. The slope of the graph of Ax + By = C (A ≠ 0 and B ≠ 0) is 

B . A

Solution Ax  By  C By   Ax  C A C y  x B B A False. m   B 138. The y-intercept of the graph of Ax + By = C

 B (A ≠ 0, B ≠ 0) is  0,  .  C

Solution Ax  By  C By   Ax  C A C y  x B B C y -intercept: B 139. y   and y   are parallel lines.

Solution Both are horizontal. True. 140. x 

11 11 and y   are perpendicular lines. 11 11

Solution

11 11   11   1; True. 11 11

141. The equation of the line passing through (–99, 99) and parallel to x = 99 is y = –99.

Solution

x  99 is vertical, so the parallel line must be vertical too x  99 . False.

142. The equation of the line passing through (99, –99) and perpendicular to y = 99 is x = 99.

Solution

x  99 is horizontal, so the perpendicular line must be vertical too x  99 . True.

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637


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

5 x  10 y 

143. The equations

15 and x  2 y  3 describe the same line.

Solution 5 x  10 y 

15

5 x  10 y

15

5 x  2y 

5 3; True.

144. To determine whether lines are parallel, perpendicular, or neither, we always graph the equations and inspect them.

Solution False. You can tell by calculating the slopes.

EXERCISES 2.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Given y  x 3  16 x. a. Let x = 0 and solve for y. b. Let y = 0 and solve for x by factoring.

Solution a.

y  03  16  0  0

b.

0  x x 2  16

0  x  x  4 x  4 x  0, 4,  4 2. Given y  x 2  5. a. Replace x with –x. Do you obtain an equivalent equation? b. Replace y with –y. Do you obtain an equivalent equation? c. Replace x with –x and y with –y. Do you obtain an equivalent equation?

Solution a.

y    x   5  x 2  5, so yes.

b.

 y  x2  5

2

y   x 2  5, so no. c.

 y  x   5 2

y   x 2  5, so no.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

638


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

3. Use a special-product formula to find  x  7  . 2

Solution x2  2 7 x   7 

2

 x 2  14 x  49

4. Complete the square on x 2  6 x and factor the resulting trinomial.

Solution  6  x2  6x     2   x2  6x  9   x  3

2

2

5. Complete the square on x 2  5 x and factor the resulting trinomial.

Solution 2

5 x  5x    2 25  x 2  5x  4 2   5  x   2  2

6. Draw a circle in the rectangular coordinate system with center (2, 2) and radius 5.

Solution

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639


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The point where a graph intersects the x-axis is called the __________.

Solution x-intercept 8. The y-intercept is the point where a graph intersects the ____________.

Solution y-axis 9. If a line divides a graph into two congruent halves, we call the line an ___________.

Solution axis of symmetry 10. If the point (–x, y) lies on a graph whenever (x, y) does, the graph is symmetric about the _________.

Solution y-axis 11. If the point (x, –y) lies on a graph whenever (x, y) does, the graph is symmetric about the _______.

Solution x-axis 12. If the point (–x, –y) lies on a graph whenever (x, y) does, the graph is symmetric about the ______.

Solution origin 13. A ___________ is the set of all points in a plane that are fixed distance from a point called its ___________.

Solution circle, center 14. A __________ is the distance from the center of a circle to a point on the circle.

Solution radius 15. The standard form of an equation of a circle with center at the origin and radius r is ____________.

Solution

x2  y 2  r 2

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

640


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

16. The standard form of an equation of a circle with center at (h, k) and radius r is _________.

Solution

 x  h   y  k   r 2

2

2

Practice Find the x- and y-intercepts of each graph. Do not graph the equation. 17. y = x2 – 4

Solution

y  x2  4

y  x2  4

x  2, x  2

y  02  4 y  4

0   x  2  x  2 

y -int:  0,  4 

x-int:  2, 0  ,  2, 0  18. y = x2 – 9

Solution y  x2  9

y  x2  9

0  x2  9

y  02  9 y  9

0   x  3  x  3  x  3, x  3

x-int:  3, 0  ,  3, 0 

y -int:  0,  9 

19. y = 4x2 – 2x

Solution y  4 x 2  2x

y  4 x2  2x

0  2 x  2 x  1

y  4 0   2 0 

x  0, x 

1 2

1  x-int:  0, 0  ,  , 0  2 

2

y 0 y -int:  0, 0 

20. y = 2x – 4x2

Solution y  2x  4x 2

y  2x  4 x2

0  2x  1  2x 

y  2 0   4 0

x  0, x 

1 2

2

y 0

1  x-int:  0, 0  ,  , 0  y -int:  0, 0  2 

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641


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

21. y = x2 – 4x – 5

Solution

y  x2  4x  5

y  x2  4x  5

x  1, x  5

y  5

0   x  1 x  5 

x-int:  1, 0  ,  5, 0 

y  02  4  0   5 y -int:  0,  5 

22. y = x2 – 10x + 21

Solution

y  x 2  10 x  21

y  x 2  10 x  21

x  3, x  7

y  21

0   x  3 x  7  x-int:  3, 0  ,  7, 0 

y  02  10  0   21 y -int:  0, 21

23. y = x2 + x – 2

Solution

y  x2  x  2

y  x2  x  2

x  2, x  1

y  02  0  2 y  2

0   x  2 x  1

x-int:  2, 0  ,  1, 0 

y -int:  0, 2 

24. y = x2 + 2x – 3

Solution

y  x 2  2x  3

y  x 2  2x  3

x  3, x  1

y  3

0   x  3 x  1

x-int:  3, 0  ,  1, 0 

y  02  2  0   3 y -int:  0,  3 

25. y = x3 – 9x

Solution

y  x 3  9x

0  x x2  9

0  x  x  3 x  3  x  0, x  3, x  3

x-int:  0, 0  ,  3, 0  ,  3, 0 

y  x 3  9x

y  03  9  0  y 0

y -int:  0, 0 

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642


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

26. y = x3 + x

Solution y  x3  x

y  x3  x

0  x x2  1

y  03  0 y 0

  x  0,  x  1  0 2

x-int:  0, 0 

y -int:  0, 0 

27. y = x4 – 1

Solution y  x4  1

y  x4  1

0  x2  1 x2  1

y  04  1 y  1

   0   x  1  x  1 x  1 x  1  0 2

2

y -int:  0,  1

x  1, x  1

x-int:  1, 0  ,  1, 0  28. y = x4 – 25x2

Solution

y  x 4  25 x 2

0  x 2 x 2  25

0  x 2  x  5 x  5 x  0, x  5, x  5

x-int:  0, 0  ,  5, 0  5, 0 

y  x 4  25 x 2 y  04  25  0 

2

y 0

y -int:  0, 0 

Graph each equation. 29. y = x2 Solution

y  x2

x-int:  0, 0 

y -int:  0, 0 

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643


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

30. y = –x2

Solution

y  x2

x-int:  0, 0 

y -int:  0, 0 

31. y = –x2 + 2

Solution y  x2  2 x-int:

 2, 0 ,   2, 0

y -int:  0, 2 

32. y = x2 – 1

Solution

y  x2  1

x-int:  1, 0  ,  1, 0  y -int:  0,  1

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644


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

33. y = x2 – 4x

Solution

y  x2  4x

x-int:  0, 0  ,  4, 0  y -int:  0, 0 

34. y = x2 + 2x

Solution

y  x 2  2x

x-int:  0, 0  ,  2, 0  y -int:  0, 0 

35. y 

1 2 x  2x 2

Solution 1 y  x 2  2x 2 x-int:  0, 0  ,  4, 0  y -int:  0, 0 

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645


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

36. y 

1 2 x 3 2

Solution 1 y  x2  3 2 x-int: none

y -int:  0, 3

Find the symmetries, if any, of the graph of each equation. Do not graph the equation. 37. y = x2 + 5

Solution x-axis y  x  5 not equivalent: no symmetry 2

y  x2  5 y -axis

origin

y  x   5

 y  x   5

y  x2  5

 y  x2  5

equivalent: symmetry

not equivalent: no symmetry

2

2

38. y = 3x + 2

Solution x-axis  y  3x  2 not equivalent: no symmetry

y  3x  2 y -axis

origin

y  3 x   2

 y  3 x   2

y  3 x  2 not equivalent: no symmetry

 y  3 x  2 y  3x  2 not equivalent: no symmetry

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646


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

39. y2 + 2 = x

Solution y2  2  x y -axis

x-axis

 y   2  x 2

y2  2  x

origin

 y   2  x

y  2  x not equivalent: no symmetry 2

2

y 2  2  x not equivalent: no symmetry

equivalent: symmetry

40. y2 + y = x

Solution y2  y  x x-axis

y -axis

origin

 y    y   x

y  y  x

 y    y   x

2

2

2

not equivalent: no symmetry

y y x 2

y 2  y  x y2  y  x not equivalent: no symmetry

not equivalent: no symmetry

41. y2 = x2

Solution y 2  x2 y -axis

x-axis

 y   x 2

origin

y  x 

2

2

 y   x 

2

2

2

y 2  x2

y 2  x2

y 2  x2

equivalent: symmetry

equivalent: symmetry

equivalent: symmetry

42. y = 3x + 8

Solution x-axis  y  3x  8 not equivalent: no symmetry

y  3x  8 y -axis

origin

y  3 x   8

 y  3 x   8

y  3 x  8 not equivalent: no symmetry

 y  3 x  8 y  3x  8 not equivalent: no symmetry

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647


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

43. y = 3x2 + 11

Solution x-axis  y  3 x 2  11 not equivalent: no symmetry

y  3 x 2  11 y -axis

origin

y  3   x   11

 y  3   x   11

y  3 x 2  11

 y  3 x 2  11 not equivalent: no symmetry

2

2

equivalent: symmetry 44. x2 + y2 = 16

Solution x-axis

x 2  y 2  16 y -axis

origin

x 2    y   16

  x   y  16

  x     y   16

x 2  y 2  16

x 2  y 2  16

x 2  y 2  16

equivalent: symmetry

equivalent: symmetry

equivalent: symmetry

2

2

2

2

2

45. y = 3x3 + 5

Solution y  3x3  5 x-axis

y -axis

origin

 y  3x  5 not equivalent: no symmetry

y  3 x   5

 y  3 x   5

y  3 x 3  5 not equivalent: no symmetry

 y  3 x 3  5

3

3

3

y  3x3  5 not equivalent: no symmetry

46. y = 3x3 + 7x

Solution x-axis  y  3x 3  7 x not equivalent: no symmetry

y  3x 3  7 x y -axis

origin

y  3 x   7x

 y  3 x   7 x 

y  3 x 3  7 x not equivalent: no symmetry

 y  3 x  7 x

3

3

y  3x 3  7 x equivalent: symmetry

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648


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

47. y2 = 3x

Solution y 2  3x y -axis

x-axis

  y   3x y 2  3x equivalent: symmetry

origin

y  3 x 

2

 y   3 x  2

2

y 2  3 x not equivalent: no symmetry

y 2  3 x not equivalent: no symmetry

48. y = 3x4 + 2

Solution x-axis  y  3x 4  2 not equivalent: no symmetry

y  3x 4  2 y -axis

origin

y  3 x   2

 y  3 x   2

y  3x 4  2

 y  3x 4  2 not equivalent: no symmetry

4

equivalent: symmetry

4

49. y  2 x

Solution

y 2x x-axis

y -axis

origin

y  2 x

y  2 x

 y  2 x

not equivalent: no symmetry

y  2 1 x

 y  2 1 x

y 2x

y  2 x

equivalent: symmetry

not equivalent: no symmetry

50. y  x  1

Solution

y  x1 x-axis

y -axis

origin

y  x  1

y  x  1

 y  x  1

not equivalent: no symmetry

not equivalent: no symmetry

not equivalent: no symmetry

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649


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

51.

y x Solution

y x x-axis

y -axis

origin

y  x

y  x

 y  x

1 y  x

not equivalent: no symmetry

1 y   x

y x

y  x not equivalent: no symmetry

equivalent: symmetry 52. y  x

Solution

y  x x-axis

y -axis

origin

y  x

y  x

 y  x

1 y  x

y  1 x

1 y  1 x

y  x

y  x

equivalent: symmetry

equivalent: symmetry

y  x equivalent: symmetry

Graph each equation. Be sure to find any intercepts and symmetries. 53. y = x2 + 4x

Solution

y  x2  4x

x-int:  0, 0  ,  4, 0  y -int:  0, 0 

symmetry: none

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650


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

54. y = x2 – 6x

Solution

y  x2  6x

x-int:  0, 0  ,  6, 0  y -int:  0, 0 

symmetry: none

55. y = x3

Solution

y  x3

x-int:  0, 0 

y -int:  0, 0 

symmetry: origin

56. y = x3 + x

Solution

y  x3  x

x-int:  0, 0 

y -int:  0, 0 

symmetry: origin

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651


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

57. y  x  2

Solution

y  x 2

x-int:  2, 0 

y -int:  0, 0 

symmetry: none

58. y  x  2

Solution

y  x 2

x-int:  2, 0  ,  2, 0  y -int:  0,  2 

symmetry: y -axis

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652


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

59. y   x  3

Solution

y   x 3

x-int:  3, 0  ,  3, 0  y -int:  0, 3

symmetry: y -axis

60. y  3 x

Solution

y 3x

x-int:  0, 0 

y -int:  0, 0 

symmetry: y -axis

61. y2 = –x

Solution

y 2  x

x-int:  0, 0 

y -int:  0, 0 

symmetry: x-axis

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653


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

62. y2 = 4x

Solution

y 2  4x

x-int:  0, 0 

y -int:  0, 0 

symmetry: x-axis

63. y2 = 9x

Solution

y 2  9x

x-int:  0, 0 

y -int:  0, 0 

symmetry: x-axis

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654


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

64. y2 = –4x

Solution y 2  4 x

x-int:  0, 0 

y -int:  0, 0 

symmetry: x-axis

Graph each equation. Be sure to find any intercepts and symmetries. 65. y  2 x

Solution

y  2x

66. y   x

Solution

y  x

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655


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

67. y 

x 1

Solution

y 

x 1

x-int:  1, 0 

y -int:  0,  1

symmetry: none

68. y  1  x

Solution

y  1 x

x-int:  1, 0 

y -int:  0, 1

symmetry: none

69. xy  4

Solution xy  4

x-int: none y -int: none symmetry: origin

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656


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

70. xy  9

Solution xy  9

x-int: none y -int: none symmetry: origin

71. y  3 x

Solution

y 3x

x-int:  0, 0 

y -int:  0, 0 

symmetry: origin

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657


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

72. y   3 x

Solution

y  3 x

x-int:  0, 0 

y -int:  0, 0 

symmetry: origin

Identify the center and radius of each circle written in standard form. 73. x 2  y 2  144

Solution x 2  y 2  144

 x  0    y  0   12 C:  0, 0  ; r  12 2

2

2

74. x 2  y 2  121

Solution x 2  y 2  121

 x  0    y  0   11 C:  0, 0  ; r  11 2

2

2

75. x 2   y  5   64 2

Solution x 2   y  5   64 2

 x  0   y  5  8 C:  0, 5  ; r  8 2

2

2

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658


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

76. x 2   y  3   8 2

Solution x 2   y  3  8 2

 x  0    y   3     8  2

2

2

 x  0    y   3     2 2  C:  0,  3  ; r  2 2 2

2

77.  x  6   y 2  2

2

1 4

Solution

 x  6   y  41 2

2

 x   6     y  0    21  2

2

C:  6, 0  ; r 

78.  x  5   y 2  2

2

1 2

16 25

Solution 16  x  5   y  25 2

2

 x  5    y  0    45  2

2

2

4 C:  5, 0  ; r  5

79.  x  4    y  1  25 2

2

Solution

 x  4    y  1  25  x  4    y  1  5 C:  4, 1 ; r  5 2

2

2

2

2

80.  x  11   x  7   169 2

2

Solution

 x  11   y  7   169  x   11    y   7    13 C:  11,  7  ; r  13 2

2

2

2

2

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659


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs 2

2  1 81.  x     y  2   45 4 

Solution 2

2  1  x     y  2   45 4  2

  45 

2  1  x    y   2   4 

2

2

 

2 2  1  x    y   2   3 5 4  1  C:  ,  2  ; r  3 5 4 

82. x  5

   y  3  225 2

2

Solution

 x  5    y  3  225  x    5    y  3   15 C:   5, 3  ; r  15 2

2

2

2

2

Write an equation in standard form of the circle with the given properties. 83. Center at the origin; r = 5

Solution

 x  0   y  0  5 2

2

2

x 2  y 2  25

84. Center at the origin; r 

3

Solution

 x  0   y  0   3  2

2

2

x2  y 2  3

85. Center at (0, –6); r = 9

Solution

 x  0    y   6    9 x   y  6   81 2

2

2

2

2

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660


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

86. Center at (0, 7); r = 14

Solution

 x  0    y  7   14 x   y  7   196 2

2

2

2

2

87. Center at (8, 0); r 

1 5

Solution  

 x  8    y  0    51  2

2

 x  8

2

2

  1  y2  25

88. Center at (–10, 0); r 

11

Solution

 x   10    y  0    11  2

2

2

 x  10   y  11 2

2

89. Center at (–2, 12), r = 20

Solution

 x   2    y  12  20 2

2

2

 x  2    y  12   400 2

2

2  90. Center at  , 5  ; r  6 7 

Solution 2

2  2 2  x    y   5   6 7 

2

2  2  x     y  5   36 7 

Write an equation in general form of the circle with the given properties. 91. Center at the origin; r = 1

Solution x 2  y 2  12  x 2  y 2  1  0 92. Center at the origin; r = 4

Solution x 2  y 2  4 2  x 2  y 2  16  0

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661


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

93. Center at (6, 8); r = 4

Solution

 x  6   y  8  4 2

2

2

x 2  12 x  36  y 2  16 y  64  16 x 2  y 2  12 x  16 y  84  0

94. Center at (5, 3); r = 2

Solution

 x  5   y  3  2 2

2

2

x 2  10 x  25  y 2  6 y  9  4 x 2  y 2  10 x  6 y  30  0

95. Center at (3, –4); r 

2

Solution

 x  3   y  4   2  2

2

2

x 2  6 x  9  y 2  8 y  16  2 x 2  y 2  6 x  8 y  23  0

96. Center at (–9, 8); r  2 3

Solution

 x  9   y  8  2 3  2

2

2

x 2  18 x  81  y 2  16 y  64  12 x 2  y 2  18 x  16 y  133  0

97. Ends of diameter at (3, –2) and (3, 8)

Solution 33 2  8  3, y  3 2 2 r  distance from center to endpoint

Center: x 

3  3  3  8  5 2

2

 x  3   y  3  5 2

2

2

x 2  6 x  9  y 2  6 y  9  25 x2  y 2  6x  6 y  7  0

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662


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

98. Ends of diameter at (5, 9) and (–5, –9)

Solution 5  5 9  9  0, y  0 2 2 r  distance from center to endpoint

Center: x 

0  5   0  9   106 2

2

 x  0    y  0    106  2

2

2

x 2  y 2  106 x 2  y 2  106  0

99. Center at (–3, 4) and passing through the origin

Solution r  distance from center to origin 

 0   3     0  4   5 2

2

 x  3   y  4  5 2

2

2

x 2  6 x  9  y 2  8 y  16  25 x2  y 2  6x  8 y  0

100. Center at (–2, 6) and passing through the origin

Solution r  distance from center to origin 

0   2   0  6   40 2

2

 x  2    y  6    40  2

2

2

x 2  4 x  4  y 2  12 y  36  40 x 2  y 2  4 x  12 y  0

Convert the general form of each circle given into standard form. 101. x2 + y2 – 6x + 4y + 4 = 0

Solution x2  y 2  6x  4 y  4  0 x 2  6 x  y 2  4 y  4 x 2  6 x  9  y 2  4 y  4  4  9  4

 x  3   y  2  9 2

2

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663


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

102. x2 + y2 + 4x – 8y – 5 = 0

Solution x2  y 2  4x  8 y  5  0 x2  4x  y 2  8 y  5 x 2  4 x  4  y 2  8 y  16  5  4  16

 x  2    y  4   25 2

2

103. x2 + y2 – 10x – 12y + 57 = 0

Solution x 2  y 2  10 x  12 y  57  0 x 2  10 x  y 2  12 y  57 x 2  10 x  25  y 2  12 y  36  57  25  36

 x  5   y  6  4 2

2

104. x2 + y2 + 2x + 18y + 57 = 0

Solution x 2  y 2  2 x  18 y  57  0 x 2  2 x  y 2  18 y  57 x 2  2 x  1  y 2  18 y  81  57  1  81

 x  1   y  9   25 2

2

105. 2x2 + 2y2 – 8x – 16y + 22 = 0

Solution 2 x 2  2 y 2  8 x  16 y  22  0 x 2  y 2  4 x  8 y  11  0 x 2  4 x  y 2  8 y  11 x 2  4 x  4  y 2  8 y  16  11  4  16

 x  2   y  4   9 2

2

106. 3x2 + 3y2 + 6x – 30y + 3 = 0

Solution 3 x 2  3 y 2  6 x  30 y  3  0 x 2  y 2  2 x  10 y  1  0 x 2  2 x  y 2  10 y  1 x 2  2 x  1  y 2  10 y  25  1  1  25

 x  1   y  5   25 2

2

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664


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Graph each circle. 107. x2 + y2 – 25 = 0

Solution x 2  y 2  25  0 x 2  y 2  25

C  0, 0  , r  5

108. x2 + y2 – 8 = 0

Solution x2  y 2  8  0 x2  y 2  8 C  0, 0  , r 

8 2 2

109. (x – 1)2 + (y + 2)2 = 4

Solution

 x  1   y  2   4 C  1,  2  , r  2 2

2

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665


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

110. (x + 1)2 + (y – 2)2 = 9

Solution

 x  1   y  2   9 C  1, 2  , r  3 2

2

111. x2 + y2 + 2x – 24 = 0

Solution x 2  y 2  2 x  24  0 x 2  2 x  y 2  24 x 2  2 x  1  y 2  24  1

 x  1  y  25 C  1, 0  , r  5 2

2

112. x2 + y2 – 4y = 12

Solution x 2  y 2  4 y  12 x  y 2  4 y  4  12  4 2

x 2   y  2   16 2

C  0, 2  , r  4

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666


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

113. x2 + y2 + 4x + 2y – 11 = 0

Solution x 2  y 2  4 x  2 y  11  0 x 2  4 x  y 2  2 y  11 x 2  4 x  4  y 2  2 y  1  11  4  1

 x  2    y  1  16 C  2,  1 , r  4 2

2

114. x2 + y2 – 6x + 2y + 1 = 0

Solution x2  y 2  6x  2 y  1  0 x 2  6 x  y 2  2 y  1 x 2  6 x  9  y 2  2 y  1  1  9  1

 x  3   y  1  9 C  3,  1 , r  3 2

2

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667


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

115. 9x2 + 9y2 – 12y = 5

Solution 9 x 2  9 y 2  12 y  5 4 5 x2  y 2  y  3 9 4 4 5 4 x2  y 2  y    3 9 9 9 2  2 x2   y    1 3 

 2 C  0,  , r  1  3

116. 4x2 + 4y2 + 4y = 15

Solution 4 x 2  4 y 2  4 y  15 15 4 1 15 1 2 2 x  y  y   4 4 4 2  1 x2   y    4 2  x2  y 2  y 

 1 C  0,   , r  2 2  

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668


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

117. 4x2 + 4y2 – 4x + 8y + 1 = 0

Solution 4x2  4 y 2  4x  8 y  1  0 1 4 1 1 1 x2  x   y 2  2 y  1     1 4 4 4 2 2  1 x      y  1  1 2   x2  y 2  x  2 y  

1  C  ,  1 , r  1 2 

118. 9x2 + 9y2 – 6x + 18y + 1 = 0

Solution 9 x 2  9 y 2  6 x  18 y  1  0 2 1 x2  y 2  x  2 y   3 9 2 1 1 1 2 2 x  x   y  2y  1     1 3 9 9 9 2 2  1  x     y  1  1 3 

1  C  ,  1 , r  1 3  

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669


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Use a graphing calculator to graph each equation. Then find the coordinates of the vertex of the parabola to the nearest hundredth. 119. y = 2x2 – x + 1

Solution y  2x 2  x  1

Vertex:  0.25, 0.88 

120. y = x2 + 5x – 6

Solution y  x 2  5x  6

Vertex:  2.50,  12.25 

121. y = 7 + x – x2

Solution y  7  x  x2

Vertex:  0.50, 7.25 

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670


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

122. y = 2x2 – 3x + 2

Solution y  2x 2  3x  2

Vertex:  0.75, 0.88 

Use a graphing calculator to solve each equation. Round to the nearest hundredth. 123. x2 – 7 = 0

Solution Graph y  x 2  7. Find the x-intercepts. x  2.65, x  2.65

124. x2 – 3x + 2 = 0

Solution Graph y  x 2  3 x  2. Find the x-intercepts. x  1.00, x  2.00

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671


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

125. x3 – 3 = 0

Solution Graph y  x 3  3. Find the x-intercepts. x  1.44

126. 3x3 – x2 – x = 0

Solution Graph y  3 x 3  x 2  x . Find the x-intercepts. x  0.43, x  0, x  0.77

Fix It In exercises 127 and 128, identify the step the first error is made and fix it. 127. Write the equation of the circle with center (3, –5) and radius 4 in general form.

Solution Step 3 was incorrect.

   y  5  16

   y  5  16  0

Step 1: x  3

Step 2: x  3

2

2

2

2

Step 3: x 2  6 x  9  y 2  10 y  25  16  0 Step 4: x 2  6x  9  y 2  10 y  9  0

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672


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

128. Convert the general form of the circle x 2  y 2  4 x  6 y  87  0 into standard form.

Solution Step 4 was incorrect. Step 1: x 2  y 2  4 x  6 y  87  0 Step 2: x 2  4 x  y 2  6 y  87

 

   y  3  100

Step 3: x 2  4 x  4  y 2  6 y  9  87  4  9 Step 4: x  2

2

2

Applications 129. Golfing Michelle Wie’s tee shot follows a path given by y = 64t – 16t2, where y is the height of the ball (in feet) after t seconds of flight. How long will it take for the ball to strike the ground?

Solution Let y  0:

y  64t  16t 2

0  16t  4  t 

t  0 or t  4 It strikes the ground after 4 seconds. 130. Golfing Halfway through its flight, the golf ball of Exercise 129 reaches the highest point of its trajectory. How high is that?

Solution From #129, the flight lasts 4 seconds. Thus, half the flight is 2 seconds. Let t  2:

y  64t  16t 2 y  64  2  16  2  128  64; The highest point is 64 feet above ground. 2

131. Stopping distances The stopping distance D (in feet) for a Ford Fusion car moving V miles per hour is given by D = 0.08V2 + 0.9V. Graph the equation for velocities between 0 and 60 mph.

Solution

D  0.08V 2  0.9V ;

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673


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

132. Stopping distances See Exercise 131. How much farther does it take to stop at 60 mph than at 30 mph?

Solution Refer to the graph for #131. The y-coordinate for x  30 is y  99. The y-coordinate for x  60 is y  342. 342  99  243

At 60 mph, 243 more feet is required to stop than at 30 mph. 133. Basketball court The center circle of the Toronto Raptors basketball court is a circle with a 12-ft diameter. If the center of the circle is located at the origin, find an equation in standard form that models the circle.

Solution

r

12 6 2

 x  0   y  0  6 2

2

2

x 2  y 2  36 134. Oil spill Oil spills from a tanker off the coast of Florida and surfaces continuously at coordinates (0, 0). If oil spreads in a circular pattern for ten hours and the circle’s radius increases at a rate of 2 inches per hour, write an equation of the circle that models the range of the spill’s effect.

Solution

r  10  2 in.   20 in.

 x  0   y  0  20 2

2

2

x 2  y 2  400 135. Super Loop The Fire Ball Super Loop is a rollercoaster ride that is shaped like a circle. Find an equation of the loop in standard form if it is positioned 5 feet off of the ground, has a diameter of 60 feet, and its center is at coordinates (0, 35).

Solution

r

60  30 2

 x  0   y  35  30 x   y  35   900 2

2

2

2

2

136. Hurricane As a hurricane strengthens, an eye begins to form at the center of the storm. At a wind speed of 80 mph the eye of a hurricane is circular when viewed from above and is 30 miles in diameter. If the eye is located at map coordinates (5, 10), find an equation, in standard form, of the circle that models the eye of the hurricane.

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674


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

r

30  15 2

 x  5   y  10  15  x  5   y  10  225 2

2

2

2

2

137. CB radios The CB radio of a trucker covers the circular area shown in the illustration. Find an equation of that circle, in general form.

Solution

r

 10  7   0  4  5 2

2

 x  7   y  4  5 2

2

2

x 2  14 x  49  y 2  8 y  16  25 x 2  y 2  14 x  8 y  40  0 138. Firestone tires Two 24-inch-diameter Firestone tires stand against a wall, as shown in the illustration. Find equations in general form of the circular boundaries of the tires.

Solution

First tire

Second tire

C  12, 12 , r  12

C  36, 12 , r  12

 x  12   y  12  12 2

2

2

 x  36   y  12  12 2

2

2

x 2  24 x  144  y 2  24 y  144  144

x 2  72 x  1296  y 2  24 y  144  144

x 2  y 2  24 x  24 y  144  0

x 2  y 2  72 x  24 y  1296  0

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675


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Discovery and Writing 139. Draw three graphs: one that is symmetric to the x-axis, one that is symmetric to the y-axis, and one that is symmetric to the origin.

Solution Answers may vary. 140. Explain how you test for symmetry with respect to the x-axis, y-axis, and origin.

Solution Answers may vary. 141. How do you recognize the equation of a circle?

Solution Answers may vary. 142. Describe the process of converting a circle in general form into standard form.

Solution Answers may vary. When converting a circle’s equation from general to standard form, it is possible to obtain a constant term on the right side that is zero or negative. If the constant term is zero, the graph is a single point. If the constant term is negative, the graph is nonexistent. Determine whether the graph of the equation is a single point or nonexistent. 143. x2 – 4x + y2 – 6y + 13 = 0

Solution

x 2  4 x  y 2  6 y  13  0 x 2  4 x  4  y 2  6 y  9  13  4  9

 x  2   y  3  0  a single point 2

2

144. x2 – 12x + y2 + 4y + 43 = 0

Solution

x 2  12 x  y 2  4 y  43  0 x 2  12 x  36  y 2  4 y  4  43  36  4

 x  6   y  2  3  nonexistent 2

2

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 145. The graphs of y = x2, y = x4, and y = x6 are symmetric with respect to the x-axis.

Solution False. The graphs are symmetric with respect to the y-axis.

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676


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

146. The graphs of y = x, y = x3, and y = x5 are symmetric with respect to the y-axis.

Solution False. The graphs are symmetric with respect to the origin. 147. If the graph of an equation is symmetric with respect to the x-axis and y-axis, then it is symmetric with respect to the origin.

Solution True. 148. If the graph of an equation is symmetric with respect to the origin, then it is symmetric with respect to the x-axis and y-axis.

Solution False. The line y  x has symmetry with respect to the origin, but not with respect to either the x- or y-axis. 149. The center of the circle 2

2 3   3 3 11   3 11  4 x    y    2 is  , .     8     8    

Solution True. 150. The radius of the circle 2

2 3   2 6    4 2 is 8 2. x     y    9 5    

Solution True. 2

2   1 1 151. The graph of the equation  x  4    y    0 is the single point  4,  . 7 7  

Solution

False. The graph is the single point  4,  71  . 152. The graph of the equation x2 + y2 = –9 is nonexistent.

Solution True.

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677


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

EXERCISES 2.6 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

True or False: If

a c  , then ac = bd. b d

Solution False, ad  bc 2. True or False: If

2x 4 , then 2 x  x  3   4  5  .  5 x3

Solution True 3. Solve: If x  x  7   4  2 

Solution

x  x  7   4  2 x2  7x  8 x2  7x  8  0

 x  8 x  1  0 x  8, x  1 4. Given y = kx2. Find the value of k if x = 6 and y = 72.

Solution

y  kx 2 72  k  62 72  36k 5. Given I 

k . Find the value for k if I = 100 and d = 30. d2

Solution

I

k d2

100 

k

302 k 100  900 90,000  k

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678


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

k . x a. Find the value of k if y = 40 and x = 2. b. Using the value of k from part a, find x if y = 10.

6. Given y 

Solution k a. y  x k 40  2

80  k b.

k x 80 10  x y 

10 x  80 x 8

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A ratio is the __________ of two numbers. Solution quotient 8. A proportion is a statement that two __________ are equal.

Solution ratios 9. In the proportion

a c  , b and c are called the __________. b d

Solution means 10. In the proportion

a c  , a and d are called the __________. b d

Solution extremes 11. In a proportion, the product of the __________ is equal to the product of the __________.

Solution extremes, means

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679


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

12. The equation y = kx indicates __________ variation.

Solution direct 13. The equation y 

k indicates __________ variation. x

Solution inverse 14. In the equation y = kx, k is called the __________ of proportionality.

Solution constant 15. The equation y = kxz represents __________ variation.

Solution joint 16. In the equation y 

kx 2 , y varies directly with __________ and inversely with z

__________.

Solution x2, z

Practice Solve each proportion. 17.

4 2  x 12

Solution 4 2  x 12 4  12  2  x 48  2 x 24  x 18.

9 x  2 6 Solution 9 x  2 6 96  x 2 54  2 x 27  x

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680


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

19.

x 3  2 x1

Solution 3 x  2 x1 x  x  1  3  2 x2  x  6 x2  x  6  0

 x  3 x  2  0 x  3 or x  2

20.

x 5 7  6 8 x Solution x 5 7  6 8 x  x  58  x   7  6  x 2  3 x  40  42 0  x 2  3x  2

0   x  2  x  1

x  1 or x  2

Set up and solve a proportion to answer each question. 21. The ratio of women to men in a mathematics class is 3:5. How many women are in the class if there are 30 men?

Solution Let x  the number of women. 3 x  5 30 3  30  5  x 90  5 x 18  x  There are 18 women.

22. The ratio of lime to sand in mortar is 3:7. How much lime must be mixed with 21 bags of sand to make mortar?

Solution Let x  the number of bags of lime. 3 x  7 21 3  21  x  7 63  7 x 9  x  9 bags of lime should be used.

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681


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Find the constant of proportionality. 23. y is directly proportional to x. If x = 30, then y = 15.

Solution y  kx

15  k  30  1 k 2

24. z is directly proportional to t. If t = 7, then z = 21.

Solution z  kt

21  k  7  3k

25. I is inversely proportional to R. If R = 20, then I = 50.

Solution k I R k 50  20 1000  k 26. R is inversely proportional to the square of I. If I = 25, then R = 100.

Solution

k R 2 I k 100  252 k 100  625 62500  k 27. E varies jointly with I and R. If R = 25 and I = 5, then E = 125.

Solution E  kIR

125  k  5  25  125  125k 1k

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682


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

28. z is directly proportional to the sum of x and y. If x = 2 and y = 5, then z = 28.

Solution

z  k x  y

28  k  2  5 28  7k 4k

Solve each problem. 29. y is directly proportional to x. If y = 15 when x = 4, find y when x 

7 . 5

Solution

y  kx

15  k  4  15 k 4

15 x 4 15 7 y   4 5 21 y  4 y 

30. w is directly proportional to z. If w = –6 when z = 2, find w when z = –3.

Solution w  kz

w  3z

3  k

w 9

6  k  2 

w  3  3

31. w is inversely proportional to z. If w = 10 when z = 3, find w when z = 5.

Solution k w z k 10  3 30  k

30 z 30 w  5 w 6 w 

32. y is inversely proportional to x. If y = 100 when x = 2, find y when x = 50.

Solution k x k 100  2 200  k y 

200 x 200 y  50 y 4 y 

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683


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

33. P varies jointly with r and s. If P = 16 when r = 5 and s = –8, find P when r = 2 and s = 10.

Solution P  krs

16  k  5  8  16  40k 16  k 40 2  k 5

2 rs 5 2 P    2  10  5 P  8 P

34. m varies jointly with the square of n and the square root of q. If m = 24 when n = 2 and q = 4, find m when n = 5 and q = 9.

Solution

m  kn2 q 24  k  2 

2

m  3n2 q

4

24  k  4  2  24  8k

m  3  5

2

9

m  3  25  3  m  225

3k Determine whether the graph could represent direct variation, inverse variation, or neither. 35.

Solution direct 36.

Solution neither 37.

Solution neither

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684


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

38.

Solution inverse Fix It In exercises 39 and 40, identify the step the first error is made and fix it. 39. Solve the proportion:

x 7  4 x 3

Solution Step 4 is incorrect. Step 1: x  x  3   7  4  Step 2: x  x  3   28 Step 3: x  x  3   28  0 or x 2  3 x  28  0 Step 4:  x  7  x  4   0 Step 5: x  7 or x  4 40. Given y is inversely proportional to x2. If y = 20 when x = 5, find y when x = 10.

Solution Step 4 is incorrect. Step 1: y 

k x2

Step 2: 20 

k 25

Step 3: k  500 Step 4: y 

500 100

Step 5: y  5

Applications Set up and solve the required proportion. 41. Cellphones A country has 221 mobile cellular telephones per 250 inhabitants. If the country’s population is about 280,000, how many mobile cellular telephones does the country have?

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685


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution Let x  the number of phones. x 221  250 280000 221  280000  250  x 61880000  25 x 247, 520  x 247,520 have cellular phones. 42. Caffeine Many convenience stores sell supersize 44-ounce soft drinks in refillable cups. For each of the products listed in the table, find the amount of caffeine contained in one of the supersize cups. Round to the nearest milligram.

Soft Drink 12 oz

Caffeine (mg)

Mountain Dew

55

Coca-Cola Classic

47

Pepsi

37

Based on data from the Los Angeles Times

Solution Let x  the amount of caffeine.

55 x  12 44 55  44  12  x 2420  12 x

47 x  12 44 47  44  12  x 2068  12 x

37 x  12 44 37  44  12  x 1628  12 x

202 mg  x

172 mg  x

136 mg  x

43. Wallpapering Read the instructions on the label of wallpaper adhesive. Estimate the amount of adhesive needed to paper 500 square feet of kitchen walls if a heavy wallpaper will be used. COVERAGE: One-half gallon will hang approximately 4 single rolls (140 square feet), depending on the weight of the wall covering and the condition of the wall.

Solution Let x  the amount of adhesive needed. 1 2

x 500

140 1  500  140  x 2 250  140 x 1.79  x About 2 gallons of adhesive will be needed.

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686


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

44. Veterinarian care For one particular breed of dog, a veterinarian administers 0.006 gram of a specific medication for each kilogram of body mass. How much medication, in milligrams, would the veterinarian administer for a 30-kilogram dog of this breed?

Solution Let x  the dosage. 0.006 x  1 30 0.006  30  1  x 0.18  x The dosage should be 0.18 g, or 180 mg. 45. Gas laws The volume of a gas varies directly with the temperature and inversely with the pressure. When the temperature of a certain gas is 330C, the pressure is 40 pounds per square inch and the volume is 20 cubic feet. Find the volume when the pressure increases 10 pounds per square inch and the temperature decreases to 300C.

Solution V  20 

kT P k  330 

40 800  330k 800 k 330 80 k 33

80 T V  33 P 80 300  33  V  50

V 

8000 11

50 160 6  14 ft 3 V  11 11

46. Hooke’s Law The force f required to stretch a spring a distance d is directly proportional to d. A force of 5 newtons stretches a spring 0.2 meter. What force will stretch the spring 0.35 meter?

Solution f  kd

f  25d

25  k

f  8.75 Newtons

5  k  0.2 

f  25  0.35 

47. Free-falling objects The distance that an object will fall in t seconds varies directly with the square of t. An object falls 16 feet in 1 second. How long will it take the object to fall 144 feet?

Solution d  kt 2 16  k  1 16  k

d  16t 2 2

144  16t 2 9  t2 3  t  3 seconds

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

687


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

48. Heat dissipation The power, in watts, dissipated as heat in a resistor varies directly with the square of the voltage and inversely with the resistance. If 20 volts are placed across a 20-ohm resistor, it will dissipate 20 watts. What voltage across a 10-ohm resistor will dissipate 40 watts?

Solution P 20 

kV 2 R

k  20 

20 400  400k 1k

V2 R V2 40  10 400  V 2 20  V  20 volts P

2

49. Period of a pendulum The time required for one complete swing of a pendulum is called the period of the pendulum. The period varies directly with the square of its length. If a 1-meter pendulum has a period of 1 second, find the length of a pendulum with a period of 2 seconds.

Solution t  kl 2 1  k  1

t  l2 2

1k

2  l2 2 l

2 meters

50. Frequency of vibration The pitch, or frequency, of a vibrating string varies directly with the square root of the tension. If a string vibrates at a frequency of 144 hertz due to a tension of 2 pounds, find the frequency when the tension is 18 pounds.

Solution f k T

144  k 2 144 2

k

f  f 

144 2 144

T

18 2 f  144 9

f  144  3   432 hertz 51. Illumination Intensity of illumination from a light source varies inversely with the square of the distance from the source. If the intensity of a light source is 60 lumens at a distance of 10 feet, find the intensity at 20 feet.

Solution

k d2 k 60  2 10 k 60  100 6000  k I

6000 d2 6000 I 202 6000 I 400 I  15  15 lumens I

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688


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

52. Illumination Intensity of illumination from a light source varies inversely with the square of the distance from the source. If the intensity of a light source is 100 lumens at a distance of 15 feet, find the intensity at 25 feet.

Solution

22500 d2 22500 I 252 22500 I 625 I  36  36 lumens

k d2 k 100  2 15 k 100  225 22500  k I

I

53. Kinetic energy The kinetic energy of an object varies jointly with its mass and the square of its velocity. What happens to the energy when the mass is doubled and the velocity is tripled?

Solution E  kmv 2  k  2m 3v 

2

 

 k  2m 9v 2  18  kmv 2

The energy is multiplied by 18. 54. Heat dissipation The power, in watts, dissipated as heat in a resistor varies jointly with the resistance, in ohms, and the square of the current, in amperes. A 10-ohm resistor carrying a current of 1 ampere dissipates 10 watts. How much power is dissipated in a 5-ohm resistor carrying a current of 3 amperes?

Solution P  kRC 2

P  RC 2

10  k  10  1 10  10k 1k

2

P  5  3

2

P  5 9 P  45 watts

55. Gravitational attraction The gravitational attraction between two massive objects varies jointly with their masses and inversely with the square of the distance between them. What happens to this force if each mass is tripled and the distance between them is doubled?

Solution G

km1m2 d2

 

k  3m1  3m2 

 2d 

2

k  9m1m2

4d 2 9 km1m2   4 d2 The force is multiplied by

9 . 4

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689


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

56. Gravitational attraction In Problem 55, what happens to the force if one mass is doubled and the other tripled and the distance between them is halved?

Solution G

km1m2 d

2

k  2m1  3m2 

  d 2

2

k  6m1m2

 24 

d2 4

km1m2

d2 The force is multiplied by 24. 57. Plane geometry The area of an equilateral triangle varies directly with the square of the length of a side. Find the constant of proportionality.

Solution Consider this figure:

h  3 can be computed using the Pythagorean Theorem. 1 1 bh   2  3  3 2 2 A  ks 2

A

3  k  2

2

3  4k 3 k 4

58. Solid geometry The diagonal of a cube varies directly with the length of a side. Find the constant of proportionality.

Solution Consider this figure:

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690


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

The diagonal is obtained by repeatedly using the Pythagorean Theorem. d  ks 3  k  1 3k

Discovery and Writing 59. Explain the terms extremes and means. Solution Answers may vary. 60. Distinguish between a ratio and a proportion.

Solution Answers may vary. 61. What is k in a variation problem?

Solution Answers may vary. 62. Describe a strategy to solve a variation problem.

Solution Answers may vary. 63. Explain why

y  k indicates that y varies directly with x. x

Solution Answers may vary. 64. Explain why xy = k indicates that y varies inversely with x.

Solution Answers may vary. 65. Explain the term joint variation and give an example.

Solution Answers may vary. 66. As temperature increases on the Fahrenheit scale, it also increases on the Celsius scale. Is this direct variation? Explain.

Solution Answers may vary.

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691


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Critical Thinking In Exercises 67–70, match each variation sentence on the left with a variation equation on the right. 67. B varies inversely as the cube of r. 68. B varies jointly as the cube of r and the square of t. 69. B varies directly as the square of t and inversely as the cube of r. 70. B varies inversely as the cube of r and jointly as s and the square of t.

a.

B

kst 2 r3

b.

B

kt 2 r3

c.

B  kr 3t 2

d.

B

k r3

Solution 67. d 68. c 69. b 70. a

In Exercises 71–74, match each variation equation on the left with a variation sentence on the right. 71. M 

kn p4

72. M  knq2 p4

a. M varies directly as the fourth power of p and inversely as the square of q. b. M varies jointly as the square of q, the fourth power of p, and inversely as n.

73. M 

kp4 q2

c. M varies directly as n and inversely as the fourth power of p.

74. M 

kq2 p4 n

d. M varies jointly as n, the square of q, and the fourth power of p.

Solution 71. c 72. d 73. a 74. b

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692


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

CHAPTER REVIEW SOLUTIONS Exercises A relation is given. (a) State the domain and the range. (b) Determine if the relation is a function. 1.

{(3, 4), (4, 5), (5, 6), (6, 7)}

Solution

D  3, 4, 5, 6 ; R  4, 5, 6, 7

Each element of the domain is paired with only one element of the range. Function. 2. {(2, 4), (2, 5), (3, 6), (–4, 3)}

Solution

D  2, 3,  4 ; R  4, 5, 6, 3

2 is both paired with 4 and 5. Not a function.

Determine whether each equation defines y to be a function of x. Assume that all variables represent real numbers. 3.

y 3 Solution y 3 Each value of x is paired with only one value of y. function

4. y + 5x2 = 2

Solution

y  5x 2  2 y  5x 2  2 Each value of x is paired with only one value of y. function 5. y2 – x = 5

Solution

y2  x  5 y2  x  5 y 

x 5

Each value of x is paired with more than one value of y. not a function 6.

y  x x Solution y  x x Each value of x is paired with only one value of y. function

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693


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Find the domain of each function. Write each answer using interval notation. 7.

f  x   3x 2  5

Solution

f  x   y  3x 2  5

domain   ,  

8.

f x 

3x x 5

Solution

3x x 5 domain   , 5    5,   f x  y 

9.

f x 

3x 4 x  16 2

Solution

f x 

3x 4 x  16 2

3x

4  x  2 x  2

x  2, x  2

domain   ,  2   2, 2   2,  

 

10. f x 

x1

Solution

f x  y 

x1

domain   1,  

 

11. f x  5  x

Solution

f x  y  5  x

domain   , 5

 

12. f x 

x2  1

Solution

f x  y 

x2  1

x 2  1  0  domain   ,  

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694


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Evaluate the functions at the given values of the independent variable. 13. f  x   5 x  4 a.

f 4

b.

f  5 

Solution

f  x   5 x  4

a.

f  4   5  4   4  20  4  16

b.

f  5   5  5   4  25  4  29

14. f  x   x 2  2 x  9 a.

f  3

b.

f  2 

Solution

f  x   x 2  2x  9

a.

f  3   32  2  3   9  969  6

b.

f  2    2   2  2   9 2

 449  1

a.

3x x 9 f 4

b.

f  2 

 

15. f x 

2

Solution

f  x  a.

3x x 9 2

f  4 

3  4

42  9 12  7

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695


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

b.

f  2  

3  2

 2  9 2

6 6  5 5

16. f  x   x 2  6 x  1 a.

f x 

b.

f x2

c.

f  x  5

 

Solution

f  x   x 2  6x  1

a.

f x   x   6 x   1 2

 x 2  6x  1

b.

     6x   1

f x2  x2

2

2

 x4  6x2  1

c.

f  x  5   x  5  6  x  5  1 2

 x 2  10 x  25  6 x  30  1  x2  4x  4

Evaluate the difference quotient for each function f (x). 17. f(x) = 5x – 6

Solution f  x  h  f  x  h

5  x  h   6   5 x  6  5 x  5h  6   5 x  6            h h 5 x  5h  6  5 x  6 5h   5 h h

18. f(x) = 2x2 – 7x + 3

Solution f  x  h  f  x  h

 2 x  h 2  7 x  h  3   2 x 2  7 x  3            h 2 x 2  4 xh  2h2  7 x  7h  3  2 x 2  7 x  3      h 2 x 2  4 xh  2h2  7 x  7h  3  2 x 2  7 x  3  h 4 xh  2h2  7h h  4 x  2h  7     4 x  2h  7 h h

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696


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

19. Target heart rate The target heart rate f (x), in beats per minute, at which a person should train to get an effective workout is a function of the person’s age x in years. If f (x) = –0.6x + 132, find the target heart rate for a 45-year-old college professor.

Solution

f  x   0.6 x  132

f  45  0.6  45   132  105 20. Concessions A concessionaire at a basketball game pays a vendor $50 per game for selling hamburgers at $3.50 each. a. Write a function that describes the income I the vendor earns for the concessionaire during the game if the vendor sells x hamburgers. b. Find the income if the vendor sells 200 hamburgers.

Solution a.

I  h   3.5h  50

b.

I  200   3.5  200   50  $650

Refer to the illustration and find the coordinates of each point.

21. A

Solution

A  2, 0 

22. B

Solution

B  2, 1

23. C

Solution

C  0,  1

24. D

Solution D  3,  1

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697


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Graph each point. Indicate the quadrant in which the point lies or the axis on which it lies. 25-28 Solution

25.  3, 5 

Solution

 3, 5 : QII

26.  5,  3

Solution

5,  3 : QIV

27.  0,  7 

Solution

0,  7  : negative y -axis

 1  28.   , 0   2  Solution  1    , 0  : negative x-axis  2 

Solve each equation for y and graph the equation. 29. 2 x  y  6

Solution 2x  y  6

 y  2 x  6 y  2x  6 x

y

0

–6

2

–2

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698


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

30. 2x  5 y  10

Solution 2 x  5 y  10

5 y  2 x  10 y 

2 x 2 5

x

y

0

–2

–5

0

Use the x- and the y-intercepts to graph each equation. 31. 3 x  5 y  15

Solution 3x  5 y  15

3 x  5 y  15

3x  5  0  15

3  0   5 y  15

3x  15

5 y  15 y  3

x 5

5, 0

0,  3

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699


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

32. x  y  7

Solution xy 7

x 0  7 x7

 7, 0

xy 7 0 y  7 y 7

0, 7 

33. x  y  7

Solution x  y  7

x  0  7 x  7

 7, 0

x  y  7 0  y  7 y  7

0,  7 

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700


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

34. x  5 y  5

Solution x  5y  5

x  5  0  5 x 5

5, 0

x  5y  5 0  5y  5  5y  5 y  1

0,  1

Graph each equation. 35. y  4

Solution y  4  horizontal

36. x  2

Solution

x  2  vertical

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701


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

37. Depreciation A Ford Mustang purchased for $18,750 is expected to depreciate according to the formula f ( x )  2200 x  18,750, where f ( x ) is the value of the Mustang after x years. Find its value after 3 years.

Solution

Let x  3: y  2200 x  18, 750  2200  3   18, 750  6600  18, 750  $12, 150 38. House appreciation A house purchased for $250,000 is expected to appreciate according to the formula f ( x )  16,500 x  250,000, where f ( x ) is the value of the house after x years. Find the value of the house 5 years later.

Solution

Let x  5: y  16,500 x  250,000  16,500  5   250,000  82,500  250,000  $332,500

Find the length of the segment PQ. 39. P  3, 7  ; Q  3,  1

Solution

x  x    y  y    3  3    7   1    6    8 

d

2

2

1

2

2

1

2

2

2

2

 36  64 

100  10

40. P  8, 6  ; Q  12, 10 

Solution

x  x    y  y    12   8     10  6    4   4

d 

2

2

2

1

2

1

2

2

41. P

2

2

16  16  32  4 2

 3, 9 ; Q  3, 7 

Solution d 

x  x    y  y 

 3  3   9  7 

2

2

1

 02   2   04 

2

2

2

1

2

2

4 2

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702


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

42. P  a,  a  ; Q  a, a 

Solution

x  x    y  y    a   a     a  a    2a    2a 

d 

2

2

2

1

2

1

2

2

2

2

 4a2  4a2  8a2  2 2 a

Find the midpoint of the segment PQ. 43. P  3, 7  ; Q  3,  1

Solution

  3  3 7    1   x  x2 y 1  y 2  0 6   M  ,   M  0, 3  M  1 , ,   M    2 2 2 2 2 2    

44. P  0, 5  ; Q  12, 10 

Solution

 0   12  5  10   x  x2 y 1  y 2   12 15   15    M M  1 , , ,   M  6,    M    2 2 2 2 2 2 2        

45. P

 3, 9 ; Q  3, 7 

Solution

 3  3 97  2 3 16   x  x2 y 1  y 2  , , , m M  1   M   M   2 2  2 2  2   2  

 3, 8

46. P  a,  a  ; Q  a, a 

Solution

 a   a  a  a   x  x2 y 1  y 2  0 0   M  ,   M  0, 0  M  1 , ,   M   2 2 2  2 2  2   

Find the slope of the line PQ, if possible. 47. P  3,  5  ; Q  1, 7 

Solution m

y2  y1 x2  x 1

7   5  13

12  6 2

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703


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

48. P  2, 7  ; Q  5,  7 

Solution y  y 1 7  7 14 m 2   2 x2  x 1 5  2 7

 1 49. P 5,  8 ; Q  5,   2

Solution m

y2  y1 x2  x 1

1 2

  8  55

2  50. P  ,  8  ; Q 1,  8 3 

Solution m

y2  y1 x2  x 1

8 21 0

: und.

8    8  1 

2 3

0 0 1 23

51. P  b, a  ; Q  a, b 

Solution y  y1 b  a m 2   1 x2  x 1 ab 52. P  a  b, b  ; Q  b, b  a 

Solution

m

y2  y 1 x2  x 1

 b  a   b  a  1 b   a  b  a

Find two points on the line and find the slope of the line. 53. y  3x  6

Solution y  3x  6 x

y

0

6

1

9

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704


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

m

y2  y 1 x2  x 1

54. y  

96 10 3  3 1 

1 x 6 5

Solution

y 

1 x 6 5

x

y

0

–6

5

–7

m

y2  y 1 x2  x 1

 7   6 

50 1 1   5 5

Determine whether the slope of each line is 0 or undefined. 55.

Solution The slope is zero. 56.

Solution The slope is undefined.

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705


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Determine whether the slope of each line is positive or negative. 57.

Solution The slope is negative. 58.

Solution The slope is positive. Determine whether the lines with the given slopes are parallel, perpendicular, or neither. 59. m1  5; m2  

1 5

Solution  1 m1m2  5     1  5 perpendicular

60. m1 

2 7 ; m2  7 2

Solution m1  m2 ; m1m2  neither

2 7   1  1 7 2

61. A line passes through (–2, 5) and (6, 10). A line parallel to it passes through (2, 2) and (10, y). Find y.

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706


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution y  y1 10  5 5 m 2   x 2  x 1 6   2  8 m

y2  y 1 x2  x 1

y 2 5  10  2 8

8  y  2  5 8 8 y  16  40 8 y  56 y 7

62. A line passes through (–2, 5) and (6, 10). A line perpendicular to it passes through (–2, 5) and (x, –3). Find x.

Solution y  y1 10  5 5   m 2 x 2  x 1 6   2  8 m

y2  y 1 x2  x 1

3  5

x   2 

8 5

5  8   8  x  2  40  8 x  16 8 x  24 x3

63. Rate of descent If an airplane descends 3000 feet in 15 minutes, what is the average rate of descent in feet per minute?

Solution

m

y 3000   200 ft per minute x 15

64. Rate of growth A small business predicts sales according to a straight-line method. If sales were $50,000 in the first year and $147,500 in the third year, find the rate of growth in dollars per year (the slope of the line).

Solution

m

y 147,500  50,000 97,500    $48, 750 per year x 31 2

65. Find the average rate of change of f  x   2 x 2 from x1  2 to x2  5.

Solution f  x2   f  5   2  5   50 2

f  x1   f  2  2  2  8 2

f  x2   f  x 1  x2  x 1

50  8 42   14 52 3

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707


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

66. Find the average rate of change of f  x   x 3  1 from x1  1 to x2  3.

Solution

f  x2   f  3  33  1  26 f  x1   f  1   1  1  2 3

f  x2   f  x 1  x2  x 1

26   2 3   1

28 7 4

Use slope-intercept form to write an equation of each line. 67. The line has a slope of

2 and a y-intercept of 3. 3

Solution y  mx  b 2 y  x3 3 68. The slope is 

3 and the line passes through (0,  5). 2

Solution y  mx  b 3 y   x 5 2 Find the slope and the y-intercept of the graph of each line. 69. 3 x  2 y  10

Solution 3 x  2 y  10 2 y  3 x  10 3 y  x 5 2 3 m  ,  0,  5  2 70. 2 x  4 y  8

Solution 2 x  4 y  8 4 y  2 x  8 1 y   x 2 2 1 m   ,  0,  2  2

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708


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

71. 2 y  3 x  10

Solution 2 y  3 x  10

y  m

3 x 5 2

3 ,  0,  5  2

72. 2x  4 y  8

Solution 2 x  4 y  8 4 y  2 x  8 1 y   x 2 2 1 m   ,  0,  2  2 73. 5 x  2 y  7

Solution 5x  2 y  7

2 y  5 x  7 5 7 y  x 2 2   5 7 m   ,  0,  2  2 74. 3x  4 y  14

Solution 3 x  4 y  14

4 y  3 x  14 3 7 y  x 4 2 3  7 m  ,  0,   4  2 Use slope-intercept form to graph each equation. 75. y 

3 x 2 5

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709


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution 3 x 2 5 3 m  , b  2 5 y 

76. y  

4 x3 3

Solution 4 y   x 3 3 4 m ,b3 3

Determine whether the graphs of each pair of equations are parallel, perpendicular, or neither. 77. y  3 x  8, 2 y  6 x  19

Solution y  3x  8

m3

2 y  6 x  19 y  3x  m3

19 2

The lines are parallel.

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710


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

78. 2 x  3 y  6, 3 x  2 y  15

Solution 2x  3 y  6

3 x  2 y  15

3 y  2 x  6 2 y  3 x  15 3 15 2 y  x y   x2 2 2 3 3 2 m m 2 3 The lines are perpendicular.

Use point-slope form to write an equation of each line. Write the answer in standard form. 79. The line passes through the origin and the point (–5, 7).

Solution

m

y2  y1 x2  x 1

70 7  5 5  0

y  y 1  m  x  x1  7  x  0 5 7 y  x 5  7  5y  5 x   5  5 y  7 x 7x  5 y  0 y 0  

80. The line passes through (–2, 1) and has a slope of –4.

Solution

y  y 1  m  x  x1  y  1  4  x  2  y  1  4 x  8

4 x  y  7 81. The line passes through (2, –1) and has a slope of  51 .

Solution

y  y 1  m  x  x1 

1  x  2 5  1  5  y  1  5     x  2    5  y 1 

5 y  5    x  2

5 y  5  x  2 x  5 y  3

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711


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

82. The line passes through (7, –5) and (4, 1).

Solution m

y2  y 1 x2  x 1

1   5  47

6  2 3

y  y 1  m  x  x1  y  5  2  x  7  y  5  2 x  14 2x  y  9 Write an equation of each line. 83. The line has a slope of 0 and passes through (–5, 17).

Solution m  0  horizontal

y  17 84. The line has no defined slope and passes through (–5, 17).

Solution m is undefined  vertical

x  5 Write an equation of each line. Write the answer in slope-intercept form. 85. The line is parallel to 3 x  4 y  7 and passes through (2, 0).

Solution 3x  4 y  7 4 y  3 x  7 3 7 y  x 4 4

m

3 4

3 . 4 y  y 1  m  x  x1 

Use m 

3  x  2 4 3 3 y  x 4 2

y 0 

86. The line passes through (7, –2) and is parallel to the line segment joining (2, 4) and (4, –10).

Solution

m

y2  y 1 x2  x1

10  4  7 42

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712


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y  y 1  m  x  x1  y  2  7  x  7  y  2  7 x  49 y  7 x  47 87. The line passes through (0, 5) and is perpendicular to the line x  3 y  4.

Solution x  3y  4 3y  x  4 1 4 y  x 3 3

1 3 Use m  3. m

y  y 1  m  x  x1  y  5  3  x  0 y  5  3x y  3x  5

88. The line passes through (7, –2) and is perpendicular to the line segment joining (2, 4) and (4, –10).

Solution

m

y2  y 1 x2  x1

10  4  7 42

1 . 7 y  y 1  m  x  x1 

Use m 

1  x  7 7 1 y 2 x1 7 1 y  x 3 7 y 2

89. Billing for services Valeria’s Painting and Decorating Service charges a fixed amount for accepting a wallpapering job and adds a fixed dollar amount for each roll hung. If the company bills a customer $177 to hang 11 rolls and $294 to hang 20 rolls, find the cost to hang 27 rolls.

Solution Let x = the number of rolls hung and let y = the total charge. Then two points on the line are given: (11, 177) and (20, 294)

m

294  177 117   13 20  11 9

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713


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

y  y 1  m  x  x1 

y  177  13  x  11 y  177  13 x  143 y  13 x  34

Let x  27:

y  13  27   34  385. The charge is $385.

90. Paying for college Wang Lei must earn $5040 for next semester’s tuition. Assume he works x hours tutoring algebra at $14 per hour and y hours tutoring Spanish at $18 per hour and makes his goal. Write an equation expressing the relationship between x and y, and graph the equation. If Wang Lei tutors algebra for 180 hours, how long must he tutor Spanish?

Solution 14 x  18 y  5040 Let x  180:

14  180   18 y  5040

2520  18 y  5040 18 y  2520 y  140 140 hours of tutoring Spanish

Find the x- and y-intercepts of each graph. Do not graph the equation. 91. y  4 x  8 x 2

Solution y  4x  8x2

0  4 x  1  2x 

x  0, x 

1 2

1  x-int :  0, 0  ,  , 0  2 

y  4 x  8x 2 y  4 0   8 0 

2

y 0

y -int:  0, 0 

92. y  x 2  10 x  24

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714


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution y  x 2  10 x  24

y  x 2  10 x  24

x  12, x  2

y  24

0   x  12  x  2 

x-int :  12, 0  ,  2, 0 

y  02  10  0   24 y -int :  0,  24 

Find the symmetries, if any, of the graph of each equation. Do not graph the equation. 93. y 2  8 x

Solution y 2  8x y -axis

x-axis

origin

y  8 x 

 y   8x 2

 y   8 x  2

2

y 2  8x equivalent: symmetry

y 2  8 x not equivalent: no symmetry

y 2  8 x not equivalent: no symmetry

94. y  3 y 4  6

Solution y  3x4  6 x-axis

y -axis

origin

 y  3x  6

y  3 x   6

 y  3 x   6

y  3x4  6

 y  3x4  6 not equivalent: no symmetry

4

not equivalent: no symmetry

4

equivalent: symmetry

4

95. y  2 x

Solution y  2 x x-axis

y -axis

origin

 y  2 x

y  2  x

 y  2  x

y 2x

y  2 1 x

y  2 x

not equivalent: no symmetry

y  2 x

y  2 1 x

equivalent: symmetry

22x not equivalent: no symmetry

96. y  x  2

Solution y  x2 x-axis

y -axis

origin

y  x  2

y  x  2

 y  x  2

not equivalent: no symmetry

not equivalent: no symmetry

not equivalent: no symmetry

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715


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Graph each equation. Find all intercepts and symmetries. 97. y  x 2  2

Solution y  x2  2

x-int : none, y -int :  0, 2  symmetry: y -axis

98. y   x 2  9

Solution y  x2  9

x-int :   3, 0  , y -int :  0, 9  symmetry: y -axis

99. y  x 3  2

Solution y  x3  2 x-int :

 2, 0  , y-int : 0,  2 3

symmetry: none

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716


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

100. y 

x 2

Solution y 

x 2

x-int : none, y -int :  0, 2  symmetry: none

101. y   x  4

Solution y   x 4

x-int :  4, 0  , y -int : none symmetry: none

102. y 

1 x 2

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717


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution 1 y  x 2 x-int :  0, 0  , y -int :  0, 0  symmetry: y -axis

103. y  x  1  2

Solution y  x1 2

x-int : none, y -int :  0, 3 

symmetry: none

104. y  3 x  1

Solution y  3x 1

x-int :  1, 0  , y -int :  0,  1 symmetry: none

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718


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Use a graphing calculator to graph each equation. 105. y  x  4  2

Solution y  x 4 2

106. y   x  2  3

Solution

y   x23

107. y  x  2 x

Solution

y  x2 x

108. y 2  x  3

Solution y2  x  3 Graph y  

x  3.

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719


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Identify the center and radius of each circle given in standard form. 109. x 2  y 2  64

Solution x 2  y 2  64

 x  0   y  0  8 C :  0, 0  ; r  8 2

2

2

110. x 2   y  6   100 2

Solution x 2   y  6   100 2

 x  0    y  6   10 C:  0, 6  ; r  10 2

111.

2

2

 x  7   y  41 2

2

Solution

 x  7   y  41 2

2

 x   7     y  0    21  2

2

C:  7, 0  ; r 

112.

2

1 2

 x  5   y  1  9 2

2

Solution

 x  5    y  1  9  x  5    y    1   3 C:  5,  1 ; r  3 2

2

2

2

2

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720


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Write an equation of each circle in standard form. 113. Center at (0, 0); r  7

Solution

 x  0   y  0  7 2

2

2

x 2  y 2  49

114. Center at (3, 0); r 

1 5

Solution  

 x  3    y  0    51  2

2

 x  3

2

2

  1  y  25 2

115. Center at (  2, 12); r  5

Solution

 x   2     y  12   5 2

2

2

 x  2    y  12   25 2

2

2  116. Center at  , 5  ; r  9 7 

Solution 2

2  2 2  x     y  5  9 7  2

2  2  x     y  5   81 7  

Write an equation of each circle in standard form and general form. 117. Center at (–3, 4); radius 12

Solution

C  3, 4  ; r  12

 x  h   y  k   r  x  3    y  4   144 2

2

2

2

2

or x 2  y 2  6 x  8 y  119  0

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721


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

118. Ends of diameter at (–6, –3) and (5, 8)

Solution 6  5 1  2 2 3  8 5 y   2 2 r  distance from center to endpoint

Center: x 

2

2

 1  5     5    8   2  2  2

121 2

2

  1 5 121 , or x    y    2 2 2   x 2  y 2  x  5 y  54  0

Convert the general form of each circle given into standard form. 119. x 2  y 2  6 x  4 y  4  0

Solution x2  y 2  6x  4 y  4  0 x 2  6 x  y 2  4 y  4 2 x  6 x  9  y 2  4 y  4  4  9  4

 x  3   y  2  9 2

2

120. 2 x 2  2 y 2  8 x  16 y  10  0

Solution 2 x 2  2 y 2  8 x  16 y  10  0 x2  y 2  4x  8 y  5  0 x2  4x  y 2  8 y  5 x  4 x  4  y 2  8 y  16  5  4  16 2

 x  2    y  4   25 2

2

Graph each circle. 121. x 2  y 2  16  0

Solution x 2  y 2  16  0

 x  0    y  0   16 C  0, 0  , r  4 2

2

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722


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

122. x 2  y 2  4 x  5

Solution x2  y 2  4x  5 x2  4x  y 2  5 x2  4x  4  y 2  5  4

 x  2  y  9 C  2, 0  , r  3 2

2

123. x 2  y 2  2 y  15

Solution x 2  y 2  2 y  15 x 2  y 2  2 y  1  15  1 x 2   y  1  16 2

C  0, 1 , r  4

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723


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

124. x 2  y 2  4 x  2 y  4

Solution x2  y 2  4x  2 y  4 x2  4x  4  y 2  2 y  1  4  4  1

 x  2    y  1  9 C  2,  1 , r  3 2

2

Use a graphing calculator to solve each equation. If an answer is not exact, round to the nearest hundredth. 125. x 2  11  0

Solution Graph y  x 2  11. Find the x-intercepts. x  3.32, x  3.32

126. x 3  x  0

Solution Graph y  x 3  x . Find the x-intercepts. x  1, x  0, x  1

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724


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

127. x 2  2  1  0

Solution Graph y  x 2  2  1. Find the x-intercepts. x  1.73, x  1, x  1, x  1.73

128. x 2  3 x  5

Solution Graph y  x 2  3 x  5. Find the x-intercepts. x  1.19, x  4.19

Solve each proportion. 129.

x3 x1  10 x

Solution x3 x1  x 10 x  x  3   10  x  1 x 2  3 x  10 x  10 x  7 x  10  0 2

 x  5  x  2   0

x  5 or x  2

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725


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

130.

x1 12  2 x1

Solution x1 12  2 x1  x  1 x  1  2  12  x 2  1  24 x 2  25 x  5

131. Pizza party If 10 medium pizzas will feed 27 people, how many medium pizzas would be required to feed a 162-member marching band?

Solution Let x = the dosage needed. 250 x  110 176 250  176  110  x 44000  110 x 400  x

The dosage is 400 mg. 132. Hooke’s Law The force required to stretch a spring is directly proportional to the amount of stretch. If a 3-pound force stretches a spring 5 inches, what force would stretch the spring 3 inches?

Solution 3 s 5 3 f  3 5 9 f  pounds 5

f  ks

f 

3  k 5  3 k 5

133. Kinetic energy A moving body has a kinetic energy directly proportional to the square of its velocity. By what factor does the kinetic energy of an automobile increase if its speed increases from 30 mph to 50 mph?

Solution E  kv 2 30 mph E  k  30  E  900k

50 mph 2

E  k  50 

2

E  2500 k

Factor of increase 

2500k 25  900k 9

134. Gas laws The volume of gas in a balloon varies directly as the temperature and inversely as the pressure. If the volume is 400 cubic centimeters when the temperature is 300 K and the pressure is 25 dynes per square centimeter, find the volume when the temperature is 200 K and the pressure is 20 dynes per square centimeter.

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726


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution V  400 

kT P k  300 

V  V 

25 10000  300k

100 3

P 100 3

T

 200 

20 1000 V  3 1 V  333 cm3 3

100 k 3

135. Area The area of a rectangle varies jointly with its length and width. Find the constant of proportionality.

Solution A  klw A  1 lw  k  1

136. Electrical resistance The resistance of a wire varies directly as the length of the wire and inversely as the square of its diameter. A 1000-foot length of wire, 0.05 inch in diameter, has a resistance of 200 ohms. What would be the resistance of a 1500-foot length of wire that is 0.08 inch in diameter?

Solution R 200 

kL 2

D k  1000 

0.05 

2

1000k 0.0025 0.0005  k 200 

R V 

0.0005L D2 0.0005  1500 

0.08 

2

V  117 ohms

CHAPTER TEST SOLUTIONS Find the domain of each function. Write each answer using interval notation. 1.

f x 

3 2x  5

Solution 3 2x  5  5 5  domain   ,    ,   2 2   f x 

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727


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

2.

f  x  x  3 Solution

f  x 

x  3: domain  3,  

Find f(–1) and f(2). 3.

x x1

f x 

Solution 1 1 1    1  1 2 2 2 2 f 2   2 21 1

f   1 

4.

f  x  x  7 Solution f  1 

1  7  6

f  2  2  7  9  3

Find the difference quotient. 5.

f  x   x2  x  5

Solution f  x  h  f  x  h

 x  h 2  x  h  5   x 2  x  5          h  x 2  2 xh  h2  x  h  5   x 2  x  5      h x 2  2 xh  h2  x  h  5  x 2  x  5  h 2 h 2 x  h  1  2 xh  h  h    2x  h  1 h h

Indicate the quadrant in which the point lies or the axis on which it lies. 6.

 3,   Solution  3,    QII

7.

0,  8 Solution

0,  8  negative y -axis

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728


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Find the x- and y-intercepts and use them to graph the equation. 8.

x  3y  6 Solution x  3y  6

x  3y  6 0  3y  6

x  3 0   6

y 2

x 6

0, 2

6, 0 

9.

2 x  5 y  10 Solution 2 x  5 y  10

2 x  5 y  10

2 x  5  0   10

2  0   5 y  10

x 5

y  2

5, 0 

0,  2 

Graph each equation. 10. 2  x  y   3 x  5

Solution

2  x  y   3x  5 2x  2 y  3x  5 2y  x  5 y 

x 0 1

1 5 x 2 2

Y 5 2 3

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729


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

11. 3 x  5 y  3  x  5 

Solution

3x  5 y  3  x  5 3 x  5 y  3 x  15  5 y   15 y 3

12.

x

Y

0

3

–2

3

1 x  2y   y  1 2

Solution 1 x  2y   y  1 2 1 x y  y 1 2 x  2y  2y  2 4 y   x  2 1 1 y  x 4 2 x

Y

0

1 2

2

1

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730


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

13.

x  y 5  3x 7

Solution x  y 5  3x 7 x  y  5  21x y  20 x  5 x

Y

0

5

1 4

0

Find the distance between points P and Q. 14. P  1,  1 ; Q  3, 4 

Solution

x  x    y  y    1   3     1  4    4    5 

d 

2

2

1

2

2

2

16  25 

2

1

2

2

41

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731


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

15. P  0,   ; Q   , 0 

Solution

x  x    y  y    0          0  2

d 

2

1

2

2

1

2

2

2 2

2 2   2  4.44

Find the midpoint of the line segment PQ. 16. P  3,  7  ; Q  3, 7 

Solution

 3   3  7  7   x  x2 y 1  y 2  0 0   M  ,   M  0, 0  , , M  1   M    2 2 2 2 2 2    

  8, 18 

17. P 0, 2 ; Q

Solution 0  8 2 2 4 2  x  x2 y 1  y 2  2  18    M M M  1 , , ,   M    2 2 2 2 2 2      

 2, 2 2 

Find the slope of the line PQ. 18. P  3,  9  ; Q  5, 1

Solution m

19. P

y2  y1 x2  x1

1   9  5  3

10 5  8 4

 3, 3 ; Q   12, 0 

Solution m

y2  y1 x2  x 1

03  12  3

3 3 3

1 3

3 3

Determine whether the two lines are parallel, perpendicular, or neither. 20. y  3 x  2; y  2 x  3

Solution y  3x  2 m3

y  2x  3

m2 neither

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732


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

21. 2 x  3 y  5; 3 x  2 y  7

Solution 2x  3 y  5 3x  2 y  7 3 y  2 x  5 2 y  3 x  7 2 5 3 7 y  x y  x 3 3 2 2 3 2 m m 2 3 perpendicular Write an equation of the line with the given properties. Your answers should be written in slope-intercept form, if possible. 22. Passing through (3,–5); m = 2

Solution

y  y 1  m  x  x1  y  5  2  x  3 y  5  2x  6 y  2 x  11

23. m  3; b 

1 2

Solution y  mx  b y  3x 

1 2

24. Parallel to 2x – y = 3; b = 5

Solution 2x  y  3  y  2 x  3 y  2x  3 m2

y  2x  5

25. Perpendicular to 2x – y = 3; b = 5

Solution 2x  y  3  y  2 x  3 y  2x  3 m2

y 

1 x 5 2

  1 3 26. Passing through  2,   and  3,  2   2

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733


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution m

y2  y1 x2  x 1

   23 

1 2

32

4

 2 2 1

y  y 1  m  x  x1  1  2  x  3 2 1 y   2x  6 2 11 y  2x  2 y

27. Parallel to the y-axis and passing through (3, –4)

Solution If the line is parallel to the y-axis, then it is a vertical line: x = 3 Find the x- and y-intercepts of each graph. 28. y  x 3  16 x

Solution y  x 3  16 x

0  x x 2  16

y  x 3  16 x

y  03  16  0 

0  x  x  4  x  4 

y 0

y -int:  0, 0 

x  0, x  4, x  4

x-int:  0, 0  ,  4, 0  ,  4, 0 

29. y  x  4

Solution y  x4

y  x 4

0  x4

y  04

0  x4

y  4

4 x

y 4

x-int:  4, 0 

y -int:  0, 4 

Find the symmetries of each graph. 30. y 2  x  1

Solution x-axis

 y   x  1 2

y2  x  1 eequivalent: symmetry

y2  x  1 y -axis y 2  x  1 not equivalent: no symmetry

origin

 y   x  1 2

y 2  x  1 not equivalent: no symmetry

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734


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

31. y  x 4  1

Solution x-axis  y  x4  1 not equivalent: no symmetry

y  x4  1 y -axis

origin

y  x   1

 y  x   1

y  x4  1

 y  x4  1 not equivalent: no symmetry

4

equivalent: symmetry

4

Graph each equation. Find all intercepts and symmetries. 32. y  x 2  9

Solution y  x2  9

x-int:  3, 0  ,  3, 0  y -int:  0,  9 

symmetry: y -axis

33. x  y

Solution x y

x-int:  0, 0 

y -int:  0, 0 

symmetry: x-axis

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735


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

34. y  2 x

Solution y 2 x

x-int:  0, 0 

y -int:  0, 0 

symmetry: none

35. x  y 3

Solution x  y3

x-int:  0, 0 

y -int:  0, 0 

symmetry: origin

Write an equation of each circle in standard form. 36. Center at (5, 7); radius of 8

Solution

C  5, 7  ; r  8

 x  h   y  k   r  x  5    y  7   64 2

2

2

2

2

37. Center at (2, 4); passing through (6, 8)

Solution r  

2  6  4  8 2

2

32

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736


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

 x  h   y  k   r  x  2    y  4   32 2

2

2

2

2

Graph each equation. 38. x 2  y 2  9

Solution x2  y 2  9

C  0, 0  , r  3

39. x 2  4 x  y 2  3  0

Solution x2  4x  y 2  3  0 x 2  4 x  y 2  3 x 2  4 x  4  y 2  3  4

 x  2  y  1 C  2, 0  , r  1 2

2

Write each statement as an equation. 40. y varies directly as the square of z.

Solution y  kz 2 41. w varies jointly with r and the square of s.

Solution w  krs 2

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737


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

42. P varies directly with Q. P = 7 when Q = 2. Find P when Q = 5.

Solution P  kQ

7  k 2

7 k 2

7 Q 2 7 P  5 2 35 P 2 P

43. y is directly proportional to x and inversely proportional to the square of z, and y = 16 when x = 3 and z = 2. Find x when y = 2 and z = 3.

Solution y  16 

kx z2 k 3

2 3k 16  4 64 k 3

2

y  2

64 3

x

z

2

64 3

x 2

3 64 18  x 3 3 3 64  18   x 64 64 3 27  x 32

Use a graphing calculator to find the positive root of each equation. Round to two decimal places. 44. x 2  7  0

Solution Graph y  x 2  7. Find any positive x-intercept. x  2.65

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738


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

45. x 2  5 x  5  0

Solution Graph y  x 2  5 x  5. Find any plosive x-intercept. x  5.85

GROUP ACTIVITY SOLUTIONS Average Velocity Real-World Example of Slope Roller coaster fans love the thrill of riding a fast, long, and high roller coaster ride at an amusement park. Engineers who design them consider rider safety, environmental safety, and even factor in “excitement” as a technical variable in their design. Knowing the roller coaster’s speed, acceleration, height, length, and duration of the ride are all important.

Group Activity We have learned in this chapter that slope m is the change in y’s divided by the change in x’s. Similarly, average velocity vavg is defined as the change in distance s divided by the change in time t.

v avg 

s2  s1 t2  t1

Calculate the average velocity of some of the fastest roller coasters in the world, from the start of the ride to the end of the ride. Round to the nearest whole number. Recall that there are 5280 feet in one mile.

Formula Rossa Unit Arab Emirates

6790

Duration in Minutes 1:32

Kingda Ka U.S.A.

3118

0:28

Roller Coaster

Length in Feet

Average Velocity in Miles per Hour

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739


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Top Thriller U.S.A.

2800

0:30

Red Force Spain

2887

0.24

Do-Dodonpa Japan

4081

0:55

Steel Dragon 2000 Japan

8133

4:00

Fury 325 U.S.A.

6602

3:25

Tower of Terror 2 Australia

1235

0:28

To approximate the instantaneous velocity of a roller coaster at a precise point in time, calculus is required.

Solution Roller Coaster

Length in feet

Length in miles

Duration in minutes

Duration in hours

Avg Velocity in miles per hour

Formula Rossa UAE

6790

1.2860

1:32

0.0256

1.286  0  50 0.0256  0

Kingda Ka USA

3118

0.5905

0:28

0.0078

0.5905  0  76 0.0078  0

Top Thriller USA

2800

0.5303

0:30

0.0083

0.5303  0  64 0.0083  0

Red Force Spain

2887

0.5468

0:24

0.0067

0.5468  0  82 0.0067  0

Do-Dodonpa Japan

4081

0.7729

0:55

0.0153

0.7729  0  51 0.0153  0

Steel Dragon 2000

8133

1.5403

4:00

0.0667

1.5403  0  23 0.0667  0

Fury 325 USA

6602

1.2504

3:25

0.0569

1.2504  0  22 0.0501  0

Tower of Terror 2

1235

0.2339

0:28

0.0078

0.2339  0  30 0.0078  0

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

740


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER 3: FUNCTIONS

TABLE OF CONTENTS End of Section Exercise Solutions .................................................................................. 741 Exercises 3.1 ............................................................................................................................. 741 Exercises 3.2 ............................................................................................................................ 775 Exercises 3.3 ........................................................................................................................... 823 Exercises 3.4 ........................................................................................................................... 850 Exercises 3.5 ........................................................................................................................... 876 Chapter Review Solutions................................................................................................ 910 Chapter Test Solutions .................................................................................................... 933 Cumulative Review Solutions ..........................................................................................940 Group Activity Solutions .................................................................................................. 950

END OF SECTION EXERCISE SOLUTIONS EXERCISES 3.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Evaluate the function f ( x )  2 x  3 at the integers –2, –1, 0, 1, and 2. Solution f ( x )  2x  3 x

y

–2

2( 2)  3  7

–1

2( 1)  3  5

0

2(0)  3  3

1

2(1)  3  1

2

2(2)  3  1

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

741


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

2. Evaluate the function f ( x )   x 2  1 at the integers –2, –1, 0, 1, and 2. Solution f ( x)  x2  1 x

y

–2

( 2)2  1  5

–1

( 1)2  1  2

0

(0)2  1  1

1

(1)2  1  2

2

(2)2  1  5

3. Evaluate the function f ( x )  2 x 3  1 at the integers –2, –1, 0, 1, and 2. Solution f ( x)  2x 3  1 x

y

–2

2( 2)3  1  15

–1

2( 1)3  1  1

0

2(0)3  1  1

1

2(1)3  1  3

2

2(2)3  1  17

4. Evaluate the function f ( x )   x  4 at the integers –6, –5, –4, –3, and –2. Solution f ( x)   x  4 x

y

–6

 6  4  2

–5

 5  4  1

–4

 4  4  0

–3

 3  4  1

–2

 2  4  2

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742


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

5. Evaluate the function f ( x ) 

x  2 at the integers 2, 3, 6, 11, and 18.

Solution f ( x) 

x 2

x

y

2

22 0

3

32  1

6

62  2

11

11  2  3

18

18  2  4

6. Evaluate the function f ( x )  2 3 x  1 at the integers –8, –1, 0, 1, and 8.

Solution f ( x )  2 3 x  1 x

y

–8

2 3 8  1  5

–1

2 3 1  1  3

0

2 3 8  1  1

1

2 3 1  1  1

8

2 3 8  1  3

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The graph of a function y = f (x) in the xy-plane is the set of all points __________ that satisfy the equation, where x is in the __________ of f and y is in the __________ of f.

Solution (x, y), domain, range 8. If every __________ line that intersects a graph does so __________, the graph represents a function.

Solution vertical, once 9. We call f(x) = x the __________ function because it pairs each real number with itself.

Solution identity

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743


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

10. We call f(x) = x2 the __________ function because it pairs each real number with its square.

Solution squaring 11. We call f(x) = x3 the __________ function because it pairs each real number with its cube.

Solution cubing 12. We call f(x) = |x| the __________ function because it pairs each real number with its absolute value.

Solution absolute value 13. We call f ( x )  x the __________ function because it pairs each real number with its principal square root.

Solution square root 14. We call f ( x )  3 x the __________ function because it pairs each real number with its cube root.

Solution cube root Practice Graph each function. Use the graph to identify the domain and range of each function. 15. f ( x )  2 x  3

Solution f ( x )  2x  3

domain = ( , ) range = ( , )

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744


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

16. f ( x )  2 x  4

Solution f ( x )  2 x  4

domain = ( , ) range = ( , ) 17. f ( x )  

3 x4 4

Solution 3 f ( x)   x  4 4

domain = ( , ) range = ( , ) 18. f ( x ) 

1 x 3 2

Solution 1 f ( x)  x  3 2

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745


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = ( , ) range = ( , ) 19. f ( x )  x 2  4

Solution f ( x)  x2  4

domain = ( , ) range = [4, ) 20. f ( x )   x 2  3

Solution f ( x)   x2  3

domain = ( , ) range = ( , 3]

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746


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

21. f ( x )  

1 2 x 5 2

Solution 1 f ( x)   x2  5 2

domain = ( , ) range = ( , 5] 22. f ( x ) 

1 2 x 2 3

Solution 1 f ( x)  x2  2 3

domain = ( , ) range = [2, ) 23. f ( x )  3( x  2)2

Solution f ( x )  3( x  2)2

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747


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = ( , ) range = [0, ) 24. f ( x )   x 2  2 x  1

Solution f ( x )   x 2  2x  1

domain = ( , ) range = [, 0) 25. f ( x )  x 3  2

Solution f ( x)  x3  2

domain = ( , ) range = ( , )

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748


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

26. f ( x )   x 3  2

Solution f ( x)   x3  2

domain = ( , ) range = ( , ) 27. f ( x )   x 3  1

Solution f ( x)  x3  1

domain = ( , ) range = ( , ) 28. f ( x ) 

1 3 x 1 4

Solution 1 f ( x)  x3  1 4

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749


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = ( , ) range = ( , ) 29. f ( x )  

1 3 x 4 2

Solution 1 f ( x)   x3  4 2

domain = ( , ) range = ( , ) 30. f ( x )  ( x  1)3

Solution f ( x )  ( x  1)3

domain = ( , ) range = ( , ) 31. f ( x )   x

Solution f ( x)   x

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750


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = ( , ) range = ( , 0] 32. f ( x )   x  3

Solution f ( x)   x  3

domain = ( , ) range = ( ,  3] 33. f ( x )  x  2

Solution f ( x)  x  2

domain = ( , ) range = [0, )

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751


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

34. f ( x )   x  2

Solution f ( x)   x  2

domain = ( , ) range = ( , 0] 35. f ( x ) 

1 x3 2

Solution f ( x) 

1 x3 2

domain = ( , ) range = [0, ) 36. f ( x )  

1 x3 2

Solution f ( x)  

1 x3 2

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752


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = ( , ) range = ( , 0] 37. f ( x )  4 x  1

Solution f ( x)  4 x  1

domain = ( , ) range = [1, ) 38. f ( x ) 

1 x 2 4

Solution 1 f ( x)  x  2 4

domain = ( , ) range = [2, )

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753


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

39. f ( x ) 

x 2

Solution f ( x) 

x 2

domain = [0, ) range = [2, ) 40. f ( x )   x  1

Solution f ( x)   x  1

domain = [1, ) range = ( , 0] 41. f ( x )  2 x  3

Solution f ( x)  2 x  3

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754


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = [0, ) range = [3, ) 42. f ( x )  

1 x 4 2

Solution 1 f ( x)   x 4 2

domain = [0, ) range = ( ,  4] 43. f ( x )  2 x  4

Solution f ( x )  2x  4

domain = [2, ) range = [0, ) 44. f ( x )   2 x  4

Solution f ( x)   2x  4

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755


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = [2, ) range = ( , 0] 45. f ( x )  3 x  2

Solution f ( x)  3 x  2

domain = ( , ) range = ( , ) 46. f ( x )   3 x  1

Solution f ( x)   3 x  1

domain = ( , ) range = ( , )

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756


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

47. f ( x )  3 3 x

Solution f ( x)  33 x

domain = ( , ) range = ( , ) 48. f ( x )  2 3 x  5

Solution f ( x )  2 3 x  5

domain = ( , ) range = ( , ) 49. f ( x )  2 3 x  1

Solution f ( x )  2 3 x  1

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757


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain = ( , ) range = ( , ) 50. f ( x )  3 x  1  7

Solution f ( x)  3 x  1  7

domain = ( , ) range = ( , ) 51. f ( x ) 

1 x

Solution 1 f ( x)  x

Domain: ( , 0)  (0, ) Range: ( , 0)  (0, )

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758


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

52. f ( x )  

1 x

Solution 1 f ( x)   x

Domain: ( , 0)  (0, ) Range: ( , 0)  (0, )

Use the Vertical Line Test to determine whether each graph represents a function. 53.

Solution function 54.

Solution not a function 55.

Solution not a function

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759


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

56.

Solution function 57.

Solution function 58.

Solution not a function 59.

Solution not a function 60.

Solution not a function

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760


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

61.

Solution function 62.

Solution function 63.

Solution function 64.

Solution function Use the graph of the function f shown to determine each of the following.

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761


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

65. f (11)

Solution The point ( 11, 2) is on the graph, so f ( 11)  2. 66. f ( 3)

Solution The point ( 3, 4) is on the graph, so f ( 3)  4. 67. f (2)

Solution The point (2, 0) is on the graph, so f (2)  0. 68. f (10)

Solution The point (10, 8) is on the graph, so f (10)  8. 69. x-intercepts

Solution The x-intercepts have y-coordinates of 0: (2, 0) and (8, 0) 70. y-intercept

Solution The y-intercept has an x-coordinate of 0: (0, 2) 71. an x-value for which f ( x )  6

Solution The point ( 6, 6) is on the graph, so f ( x )  6 when x  6. 72. the x-value for which f ( x )  4

Solution The point (5,  4) is on the graph, so f ( x )  4 when x  5. Use the graph of the function f shown to determine each of the following.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

762


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

73. f ( 3)

Solution The point (3,  25) is on the graph, so f ( 3)  25. 74. f (3)

Solution The point (3,  25) is on the graph, so f (3)  25. 75. x-intercepts

Solution The x-intercepts have y-coordinates of 0: ( 2, 0) and (2, 0) 76. y-intercept

Solution The y-intercept has an x-coordinate of 0: (0,  16) 77. an x-value for which f ( x )  16

Solution The point (0,  16) is on the graph, so f ( x )  16 when x  0. 78. the x-values for which f ( x )  25

Solution The point (3,  25) and (3, 25) are on the graph, so f ( x )  25 when x  3 and 3. Use the graph of the function f shown to determine each of the following.

79. f ( 6)

Solution The point ( 6, 2) is on the graph, so f ( 6)  2. 80. f ( 2)

Solution The point  2,  2  is on the graph, so f  2   2.

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763


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

81. f ( 1)

Solution The point  1,  3  is on the graph, so f  1  3.  10  82. f    3

Solution The point

 , 4 is on the graph, so f    4. 10 3

10 3

83. x-intercept

Solution The x-intercepts has a y-coordinates of 0: ( 4, 0) 84. y-intercept

Solution The y-intercept has an x-coordinate of 0: (0, 4) 85. the x-value for which f ( x )  1

Solution The point ( 5, 1) is on the graph, so f ( x )  1 when x  5. 86. the x-values for which f ( x )  4

Solution All points with a y-coordinate of 4 have x  1, so (  1,  ). Use the graph to determine each function’s domain and range. 87.

Solution Domain: (–∞, ∞), Range: (–∞, ∞) 88.

Solution Domain: (–∞, ∞), Range: (–∞, ∞)

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764


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

89.

Solution Domain: (–∞, ∞), Range: [–4, ∞) 90.

Solution Domain: (–∞, ∞), Range: (–∞, 9] 91.

Solution Domain: (–∞, ∞), Range: (–∞, ∞) 92.

Solution Domain: (–∞, ∞), Range: (–∞, ∞)

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765


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

93.

Solution Domain: (–∞, ∞), Range: (–∞, 1] 94.

Solution Domain: (–∞, ∞), Range: [0, ∞) 95.

Solution Domain: [–2, 1), Range: (–3, 3] 96.

Solution Domain: (–3, 2], Range: [–1, 3)

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766


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

97.

Solution Domain: (  , 0)  (0,  ), Range: (  , 0)  (0,  ) 98.

Solution Domain: (–∞, ∞), Range: (–∞, ∞) 99.

Solution Domain: (–∞, 0], Range: [2, ∞) 100.

Solution Domain: [0, ∞), Range: (–∞, 1] 101.

Solution Domain: (–∞, ∞), Range: {–2, 2}

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767


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

102.

Solution Domain: (–∞, ∞), Range: (–∞, ∞) Use a graphing calculator to graph each function. Then determine the domain and range of the function. 103. f ( x )  3 x  2

Solution f ( x )  3x  2

domain: (–∞, ∞); range: [0, ∞) 104. f ( x )  2 x  5

Solution f ( x )  2x  5

domain:  5 ,  ; range: [0, ∞) 2 105. f ( x )  3 5 x  1

Solution f ( x )  3 5x  1

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768


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

domain: (–∞, ∞); range: (–∞, ∞) 106. f ( x )   3 3 x  2

Solution f ( x)   3 3x  2

domain: (–∞, ∞); range: (–∞, ∞)

Fix It In exercises 107 and 108, identify the step the first error is made and fix it. 107. Given the f (x) = –3|x| + 4. Graph the function by completing a table of five values and plotting the points. Then determine the function’s domain and range.

Solution Step 4 was incorrect Step 1: x

y

−2

−2

−1

1

0

4

1

1

2

−2

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

769


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Step 2:

Step 3: domain: ( , ) Step 4: range: ( , 4] 108. Given the f ( x )  2 x  5 . Graph the function by completing a table of four values and plotting the points. Then determine the function’s domain and range.

Solution Step 2 was incorrect. Step 1: x

y

−5

0

−4

−2

−1

−4

4

−6

11

−8

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

770


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Step 2:

Step 3: domain: [5, ) Step 4: range: ( , 0]

Applications 109. Rain in Dallas, Texas The graph shows the average number of inches of rain, per month, in Dallas, Texas, for the months of May through October.

Use the graph to approximate the following. a. Domain and range b. Identify the average number of inches of rain in May. c. Identify the average number of inches of rain in June. d. What month is the average number of inches of rain 2?

Solution a. domain: [1, 6] ; range: [2, 5] b. The point (1, 5) is on the graph ⇒ 5 in. c. The point (2, 4) is on the graph ⇒ 4 in. d. The point (4, 2) is on the graph ⇒ August

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771


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

110. Height of terrain The graph of the function displays the height in yards of a terrain traveled by a motorcyclist at specific mile markers along a high-way. Use the graph to find each of the following.

a. b. c. d. e. f.

Determine the domain and range. Determine the height at mile marker 2. Determine the height at mile marker 8. At what mile markers is the terrain flat? Determine the y-intercept. Between mile markers 4 and 12, how many times is the height of the terrain 200 yards?

Solution a. domain: [0, 16] ; range: [0, 900] b. The point (2, 300) is on the graph ⇒ 300 yd c. The point (8, 500) is on the graph ⇒ 500 yd d. The points (4, 0) , (12, 0) and (16, 0) are on the graph ⇒ mile markers 4, 12, and 16 e. The y-intercept has an x-coordinate of 0: (0, 900) f.

The graph crosses the y-coordinate of 200 yd twice between x = 4 and x = 12.

Discovery and Writing 111. Describe a strategy for sketching the graph of a function.

Solution Answers may vary. 112. Describe how to use the Vertical Line Test to determine whether a graph represents a function.

Solution Answers may vary. 113. Explain how to determine the domain and range of a function’s graph.

Solution Answers may vary.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

772


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

114. Draw a graph that has the following characteristics:

    

Domain: [–5, 10] Range: [–10, 10] x-intercepts: (–5, 0) and (5, 0) y-intercept: (0, –10) passes through (–2, 3) and (10, 10)

Solution Answers may vary. 115. Use a graphing calculator to graph the function f ( x )  to find

x , and use TRACE and ZOOM

5 to three decimal places.

Solution

5  2.236 116. Use a graphing calculator to graph the function f ( x )  3 x and use the TRACE and ZOOM features to find 3 2 to three decimal places.

Solution 3

2  1.260

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 117. Functions always pass the Horizontal Line Test.

Solution False. Functions always pass the Vertical Line Test. 118. A function can have at most one y-intercept.

Solution True. 119. If the domain of a function is all real numbers, then the range is all real numbers.

Solution False. The range does not have to be the same as the domain. 120. The domain of the square root function and cube root function is all real numbers.

Solution False. The domain of the square root function is [0, ).

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

773


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Match the graphs of the functions on the left with the graph of the function on the right so that both have identical domains and ranges. 121.

a.

122.

b.

123.

c.

124.

d.

Solution 121. b 122. d 123. a 124. c

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774


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

EXERCISES 3.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Use f(x) and write the function whose equation is represented by each graph. a.

b.

c.

d.

e.

Solution a.

f ( x)  3 x

b.

f ( x)  x

c.

f ( x)  x2

d.

f ( x) 

e.

f ( x)  x3

x

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775


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

2. Graph f(x) = x2 and g(x) = x2 + 2 on the same coordinate axes by plotting points. Describe how the graph of g(x) compares to the graph of f(x).

Solution It is the graph of f ( x ) translated 2 units upward.

3. Graph f(x) = x3 and g(x) = (x – 2)3 on the same coordinate axes by plotting points. Describe how the graph of g(x) compares to the graph of f(x).

Solution It is the graph of f ( x ) translated two units right.

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776


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

4. Use plotting points to graph f(x) = |x| and g(x) = 2|x| on the same coordinate axes. Describe how the graph of g(x) compares to the graph of f(x).

Solution It is the graph of f ( x ) vertically stretched by a factor of 2.

5. Graph f ( x )  3 x and g( x )   3 x on the same coordinate axes by plotting points. Describe how the graph of g(x) compares to the graph of f(x).

Solution It is the graph of f ( x ) reflected about the x-axis.

6. Graph f ( x )  x and g( x )   x on the same coordinate axes by plotting points. Describe how the graph of g(x) compares to the graph of f(x).

Solution It is the graph of f ( x ) reflected about the y-axis.

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

777


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Fill in the blanks. 7. The graph of y  f ( x )  5 is identical to the graph of y  f ( x ) except that it is translated 5 units __________.

Solution upward 8. The graph of __________ is identical to the graph of y  f ( x ) except that it is translated 7 units downward.

Solution f ( x)  7 9. The graph of y  f ( x  3) is identical to the graph of y  f ( x ) except that it is translated 3 units __________.

Solution to the right 10. The graph of y  f ( x  2) is identical to the graph of y  f ( x ) except that it is translated 2 units __________.

Solution to the left 11. To draw the graph of y  ( x  2)2  3 , translate the graph of y  x 2 _____ units to the left and 3 units __________.

Solution 2, downward 12. To draw the graph of y  ( x  3)3  1 , translate the graph of y  x 3 3 units to the _______ and 1 unit _________.

Solution right, upward 13. The graph of y  f (  x ) is a reflection of the graph of y  f ( x ) about the __________

Solution y-axis 14. The graph of __________ is a reflection of the graph of y  f ( x ) about the x-axis.

Solution y  f ( x ) 15. The graph of y = f(4x) shrinks the graph of y  f ( x ) __________ by multiplying each

x-value of f ( x ) by 1 . 4

Solution horizontally

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

778


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

16. The graph of y  8f ( x ) stretches the graph of y  f ( x ) __________ by a factor of 8.

Solution vertically

Practice The graph of each function is a translation of the graph of f ( x )  x 2 . Graph each function. 17. g( x )  x 2  2

Solution

g( x )  x 2  2 Shift f ( x )  x 2 D 2

18. g( x )  ( x  2)2

Solution

g( x )  ( x  2)2 Shift f ( x )  x 2 R 2

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

779


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

19. g( x )  ( x  3)2

Solution g( x )  ( x  3)2 Shift f ( x )  x 2 L 3

20. g( x )  x 2  3

Solution g( x )  x 2  3 Shift f ( x )  x 2 U3

21. h( x )  ( x  1)2  2

Solution h( x )  ( x  1)2  2 Shift f ( x )  x 2 U2, L 1

22. h( x )  ( x  3)2  1

Solution h( x )  ( x  3)2  1

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780


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x )  x 2 D 1, R 3

2

 1 1 23. h( x )   x    2 2  Solution 2

 1 1 h( x )   x    2 2  Shift f ( x )  x 2 D 1 , L 1 2

2

2

 3 5 24. h( x )   x    2 2  Solution 2

 3 5 h( x )   x    2 2   Shift f ( x )  x 2 U 5 , R 3 2

2

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781


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

The graph of each function is a translation of the graph of f ( x )  x 3 . Graph each function. 25. g( x )  x 3  1

Solution g( x )  x 3  1 Shift f ( x )  x 3 U 1

26. g( x )  x 3  3

Solution g( x )  x 3  3 Shift f ( x )  x 3 D 3

27. g( x )  ( x  2)3

Solution g( x )  ( x  2)3 Shift f ( x )  x 3 R 2

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782


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

28. g( x )  ( x  3)3

Solution g( x )  ( x  3)3 Shift f ( x )  x 3 L 3

29. h( x )  ( x  2)3  3

Solution h( x )  ( x  2)3  3 Shift f ( x )  x 3 D 3, R 2

30. h( x )  ( x  1)3  4

Solution h( x )  ( x  1)3  4 Shift f ( x )  x 3 U4, L 1

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783


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

The graph of each function is a translation of the graph of f ( x )  x 3 . Graph each function. 31. h( x )  ( x  5)3  1

Solution h( x )  ( x  5)3  1 Shift f ( x )  x 3 U 1, R 5

32. h( x )  ( x  4)3  3

Solution h( x )  ( x  4)3  3 Shift f ( x )  x 3 D 3, L 4

The graph of each function is a translation of the graph of f ( x )  x . Graph each function. 33. g( x )  x  2

Solution g( x )  x  2

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784


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x )  x U2

34. g( x )  x  2

Solution g( x )  x  2 Shift f ( x )  x D 2

35. g( x )  x  5

Solution g( x )  x  5 Shift f ( x )  x R 5

36. g( x )  x  4

Solution g( x )  x  4

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785


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x )  x L 4

37. h( x )  x  2  1

Solution h( x )  x  2  1 Shift h( x )  x D 1, L 2

38. h( x )  x  3  3

Solution h( x )  x  3  3 Shift f ( x )  x U3, R 3

39. h( x )  x  6  3

Solution h( x )  x  6  3

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786


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x )  x D 3, R 6

40. h( x )  x  4  5

Solution h( x )  x  4  5 Shift f ( x )  x U 5, L 4

The graph of each function is a translation of the graph of f ( x )  41. g( x ) 

x . Graph each function.

x 1

Solution g( x ) 

x 1

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787


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x ) 

42. g( x ) 

x U1

x 3

Solution g( x ) 

x 3

Shift f ( x ) 

43. g( x ) 

x D3

x2

Solution g( x ) 

x2

Shift f ( x ) 

44. g( x ) 

x L2

x 4

Solution g( x ) 

x 4

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788


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x ) 

45. h( x ) 

x R4

x 2 1

Solution

h( x ) 

x 2 1

Shift f ( x ) 

46. h( x ) 

x D 1, R 2

x23

Solution

h( x ) 

x23

Shift f ( x ) 

47. h( x ) 

x U3, L 2

x 5 2

Solution

h( x ) 

x 5 2

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789


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x ) 

48. h( x ) 

x U 2, R 5

x 6 2

Solution

h( x ) 

x 6 2

Shift f ( x ) 

x D 2, L 6

The graph of each function is a translation of the graph of f ( x )  3 x . Graph each function. 49. g( x )  3 x  4 Solution g( x )  3 x  4

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790


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x )  3 x D 4

50. g( x )  3 x  3 Solution g( x )  3 x  3

Shift f ( x )  3 x U3

51. g( x )  3 x  2 Solution g( x )  3 x  2

Shift f ( x )  3 x R 2

52. g( x )  3 x  5 Solution g( x )  3 x  5

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791


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x )  3 x L 5

53. h( x )  3 x  1  1 Solution

h( x )  3 x  1  1 Shift f ( x )  3 x D 1, L 1

54. h( x )  3 x  1  1 Solution

h( x )  3 x  1  1 Shift f ( x )  3 x D 1, R 1

55. h( x )  3 x  3  4 Solution

h( x )  3 x  3  4

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792


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift f ( x )  3 x U 4, R 3

56. h( x )  3 x  1  7 Solution

h( x )  3 x  1  7 Shift f ( x )  3 x U 7, L 1

The graph of each function is a reflection of the graph of y  x 2 , y  x 3 , y  x , y

x , or y  3 x . Graph each function.

57. f ( x )   x 2 Solution f ( x)  x2

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793


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Reflect y  x 2 about x

58. g( x )  (  x )2 Solution g( x )  (  x )2 Reflect y  x 2 about y

59. h( x )   x 3 Solution h( x )   x 3 Reflect y  x 3 about x

60. g( x )  (  x )3 Solution g( x )  (  x )3

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794


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Reflect y  x 3 about y

61. f ( x )   x Solution f ( x)   x Reflect y  x about x

62. f ( x )   x Solution f ( x)  x Reflect y  x about y

63. f ( x )   x Solution

f ( x)   x

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795


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Reflect y 

64. f ( x ) 

x about x

x

Solution

f ( x) 

x

Reflect y 

x about y

65. f ( x )   3 x Solution

f ( x)   3 x Reflect y  3 x about x

66. g( x )  3  x Solution g( x )  3  x

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796


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Reflect y  3 x about y

The graph of each function is a horizontal stretching or shrinking of the graph of y  x2, y  x3, y  x , y 

x , or y  3 x . Graph each function.

67. f ( x )  2 x 2 Solution f ( x )  2 x 2 : Stretch

y  x 2 vert. by a factor of 2

68. g( x ) 

1 2 x 2

Solution 1 g( x )  x 2 : Shrink 2

y  x 2 vert. by a factor of 1

2

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797


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

69. f ( x ) 

1 3 x 2

Solution f ( x) 

1 3 x : Shrink 2

y  x 3 vert. by a factor of 1

2

70. g( x )  2 x 3 Solution

g( x )  2 x 3 : Stretch y  x 3 vert. by a factor of 2

71. h( x )  4 x Solution h( x )  4 x

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798


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Stretch y  x vert. by a factor of 4.

72. f ( x ) 

1 x 3

Solution 1 f ( x )  x : Shrink 3 y  x vert. by a factor of 1

3

73. f ( x )  3 x Solution

f ( x )  3 x : Stretch y 

x vert. by a factor of 3

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799


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

74. f ( x ) 

1 x 4

Solution 1 f ( x)  x : Shrink 4 y 

x vert. by a factor of 1

75. f ( x ) 

4

13 x 2

Solution 1 f ( x )  3 x : Shrink 2 y  3 x vert. by a factor of 1

2

76. f ( x )  4 3 x Solution

f ( x )  4 3 x : Stretch

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800


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

y  3 x vert. by a factor of 4

The graph of each function is a horizontal stretching or shrinking of the graph of y  x2, y  x3, y  x , y 

x , or y  3 x . Graph each function.

77. f ( x )  (2 x )2 Solution f ( x )  (2 x )2 : Shrink

y  x 2 hor. by a factor of 2

1  78. g( x )   x  4 

2

Solution

1  g( x )   x  4 

2

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801


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shrink y  x 2 vert. by a factor of

1  79. f ( x )   x  2 

1 4

3

Solution

1  f ( x)   x  2 

3

Shrink y  x 3 vert. by a factor of

1 2

80. f ( x )  (2 x )3 Solution f ( x )  (2 x )3

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802


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Stretch y  x 3 vert. by a factor of 2.

81. f ( x ) 

1 x 5

Solution

f ( x) 

1 x 5

Stretch y  x hor. by a factor of 5.

82. f ( x )  3 x Solution f ( x)  3x

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803


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shrink y  x hor. by a factor of 1

3

83. g( x )  6 x Solution g( x )  6 x

Shrink y 

84. g( x ) 

x hor. by a factor of 1

6

1 x 3

Solution g( x ) 

1 x 3

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804


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Stretch y 

85. f ( x )  3

x hor. by a factor of 3.

1 x 5

Solution f ( x)  3

1 x 5

Stretch y  3 x hor. by a factor of 5.

86. f ( x )  3 4 x Solution

f ( x)  3 4x

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805


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shrink y  3 x hor. by a factor of 1 . 4

Graph each function using a combination of transformations applied to the graph of a basic function. 87. g( x )  3( x  2)2  1 Solution g( x )  3( x  2)2  1

Start with y  x 2 Shift L 2, Stretch vert. by a factor of 3, Shift D 1

1 88. g( x )   ( x  1)2  1 3 Solution

1 g( x )   ( x  1)2  1 3

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806


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Start with y  x 2 , Shift L 1, Shrink vert. by a factor of 1 , Reflect x, Shift U 1 3

89. h( x )  2 x  3 Solution h( x )  2 x  3

Start with y  x Stretch vert. by a factor of 2; Reflect x; Shift U 3

90. f ( x )  2 x  3 Solution f ( x )  2 x  3

Start with y  x Shift L 3, Stretch vert. by a factor of 2, Reflect x

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807


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

91. f ( x )  2 x  2  1 Solution f ( x)  2 x  2  1

Start with y  x Shift R 2, Stretch vert. by a factor of 2, Shift U 1

92. f ( x )  3 x  5  2 Solution f ( x )  3 x  5  2

Start with y  x Shift L 5, Stretch vert. by a factor of 3, Reflect x, Shift D 2

93. f ( x )  2 x  3 ( x  0) Solution

f ( x)  2 x  3 Start with y 

x

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808


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Stretch vert. by a factor of 2, Shift U 3

94. g( x )  2 x  3 ( x  3) Solution g( x )  2 x  3

Start with y 

x

Shift L 3, Stretch vert. by a factor of 2

95. h( x )  2 x  2  1 ( x  2) Solution

h( x )  2 x  2  1 Start with y 

x

Shift R 2, Stretch vert. by a factor of 2, Shift U 1

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809


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

1 x 5 2 2 ( x  5)

96. h( x ) 

Solution 1 h( x )  x 5 2 2

Start with y 

x

Shift L 5, Shrink vert. by a factor of 1 , Shift D 2 2

97. g( x )  2( x  2)3  1 Solution g( x )  2( x  2)3  1

Start with y  x 3 Shift L 2, Stretch vert. by a factor of 2, Reflect x, Shift D 1

98. g( x ) 

1 ( x  1)3  1 3

Solution 1 g( x )  ( x  1)3  1 3

Start with y  x 3

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810


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Shift L 1, Shrink vert. by a factor of 1 , Shift D 1 2

99. f ( x )  2 3 x  4 Solution

f ( x)  23 x  4 Start with y  3 x Stretch vert. by a factor of 2, Shift U 4

100. f ( x )  2 3 x  1 Solution

f ( x )  2 3 x  1 Start with y  3 x Shift L 1, Stretch vert. by a factor of 2, Reflect x

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811


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Use the following graph and transformations to graph g( x ). Sketch the graph of each function.

101. y  f ( x )  1 Solution Shift y  f ( x ) U 1

102. y  f ( x  1) Solution Shift y  f ( x ) L 1

103. y  2f ( x ) Solution Stretch y  f ( x ) vert. by a factor of 2

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812


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x 104. y  f   2 Solution Stretch y  f ( x ) hor. by a factor of 2

105. y  f ( x  2)  1 Solution Shift y  f ( x ) U 1, R 2

106. y  f ( x ) Solution Reflect y  f ( x ) about x

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813


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

107. y  2f (  x ) Solution Stretch y  f ( x ) vert. by a factor of 2, reflect about y

108. y  f ( x  1)  2 Solution Shift y  f ( x ) D 2, L 1

The figure shows the graph of f(x). Use the given graph and transformations to graph each function.

109. y  f ( x )  2 Solution Shift y  f ( x ) U2

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814


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

110. y  f ( x )  2 Solution Shift y  f ( x ) D 2

111.

y  f ( x  2) Solution Shift y  f ( x ) L 2

112. y  f ( x  2) Solution Shift y  f ( x ) R 2

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815


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

113. y  f ( x  4)  2 Solution Shift y  f ( x ) D 2, R 4

114. y  f ( x  4)  2 Solution Shift y  f ( x ) D 2, R 4

115. y  f (  x ) Solution Reflect y  f ( x ) about y

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816


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

116. y  f ( x ) Solution Reflect y  f ( x ) about x

117. y  4f ( x ) Solution Stretch y  f ( x ) vert. by a factor of 4

118. y 

1 f ( x) 4

Solution Shrink y  f ( x ) vert. by a factor of 1

4

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817


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

119. y  f (4 x ) Solution Shrink y  f ( x ) horiz. by a factor of 4

1  120. y  f  x  4  Solution Shrink y  f ( x ) horiz. by a factor of 1

4

Fix It In exercises 121 and 122, identify the step the first error is made and fix it. 121. Use the graph of f ( x )  x to graph g( x )  2 x  1  3 . Use the following sequence of transformations: translate the graph horizontally, vertically stretch or shrink the graph, reflect the graph, and translate the graph vertically. Solution Step 4 was incorrect: Step 1:

Step 2:

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818


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Step 3:

Step 4:

122. Use the graph of f ( x )  x to graph g( x )  3  x  2  4 . Use the following sequence of transformations: reflect the graph, vertically stretch or shrink the graph, translate the graph horizontally, and translate the graph vertically. Solution Step 3 was incorrect. Step 1:

Step 2:

Step 3:

Step 4:

Discovery and Writing Use a graphing calculator to perform each experiment. Write a brief paragraph describing your findings. 123. Investigate the translations of the graph of a function by graphing the parabola y  ( x  k )2  k for several values of k. What do you observe about successive positions of the vertex? Solution Answers may vary.

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819


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

124. Investigate the translations of the graph of a function by graphing the parabola y  ( x  k )2  k 2 for several values of k. What do you observe about successive positions of the vertex? Solution Answers may vary. 125. Investigate the horizontal stretching of the graph of a function by graphing y  ax for several values of a. What do you observe? Solution Answers may vary. 126. Investigate the vertical stretching of the graph of a function by graphing y  b x for several values of b. What do you observe? Are these graphs different from the graphs in Exercise 107? Solution Answers may vary. Write a paragraph using your own words. 127. a. Describe the change that must be made to the equation of a function to translate it vertically upward. b. Describe the change that must be made to the equation of a function to translate it vertically downward. Solution Answers may vary. 128. a. Describe the change that must be made to the equation of a function to translate it horizontally to the right. b. Describe the change that must be made to the equation of a function to translate it horizontally to the left. Solution Answers may vary. 129. a. Describe the change that must be made to the equation of a function to reflect it about the x-axis. b. Describe the change that must be made to the equation of a function to reflect it about the y-axis. Solution Answers may vary. 130. a. Describe the change that must be made to the equation of a function to stretch it vertically. b. Describe the change that must be made to the equation of a function to stretch it horizontally. Solution Answers may vary.

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820


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

131. a. Describe the change that must be made to the equation of a function to shrink it vertically. b. Describe the change that must be made to the equation of a function to shrink it horizontally. Solution Answers may vary. Critical Thinking Write the equation for the graph of the function shown. 132.

Solution Shift the function f ( x )  x 3 R 2.

g( x )  ( x  2)3

133.

Solution Reflect the function f ( x ) 

x about x and shift U 1. g( x )   x  1

134.

Solution Reflect the function f ( x )  x about x and shift L 3. g( x )   x  3

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821


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

135.

Solution Reflect the function f ( x )  3 x about x and stretch vert. by a factor of 2. g( x )  2 3 x 136.

Solution Reflect the function f ( x ) 

x about y and shift R 3. g( x ) 

 x 3

137.

Solution Reflect the function f ( x )  x 2 about x, shrink vert. by a factor of 1 , shift U 8. 2

g( x )   1 x 2  8 2 138.

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822


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution Reflect the function f ( x )  x 3 about x and shift U 2 and R 2. g( x )  ( x  2)3  2

EXERCISES 3.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Given f ( x )  2 x 4  4 x 2  6. Determine f (  x ). Solution f ( x)  2x 4  4x 2  6 f (  x )  2(  x )4  4(  x )2  6 f ( x)  2x 4  4x 2  6

2. Given f ( x )  3 x 5  5 x 3 . Determine f (  x ). Solution f ( x)  3x5  5x 3 f (  x )  3(  x )5  5(  x )3  3 x 5  5 x 3

3. If x > –4, then f ( x )  2 x 3 . Find f ( 2). Solution f ( 2)  2( 2)3  16 4. If x

–4, then f ( x )  3 x . Find f ( 6).

Solution f ( 6)  3 6  18

5. What is the greatest integer less than or equal to 4 2 ? 3

Solution –5 6. What is the greatest integer less than or equal to π ? Solution 3 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

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823


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Fill in the blanks. 7. If the graph of a function is symmetric about the __________, it is called an even function. Solution y-axis 8. If the graph of a function is symmetric about the origin, it is called an __________ function. Solution odd 9. If a function is even, then f (  x )  __________. Solution f ( x) 10. If a function is odd, then f (  x )  __________. Solution f ( x ) 11. If the values of f ( x ) get larger as x increases on an interval, we say that the function is __________ on the open interval. Solution increasing 12. If the values of f ( x ) get smaller as x increases on an interval, we say that the function is __________ on the open interval. Solution decreasing 13. If the values of f ( x ) do not change as x increases on an interval, we say that the function is __________ on the open interval. Solution constant 14. A local __________ occurs where a function changes from increasing to decreasing. Solution maximum 15. A local __________ occurs where a function changes from decreasing to increasing. Solution minimum 16. __________ functions are defined by different equations for different intervals in their domains. Solution Piecewise-defined

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824


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Practice Determine whether each function is even, odd, or neither. 17.

Solution symmetric about y-axis ⇒ even 18.

Solution symmetric about y-axis ⇒ even 19.

Solution symmetric about origin ⇒ odd 20.

Solution symmetric about origin ⇒ odd 21.

Solution no symmetry ⇒ neither © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

825


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

22.

Solution no symmetry ⇒ neither Determine algebraically whether each function is even, odd, or neither. 23. f ( x )  x 6  x 2 Solution f ( x)  x4  x2

f (  x )  (  x )4  (  x )2  x 4  x 2  f ( x )  even 24. f ( x )  2 x 3  2 x Solution f ( x)  2x 3  2x

f (  x )  2(  x )3  2(  x )  2 x 3  2 x  f ( x )  odd 25. f ( x )  x 5  x 2 Solution f ( x)  x5  x2 f (  x )  (  x )5  (  x )2   x 5  x 2  neither

26. f ( x )  3 x 6  x 2 Solution f ( x )  3 x 6  x 2

f (  x )  3(  x )6  (  x )2  3 x 6  x 2  f ( x )  even 27. f ( x )  4 x 7  x 3 Solution f ( x )  4 x 7  x 3

f (  x )  4(  x )7  (  x )3  4 x 7  x 3  f ( x )  odd

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826


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

28. f ( x )  9 x 3  x 2 Solution f ( x )  9x 3  x 2 f (  x )  9(  x )3  (  x )2  9 x 3  x 2  neither

29. f ( x )  5 x 3  3 x Solution f ( x)  5x 3  3x

f (  x )  5(  x )3  3(  x )  5 x 3  3 x  f ( x )  odd 30. f ( x )  4 x 2  5 Solution f ( x)  4x2  5

f (  x )  4(  x )2  5  4 x 2  5  f ( x )  even 31. f ( x ) 

x x 1 2

Solution

x x2  1 x f ( x )  (  x )2  1 x  2  f ( x )  odd x 1 f ( x) 

32. f ( x ) 

2x x2  9

Solution 2x f ( x)  2 x 9 2(  x ) f ( x )  (  x )2  9 2 x  2  f ( x )  odd x 9

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827


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

33. f ( x ) 

1 x6

Solution 1 f ( x)  6 x 1

f ( x ) 

( x )

6

1 x6

 f ( x )  even

2 x4

34. f ( x )  

Solution

2 x4

f ( x)  

f ( x )  

35. f ( x ) 

2 2   4  f ( x )  even 4 ( x ) x

x1

Solution

f ( x) 

x1

f ( x ) 

 x  1  neither

36. f ( x )  2 x  5 Solution

f ( x )  2x  5

f (  x )  2(  x )  5

 2 x  5  neither x

37. f ( x ) 

x

Solution x f ( x)  x f ( x )  

x x x x

 f ( x )  odd

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828


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

38. f ( x )  2 x  x Solution f ( x)  2x  x f (  x )  2(  x )   x  2 x  x  neither

State the open intervals where each function is increasing, decreasing, or constant. 39.

Solution decreasing: ( , 0) ; increasing: (0, )

40.

Solution constant: ( , 0) ; decreasing: (0, ) 41.

Solution increasing: ( , 0) ; decreasing: (0, ) 42.

Solution decreasing: ( , 0) ; constant: (0, )

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829


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

43.

Solution decreasing: ( ,  2) ; constant: ( 2, 2) ; increasing: (2, ) 44.

Solution increasing: ( , 0) ; decreasing: (0, 3) ; constant: (3, ) Use the graph to identify any local maxima and local minima. 45.

Solution local min. is 2 46.

Solution local max. is –2

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830


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

47.

Solution local max. is 5 48.

Solution local min. is –4 49.

Solution local max. is 2, local min. is 1 50.

Solution local max. is –1, local min. is –2 51.

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831


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution local max. is 1, local min. is 0 52.

Solution local max. is 0, local min. is  1

2

53.

Solution local max. is 3, local min. is –3 54.

Solution local max. is –2, local min. is –4 Evaluate each piecewise-defined function.

2 x  2 if x  0 55. f ( x )   if x  0 3 a.

f ( 2)

b.

f (0)

Solution a.

f ( 2)  2( 2)  2  2

b.

f (0)  3

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832


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

 x  2 if x  1 56. f ( x )   2 if x  1  x a.

f ( 1)

b.

f (5)

Solution a.

f ( 1)  1  2  3

b.

f (5)  52  25

2 x 3 if x  1 57. g( x )    x if x  1

a.

g( 3)

b.

g( 1)

c.

 1 g  4

Solution a. g(3)  2(3)3  54 b.

g( 1)  2( 1)3  2

c.

 1 g   4

1 1  4 2

  x 2  3 x  1 if x  1 58. g( x )   3 if x  1  x  2

a.

g( 1)

b.

g(1)

c.

 27  g   8 

Solution a. g( 1)  ( 1)2  3( 1)  1  3 b.

g(1)  3 1  2  1  2  1

c.

 27  3 27 3 1 g 2 2  8 8 2 2  

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833


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

2 if x  0  59. f ( x )  2  x if 0  x  2  x  1 if x  2 

a.

f ( 1)

b.

f (1)

c.

f (2)

Solution a. f ( 1)  2 b.

f (1)  2  1  1

c.

f (2)  2  1  3

2 x if x  0  60. f ( x )  3  x if 0  x  2 x if x  2  a.

f ( 0.5)

b.

f (0)

c.

f (2)

Solution a. f ( 0.5)  2( 0.5)  1 b.

f (0)  3  0  3

c.

f (2)  2  2

x3 if x  2  61. g( x )  2 x  5 if  2  x  2 5 if x  2  a.

g( 4)

b.

g( 1)

c.

9 g  4

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834


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution a. g( 4)  ( 4)3  64 b.

g( 1)  2( 1)  5  7

c.

9 g   5 4

2 x  3 if x  4  62. g( x )   x 2  12 if  4  x  4   x  2 if x  4 a.

g( 4)

b.

g(3)

c.

g(8)

Solution a. g( 4)  2( 4)  3  5 b.

g(3)  32  12  3

c.

g(8)  8  2  2 2  2

Graph each piecewise-defined function.

 x  2 if x  0 63. f ( x )   if x  0 2 Solution  x  2 if x  0 f ( x)   if x  0 2

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835


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

2 x if x  0 64. f ( x )   2 if x  0 x   Solution if x  0 2 x f ( x)   2 x if x  0

 x if x  0 65. f ( x )   2 if x  0 Solution  x if x  0 f ( x)   2 if x  0

 x  66. f ( x )   1  x 2

if x  0 if x  0

Solution  x if x  0  f ( x)   1 if x 0  x 2

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836


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

  4  x if x  1 67. f ( x )   if x  1 3 Solution   4  x if x  1 f ( x)   if x  1 3

  5  x if x  1 68. f ( x )   if x  1 3 Solution   5  x if x  1 f ( x)   if x  1 3

  x if x  0 69. f ( x )   2 if x  0  x Solution   x if x  0 f ( x)   2 if x  0  x

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837


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

 x 70. f ( x )    x Solution  x f ( x)    x

if x  0 if x  0

if x  0 if x  0

0 if x  0  if 0  x  2 71. f ( x )   x 2 4  2 x if x  2  Solution 0 if x  0  2 if 0  x  2 f ( x)  x 4  2 x if x  2 

2 if x  0  72. f ( x )  2  x if 0  x  2 x if x  2 

Solution 2 if x  0  f ( x )  2  x if 0  x  2 x if x  2 

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838


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

2 x 2 if x   1  if  1  x  1 73. f ( x )  2 4 x  1 if x  1  Solution 2 x 2 if x   1  if  1  x  1 f ( x )  2 4 x  1 if x  1 

 x 2  9 if x   1  74. f ( x )  8 if  1  x  1 2 x  2 if x  1  Solution  x 2  9 if x   1  if  1  x  1 f ( x )  8 2 x  2 if x  1 

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839


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Evalute each function at the indicated x-values. 75. f ( x )   x  a.

f (3)

b.

f ( 4)

c.

f ( 2.3)

Solution a. f (3)   3  3 b.

f ( 4)   4   4

c.

f ( 2.3)   2.3  3

76. f ( x )   3 x  a.

f (4)

b.

f ( 2)

c.

f ( 1.2)

Solution a. f (4)   3(4)   12  12 b.

f ( 2)   3( 2)   6  6

c.

f ( 1.2)   3( 1.2)   3.6  4

x 77. g( x )     2  a.

g(7)

b.

g( 3)

c.

 1 g  3

Solution a.

7 g(7)     3  2 

b.

 3  g( 3)     2  2 

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840


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

c.

 1   1   3   1  g     0  3   2   6   

 x  78. g( x )     3  a.

g(5)

b.

g( 3)

c.

g( 10)

Solution a.

 5  g(5)     2  3 

b.

3 g( 3)     1  3 

c.

 10  g( 10)     3  3 

79. f ( x )   x  3 a.

f ( 1)

b.

2 f  3

c.

f (1.3)

Solution a. f ( 1)   1  3   2  2

     3  3   3 2 3

2 3

2 3

b.

f

c.

f (1.3)   1.3  3   4.3  4

80. f ( x )   4 x   1 a.

f ( 3)

b.

f (0)

c.

f ( )

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841


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution a. f ( 3)   4( 3)  1   12  1  12  1  13 b.

f (0)   4(0)  1  0  1  0  1  1

c.

f ( )   4( )  1  12  1  11

Graph each function. 81. y   2 x  Solution y   2 x 

82. f ( x )  2  x  Solution f ( x )  2  x 

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842


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x 83. f ( x )     2  Solution x f ( x )     2 

1  84. y   x  3  3  Solution 1  y   x  3  3 

85. y   x   1 Solution y   x   1

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843


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

86. y   x  2 Solution y   x  2

Fix It In exercises 87 and 88, identify the step the first error is made and fix it. 87. Determine whether the function f ( x )  x 3  12 x is even, odd, or neither. Solution Step 4 was incorrect. Step 1: f (  x )  (  x )3  12(  x ) Step 2: f (  x )  (  x )(  x )(  x )  12(  x ) Step 3: f (  x )   x 3  12 x Step 4: f (  x )  ( x 3  12 x ) Step 5: This function is odd. 88. Determine whether the function f ( x )  x 4  5 x 2 even, odd, or neither. Solution Step 3 was incorrect. Step 1: f (  x )  (  x )4  5(  x )2 Step 2: f (  x )  (  x )(  x )(  x )(  x )  5(  x )(  x ) Step 3: f (  x )  x 4  5 x 2 Step 4: f (  x )  f ( x ) Step 5: The function is even.

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844


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Applications 89. Grading scales A mathematics instructor assigns letter grades according to the following scale. From

Up to but Less Than

Grade

60%

70%

D

70%

80%

C

80%

90%

B

90%

100% (including 100%)

A

Graph the ordered pairs (p, g), where p represents the percent and g represents the grade. Find the final semester grade of a student who has test scores of 67%, 73%, 84%, 87%, and 93%. Solution

67  73  84  87  93 404   80.8 5 5 The student's grade is B. 90. Calculating grades See Exercise 71 and find the final semester grade of a student who has test scores of 53%, 65%, 64%, 73%, 89%, and 82%. Solution Refer to #71.

53  65  64  73  89  82 426   71 6 6 The student's grade is C. 91. Renting a jeep A rental company charges $50 to rent a Jeep for one day, plus $4 for every 100 miles (or portion of 100 miles) that it is driven. Graph the ordered pairs (m, C), where m represents the miles driven and C represents the cost. Find the cost if the Jeep is driven 275 miles in one day.

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845


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution $32 for 275 miles

92. Riding in a taxi A taxicab company charges $5 for a trip up to 1 mile, and $2 for every extra mile (or portion of a mile). Graph the ordered pairs (m, C), where m represents the miles traveled and C represents the cost. Find the cost to ride 3.5 miles.

Solution $23 for 10 1 miles 4

93. Computer communications An online information service charges for connect time at a rate of $12 per hour, computed for every minute or fraction of a minute. Graph the points (t, C), where C is the cost of t minutes of connect time. Find the cost of 7 1 minutes. 2

Solution $12 per hour ⇒ $0.20 per minute $1.60 for 7 1 minutes 2

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846


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

94. iPad repair There is a charge of $30, plus $40 per hour (or fraction of an hour), to repair an iPad. Graph the points (t, C), where t is the time it takes to do the job and C is the cost. If it takes 4 hours to repair the iPad, how much did it cost? Solution $190 for 4 hours

95. Rounding numbers Measurements are rarely exact; they are often rounded to an appropriate precision. Graph the points (x, y), where y is the result of rounding the number x to the nearest ten. Solution

96. Signum function Computer programmers often use the following function, denoted by y = sgn x. Graph this function and find its domain and range. 1 if x  0  y  0 if x  0  1 if x  0 

Solution y  sgn x

Domain = ( , ) range = {1, 0, 1}

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847


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

97. Graph the function defined by y 

x x

and compare it to the graph in Exercise 78. Are

the graphs the same? Solution y 

x x

, Not defined at x  0, so not the same

98. Graph: y  x  x . Solution y x x

Discovery and Writing 99. If you are given a function’s graph, how do you determine whether the function is even, odd, or neither? Solution Answers may vary. 100. If you are given a function’s equation, how do you determine whether the function is even, odd, or neither? Solution Answers may vary. 101. What does it mean for a function to be increasing on an interval? Give two examples of real-life functions that are increasing. Solution Answers may vary. 102. What does it mean for a function to be decreasing on an interval? Give two examples of real-life functions that are decreasing. Solution Answers may vary. 103. Describe what happens at the point where the graph of a function changes from increasing to decreasing.

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848


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution Answers may vary. 104. Describe what happens at the point where the graph of a function changes from decreasing to increasing. Solution Answers may vary. 105. In this section, we discussed maximum and minimum values; state two real-life situations where a maximum or minimum value is important. Solution Answers may vary. 106. Use postal rates for a first-class postage stamp to create a step function for calculating costs of mailing a first-class letter. Solution Answers may vary. 107. A water park charges the following prices for daily admission to the park: general admission, $39.99; children under 48 inches, $34.99. Describe how this information can be represented by a piecewise-defined function Solution Answers may vary. 108. Construct a piecewise-defined function that occurs in everyday life. Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 109. The function f ( x )  x 100  x 50 is an even function. Solution

f (  x )  (  x )100  (  x )50  x 100  x 50  f ( x ) : True. 110. The function g( x )  x 101  x 51 is an odd function. Solution

f (  x )  (  x )101  (  x )51   x 101  x 51  f ( x ) : True. 111. The function f ( x )  7 x is an odd function. Solution

f (  x )  7  x   7 x  f ( x ) : True.

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849


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

112. The function f ( x )  8 x is an even function. Solution

f (  x )  8  x . This is not defined for x  0, so it is not even. False. 113. The quotient of two odd functions is an even function. Solution True. 114. All functions have a local maximum value and a local minimum value. Solution False. The function y  2 x  3 has no local minimum or maximum values. 115. Local maximum and minimum values can be the same. Solution True. 116. If function f decreases on the interval ( , x1 ) and increases on the interval ( x1 , ) , then f ( x1 ) is a local maximum value. Solution False. f ( x1 ) is a local minimum value. 117. If function f increases on the interval ( , x1 ) and decreases on the interval ( x1 , ) , then f ( x1 ) is a local minimum value. Solution False. f ( x1 ) is a local maximum value. 118.   [ ]    4 Solution True.

EXERCISES 3.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Add and simplify: ( 5 x 2  4 x  1)  (2 x 2  5 x  2) Solution 3 x 2  x  1

2. Subtract and simplify: ( 4 x 2  3 x  1)  (2 x 2  2 x  1) Solution 4 x 2  2 x 2  3 x  2 x  1  1  6 x 2  5 x  2

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850


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

3. Multiply and simplify: (  x 2  2 x  3)(2 x  1) Solution 2 x 3  x 2  4 x 2  2 x  6 x  3  2 x 3  5 x 2  8 x  2 4. Divide and simplify:

6 x 2 6 x 3

Solution 6 x3 x3   x 2 6 x 2 5. Find the domain of f ( x ) 

x  3.

Solution x  3  0, x  3 ; domain: [3, ) 6. If f ( x )  x  5 , find and simplify [f ( x )]2  2f ( x )  20. Solution [f ( x )]2  2f ( x )  20  ( x  5)2  2( x  5)  20  x 2  10 x  25  2 x  10  20  x 2  8x  5

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7.

(f  g)( x )  __________ Solution f ( x )  g( x )

8.

(f  g)( x )  __________ Solution f ( x )  g( x )

9.

(f  g)( x )  __________ Solution f ( x )g( x )

10. (f g)( x )  __________, where g( x )  0 Solution f ( x) g( x )

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851


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

11. The domain of f + g is the __________ of the domains of f and g. Solution intersection 12. (f  g)( x )  __________ Solution f ( g( x )) 13. ( g  f )( x )  __________ Solution g(f ( x )) 14. To determine (f  g)( 5) , first find __________. Solution g( 5) 15. Composition of functions is not __________. Solution commutative 16. To be in the domain of the composite function f  g , a number x has to be in the __________ of g, and the output of g must be in the __________ of f. Solution domain; domain Practice Let f ( x )  2 x  1 and g( x )  3 x  2. Find each function and its domain. 17. f  g Solution (f  g)( x )  f ( x )  g( x )  (2 x  1)  (3 x  2)  5 x  1; domain  ( , ) 18. f  g Solution (f  g)( x )  f ( x )  g( x )  (2 x  1)  (3 x  2)   x  3; domain  ( , ) 19. f  g Solution (f  g)( x )  f ( x )g( x )  (2 x  1)(3 x  2)  6 x 2  x  2; domain  ( , ) 20. f g Solution (f g)( x ) 

  f ( x ) (2 x  1) 2 2  ; domain   ,    ,  g( x ) (3 x  2) 3 3    

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852


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Let f ( x )  x 2  x and g( x )  x 2  1 . Find each function and its domain. 21. f  g Solution (f  g)( x )  f ( x )  g( x )  ( x 2  x )  ( x 2  1)  x  1; domain  ( , ) 22. f  g Solution (f  g)( x )  f ( x )  g( x )  ( x 2  x )  ( x 2  1)  2 x 2  x  1; domain  ( , ) 23. f g Solution (f g)( x ) 

f ( x) x2  x x ( x  1) x  2   ; domain  ( ,  1)  ( 1, 1)  (1, ) g( x ) x  1 ( x  1)( x  1) x  1

24. f  g Solution (f  g)( x )  f ( x )g( x )  ( x 2  x )( x 2  1)  x 4  x 3  x 2  x; domain  ( , ) Let f ( x )  x 2  7 x  3 and g( x )  x 2  5 x  6. Find each function and its domain. 25. f  g Solution (f  g)( x )  f ( x )  g( x )  ( x 2  7 x  3)  ( x 2  5 x  6)  2 x 2  12 x  9; domain  ( , ) 26. f  g Solution (f  g)( x )  f ( x )  g( x )  ( x 2  7 x  3)  ( x 2  5 x  6)  2 x  3; domain  ( , ) 27. f  g Solution (f  g)( x )  f ( x )g( x )  ( x 2  7 x  3)( x 2  5 x  6)  x 4  12 x 3  44 x 2  57 x  18; domain  ( , ) 28. f g Solution (f g)( x ) 

f ( x) x2  7 x  3 x2  7x  3 ; domain  ( , 2)  (2, 3)  (3, )  2  g( x ) x  5 x  6 ( x  2)( x  3)

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853


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Let f ( x )  x 3  2 x 2  x  5 and g( x )  x 2  4 x  5. Find each function and its domain. 29. f  g Solution (f  g)( x )  x 3  2 x 2  x  5  ( x 2  4 x  5)

 x 3  2x 2  x  5  x 2  4x  5  x 3  x 2  5 x; domain  ( , ) 30. f  g Solution (f  g)( x )  x 3  2 x 2  x  5  x 2  4 x  5

 x 3  3 x 2  3 x  10; domain  ( , ) 31.

f g

Solution f  x 3  2x 2  x  5 ; domain = ( ,  5)  ( 5, 1)  (1,  )   ( x)  x2  4x  5 g 32. f  g Solution (f  g)( x )  ( x 2  2 x 2  x  5)( x 2  4 x  5)

 x 5  6 x 4  2 x 3  19 x 2  15 x  25 ; domain = ( ,  ) Let f ( x ) 

x1 x 8 and g( x )  . Find each function and its domain. x 3 x 3

33. f  g Solution

(f  g)( x ) 

x  1 x  8 2x  7   ; domain = ( , 3)  (3,  ) x3 x3 x3

34. f  g Solution

(f  g)( x ) 

x 1 x 8 9   ; domain = ( , 3)  (3,  ) x 3 x 3 x 3

35. f  g Solution (f  g)( x ) 

( x  1)( x  8) x 2  7 x  8 ; domain = ( , 3)  (3,  )  ( x  3)2 ( x  3)2

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854


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

36.

f g

Solution f   x  1  x  3 x  1 ; domain = ( ,  8)  ( 8,  3)  ( 3,  )   ( x)     g  x  3 x  8 x  8 Let f ( x )  x 2  7 and g( x ) 

x . Find each function and its domain.

37. f  g Solution

(f  g)( x )  f ( x )  g( x )  ( x 2  7) 

 x   x  x  7; domain  (0, ) 2

38. f  g Solution

(f  g)( x )  f ( x )  g( x )  ( x 2  7) 

 x   x  x  7; domain  (0, ) 2

39. f g Solution (f g)( x ) 

f ( x ) ( x 2  7)  ; domain  (0, ) g( x ) x

40. f  g Solution

(f  g)( x )  f ( x )g( x )  ( x 2  7)

 x   x x  7 x ; domain  (0, ) 2

Let f ( x )  x 2  1 and g( x )  3 x  2. Find each value, if possible. 41. (f  g)(2) Solution (f  g)(2)  f (2)  g(2)  [(2)2  1]  [3(2)  2]  3  4  7 42. (f  g)( 3) Solution (f  g)( 3)  f ( 3)  g( 3)  [( 3)2  1]  [3( 3)  2]  8  ( 11)  3 43. (f  g)(0) Solution (f  g)(0)  f (0)  g(0)  [(0)2  1]  [3(0)  2]  1  ( 2)  1

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855


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

44. (f  g)( 5) Solution (f  g)( 5)  f ( 5)  g( 5)  [( 5)2  1]  [3( 5)  2]  24  ( 17)  41 45. (f  g)(2) Solution (f  g)(2)  f (2)  g(2)  [(2)2  1]  [3(2)  2]  (3)(4)  12 46. (f  g)( 1) Solution (f  g)( 1)  f ( 1)  g( 1)  [( 1)2  1]  [3( 1)  2]  (0)( 5)  0

2 47. (f g)   3 Solution  2 2  2   1    f   3   5 3 2 (f g)         9  undefined 0  3 g2  2   3    2  3  3 

48. (f g)(0) Solution (f g)(0) 

1 f (0) [(0)2  1] 1    g(0) [3(0)  2] 2 2

Let f ( x )  2 x  5 and g( x )  3 x . Find each value. 49. (f  g)(8) Solution (f  g)(8)  f (8)  g(8)  [2(8)  5]  3 8  11  2  13

50. (f  g)( 8) Solution (f  g)( 8)  f ( 8)  g( 8)  [2( 8)  5]  3 8  21  ( 2)  23

51. (f  g)( 27) Solution (f  g)( 27)  f ( 27)  g( 27)  [2( 27)  5]  3 27  59  ( 3)  56

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856


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

52. (f  g)(8) Solution (f  g)(8)  f (8)  g(8)  [2(8)  5]  3 8  11  2  9

53. (f  g)( 1) Solution (f  g)( 1)  f ( 1)  g( 1)  [2( 1)  5]  3 1  ( 7)( 1)  7

54. (f  g)(1) Solution (f  g)(1)  f (1)  g(1)  [2(1)  5]  3 1  ( 3)(1)  3

 1 55. (f g)   8 Solution

  1   1 f   2    5  19 8 8  1     4   19 (f g)        8 1 2     g 1  1 3     2 8 8  1 56. (f g)     8 Solution

  1   1 f    2     5  21 8 8    1    4  21  (f g)      8 1 2   1 1   g  3     2 8  8 Find two functions f and g such that h(x) can be expressed as the function indicated. Several answers are possible. 57. h( x )  3 x 2  2 x; f  g Solution Let f ( x )  3 x 2 and g( x )  2 x. Then (f  g)( x )  3 x 2  2 x  h( x ). 58. h( x )  3 x 2 ; f  g Solution Let f ( x )  3 and g( x )  x 2 . Then (f  g)( x )  3 x 2  h( x ).

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857


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

59. h( x ) 

3x 2 ; f g x2  1

Solution Let f ( x )  3 x 2 and g( x )  x 2  1. Then (f g)( x ) 

3x 2  h( x ). x2  1

60. h( x )  5 x  x 2 ; f  g Solution Let f ( x )  5 x and g( x )   x 2 . Then (f  g)( x )  5 x  x 2  h( x ). 61. h( x )  x(3 x 2  1); f  g Solution Let f ( x )  3 x 3 and g( x )   x . Then (f  g)( x )  3 x 3  x

 x(3 x 2  1)  h( x ). 62. h( x )  (3 x  2)(3 x  2); f  g Solution Let f ( x )  9 x 2 and g( x )  4. Then (f  g)( x )  9 x 2  4  (3 x  2)(3 x  2)  h( x ).

63. h( x )  x 2  7 x  18; f  g Solution Let f ( x )  x  9 and g( x )  x  2. Then (f  g)( x )  ( x  9)( x  2)

 x 2  7 x  18  h( x ). 64. h( x )  5 x 5 ; f g Solution Let f ( x )  5 x 6 and g( x )  x . Then (f g)( x ) 

5x6  5 x 5  h( x ). x

Let f ( x )  3 x and g( x )  x  1. Determine the domain of each composite function and then find the composite function.

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858


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

65. f  g Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of f ( x ) . Domain of g( x ) : ( ,  ). Domain of f ( x )  ( ,  ). Thus, all values of g( x ) are in the domain of f ( x ) . Domain of f  g : (  ,  ) (f  g)( x )  f ( g( x ))  f ( x  1)  3( x  1)  3 x  3 66. g  f Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ) . Domain of f ( x ) : ( ,  ). Domain of g( x )  ( , ). Thus, all values of f ( x ) are in the domain of g( x ) . Domain of g  f : (  ,  ) ( g  f )( x )  g(f ( x ))  g(3 x )  3 x  1 67. f  f Solution The domain of f  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of f ( x ) . Domain of f ( x ) : ( ,  ). Thus, all values of f ( x ) are in the domain of f ( x ) . Domain of f  f : (  ,  ) (f  f )( x )  f (f ( x ))  f (3 x )  3(3 x )  9 x 68. g  g Solution The domain of g  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of g( x ) . Domain of g( x ) : ( ,  ). Thus, all values of g( x ) are in the domain of g( x ) . Domain of g  g : (  ,  ) ( g  g)( x )  g( g( x ))  g( x  1)  x  1  1  x  2 Let f ( x )  x 2 and g( x )  2 x . Determine the domain of each composite function and then find the composite function. 69. g  f Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ) . Domain of f ( x ) : ( ,  ). Domain of g( x )  ( , ). Thus, all values of f ( x ) are in the domain of g( x ) . Domain of g  f : (  ,  )

( g  f )( x )  g(f ( x ))  g( x 2 )  2 x 2

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859


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

70. f  g Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of f ( x ) . Domain of g( x ): ( ,  ). Domain of f ( x )  ( ,  ). Thus, all values of g( x ) are in the domain of f ( x ) . Domain of f  g : (  ,  )

(f  g)( x )  f ( g( x ))  f (2 x )  (2 x )2  4 x 2 71. g  g Solution The domain of g  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of g( x ) . Domain of g( x ): ( ,  ). Thus, all values of g( x ) are in the domain of g( x ) . Domain of g  g : (  ,  ) ( g  g)( x )  g( g( x ))  g(2 x )  2(2 x )  4 x 72. f  f Solution The domain of f  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of f ( x ) . Domain of f ( x ) : ( ,  ). Thus, all values of f ( x ) are in the domain of f ( x ) . Domain of f  f : (  ,  ) (f  f )( x )  f (f ( x ))  f ( x 2 )  ( x 2 )2  x 4 Let f ( x )  2 x 2  3 x  7 and g( x )  4 x  1. Determine the domain of the composite function and then find the composite function. 73. f  g Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of f ( x ) . Domain of g( x ) : ( ,  ). Domain of f ( x )  ( ,  ). Thus, all values of g( x ) are in the domain of f ( x ) . Domain of f  g : (  ,  ) (f  g)( x )  f ( g( x ))  f (4 x  1)  2(4 x  1)2  3(4 x  1)  7

 2(16 x 2  8 x  1)  12 x  3  7  32 x 2  16 x  2  12 x  10  32 x 2  28 x  12 74. g  f Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ) . Domain of f ( x ) : ( ,  ). Domain of g( x )  ( ,  ). Thus, all values of f ( x ) are in the domain of g( x ) . Domain of g  f : (  , ) ( g  f )( x )  g(f ( x ))  g(2 x 2  3 x  7)  4(2 x 2  3 x  7)  1  8 x 2  12 x  28  1  8 x 2  12 x  27

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860


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

75. f  f Solution The domain of f  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of f ( x ) . Domain of f ( x ) : ( ,  ). Thus, all values of f ( x ) are in the domain of f ( x ) . Domain of f  f : (  ,  ) (f  f )( x )  f (f ( x ))  f (2 x 2  3 x  7)  2(2 x 2  3 x  7)2  3(2 x 2  3 x  7)  7  2(4 x 4  12 x 3  37 x 2  42 x  49)  6 x 2  9 x  21  7  8 x 4  24 x 3  74 x 2  84 x  98  6 x 2  9 x  14  8 x 4  24 x 3  68 x 2  75 x  84

76. g  g Solution The domain of g  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of g( x ) . Domain of g( x ) : ( ,  ). Thus, all values of g( x ) are in the domain of g( x ) . Domain of g  g : (  ,  ) ( g  g)( x )  g( g( x ))  g(4 x  1)  4(4 x  1)  1  16 x  4  1  16 x  5 Let f ( x )  x 3 and g( x )  x  4. Find each function and its domain. 77. g  f Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ). Domain of f ( x ) : (  ,  ). Thus , all values of f ( x ) are in the domain of g( x ). Domain of g  f : (  ,  ) ( g  f )( x )  g(f ( x ))  g( x 3 )  x 3  4 78. f  g Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of f ( x ). Domain of g( x ): (  ,  ). Thus , all values of g( x ) are in the domain of f ( x ). Domain of f  g : (  ,  )

(f  g)( x )  f ( g( x ))  f ( x  4)  ( x  4)3  x 3  12 x 2  48x  64 79. g  g Solution The domain of g  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of g( x ). Domain of g( x ): (  ,  ). Thus , all values of g( x ) are in the domain of g( x ). Domain of Domain of g  g : (  , ) ( g  g)( x )  g( g( x ))  ( x  4)  4  x  8

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861


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

80. f  f Solution The domain of f  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of f ( x ). Domain of f ( x ) : (  ,  ). Thus , all values of f ( x ) are in the domain of f ( x ). Domain of f  f : (  ,  ) (f  f )( x )  f (f ( x ))  f ( x 3 )  x 9 81. f  f  f Solution Domain of f: (  ,  ), Domain of f  f : (  ,  ), Domain of f  f  f : (  ,  ) (f  f  f )( x )  f (f (f ( x )))  f (f ( x 3 ))  f ( x )9  x 27

82. g  g  g Solution Domain of g: (  ,  ) , Domain of g  g : (  ,  ), Domain of g  g  g : (  ,  ) ( g  g  g)( x )  g( g( g( x )))  g( g( x  4))  g( x  8)  x  12 Let f ( x )  x and g( x )  x  1. Determine the domain of each composite function and then find the composite function. 83. f  g Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of f ( x ) . Domain of g( x ) : ( ,  ). Domain of f ( x )  [0,  ). Thus, we must have g( x )  0  x  1  0  x  1. Domain of f  g: [  1, ) (f  g)( x )  f ( g( x ))  f ( x  1) 

x1

84. g  f Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ) . Domain of f ( x ) : [0,  ). Domain of g( x )  ( , ). Thus, all values of f ( x ) are in the domain of g( x ). Domain of g  f : [0, )

( g  f )( x )  g(f ( x ))  g

 x  x  1

85. f  f Solution The domain of f  f is the set of all real numbers in the domain of f ( x ) such that

f ( x ) is in the domain of f ( x ) . Domain of f ( x ) : [0,  ). Thus, we must have

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862


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

f ( x)  0 

x  0. This is true for all real values of x . Domain of f  f : [0, )

(f  f )( x )  f (f ( x ))  f

 x 

x  ( x)

12

 x 12

14

4x

86. g  g Solution The domain of g  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of g( x ) . Domain of g( x ) : ( ,  ). Thus, all values of g( x ) are in the domain of g( x ). Domain of g  g: ( , ) ( g  g)( x )  g( g( x ))  g( x  1)  ( x  1)  1  x  2 Let f ( x )  x  1 and g( x )  x 2  1. Determine the domain of each composite function and then find the composite function. 87. g  f Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ) . Domain of f ( x ) : [ 1,  ). Domain of g( x )  ( , ). Thus, all values of f ( x ) are in the domain of g( x ). Domain of g  f : [1, )

( g  f )( x )  g(f ( x ))  g

 x  1   x  1  1  x 2

88. f  g Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of f ( x ) . Domain of g( x ) : ( ,  ). Domain of f ( x )  [ 1,  ). Thus, we must have g( x )  1  x 2  1  1  x 2  0. This is true for all real values of x . Domain of f  g: ( , ) (f  g)( x )  f ( g( x ))  f ( x 2  1) 

x2  1  1 

x2  x

89. g  g Solution The domain of g  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of g( x ) . Domain of g( x ) : ( ,  ). Thus, all values of g( x ) are in the domain of g( x ). Domain of g  g: ( , )

( g  g)( x )  g( g( x ))  g( x 2  1)  ( x 2  1)2  1  x 4  2 x 2

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863


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

90. f  f Solution The domain of f  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of f ( x ) . Domain of f ( x ) : [ 1,  ). Thus, we must have

f ( x )  1 

x  1  1. This is true for all real values of x . Domain of f  f : [1, )

(f  f )( x )  f (f ( x ))  f

Let f ( x ) 

1 x 1

and g( x ) 

 x  1 

x11

1 . Determine the domain of each composite function and then x 2

find the composite function. 91. f  g Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of f ( x ) . Domain of g( x ) : ( , 2)  (2,  ). Domain of f ( x )  ( , 1)  (1,  ). Thus, we must have g( x )  1 

1  1 1 x 2 x  3 x 2

Domain of f  g: ( , 2)  (2, 3)  (3, )

 1  1 1 x 2 x 2 x 2     (f  g)( x )  f ( g( x ))  f   1 1  x  2  x  2  1 x  2  1 x  2 1  ( x  2) 3  x 92. g  f Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ) . Domain of f ( x ): ( , 1)  (1,  ). Domain of

g( x )  ( , 2)  (2, ). Thus, we must have f ( x)  2 

1 3  2  1  2( x  1)  1  2 x  2  2 x  3  x  x1 2

 

Domain of f  g: ( , 1)  1, 3  3 ,  2

2

 1  1 1 x1 x1 x1     ( g  f )( x )  g(f ( x ))  g   1 1 2 x 1 x x 1 1 2( 1) 3     2x 2    x 1 x 1 93. f  f Solution The domain of f  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of f ( x ) . Domain of f ( x ) : ( , 1)  (1,  ). Thus, we must have

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864


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

f ( x)  1 

1  1  x  1  x  2 Domain of f  f : ( , 1)  1, 2  2,  x1

 

 1  1 1 x1 x1 x1     (f  f )( x )  f (f ( x ))  f   1 1  x  1  x  1  1 x  1  1 x  1 1  ( x  1) 2  x 94. g  g Solution The domain of g  g is the set of all real numbers in the domain of g( x ) such that

g( x ) is in the domain of g( x ) . Domain of g( x ) : ( , 2)  (2,  ). Thus, we must have g( x )  2 

 

1 5 Domain of g  g: ( , 2)  2, 5  5 ,   2  1  2x  4  x  2 2 x 2 2

 1  1 1 x 2 x 2 x 2     ( g  g)( x )  g( g( x ))  g   1 1  x  2  x  2  2 x  2  2 x  2 1  2( x  2) 5  2 x Let f ( x )  2 x  5 and g( x )  5 x  2. Find each value. 95. (f  g)(2) Solution (f  g)(2)  f ( g(2))  f (5(2)  2)  f (8)  2(8)  5  11 96. (f  g)( 2) Solution (f  g)( 2)  f ( g( 2))  f (5( 2)  2)  f ( 12)  2( 12)  5  29 97. ( g  f )( 3) Solution ( g  f )( 3)  g(f ( 3))  g(2( 3)  5)  g( 11)  5( 11)  2  57 98. ( g  f )(3) Solution ( g  f )(3)  g(f (3))  g(2(3)  5)  g(1)  5(1)  2  3

 1 99. (f  f )     2 Solution

      f 2     5  f (6)  2(6)  5  17

(f  f )  1  f f  1 2

1 2

2

3 100. ( g  g)   5 Solution



    g 5    2  g(1)  5(1)  2  3

( g  g) 3  g g 3 5

5

3 5

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865


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Let f ( x )  3 x 2  2 and g( x )  4 x  4. Find each value. 101. (f  g)( 3) Solution (f  g)( 3)  f ( g( 3))  f (4( 3)  4)  f ( 8)  3(8)2  2  190

 1 102. (f  g)   4 Solution

(f  g)

   f  g    f 4    4  f (5)  3(5)  2  73 1 4

1 4

1 4

2

103. ( g  f )(3) Solution ( g  f )(3)  g(f (3))  g(3(3)2  2)  g(25)  4(25)  4  104

 1 104. ( g  f )   3 Solution



    g  3    2   g     4     4  

(g  f ) 1  g f 3

105. (f  f )

1 3

1 3

2

5 3

5 3

8 3

 3

Solution (f  f )

 3   f f  3   f  3  3   2   f (7)  3(7)  2  145 2

2

106. ( g  g)( 4) Solution ( g  g)( 4)  g( g( 4))  g(4( 4)  4)  g( 12)  4( 12)  4  44 Let f ( x ) 

2 and g( x )  x

x . Find each value.

107. (f  g)(100) Solution

(f  g)(100)  f ( g(100))  f

 100   f (10)  102  51

108. (f  g)(8) Solution (f  g)(8)  f ( g(8))  f

 8   28  28  22  2 162  242  22

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866


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

 1  109. ( g  f )    32  Solution

 2   1   1  ( g  f )    g  f     g    g(64)  64  8  1   32    32    32  110. ( g  f )(8) Solution 2  1 ( g  f )(8)  g(f (8))  g    g    8 4

1 1  4 2

 81  111. ( g  g)    256  Solution  81    81    81  9   g   ( g  g)    g  g     g    256   16    256    256 

9 3  16 4

 3 112. (f  f )     5 Solution

 2    3   3   f   10   2   6   3 (f  f )     f  f      f  3  5 5 10 5       3   10   3  5 Find two functions f and g such that the composition f  g  h expresses the given correspondence. Several answers are possible. 113. h( x )  3 x  2 Solution Let f ( x )  x  2 and g( x )  3 x. Then (f  g)( x )  f ( g( x ))

 f (3 x )  3 x  2. 114. h( x )  7 x  5 Solution Let f ( x )  x  5 and g( x )  7 x. Then (f  g)( x )  f ( g( x ))

 f (7 x )  7 x  5.

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867


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

115. h( x )  x 2  2 Solution Let f ( x )  x  2 and g( x )  x 2 . Then (f  g)( x )  f ( g( x ))

 f ( x 2 )  x 2  2. 116. h( x )  x 3  3 Solution Let f ( x )  x  3 and g( x )  x 3 . Then (f  g)( x )  f ( g( x ))

 f ( x 3 )  x 3  3. 117. h( x )  ( x  2)2 Solution Let f ( x )  x 2 and g( x )  x  2. Then (f  g)( x )  f ( g( x ))

 f ( x  2)  ( x  2)2 . 118. h( x )  ( x  3)3 Solution Let f ( x )  x 3 and g( x )  x  3. Then (f  g)( x )  f ( g( x ))

 f ( x  3)  ( x  3)3 . 119. h( x ) 

x2

Solution

x and g( x )  x  2.

Let f ( x ) 

Then (f  g)( x )  f ( g( x ))

 f ( x  2)  120. h( x ) 

x  2.

1 x 5

Solution Let f ( x ) 

1 and g( x )  x  5. x

Then (f  g)( x )  f ( g( x ))

 f ( x  5) 

1 . x 5

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868


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

121. h( x ) 

x 2

Solution Let f ( x )  x  2 and g( x ) 

x.

Then (f  g)( x )  f ( g( x )) f

122. h( x ) 

 x   x  2.

1 5 x

Solution Let f ( x )  x  5 and g( x ) 

1 . x

Then (f  g)( x )  f ( g( x ))  1 1  f     5. x x

123. h( x )  x Solution Let f ( x )  x and g( x )  x . Then (f  g)( x )  f ( g( x ))

 f ( x )  x. 124. f ( x )  3 Solution Let f ( x )  3 and g( x )  x . Then (f  g)( x )  f ( g( x ))

 f ( x )  3. Use the graphs of functions f and g to answer each problem.

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869


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

125. (f  g)( 4) Solution (f  g)( 4)  f ( 4)  g( 4)  2  2  0 126. (f  g)(1) Solution (f  g)(1)  f (1)  g(1)  1  3  2 127. (f  g)(5) Solution (f  g)(5)  f (5)  g(5)  2(0)  0 128. (f g)( 1) Solution 1

(f g)( 1) 

f ( 1) 2 1   g( 1) 2 4

129. (f  g)(3) Solution (f  g)(3)  f ( g(3))  f (2)  1 130. ( g  f )(2) Solution ( g  f )(2)  g(f (2))  g(1)  3 131. (f  f )( 2) Solution (f  f )( 2)  f (f ( 2))  f (0)  1 132. ( g  g)( 5) Solution ( g  g)( 5)  g( g( 5))  g(1)  3 Use the tables of values of f and g to answer each problem.

x

f(x)

x

g(x)

2

4

0

0

4

9

2

4

6

13

3

9

13

17

4

16

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870


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

133. (f  g)(2) Solution (f  g)(2)  f (2)  g(2)  4  4  8 134. (f g)(4) Solution

(f g)(4) 

f (4) 9  g(4) 16

135. (f  g)(2) Solution (f  g)(2)  f ( g(2)  f (4)  9 136. ( g  f )(2) Solution ( g  f )(2)  g(f (2))  g(4)  16 Fix It In exercises 137 and 138, identify the step the first error is made and fix it. 137. f ( x )  3 x 2  4 x  11 and g( x )  3 x 2  2 x  5. Find (f  g)( x ). Solution Step 3 was incorrect. Step 1: (f  g)( x )  f ( x )  g( x ) Step 2: (f  g)( x )  (3 x 2  4 x  11)  ( 3 x 2  2 x  5) Step 3: (f  g)( x )  3 x 2  4 x  11  3 x 2  2 x  5 Step 4: (f  g)( x )  6 x 2  2 x  6 138. f ( x )  x 2  7 x  3 and g( x )  x  5. Find (f  g)( x ). Solution Step 4 was incorrect. Step 1: (f  g)( x )  f ( g( x )) Step 2: (f  g)( x )  f ( x  5) Step 3: (f  g)( x )  ( x  5)2  7( x  5)  3 Step 4: x 2  10 x  25  7 x  35  3 Step 5: x 2  17 x  63

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871


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Applications 139. Camcorder Suppose that the functions R( x )  300 x and C( x )  60,000  40 x model a company’s monthly revenue and cost for producing and selling camcorders. a. Find (R  C )( x ), the function that models the monthly profit, P( x ). b. Find the company’s profit if 500 camcorders are produced and sold in one month. Solution a. (R  C)( x )  300x  (60,000  40x )

 260x  60,000 b.

(R  C)(500)  260(500)  60,000  70,000

140. TV screen The height of the television screen shown is 13 inches.

a. Write a formula to find the area of the viewing screen. b. Use the Pythagorean Theorem to write a formula to find the width w of the screen. c. Write a formula to find the area of the screen as a function of the diagonal d. Solution a. A  13w b.

w 2  132  d 2 w 2  d 2  132 w  d 2  169

c.

A  13w  13 d 2  169

141. Missing hiker’s dog A hiker’s dog is missing and a search-and-rescue team is determining the circular area in which to search. On the given terrain, the hiker’s dog can travel at a rate of 2 miles hour. a. Write the function s(t ) which would represent the distance in miles the hiker’s dog could travel in t hours. b. Write the area function A(r ) which would give the circular search area, in square miles, given the hiker traveled r miles. c. Determine the composite function ( A  s)(t ). d. If 2 hours have passed, how much area should the rescue team search?

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872


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution a. s(t )  2t b.

A( r )   r 2

c.

( A  s )(t )  A( s(t ))  A(2t )   (2t )2  4 t 2

d.

4 (2)2  16 mi2

142. Area of an oil spill Suppose an oil spill from a tanker is spreading in the shape of a circular ripple. If the function d(t )  3t represents the diameter of the spill in inches at time t minutes, express the area, A, of the oil spill as a function of time. Find the area of the oil spill after 2 hours. Round to one decimal place. Solution 2

r (t ) 

 3t  d(t ) 3t 9  ; A(t )   (r (t ))2       t 2 2 2 2 4  

A(120) 

9  (120)2  101, 787.6 square inches 4

143. Hot air balloon The surface area, S(r ) , of a spherical-shaped hot air balloon with radius r in feet, is given by the formula S( r )  4 r 2 . If the balloon’s radius increases with time t in seconds, represented by r (t )  3 t 3 , find (S  r )(t ), the surface area as a 2

function of time. Solution 2

3  3  9  (S  r )(t )  S  t 3   4  t 3   4  t 6   9 t 6 2  2  4  144. Area of a square Write a formula for the area A of a square in terms of its perimeter P. Solution 2

If the perimeter is P, then each side is s 

P P2 P . . Area  s2     16 4 4

145. Perimeter of a square Write a formula for the perimeter P of a square in terms of its area A. Solution If the area is A and the length of a side is s, then s2  A  s 

A. Then P  4s  4 A.

146. Ceramics When the temperature of a pot in a kiln is 1200°F, an artist turns off the heat and leaves the pot to cool at a controlled rate of 81°F per hour. Express the temperature of the pot in degrees Celsius as a function of the time t (in hours) since the kiln was turned off.

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873


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution Use the relationship C 

F  mt  b

5 (F  32). 9 5 (( 81t  1200)  32) 9 5  ( 81t  1168) 9 5840  45t  9

C  F  C(F ) 

F  81t  1200

147. Calories burned The function C( x ) gives the number of calories burned for hiking x miles. The function M(t ) gives the average number of miles a person can hike in time t minutes. Identify the composite function that would be used to determine the number the calories a person will burn if hiking 20 minutes. Solution (C  M )(20) 148. Commission Gisela works as a sales representative and receives a monthly salary of $2000 plus a 4% commission on sales over $12,500. Given the two f ( x )  x  12500 and g( x )  0.04 x. Identify the composite function that would be used to determine her commission this month if her sales exceeded $12,500. Solution ( g  f )( x ) Discovery and Writing 149. Describe how to determine the composition of two functions. Solution Answers may vary. 150. Explain how to determine the domain of the composition of two functions. Solution Answers may vary. 151. Let f ( x )  3 x. Show that (f  f )( x )  f ( x  x ). Solution (f  f )( x )  f ( x )  f ( x )  3x  3x  6x

f ( x  x )  f (2x )  3(2x )  6x 152. Let g( x )  x 2 . Show that ( g  g)( x )  g( x  x ). Solution

( g  g)( x )  g( x )  g( x )  x 2  x 2  2x 2 g( x  x )  g(2x )  (2x )2  4 x 2

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874


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

153. Let f ( x ) 

x1 x1

. Find (f  f )( x ).

Solution x 1 x 1  x  1  x  1  1 ( x  1) x  1  1 x  1  ( x  1) (f  f )( x )  f (f ( x ))  f      x  1 x  1  x  1  xx  11  1 ( x  1) xx  11  1

 154. Let g( x ) 

x . x1

1 2  2x x

Find ( g  g)( x ).

Solution x ( x  1) x  x  x x x 1 x 1 ( g  g)( x )  g( g( x ))  g     x  x x  ( x  1) 1  x  1  x  1  1 ( x  1) xx 1  1

Let f ( x )  x 2  x , g( x )  x  3, and h( x )  3 x . Use a graphing calculator to graph both functions on the same axes. Write a brief paragraph summarizing your observations. 155. f and f  g Solution Answers may vary. 156. f and g  f Solution Answers may vary. 157. f and f  h Solution Answers may vary. 158. f and h  f Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 159. (f  g)( x )  ( g  f )( x ) Solution False. (f  g)( x )  ( f  g)( x )  ( g  f )( x ) 160. (f  g)( x )  ( g  f )( x ) Solution True.

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875


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

161. (f  g)( x ) sometimes equals ( g  f )( x ) Solution True. 162. If f ( x )  x 2 , then (f  f  f )( x )  x 6 Solution

False. (f  f  f )( x )  f (f (f ( x )))  f (f ( x 2 ))  f ( x 2 )2  f ( x 4 )  ( x 4 )2  x 8 163. If g( x )   x 3 , then ( g  g  g)( x )   x 9 Solution

False. ( g  g  g)( x )  g( g( g( x )))  g( g(  x 3 ))  g (  x 3 )3  g( x 9 )  ( x 9 )3   x 27 164. If f ( x ) 

5 , then (f  f  f  f )( x )  x x

Solution True.       5  5 (f  f  f  f )( x )  f (f (f (f ( x ))))  f  f  f      f  f  5    f (f ( x ))  f    5  x 5      x  x  5x       x 

165. If f ( x )  x 975 and g( x )  x 864 then (f  g)( 1)  1 Solution

True. (f  g)( 1)  f ( g( 1))  f ( 1)864  f (1)  1975  1. 166. If f ( x )  99 x and g( x )  77 x then (f  g)( 1)  1 Solution

 

False. (f  g)( 1)  f ( g( 1))  f 77 1  f ( 1)  99 1  1.

EXERCISES 3.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Given f(x) = 2x – 15 and g( x ) 

x  15 2

. Determine (f  g)( x ) and ( g  f )( x ).

Solution (f  g)( x )  f ( g( x ))

 x  15   x  15  f   2   15  x 2    2 

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876


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

( g  f )( x )  g(f ( x ))  g(2 x  15) 

2 x  15  15 x 2

2. Given f ( x )  x 3  27 and g( x )  3 x  27. Determine (f  g)( x ) and ( g  f )( x ). Solution (f  g)( x )  f ( g( x ))

f

 x  27    x  27   27  x 3

3

3

( g  f )( x )  g(f ( x ))

g( x 3  27)  3 x 3  27  27  x 3. Solve x 

y 5 3

for y.

Solution y 5 x 3 3x  y  5 y  3x  5 4. Solve x  2 y 3  4 for y. Solution x  2y3  4

x  4  2y3 x4  y3 2 x4 y 3 2 5. How many times would the horizontal line y = 2 intersect the graph of each of the given functions? a.

f ( x)  x

b.

f ( x)  x2

c.

f ( x)  x3

d.

f ( x)  x

f.

f ( x)  3 x

Solution a. 1 b. 2

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877


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

c. 1 d. 2 e. 1 6. If you graph f ( x )  x  2 and g( x )  x 2  2, x  2 on the same coordinate axes, the graphs of f(x) and g(x) are reflections of each other about the graph of what common function? Solution identity function y  x or f ( x )  x Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If different numbers in the domain of a function have different outputs, the function is called a __________ function. Solution one-to-one 8. If every __________ line intersects the graph of a function only once, the function is one-to-one. Solution horizontal 9. Two functions f and g are inverses if their composition in either order is the __________ function. Solution identity 10. The graph of a function and its inverse are reflections of each other about the line __________. Solution y=x Practice Determine whether each function is one-to-one. 11. f ( x )  5 Solution f ( x)  5

f (1)  f ( 1)  5 not one-to-one

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878


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

12. f ( x )  11 Solution f ( x )  11

f (1)  f ( 1)  11 not one-to-one 13. f ( x )  3 x Solution f ( x )  3x one-to-one 14. f ( x ) 

1 x 2

Solution 1 f ( x)  x 2 one-to-one 15. f ( x )  x 2  6 Solution f ( x)  x2  6 f (1)  f ( 1)  7

not one-to-one 16. f ( x )  x 4  x 2 Solution f ( x)  x4  x2 f (1)  f ( 1)  0

not one-to-one 17. f ( x )  x 3  4 Solution f ( x)  x3  4 one-to-one 18. f ( x )  ( x  1)3 Solution f ( x )  ( x  1)3 one-to-one

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879


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

19. f ( x )  x 3  x Solution f ( x)  x3  x f (1)  f ( 1)  0

not one-to-one 20. f ( x )  x 2  x Solution f ( x)  x2  x f (1)  f (0)  0

not one-to-one 21. f ( x )  x Solution f ( x)  x f (1)  f ( 1)  1

not one-to-one 22. f ( x )  x  3 Solution f ( x)  x  3 f (4)  f (2)  1

not one-to-one 23. f ( x ) 

x

Solution

f ( x) 

x

one-to-one 24. f ( x ) 

x 6

Solution

f ( x) 

x 6

one-to-one

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880


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

25. f ( x )  3 x Solution

f ( x)  3 x one-to-one 26. f ( x )  3 x  6 Solution

f ( x)  3 x  6 one-to-one 27. f ( x )  ( x  2)2 ; x  2 Solution f ( x )  ( x  2)2 ; x  2 one-to-one 28. f ( x ) 

1 x

Solution 1 f ( x)  x one-to-one Use the Horizontal Line Test to determine whether each graph represents a one-to-one function. 29.

Solution one-to-one 30.

Solution not one-to-one

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881


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

31.

Solution not one-to-one (not a function) 32.

Solution one-to-one 33.

Solution one-to-one 34.

Solution not one-to-one Verify that the functions are inverses by showing that (f  g )( x ) and ( g  f )( x ) are the identity function. 35. f ( x )  5 x and g( x ) 

1 x 5

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882


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution

1  1  (f  g)( x )  f ( g( x ))  f  x   5  x   x 5   5  1 ( g  f )( x )  g(f ( x ))  g(5 x )  (5 x )  x 5 36. f ( x )  4 x  5 and g( x ) 

x 5 4

Solution

 x 5  x  5 (f  g)( x )  f ( g( x ))  f    4 5  x 55  x  4   4  (4 x  5)  5 4 x ( g  f )( x )  g(f ( x ))  g(4 x  5)   x 4 4 37. f ( x )  x 3  8 and g( x )  3 x  8 Solution (f  g)( x )  f ( g( x ))  f

 x  8   x  8  8  x  8  8  x 3

3

3

( g  f )( x )  g(f ( x ))  g( x 3  8)  3 x 3  8  8  3

38. f ( x )  8 x 3 and g( x ) 

3

x3  x

x 2

Solution 3

3x 3x x   8   8   x (f  g)( x )  f ( g( x ))  f   2   2  8     3

( g  f )( x )  g(f ( x ))  g(8 x 3 ) 

8x 3 2x  x 2 2

39. f ( x )  5 x  1 and g( x )  ( x  1)5 Solution

  ( x  1)   1  x  1  1  x

(f  g)( x )  f ( g( x ))  f ( x  1)5 

5

5

  x  1  1   x   x

( g  f )( x )  g(f ( x ))  g 5 x  1 

5

5

5

5

40. f ( x )  x 5  2 and g( x )  5 x  2 Solution

(f  g)( x )  f ( g( x ))  f

 x  2   x  2  2  x  2  2  x 5

5

( g  f )( x )  g(f ( x ))  g x 5  2 

5

5

x5  2  2 

5

x5  x

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883


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

41. f ( x ) 

x1 1 and g( x )  x x1

Solution

1 1  1  x  1  1 ( x  1) x  1  1 1 x  1 (f  g)( x )  f ( g( x ))  f    x  1 1 1 ( x  1)  x  1 x 1 x 1

 x  1 1 x(1) x x ( g  f )( x )  g(f ( x ))  g     x   x1 x  1 x 1  1 x x1  1  1  x x

42. f ( x ) 

x1 x  11 and g( x )  x1 x1

Solution

 

 

x1 x1  x  1  x  1  1 ( x  1) x  1  1 x  1 x  1 (f  g)( x )  f ( g( x ))  f     x  x 1 x  1 1 x  x  1  ( x  1)   x  1  1 ( x  1) x  1  1

Note that ( g  f )( x ) will involve exactly the same calculations. 1 -

The function f(x) defined by the given equation is one-to-one. Find f (x). 43. f ( x )  11x Solution y  f ( x )  11x x  11 y x  y 11 x f 1 ( x )  11

44. f ( x )  5 x Solution y  f ( x )  5 x

x  5 y x   y 5 x 1 f ( x)   5 45. f ( x )  

x 7

Solution y  f ( x)  

x 7

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884


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x 7 x  y

y 7

1

f ( x )  7 x 46. f ( x ) 

x 15

Solution

x 15 y x 15 15 x  y

y  f ( x) 

f 1 ( x )  15 x 47. f ( x )  2 x  7 Solution y  f ( x)  2x  7 x  2x  7 x  7  2y x 7  y 2 x 7 f 1 ( x )  2 48. f ( x )  4 x  13 Solution y  f ( x )  4 x  13 x  4 y  13 x  13  4 y x  13  y 4 x  13 f 1 ( x )  4 49. f ( x ) 

x 6 8

Solution

y  f ( x) 

x 6 8

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885


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

y 6 8 8x  y  6 8x  6  y x

f 1 ( x )  8 x  6

50. f ( x ) 

2x  9 3

Solution

2x  9 3 2y  9 x 3 3x  2 y  9 3x  9  2 y 3x  9  y 2 3x  9 f 1 ( x )  2

y  f ( x) 

51. f ( x )  x 3  6 Solution y  f ( x)  x3  6 x  y3  6 x  6  y3 3

x 6  y f 1 ( x )  3 x  6

52. f ( x )  x 3  12 Solution y  f ( x )  x 3  12 x  y 3  12 x  12  y 3 3

x  12  y f 1 ( x )  3 x  12

53. f ( x )  ( x  9)3 Solution y  f ( x )  ( x  9)3

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886


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x  ( y  9)3 3 3

x  y 9

x 9  y f 1 ( x )  3 x  9

54. f ( x )  ( x  13)3 Solution y  f ( x )  ( x  13)3

x  ( y  13)3 3 3

x  y  13

x  13  y f 1 ( x )  3 x  13

55. f ( x )  2 3 x  7 Solution y  f ( x)  23 x  7

x  23 y  7 x  7  23 y x7 3  y 2 3  x 7    y  2   x 7 f ( x)     2 

3

1

56. f ( x )  4 3 x  15 Solution y  f ( x )  4 3 x  15

x  4 3 y  15 x  15  4 3 y x  15 3  y 4 3  x  15     y  4   x  15  f ( x)     4 

3

1

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887


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

57. f ( x ) 

6 x

Solution y  f ( x) 

6 x

6 y xy  6 6 y  x 6 1 f ( x)  x x

58. f ( x )  

10 x

Solution 10 x 10 x y xy  10

y  f (x)  

10 x 10 1 f ( x)  x y 

59. f ( x ) 

2 x 3

Solution

2 x 3 2 x y 3 ( y  3) x  2 xy  3 x  2 xy  2  3 x 2  3x y  x 2  3x 2 1 f ( x)   3 x x y  f ( x) 

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888


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

60. f ( x ) 

5 x 8

Solution

5 x 8 5 x y 8 ( y  8) x  5 xy  8 x  5 xy  5  8 x 5  8x y  x 5 1 f ( x)   8 x

y  f ( x) 

61. f ( x ) 

x 6 x 2

Solution x 6 x 2 y 6 x y 2 ( y  2) x  y  6 y  f ( x) 

xy  2 x  y  6 xy  y  2 x  6 y ( x  1)  2 x  6 2x  6 x1 2x  6 1 f ( x)  x1 y 

62. f ( x ) 

x 7 x2

Solution y  f ( x) 

x 7 x2

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889


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

y 7 y 2 x( y  2)  y  7 xy  2 x  y  7 x

xy  y  7  2 x y ( x  1)  7  2 x 7  2 x x1 2 x  7 1 f ( x)  x1 y 

Each equation defines a one-to-one function f. Determine f 1 ( x ) and verify that (f  f 1 )( x ) and (f 1  f )( x ) are both the identity function. 63. f ( x )  3 x

Solution y  f ( x )  3x x  3y

f  f  ( x )  f f ( x ) 1

f  f  ( x )  f f ( x ) 1

1

 f 1  3 x 

x f  3 x  3  3 x

x  3y 3 x 1 f 1 ( x )   x 3 3

64. f ( x ) 

1

3x 3 x 

1 x 3

Solution

1 x 3 1 x y 3 3x  y

y  f ( x) 

f  f  ( x )  f f ( x ) 1

1

f  f  ( x )  f f ( x ) 1

 f (3 x )

1   f 1  x  3  1   3 x  3  x

1 (3 x ) 3 x 

f 1 ( x )  3 x

1

65. f ( x )  3 x  2

Solution y  f ( x)  3x  2 x  3y  2 x  2  3y

x 2  y 3 x 2 f 1 ( x )  3

f  f  ( x )  f f ( x )  1

1

 x  2 f   3   x  2  3 2  3   x 22  x

f  f  ( x )  f f ( x ) 1

1

 f 1 (3 x  2) (3 x  2)  2 3 3x  x 3 

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890


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

66. f ( x )  2 x  5

Solution y  f ( x)  2x  5 x  2y  5

f  f  ( x )  f f ( x )  1

f  f  ( x )  f f ( x ) 1

1

1

 x  5 f   2   x  5  2 5  2   x 55  x

x  5  2y x 5  y 2 x 5 f 1 ( x )  2

 f 1 (2 x  5) (2 x  5)  5  2 2x  x 2

67. f ( x )  x 3  2

Solution y  f ( x)  x3  2

f  f  ( x )  f f ( x )  1

x 2  y

 x  2   x  2  2

f 1 ( x )  3 x  2

 x 22  x

x  y3  2 x 2  y 3

1

1

 f 1 ( x 3  2)

3

f

3

f  f  ( x )  f f ( x )

1

 3 ( x 3  2)  2

3

3

3

x3  x

68. f ( x )  ( x  2)3

Solution y  f ( x )  ( x  2)3

f  f  ( x )  f f ( x ) 1

 x  2    x  2  2   x  x

x  ( y  2)3 3 3

f

x  y 2

f 1 ( x )  3 x  2

1

3

 3 ( x  2)3  2

3

 x 22  x

3

3

1

1

3

3

x 2  y

f  f  ( x )  f f ( x )   f  ( x  2) 

1

69. f ( x )  5 x

Solution

y  f ( x)  5 x x5y x5  y 1

f ( x)  x

f  f  ( x )  f f ( x ) 1

1

 f ( x5 ) 

5

5

x x 5

f  f  ( x )  f f ( x ) 1

1

    x  x  f 1 5 x 5

5

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891


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

70. f ( x )  5 x  4

Solution

y  f ( x)  5 x  4 x  5 y 4 x 4 

5

f  f  ( x )  f f ( x )   f  ( x  4)  1

f  f  ( x )  f f ( x ) 1

y

 f 1 5 x  4

 ( x  4)  4 5

5

( x  4)  y

 x 44  x

f ( x )  ( x  4)

71. f ( x ) 

5

 x  4  4 5

5

 x  x 5

5

1 x3

Solution

1 x3 1 x y 3 x( y  3)  1 1 y 3 x 1 1 f ( x)   3 x

y  f ( x) 

72. f ( x ) 

1

5

5

1

1

f  f  ( x )  f f ( x )  1

1

f  f  ( x )  f f ( x )  1

1   f   3 x  1  1 33

 1   f 1    x  3 1  3 1 x 3

x

1

1

 x 33

1 x

x

x

1 x 2

Solution

1 x 2 1 x y 2 x( y  2)  1 1 y 2  x 1 f 1 ( x )   2 x

y  f ( x) 

f  f  ( x )  f f ( x )  1

1

1   f   2 x   1  1 22 x

1 1 x

f  f  ( x )  f f ( x )  1

1

 1   f 1    x  2 1  2 1 x 2

 x 22 x

x

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892


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

73. f ( x ) 

1 2x

Solution

1 2x 1 x 2y x(2 y )  1 2 xy  1 1 y  2x 1 f 1 ( x )  2x

y  f ( x) 

74. f ( x ) 

f  f  ( x )  f f ( x ) 1

1

f  f  ( x )  f f ( x ) 1

 1  f   2x  1   1  2   2x  1 

1

 1   f 1    2x  1  2 1

 2x 

1 x

1 1 x

x

x

1 x3

Solution

1 x3 1 x 3 y xy 3  1 1 y3  x 1 y 3 x 1 f 1 ( x )  3 x

y  f ( x) 

f  f  ( x )  f f ( x ) 1

1

 1  f 3   x   1  3  1 3   x   1  1 x

f  f  ( x )  f f ( x ) 1

1

 1   f 1  3  x  1 3 1

x3

3 x

x3 1

x

Find the inverse of each one-to-one function and graph both the function and its inverse on the same set of coordinate axes. 75. f ( x )  5 x

Solution y  f ( x )  5x x  5y

x  y 5 1 f 1 ( x )  x 5

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893


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

76. f ( x ) 

3 x 2

Solution

3 x 2 3 x y 2

y  f ( x) 

2 x y 3 2 f 1 ( x )  x 3 77. f ( x )  2 x  4

Solution y  f ( x )  2x  4 x  2y  4 x  4  2y x4  y 2 x4 f 1 ( x )  2 78. f ( x ) 

3 x 2 2

Solution

3 x 2 2 3 x  y 2 2 3 x2 y 2

y  f ( x) 

2 ( x  2)  y 3 2 f 1 ( x )  ( x  2) 3 79. f ( x )  2 x  10

Solution y  f ( x )  2x  4

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894


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x  2y  4 x  4  2y x4  y 2 x4 f 1 ( x )  2

80. f ( x )  3 x  3

Solution y  f ( x )  3 x  3 x  3 y  3

x  3  3 y x3  y 3 x  3 f 1 ( x )  3

81. f ( x )  ( x  2)3

Solution y  f ( x )  ( x  2)3 x  ( y  2)3 3

x  y 2

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895


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

3

x 2  y f 1 ( x )  3 x  2

82. f ( x )  ( x  6)3

Solution y  f ( x )  ( x  6)3 x  ( y  6)3 3 3

x  y 6

x 6  y f 1 ( x )  3 x  6

83. f ( x )   x 3  4

Solution y  f ( x)   x3  4 x  y3  4 y 3  x  4 y  3 ( x  4) f 1 ( x )   3 x  4

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896


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

84. f ( x )   x 3  5

Solution y  f ( x)   x3  5

x  y3  5 x  5  y3 ( x  5)  y 3 3

( x  5)  y f 1 ( x )   3 x  5

85. f ( x ) 

3

x 4

Solution

y  f ( x)  3 x  4 x  3 y 4 x3  y  4 x3  4  y f 1 ( x )  x 3  4

86. f ( x )  3 x  3

Solution

y  f ( x)  3 x  3

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897


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x  3 y 3 x3  y  3 x3  3  y f 1 ( x )  x 3  3

87. f ( x ) 

x 4

Solution y  f ( x) 

x 4

x

y 4

x  y 4 2

x 4 y 2

f 1 ( x )  x 2  4 (x  0)

88. f ( x ) 

x2

Solution y  f ( x) 

x

x2

y 2

x  y 2 2

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898


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x2  2  y f 1 ( x )  x 2  2 (x  0)

89. f ( x )  x 2  9 ( x  0)

Solution y  f ( x)  x2  9 x  y2  9 x  9  y2 x 9  y f 1 ( x ) 

90. f ( x ) 

x 9

1 2 x  6 ( x  0) 2

Solution y  f ( x) 

1 2 x 6 2

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899


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

1 2 y 6 2 1 x  6  y2 2 2( x  6)  y 2 x

2( x  6)  y f 1 ( x )  2( x  6)

The function f defined by the gven equation is one-to-one on the given domain. Find f 1 ( x ). 91. f ( x )  x 2  5 ( x  0)

Solution f ( x)  x2  5

x0

2

y  x 5

x0

2

x  y 5

y 0

2

y 0

x 5  y  x 5  y

y 0

Thus, f 1 ( x ) 

x  5 ( x  5).

92. f ( x )  x 2  5 ( x  0)

Solution f ( x)  x2  5

x0

2

y  x 5

x0

2

x  y 5

y 0

2

y 0

x 5  y  x 5  y

y 0

Thus, f 1 ( x ) 

x  5 ( x  5).

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900


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

93. f ( x )  4 x 2 ( x  0)

Solution f ( x)  4x2

x0

y  4x

2

x0

x  4y

2

y 0

x  y2 4 x   y 4 x  y 2

y 0 y 0 y 0

Thus, f 1 ( x ) 

x ( x  0). 2

94. f ( x )  4 x 2 ( x  0)

Solution f ( x )  4 x 2

x0

y  4 x

2

x0

x  4 y

2

y 0

x  y2 4 x    y 4 x  y 2 

y 0 y 0 y 0

Thus, f 1 ( x ) 

x ( x  0). 2

95. f ( x )  x 2  3 ( x  0)

Solution

f ( x)  x2  3

x0

2

y  x 3

x0

2

x  y 3

y 0

2

y 0

x3 y  x3  y

y 0

Thus, f 1 ( x )   x  3 ( x  3).

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901


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

96. f ( x ) 

1 ( x  0) x2

Solution 1 f ( x)  2 x 1 y  2 x 1 x 2 y 2 xy  1

x 0 x 0 y 0 y 0

1 y2  x y 

y 0 1 x

y 0

Thus, f 1 ( x ) 

1 x ( x  0).  x x

97. f ( x )  x 4  8 ( x  0)

Solution f ( x)  x4  8

x0

4

y  x 8

x0

4

x  y 8

y 0

4

y 0

x 8  y  x 8  y

y 0

4

Thus, f 1 ( x )  4 x  8 ( x  8). 98. f ( x ) 

1 ( x  0) x4

Solution 1 f ( x)  4 x 1 y  4 x 1 x 4 y 4 xy  1

x0 x0 y 0 y 0

1 y4  x y  4

y 0 1 x

y 0

Thus, f 1 ( x )   4

4 1 x3  ( x  0). x x

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902


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

99. f ( x )  4  x 2 (0  x  2)

Solution f ( x)  4  x2

0 x2

y  4x

2

0 x2

x  4  y2

0 y 2

x 4 y

2

0 y 2

y 4x

2

2 2

y   4x

0 y 2 2

0 y 2

Thus, f 1 ( x )  4  x 2 (0  x  2). 100. f ( x ) 

x 2  1 ( x  1)

Solution f ( x) 

x2  1

x  1

y 

x 1

x  1

x

y 1

y  1

2

2

x  y 1

y  1

y 2  x2  1

y  1

y   x2  1

y  1

2

2

Thus, f 1 ( x )   x 2  1 ( x  0).

Find the domain and the range of f. Find the range by finding the domain of f 1 . 101. f ( x ) 

x x 2

Solution x f ( x)  x 2

Domain of f  ( , 2)  (2, ) f 1 ( x ) 

2x x1

Range of f  Domain of f 1  ( , 1)  (1, )

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903


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

102. f ( x ) 

x 2 x3

Solution x 2 f ( x)  x3

Domain of f  ( ,  3)  (  3, ) f 1 ( x ) 

3x  2 1 x

Range of f  Domain of f 1  ( , 1)  (1, ) 103. f ( x ) 

1 2 x

Solution 1 f ( x)   2 x

Domain of f  ( , 0)  ( 0,  ) f 1 ( x ) 

1 x2

Range of f  Domain of f 1  ( ,  2)  (  2, )

104. f ( x ) 

3 1  x 2

Solution 3 1 f ( x)   x 2

Domain of f  ( , 0)  ( 0,  )

f 1 ( x ) 

3 x 1

2

Range of f  Domain of f 1

 

 ,  1   1 ,  2

2

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904


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

105. Use the graph of the function f to graph its inverse f 1 .

Solution

106. Use the graph of the function f to graph its inverse f 1 .

Solution

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905


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

107. Use the graph of the function f to graph its inverse f 1 .

Solution

108. Use the graph of the function f to graph its inverse f 1 .

Solution

Fix It In exercises 109 and 110, identify the step the first error is made and fix it. 109. Given the one-to-one function f ( x )  x 3  27, find f 1 ( x ).

Solution Step 4 was incorrect. Step 1: y  x 3  27

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906


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Step 2: x  y 3  27 Step 3: x  27  y 3 Step 4: 3 x  27  y Step 5: f 1 ( x )  3 x  27 110. Given the one-to-one function f ( x ) 

x 4 , find f 1 ( x ). 3

Solution Step 5 was incorrect. Step 1: y 

x 4 3

Step 2: x 

y 4 3

Step 3: 3 x  y  4 Step 4: 3 x  4  y Step 5: f 1 ( x )  3 x  4

Applications 111. Buying pizza A pizzeria charges $8.50 plus 75¢ per topping for a medium pizza. a. Find a linear function that expresses the cost f(x) of a medium pizza in terms of the number of toppings x. b. Find the cost of a pizza that has four toppings. c. Find the inverse of the function found in part (a) to find a formula that gives the number of toppings f 1 ( x ) in terms of the cost x. d. If Josh has $10, how many toppings can he afford?

Solution a.

y  0.75 x  8.50

b.

y  0.75(4)  8.50  $11.50

c.

y  0.75 x  8.50 x  0.75 y  8.50 x  8.50  0.75 y x  8.50  y 0.75

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907


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

d.

x  8.50 0.75 10  8.50  0.75 1.50  2 0.75

y 

112. Cell phone bills An international phone company charges $11 per month plus a nickel per call. a. Find a rational function that expresses the average cost f(x) of a call in a month when x calls were made. b. To the nearest tenth of a cent, find the average cost of a call in a month when 68 calls were made. c. Find the inverse of the function found in part (a) to find a formula that gives the number of calls f 1 ( x ) that can be made for an average cost x. d. How many calls need to be made for an average cost of 15¢ per call?

Solution a.

y 

b.

y 

0.05 x  11 x

0.05(68)  11 68  $0.212  21.2c 0.05 x  11 x 0.05 y  11 x y xy  0.05 y  11 y 

c.

xy  0.05 y  11 y ( x  0.05)  11 y 

d.

11 x  0.05

11 x  0.05 11  0.15  0.05 11   110 0.10

y 

Discovery and Writing 113. Describe what makes a function a one-to-one function.

Solution Answers may vary. © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

908


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

114. Explain the strategy used to determine the inverse of a one-to-one function.

Solution Answers may vary. 115. Write a brief paragraph to explain why the range of f is the domain of f 1 .

Solution Answers may vary. 116. Write a brief paragraph to explain why the graphs of a function and its inverse are reflections about the line y = x.

Solution Answers may vary. 117. Let f ( x )  x 5  x 3  x  3. Find f 1 (3). (Hint: Do not find f 1 ( x ) . Use observation and the fact that if f (a)  b, then f 1 (b)  a.)

Solution f (0)  3, so f 1 (3)  0. 118. Let f ( x )  x 5  x 3  x  3. Find f 1 ( 3). (Hint: Do not find f 1 ( x ) . Use the fact that if f (a)  b, then f 1 (b)  a.)

Solution f (0)  3, so f 1 ( 3)  0. Use a graphing calculator to graph each function for various values of a. 119. For what values of a is f ( x )  x 3  ax a one-to-one function?

Solution a0 120. For what values of a is f ( x )  x 3  ax 2 a one-to-one function?

Solution a0 Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 121. All functions have inverses.

Solution False. Only one-to-one functions have inverses. 122. If f ( x )  x 3  7, then f 1 ( x ) 

1 . x 7 3

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909


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution False.

y  f ( x)  x3  7 x  y3  7 x  7  y3 3

x 7  y f 1 ( x )  3 x  7

123. The inverse of the squaring function is the cube root function.

Solution False. The squaring function is not one-to-one and does not have an inverse. 124. The inverse of the cube root function is the cubing function.

Solution True. 125. If f ( x )  x 123 then f 1 ( x )  123 x .

Solution True. 126. f ( x )  x 888 is not a one-to-one function.

Solution True. 127. The graph of a function and its inverse are symmetric about the y-axis.

Solution False. The graph of a function and its inverse are symmetric about the line y  x 128. Functions that are either increasing or decreasing on their domains have inverses.

Solution True.

CHAPTER REVIEW SOLUTIONS Exercises Use the graph to determine the function’s domain and range. 1.

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910


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution 1 f ( x )  ( x  2)2 2

domain  ( , ) range  [0, ) Graph each function. Use the graph to identify the domain and range of each function. 2.

f ( x)  x2  4

Solution f ( x)  x2  4

domain  ( , ) range  ( , 4] 3.

f ( x)  3 x  2

Solution f ( x)  3 x  2

domain  ( , ) range  [0, )

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911


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

4.

f ( x)  

1 x 3 2

Solution 1 f ( x)   x  3 2

domain  ( , ) range  ( , 3] 5.

f ( x)  2x 3  2

Solution f ( x)  2x 3  2

domain  ( , ) range  ( , ) 6.

f ( x )  ( x  4)3

Solution f ( x )  ( x  4)3

domain  ( , ) range  ( , )

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912


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

7.

f ( x) 

x 5 1

Solution f ( x) 

x 5 1

domain  [5, ) range  [1, ) 8.

f ( x)   x  4

Solution f ( x)   x  4

domain  [0, ) range  ( ,  4)

9.

f ( x)  23 x

Solution f ( x)  23 x

domain  ( , ) range  ( , )

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913


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

10. f ( x )   3 x  1

Solution f ( x)   3 x  1

domain  ( , ) range  ( , )

Use the Vertical Line Test to determine whether each graph represents a function. 11.

Solution function 12.

Solution not a function Use the graph of the function f shown to determine each of the following.

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914


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

13. Domain and range

Solution domain  ( , ); range  ( , 4] 14. f (2)

Solution The point (2, 0) is on the graph, so f (2)  0. 15. f ( 1)

Solution The point ( 1, 3) is on the graph, so f ( 1)  3. 16. the x-values for which f ( x )  0

Solution The points ( 2, 0), (0, 0) and (2, 0) are on the graph, so f ( x )  0 when x  2, 0, and 2.

Use the graph of the function f shown to determine each of the following.

17. Domain and range

Solution domain  ( ,  ); range  ( ,  2] 18. f ( 1)

Solution The point ( 1,  6) is on the graph, so f ( 1)  6. 19. f ( 4)

Solution The point ( 4,  4) is on the graph, so f ( 4)  4. 20. the x-values for which f ( x )  8

Solution The points ( 6,  8) and (0,  8) are on the graph, so f ( x )  8 when x  6 and 0. 21. Target heart rate The target heart rate f(x), in beats per minute, at which a person should train to get an effective workout is a function of their age x in years. If

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915


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

f ( x )  0.6 x  132 , find the target heart rate for a 19-year-old college student. Round to the nearest whole number. Solution f ( x )  0.6 x  132

f (19)  0.6(19)  132  121 bpm 22. Cliff divers The height s, in feet, of a cliff diver is a function of the time t in seconds she has been falling. If s as a function of t can be expressed as s(t )  16t 2  10t  300, what is the height of the diver at 2.5 seconds?

Solution s(t )  16t 2  10t  300 s(2.5)  16(2.5)2  10(2.5)  300  225 ft

Each function is a translation of a basic function. Graph both on one set of coordinate axes. 23. g( x )  x 2  5

Solution g( x )  x 2  5 Shift y  x 2 U5

24. g( x )  ( x  7)3

Solution g( x )  ( x  7)3 Shift y  x 3 R 7

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916


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

25. g( x ) 

x23

Solution g( x ) 

x23

Shift y 

x U3, L 2

26. g( x )  x  4  2

Solution g( x )  x  4  2 Shift y  x U2, R 4

Each function is a stretching of f ( x )  x 3 . Graph both on one set of coordinate axes. 27. g( x ) 

1 3 x 3

Solution 1 g( x )  x 3 : Shrink y  x 3 vert. by a factor of 1 3 3

28. g( x )  ( 5 x )3

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917


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution

g( x )  ( 5 x )3 Shrink y  x 3 horiz. by a factor of 1 . Reflect about y. 5

Graph each function using a combination of translations, stretchings, and reflections. 29. f ( x )  2( x  6)2  8

Solution f ( x )  2( x  6)2  8 Start with y  x 2 . Shift R 6, Stretch vert. by a factor of 2, Shift D 8

30. f ( x ) 

1 ( x  2)2  6 2

Solution 1 f ( x )  ( x  2)2  6 2 Start with y  x 2 . Shift L 2, Shrink vert. by a factor of 1 , Shift U 6 2

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918


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

31. g( x )   x  4  3

Solution g( x )   x  4  3 Start with y  x Shift R 4, Reflect x, Shift U 3

32. g( x ) 

1 x 4 1 4

Solution 1 g( x )  x  4  1 4 Start with y  x Shift R 4, Shrink vert. by a factor

1, 4

of Shift U 1

33. g( x )  3 x  3  2

Solution g( x )  3 x  3  2

Start with y 

x

Shift L 3, Stretch vert. by a factor of 3, Shift U 2

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919


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

34. g( x ) 

1 ( x  3)3  2 3

Solution 1 g( x )  ( x  3)3  2 3 Start with y  x 3 Shift L 3, Shrink vert. by a factor of 1 , Shift U 2 3

35. f ( x ) 

x  3

Solution f ( x) 

x  3

Start with y 

x . Reflect y, Shift U 3

36. g( x )  2 3 x  5

Solution g( x )  2 3 x  5

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920


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Start with y  3 x . Stretch vert. by a factor of 2, Shift D 5

Determine whether each function is even, odd, or neither. 37.

Solution symmetric about y-axis  even 38.

Solution symmetric about origin  odd 39.

Solution no symmetry  neither

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921


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

40.

Solution symmetric about origin  odd 41. y  3 x 3  2 x

Solution f (  x )  3(  x )3  2(  x )  3 x 3  2 x  (3 x 3  2 x )

f (  x )  f ( x ) , so odd 42. y  x 2  4 x

Solution f ( x)  x2  4x f (  x )  (  x )2  4(  x )  x2  4x

 neither even nor odd 43. y  5 x 3  4 x 2

Solution f (  x )  5(  x )3  4(  x )2  5 x 3  4 x 2

f (  x )  f ( x )  f ( x ) , so neither 44. y  2  x 4

Solution f (  x )  2  (  x )4  2  x 4

f (  x )  f ( x ) , so even Determine the open intervals on which the graph of the function is increasing, decreasing, or constant. 45.

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922


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution inc : ( , 4); dec: (4,  ) 46.

Solution inc : ( ,  2)  (2,  ); dec: ( 2, 2) State the values of any local maxima or minima. 47.

Solution local max. is 2, local min. is 0 48.

Solution local max. is 2, local min. is –2 Evaluate each piecewise-defined function.

 x  2 if x  3 49. f ( x )   2 if x  3  x a.

f ( 2)

b.

f (3)

Solution a.

f ( 2)  2  2  4

b.

f (3)  32  9

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923


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

2 if x  3  50. f ( x )  2  x if 0  x  2  x  1 if x  2 

a.

3 f  2

b.

f (2)

Solution a.

3 3 1 f   2  2 2 2  

b.

f (2)  2  1  3

Graph each piecewise-defined function and determine the open intervals on which it is increasing, decreasing, or constant.  x  3 if x  0 51. f ( x )   if x  0 3

Solution  x  3 if x  0 f (x)   if x  0 3

inc : (  , 0); const : ( 0,  )

 x  5 if x  0 52. f ( x )   5  x if x  0

Solution  x  5 if x  0 f (x)   5  x if x  0

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924


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

inc : (  , 0); dec : ( 0,  )

3 x  1 if x  0  53. f ( x )   1 2  x  4 if x  0 3

Solution 3 x  1 if x  0  f ( x)   1 2  x  4 if x  0 3

dec : (  , 0); inc : ( 0,  )

1 2  ( x  1)  2 2  54. f ( x )  2  1 2 x  2 

if x  1 if  1  x  1 if x  1

Solution

1 2  ( x  1)  2 2  f ( x )  2  1 2 x  2 

if x  1 if  1  x  1 if x  1

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925


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

dec : (  ,  1); const : (  1, 1); inc : ( 1,  )

Evaluate each function at the indicated x-values. 55. f ( x )   2 x  Find f (1.7).

Solution f (1.7)   2(1.7)   3.4  3 56. f ( x )   x  5 Find f (4.99).

Solution f (4.99)  4.99  5   0.01  1 Graph each function. 57. f ( x )   x   2

Solution f ( x )   x   2

58. f ( x )   x  1

Solution f ( x )   x  1

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926


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

59. Renting an economy car A rental company charges $20 to rent economy car for one day, plus $8 for every 100 miles (or portion of 100 miles) that it is driven. Find the cost if the economy car is driven 295 miles in one day.

Solution 20  3(8)  $44 60. Riding with Uber Uber charges $4 for a trip up to 1 mile, and $2 for every extra mile (or portion of a mile). Find the cost to ride 11 1 miles. 2

Solution 4  11(2)  $26 Let f ( x )  x 2  1 and g( x )  2 x  1. Find each function and its domain. 61. f  g

Solution (f  g)( x )  f ( x )  g( x )  ( x 2  1)  (2x  1)  x 2  2x; domain  (, ) 62. f  g

Solution (f  g)( x )  f ( x )g( x )  ( x 2  1)(2 x  1)  2 x 3  x 2  2x  1; domain  (, ) 63. f  g

Solution (f  g)( x )  f ( x )  g( x )  ( x 2  1)  (2x  1)  x 2  2 x  2; domain  ( , ) 64. f g

Solution (f g)( x ) 

  f ( x) x2  1 1  1  ; domain   ,      ,   g( x ) 2 x  1 2  2  

Let f ( x )  2 x 2  1 and g( x )  2 x  1. Find each value, if possible. 65. (f  g)( 3)

Solution (f  g)( 3)  f ( 3)  g( 3)  [2( 3)2  1]  [2( 3)  1]  17  ( 7)  10 66. (f  g)( 5)

Solution (f  g)( 5)  f ( 5)  g( 5)  [2( 5)2  1]  [2( 5)  1]  49  ( 11)  60 67. (f  g)(2)

Solution (f  g)(2)  f (2)  g(2)  [2(2)2  1]  [2(2)  1]  7  3  21

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927


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

 1 68. (f g)   2

Solution

 

 

2

1 2 1 1  1  1 f 2 2 (f g)      2  undefined 1 1 2 0 2 1   g 2 2

Let f ( x )  x 2  1 and g( x )  2 x  1. Find each function and its domain. 69. f  g

Solution The domain of f  g is the set of all real numbers in the domain of g( x ) such that g( x ) is in the domain of f ( x ). Domain of g( x ) : (  ,  ). Domain of f ( x )  (  ,  ). Thus, all values of g( x ) are in the domain of f ( x ). Domain of f  g : (  , )

(f  g)( x )  f ( g( x ))  f (2 x  1)  (2 x  1)2  1  4 x 2  4 x  1  1  4 x 2  4 x 70. g  f

Solution The domain of g  f is the set of all real numbers in the domain of f ( x ) such that f ( x ) is in the domain of g( x ). Domain of f ( x ) : (  ,  ). Domain of g( x )  (  ,  ). Thus, all values of f ( x ) are in the domain of g( x ). Domain of g  f : ( , )

( g  f )( x )  g(f ( x ))  g( x 2  1)  2( x 2  1)  1  2 x 2  2  1  2 x 2  1 Let f ( x )  x 2  5 and g( x )  3 x  1. Find each value. 71. (f  g)( 2)

Solution (f  g)( 2)  f ( g( 2))  f (3( 2)  1)  f ( 5)  ( 5)2  5  20 72. ( g  f )( 2)

Solution

( g  f )( 2)  g(f ( 2))  g ( 2)2  5  g( 1)  3( 1)  1  2

Find two functions f and g such that the composition f  g  h expresses the given correspondence. Several answers are possible. 73. h( x ) 

x 5

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928


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution

Let f ( x ) 

x and g( x )  x  5.

Then (f  g)( x )  f ( g( x ))

 f ( x  5) 

x  5.

74. h( x )  ( x  6)3

Solution Let f ( x )  x 3 and g( x )  x  6. Then (f  g)( x )  f ( g( x ))  f ( x  6)  ( x  6)3 .

Determine whether each function is one-to-one. 75. f ( x )  x 2  8

Solution f ( x )  x 2  8 is not one-to-one, since f (2)  f ( 2)  12 76. f ( x )  2 x 3

Solution f ( x )  2 x 3 is one-to-one, since every x-value produces a different y-value. Use the Horizontal Line Test to determine whether each graph represents a one-to-one function. 77.

Solution one-to-one 78.

Solution not one-to-one Verify that the functions are inverses by showing that (f  g )( x ) and ( g  f )( x ) are the identity function. © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

929


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

79. f ( x )  8x  3 and g( x ) 

x3 8

Solution

 x  3  x  3 (f  g)( x )  f ( g( x ))  f    8 3  x 33  x  8   8  (8 x  3)  3 8 x ( g  f )( x )  g(f ( x ))  g(8 x  3)   x 8 8 80. f ( x ) 

1 2x  1 and g( x )  2 x x

Solution  1 1 1  x (f  g)( x )  f ( g( x ))  f  2    1 1 x   2 2 x x

 1  1 ( g  f )( x )  g(f ( x ))  g    2  1  2  (2  x )  x 2 x 2 x

Each equation defines a one-to-one function. Find f 1 ( x ) and verify that (f  f 1 )( x ) and (f 1  f )( x ) are the identity function.

81. f ( x )  7 x  1

Solution y  f (x)  7x  1 x  7y  1 x  1  7y x1  y 7 x1 f 1 ( x )  7 82. f ( x )  5x  8

Solution y  f ( x )  5x  8

x  5y  8 x  8  5y x 8  y 8 x 8 f 1 ( x )  5 83. f ( x )  x 3  10

Solution y  f ( x )  x 3  10

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930


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

x  y 3  10 x  10  y 3 3

x  10  y f 1 ( x )  3 x  10

84. f ( x )  3 x  5

Solution y  f ( x)  3 x  5

x  3 y 5 x3  y  5 x3  5  y f 1 ( x )  x 3  5 85. f ( x ) 

5 x

Solution 5 x 5 x y xy  5

y  f ( x) 

5 x 5 1 f ( x)  x y 

86. f ( x ) 

1 2 x

Solution 1 2 x 1 x 2 y x(2  y )  1 y  f ( x) 

2 y  2

1 x

1  y x

f 1 ( x )  2 

1 x

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931


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

87. f ( x ) 

x 1 x

Solution x 1 x y x 1 y x(1  y )  y

y  f ( x) 

x  xy  y x  xy  y x  y ( x  1) x  y x1 f 1 ( x ) 

88. f ( x ) 

x x1

3 x3

Solution

3 x3 3 x 3 y xy 3  3 3 y3  x 3 y 3 x

y  f ( x) 

f 1 ( x )  3

3 3 3x 2  x x

89. Find the inverse of the one-to-one function f ( x )  2x  5 and graph both the function and its inverse on the same set of coordinate axes.

Solution y  f ( x)  2x  5 x  2y  5 x  5  2y x 5  y 2 x 5 f 1 ( x )  2

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932


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

90. Find the range of f ( x ) 

2x  3 5 x  10

by finding the domain of f 1 .

Solution 2x  3 5 x  10 2y  3 x 5 y  10 x (5 y  10)  2 y  3 y 

5 xy  10 x  2 y  3 5 xy  2 y  10 x  3 y (5 x  2)  10 x  3 10 x  3 5x  2 Range of f  Domain of f 1 y 

  2 2   ,    ,   5 5  

CHAPTER TEST SOLUTIONS Graph each function by plotting points. 1.

f ( x)  2 x  1  2

Solution f ( x)  2 x  1  2

2.

f ( x )  2 x 3  4

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933


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution f ( x )  2 x 3  4

Use the graph of the function shown to determine the following.

3. domain and range

Solution domain: ( , )

range: ( , 5) 4.

f (1) Solution The point (1, 4) is on the graph, so f (1)  4.

Use transformations to graph each function. 5.

f ( x )  ( x  3)2  1

Solution f ( x )  ( x  3)2  1

Shift y  x 2 U 1, R 3

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934


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

6.

f ( x) 

x 15

Solution f ( x) 

x 15

Shift y 

7.

x U5, R 1

f ( x )  ( x  1)3  3

Solution f ( x )  ( x  1)3  3

Shift y  x 3 R 1, reflect about x, shift U 3.

8.

f ( x)   1 x  5  2 2

Solution f ( x)   1 x  5  2 2

Shift y  x L 5, reflect about x, shrink vert. by a factor of 1 , shift D 2 2

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935


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

9.

f ( x)  23 x  6  1

Solution f ( x)  23 x  6  1

Shift y  3 x R 6, stretch vert. by a factor of 2, shift D 1.

Determine whether the functions are even, odd, or neither. 10.

Solution symmetric about origin  odd 11. f ( x )  2 x 4  3 x 2  7

Solution y  f ( x )  2x 4  3x 2  7 f (  x )  2(  x )4  3(  x )2  7  2 x 4  3 x 2  7  f ( x )  even

Use the graph to determine any local maxima or minima. 12.

Solution local max. is 5, local min. is 4

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936


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Use the piecewise-defined function shown to find each value.

2 x  f ( x )  3  x x 

if x  0 if 0  x  2 if x  2

3 13. f   2

Solution 3 3 3 f   3  2 2 2 14. f (5)

Solution f (5)  5  5  x  1 if x  1 15. Graph f ( x )   . if x  1 4

Solution   x  1 if x  1 f (x)   if x  1 4

Let f ( x )  3 x and g( x )  x 2  2. Find each function. 16. f  g

Solution (f  g)( x )  f ( x )  g( x )  (3 x )  ( x 2  2)  x 2  3 x  2 17. f g

Solution (f g)( x ) 

f ( x) 3x  2 g( x ) x  2

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937


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

18. g  f

Solution ( g  f )( x )  g(f ( x ))  g(3 x )  (3 x )2  2  9x 2  2 19. f  g

Solution (f  g)( x )  f ( g( x ))  f ( x 2  2)  3( x 2  2)  3 x 2  6 Let f ( x )  2 x 2  5 x  1 and g( x )  5 x  1. Find each function value. 20. (f  g)( 2)

Solution (f  g)( 2)  f ( 2)  g( 2)  [2( 2)2  5( 2)  1]  [5( 2)  1]  19  ( 9)  10 21. (f  g)(2)

Solution

(f  g)(2)  f (2)  g(2)  [2(2)2  5(2)  1]  [5(2)  1]  1  11  12 22. (f  g)( 1)

Solution

(f  g)( 1)  f ( 1)  g( 1)  [2( 1)2  5( 1)  1]  [5( 1)  1]  8  ( 4)  32 23. (f g)(0)

Solution (f g)(0) 

f (0) 2(0)2  5(0)  1 1   1 g(0) 5(0)  1 1

24. (f  g)( 1)

Solution

(f  g)( 1)  f ( g( 1))  f (5( 1)  1)  f ( 4)  2( 4)2  5( 4)  1  53 25. ( g  f )( 3)

Solution

( g  f )( 3)  g(f ( 3))  g(2( 3)2  5( 3)  1)  g(34)  5(34)  1  171 Assume that f ( x ) is one-to-one. Find f 1 ( x ). 26. f ( x )  5x  2

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938


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution y  f ( x )  5x  2

x  5y  2 x  2  5y x2  y 5 x2 f 1 ( x )  5 27. f ( x ) 

x1 x1

Solution x1 x1

y  f ( x) 

y1 y 1 x( y  1)  y  1 x

xy  x  y  1 xy  y  x  1 y ( x  1)  x  1 f 1 ( x ) 

x1 x1

28. f ( x )  x 3  3

Solution

y  x3  3 x  y3  3 x  3  y3 3

x3  y f 1 ( x )  3 x  3

Find the range of f ( x ) by finding the domain of f 1 ( x ). 29. f ( x ) 

3 2 x

Solution 3 3 f ( x )   2; f 1 ( x )  x x2 Range of f  Domain of f 1  ( ,  2)  ( 2, )

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939


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

30. f ( x ) 

3x  1 x 3

Solution 3 x  1 1 3x  1 f ( x)  ; f ( x)  ; Range of f  Domain of f 1  ( , 3)  (3, ) x 3 x 3

CUMULATIVE REVIEW SOLUTIONS Use the x- and y-intercepts to graph each equation. 1.

5 x  3 y  15 Solution 5 x  3 y  15

5 x  3 y  15

5 x  3(0)  15

5(0)  3 y  15

5 x  15

3 y  15

x3

y  5

(3, 0)

2.

(0,  5)

3 x  2 y  12 Solution 3 x  2 y  12

3 x  2 y  12

3 x  2(0)  12

3(0)  2 y  12

3 x  12

2 y  12

x4

y 6

(4, 0)

(0, 6)

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940


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Find the length, the midpoint, and the slope of the line segment PQ. 3.

  7 1 P  2,  ; Q  3,   2 2    

Solution

4.

    25  16  41

a.

d  ( x2  x1 )2  ( y 2  y 1 )2  ( 2  3)2  7   1

b.

x

c.

m

2

7

x 1  x2 2 17 2

2

3  ( 2)

 1

  2  3 1 3 2  ; y  2  2  2 2 2 2 2

8 2

5



6

2

2

 1 3  ,  2 2

4 5

P(3, 7); Q( 7, 3)

Solution a.

d  ( x2  x1 )2  ( y 2  y 1 )2  (3  ( 7))2  (7  3)2 

b.

x

c.

m

x 1  x2 2

3  ( 7) 4 7  3 10   2; y   5 2 2 2 2

100  16 

116  2 29

 2, 5

37 4 2   7  3 10 5

Find the slope of the line passing through the two given points. 5.

P( 1, 9) and Q( 4,  6)

Solution y  y1 6  9 15   5 m 2 4  ( 1) 3 x2  x 1 6.

  1 1 P  2,   and Q  5,   3 3  

Solution

m

y2  y 1 x2  x 1

 3  0  0

1 1 3

52

3

Write the equation of the line with the given properties. Give the answer in slope-intercept form. 7. The line passes through (–3, 5) and (3, –7).

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941


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution y  y1 7  5 12    2 m 2 x2  x 1 3  ( 3) 6 y  y 1  m( x  x 1 ) y  5  2( x  3) y  2 x  1

3 5 7 8. The line passes through  ,  and has a slope of . 2 2 2  

Solution y  y 1  m( x  x 1 ) y  5  7 (x  3) 2

2 2 7 21 y  x 5 2 4 2 y  7 x  11 2 4

9. The line is parallel to 3x – 5y = 7 and passes through (–5, 3).

Solution 3x  5 y  7  5 y  3 x  7 y  3x7 5 5 m 3 5

use m  3 .

y  y 1  m( x  x 1 ) y  3  3 ( x  5) 5

y 3 3x 3 5

y  3x 6 5

5

10. The line is perpendicular to x – 4y = 12 and passes through the origin.

Solution x  4 y  12  4 y   x  12

y  y 1  m( x  x1 )

y  1 x 3 4 m 1 4

y  4 x

y  0  4( x  0)

use m  4.

Graph each equation. Make use of intercepts and symmetries. 11. x 2  y  2

Solution x2  y  2

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942


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

symmetry: y-axis x-int: none, y-int: (0, 2)

12. y 2  x  2

Solution y2  x  2

symmetry: x-axis x-int: (2, 0), y-int: none

Identify the center and radius of the circles. 13. x 2  ( y  7)2 

1 4

Solution x 2  ( y  7)2  C(0, 7); r 

1 4

1 1  4 2

14. ( x  5)2  ( y  4)2  144

Solution ( x  5)2  ( y  4)2  144

C(5,  4); r  144  12 Graph each circle. 15. x 2  y 2  100

Solution x 2  y 2  100

Circle: C (0, 0); r = 10

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943


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

16. x 2  2 x  y 2  8

Solution x 2  2x  y 2  8 ( x  1)2  y 2  9

Circle: C (1, 0); r = 3

Solve each proportion. 17.

x 2 x 6  x 5 Solution x 2 x 6  x 5 5( x  2)  x ( x  6) 5 x  10  x 2  6 x 0  x 2  11x  10 0  ( x  10)( x  1)

x  10 or x  1 18.

x  2 3x  1  x  6 2 x  11

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944


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution x  2 3x  1  x  6 2 x  11 ( x  2)(2 x  11)  (3 x  1)( x  6) 2 x 2  7 x  22  3 x 2  17 x  6 0  x 2  10 x  16 0  ( x  8)( x  2)

x  8 or x  2 19. Dental billing The billing schedule for dental X-rays specifies a fixed amount for the office visit plus a fixed amount for each X-ray exposure. If 2 X-rays cost $37 and 4 cost $54, find the cost of 5 exposures.

Solution 54  37 17 m   8.5 42 2

y  mx  b

y  8.5 x  20

y  8.5 x  b

y  8.5(5)  20

37  8.5(2)  b

y  62.50

37  17  b

It will cost $62.50.

20  b 20. Automobile collisions The energy dissipated in an automobile collision varies directly with the square of the speed. By what factor does the energy increase in a 50-mph collision compared with a 20-mph collision?

Solution E  ks2 E  k (50)2

E  k (20)2

E  2500k

E  400k

50 mph E 2500k 25   20 mph E 400k 4

Graph each function. 21. f ( x )  2 x  2  1

Solution y  2 x  2  1

Shift y  x R 2, reflect about x, stretch vert. by a factor of 2, shift D 1.

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945


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

22. f ( x )  x 2  4

Solution y  x2  4

Shift y  x 2 D 4.

23. f ( x )   x 2  4

Solution y   x2  4 Reflect y  x 2 about x, shift U 4.

24. f ( x )   x 3  5

Solution y  x3  5 Reflect y  x 3 about x, shift D 5.

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946


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

25. f ( x )  2 x  4  1

Solution y 2 x4 1

Shift y 

x L 4, stretch vert. by a factor of 2, shift D 1.

26. f ( x )  3 x  1  3

Solution y  3 x 13

Shift y  3 x R 1 and D 3.

Find the domain and range of the function. 27.

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947


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution domain: ( ,  ); range: [0,  ) Let f ( x )  3 x  4 and g( x )  x 2  1. Find each function and its domain. 28. (f  g)( x )

Solution (f  g)( x )  f ( x )  g( x )  (3 x  4)  ( x 2  1)   x 2  3 x  5; domain  ( , ) 29. (f  g)( x )

Solution (f  g)( x )  f ( x )g( x )  (3 x  4)( x 2  1)  3 x 3  4 x 2  3 x  4; domain  ( , ) 30. (f g)( x )

Solution (f g)( x ) 

f ( x ) 3x  4  2 ; domain  ( , ) g( x ) x 1

Let f ( x )  3 x  4 and g( x )  x 2  1. Find each value. 31. (f  g)(2)

Solution (f  g)(2)  f ( g(2))  f ((22  1))  f (5)  3(5)  4  11 32. ( g  f )(2)

Solution ( g  f )(2)  g(f (2))  g(3(2)  4)  g(2)  22  1  5 33. (f  g)( x )

Solution (f  g)( x )  f ( g( x ))  f ( x 2  1)  3( x 2  1)  4  3 x 2  1 34. ( g  f )( x )

Solution ( g  f )( x )  g(f ( x ))  g(3 x  4)  (3 x  4)2  1  9 x 2  24 x  17 Find the inverse of the function defined by each equation. 35. f ( x )  3 x  2

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948


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Solution y  3x  2

x  3y  2 x  2  3y x 2  y 3

f 1 ( x ) 

36. f ( x ) 

x 2 3

1 x 3

Solution 1 x 3 1 x y 3 x( y  3)  1 y 

1 x 1 y  3 x

y 3 

f 1 ( x ) 

1 3 x

37. y  x 2  5 (x  0)

Solution y  x2  5 x  y2  5 x  5  y2  x 5  y x  5  y ( y  0) f 1 ( x ) 

x 5

38. 3 x  y  1

Solution 3x  y  1

3y  x  1 3y  x  1 x1 y  3

f 1 ( x ) 

x1 3

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949


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

Write each sentence as an equation. 39. y varies directly with the product of w and z.

Solution y = kwz 40. y varies directly with x and inversely with the square of t.

Solution kx y  2 t

GROUP ACTIVITY SOLUTIONS Cryptography and Cybersecurity Real-World Example of Cryptography and Cybersecurity Cybersecurity is one of the most significant challenges in our contemporary world. To protect computer systems, networks, and people from identity theft, secure communication techniques are needed. Cryptography is the study of these techniques and it is closely associated with encryption. Encryption uses an algorithm to encode a message or data and then a secure key is used to decrypt it.

Group Activity We have learned about inverse functions in this chapter. Let’s encrypt a one-word message with a function and then decode it using the inverse of the function as out security key. Consider the two tables shown. Note that each letter of the alphabet corresponds to one number.

A

B

C

D

E

F

G

H

I

J

K

L

M

1

2

3

4

5

6

7

8

9

10

11

12

13

N

0

P

Q

R

S

T

U

V

W

X

Y

Z

14

15

16

17

18

19

20

21

22

23

24

25

26

A word has been encrypted using the function, f(x) = 3x + 5, where x represents the number corresponding to the letter and f(x) represents its output or encryption. a. Determine f–1 (x). b. Use f–1 (x) as the security key and decrypt the series of numbers showing in the table. Identify the word obtained.

44

8

65

29

20

44

8

65

32

14

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62

950


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 3: Functions

c. Choose a one-word message and a function to encrypt the word, but keep the function a secret. Share the series of numbers with a classmate and see if your classmate can decode the list of numbers and identify the word.

Solution a. y  f ( x )  3 x  5

x  3y  5 x  5  3y x 5  y 3 x 5 f 1 ( x )  3 x

44

8

65

29

20

44

8

65

32

14

62

x 5 3

13

1

20

8

5

13

1

20

9

3

19

M

A

T

H

E

M

A

T

I

C

S

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951


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER 4: POLYNOMIAL AND R ATIONAL F UNCTIONS

TABLE OF CONTENTS End of Section Exercise Solutions .................................................................................. 952 Exercises 4.1 ............................................................................................................................ 952 Exercises 4.2 ........................................................................................................................... 998 Exercises 4.3 .......................................................................................................................... 1035 Exercises 4.4 .......................................................................................................................... 1063 Exercises 4.5 .......................................................................................................................... 1078 Exercises 4.6 ........................................................................................................................... 1124 Chapter Review Solutions............................................................................................... 1157 Chapter Test Solutions ................................................................................................... 1186 Group Activity Solutions ................................................................................................. 1195

END OF SECTION EXERCISE SOLUTIONS EXERCISES 4.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Solve by using the Square Root Property. 2

0 = 2(x +5) –18 Solution 0 = 2  x + 5  – 18 2

18 = 2  x + 5  9   x + 5

2

2

3   x + 5  x  5  3 x  2,  8

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952


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2. Solve by completing the square. x 2 + 6x – 15 = 0

Solution

x 2  6 x – 15  0 x 2  6 x  15 x 2  6 x  9  15  9

 x  3  24 2

x  3   24 x  3  2 6

3. Solve by using the Quadratic Formula.

–x 2 – 2 x  2  0 Solution –x 2 – 2 x  2  0 a  –1, b  –2, c  2

x

 2  4  1 2 2  1 2

2

2  12 2  2 3  2 2 x  1  3 x

4.

f x 

1 2 x – 5 x  7 Find f 0 . 2

 

Solution 2 1 f 0   0  – 5 0   7  7 2

 

 

5. Given g x  x 2  8 . For what values of x is g x  0 ?

Solution

0  x2  8 8  x2  8x 2 2  x

 

6. Use the graph of f x  x 2 to graph f  x    x  2  – 4.

Solution

2

 

Shift the graph of f x  x 2 down 4.

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953


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A quadratic function is defined by the equation __________ (a ≠ 0).

Solution f (x) = ax2 + bx + c 8. The standard form for the equation of a parabola is __________ (a ≠ 0).

Solution f (x) = a(x – h)2 + k 9. The vertex of the parabolic graph of the equation y = 2(x – 3)2 + 5 will be at __________.

Solution (3, 5) 10. The vertical line that intersects the parabola at its vertex is the __________.

Solution axis of symmetry 11. If the parabola opens __________ the vertex will be a minimum point.

Solution upward 12. If the parabola opens __________ the vertex will be a maximum point.

Solution downward 13. The x-coordinate of the vertex of the parabolic graph of f(x) = ax2 + bx + c is __________.

Solution b 2a

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954


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

14. The y-coordinate of the vertex of the parabolic graph of f(x) = ax2 + bx + c is __________.

Solution b2 c4a Practice Determine whether the graph of each quadratic function opens upward or downward. State whether a maximum or minimum point occurs at the vertex of the parabola. 15. f ( x ) =

1 2 x +5 2

Solution 1 f ( x) = x2 + 3 2 a = 21  a > 0 up, minimum 16. f ( x ) = 2 x 2 - 7 x

Solution

f ( x ) = 2x 2 - 3x a=2a>0 up, minimum 2

17. f ( x ) = - 3 ( x + 1) + 7

Solution 2

f ( x ) = -3 ( x + 1) + 2 a = -3  a < 0 down, maximum 2

18. f ( x ) = - 5 ( x - 1) - 3

Solution 2

f ( x ) = -5 ( x - 1) - 1 a = -5  a < 0 down, maximum 19. f ( x ) = -2 x 2 + 5 x - 4

Solution f ( x ) = -2 x 2 + 5 x - 1 a = -2  a < 0 down, maximum

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955


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

20. f ( x ) = 2 x 2 - 3 x + 6

Solution

f ( x ) = 2x 2 - 3x + 1 a=2a>0 up, minimum Find the vertex of each parabola. 21. f ( x ) = x 2 - 1

Solution 2

y = x 2 - 1 = ( x - 0) - 1 Vertex: (0, -1) 22. f ( x ) = -x 2 + 2

Solution 2

y = -x 2 + 2 = - ( x - 0) + 2 Vertex: (0, 2) 2

23. f ( x ) = ( x - 3) + 5

Solution 2

f ( x ) = ( x - 3) + 5 Vertex: (3, 5) 2

24. f ( x ) = - 2 ( x - 3) + 4

Solution 2

f ( x ) = -2 ( x - 3) + 4 Vertex: (3, 4) 2

25. f ( x ) = - 2 ( x + 6) - 4

Solution 2

f ( x ) = -2 ( x + 6) - 4 Vertex: (-6, - 4) 26. f ( x ) =

2 1 x + 1) - 5 ( 3

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956


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution 2

f ( x ) = 31 ( x + 1) - 5 Vertex: (-1, - 5)

27. f ( x ) =

2 2 x - 3) ( 3

Solution 2

f ( x ) = 23 ( x - 3) Vertex: (3, 0)

2

28. f ( x ) = 7 ( x + 2) + 8

Solution 2

f ( x ) = 7 ( x + 2) + 8 Vertex: (-2, 8) 29. f ( x ) = x 2 - 4 x + 4

Solution

f ( x ) = x 2 - 4 x + 4; a = 1, b = -4, c = 4 x =-

b -4 ==2 2a 2 (1)

y = x 2 - 4 x + 4 = 22 - 4 (2) + 4 = 0 Vertex: (2,0) 30. f ( x ) = x 2 - 10 x + 25

Solution

y = x 2 - 10 x + 25; a = 1, b = -10, c = 25 x =-

b -10 ==5 2a 2 (1)

y = x 2 - 10 x + 25 = 52 - 10 (5) + 25 = 0 Vertex: (5,0) 31. f ( x ) = x 2 + 6 x - 3

Solution y = x 2 + 6 x - 3; a = 1, b = 6, c = -3 x =-

b 6 == -3 2a 2 (1)

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957


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2

y = x 2 + 6 x - 3 = (-3) + 6 (-3) - 3 = -12 Vertex: (-3, - 12)

32. f ( x ) = -x 2 + 9 x - 2

Solution

y = -x 2 + 9 x - 2; a = -1, b = 9, c = -2 x =-

b 9 9 == 2a 2 (-1) 2 2

æ9ö æ9ö y = -x 2 + 9 x - 2 = - ççç ÷÷÷ + 9 ççç ÷÷÷ - 2 è 2 ø÷ è 2 ÷ø = æ 9 73 ÷ö ÷÷ Vertex: ççç , è 2 4 ÷ø

73 4

33. f ( x ) = -2 x 2 + 12 x - 17

Solution

y = -2 x 2 + 12 x - 17; a = -2, b = 12, c = -17 x =-

b 12 ==3 2a 2 (-2)

y = -2 x 2 + 12 x - 17 2

= -2 (3) + 12 (3) - 17 = 1 Vertex: (3, 1) 34. f ( x ) = 2 x 2 + 16 x + 33

Solution

y = 2 x 2 + 16 x + 33; a = 2, b = 16, c = 33 x =-

b 16 == -4 2a 2 (2)

y = 2 x 2 + 16 x + 33 2

= 2 (-4) + 16 (-4) + 33 = 1 Vertex: (-4, 1) 35. f ( x ) = 3 x 2 - 4 x + 5

Solution

y = 3 x 2 - 4 x + 5; a = 3, b = -4, c = 5

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958


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x =-

b -4 4 2 == = 2a 2 (3) 6 3 2

æ 2ö æ 2ö y = 3 x - 4 x + 5 = 3 çç ÷÷÷ - 4 çç ÷÷÷ + 5 çè 3 ÷ø çè 3 ÷ø 2

= æ 2 11ö Vertex: çç , ÷÷÷ çè 3 3 ÷ø

11 3

36. f ( x ) = -4 x 2 + 3 x + 4

Solution

y = -4 x 2 + 3 x + 4; a = -4, b = 3, c = 4 3 3 b x === 2a 2 (-4) 8 2

æ 3ö æ 3ö y = -4 x + 3 x + 4 = -4 çç ÷÷÷ + 3 çç ÷÷÷ + 4 çè 8 ÷ø çè 8 ÷ø 2

= æ 3 73 ÷ö ÷ Vertex: çç , çè 8 16 ÷÷ø 37. f ( x ) =

73 16

1 2 x + 4x - 3 2

Solution 1 y = x 2 + 4 x - 3; 2 1 a = , b = 4, c = -3 2 4 b x === -4 2a 2 ( 21 ) 1 2 x + 4x - 3 2 2 1 = (-4) + 4 (-4) - 3 = -11 2 Vertex: (-4, - 11) y=

38. f ( x ) = -

2 2 x + 3x - 5 3

Solution

y =-

2 2 x + 3 x - 5; 3

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959


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2 a = - , b = 3, c = -5 3 3 3 9 b x === 4 = 2a 4 2 (- 23 ) 3 2 2 x + 3x - 5 3 2 æ9ö 2 æ9ö 13 = - çç ÷÷÷ + 3 çç ÷÷÷ - 5 = çè 4 ÷ø 3 çè 4 ÷ø 8

y =-

æ9 13 ö Vertex: çç , - ÷÷÷ çè 4 8 ÷ø

Use the graph to identify the vertex, x-intercepts, y-intercept, axis of symmetry, domain, range, and minimum or maximum point. 39.

Solution vertex: (2, 4) x-intercepts: (0, 0), (4, 0) y-intercept: (0, 0) axis of symmetry: x = 2 domain: (–, ) range: (–, 4] maximum point: (2, 4) 40.

Solution vertex : (3, –4) x-intercepts: (1, 0), (5, 0) y-intercept: (0, 5) axis of symmetry: x = 3

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960


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

domain: (–, ) range: [–4, ) minimum point: (3, –4)

Graph each quadratic function given in standard form. Identify the vertex, intercepts, and axis of symmetry. 41. f ( x ) = x 2 - 4

Solution 2

f ( x ) = x 2 - 4 = ( x - 0) - 4 a = 1  up, vertex: (0, - 4) 0 = x2 - 4 0 = ( x + 2)( x - 2) x = -2, x = 2  (-2, 0) , (2, 0) f (0) = -4  (0, -4) axis of symmetry: x = 0 f (1) = -3  (1, - 3)

(-1, - 3) on graph by symmetry

42. f ( x ) = x 2 + 1

Solution 2

f ( x ) = x 2 + 1 = ( x - 0) + 1 a = 1  up, vertex: (0, 1) 0 = x 2 + 1  impossible no x-intercepts f (0) = 1  (0, 1) axis of symmetry: x = 0 f (1) = 2  (1, 2)

(-1, 2) on graph by symmetry

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961


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

43. f ( x ) = -3 x 2 + 6

Solution 2

f ( x ) = -3 x 2 + 6 = -3 ( x - 0) + 6 a = -3  down, vertex: (0, 6) 0 = -3 x 2 + 6 x2 = 2

(

) ( 2, 0)

x =  2  - 2, 0 , f (0) = 6  (0, 6)

axis of symmetry: x = 0 f (1) = 3  (1, 3)

(-1, 3) on graph by symmetry

44. f ( x ) = -4 x 2 + 4

Solution 2

f ( x ) = -4 x 2 + 4 = -4 ( x - 0) + 4 a = -4  down, vertex: (0, 4) 0 = -4 x 2 + 4 x2 = 1 x =  1  (-1, 0) , (1, 0) f (0) = 4  (0, 4) axis of symmetry: x = 0

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962


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

f (2) = -12  (2, - 12)

(-2, - 12) on graph by symmetry

45. f ( x ) = -

1 2 x +8 2

Solution 2

f ( x ) = - 21 x 2 + 8 = - 21 ( x - 0) + 8 a = - 21  down, vertex: (0, 8) 0 = - 21 x 2 + 8 x 2 = 16 x =  4  (-4, 0) , (4, 0) f (0) = 8  (0, 8) axis of symmetry: x = 0 f (2) = 6  (2, 6)

(-2, 6) on graph by symmetry

46. f ( x ) =

1 2 x -2 2

Solution 2

f ( x ) = 21 x 2 - 2 = 21 ( x - 0) - 2 a = 21  up, vertex: (0, - 2)

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963


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

0 = 21 x 2 - 2 4 = x2 x =  2  (-2, 0) , (2, 0) f (0) = -2  (0, - 2) axis of symmetry: x = 0 f (4) = 6  (4, 6)

(-4, 6) on graph by symmetry

2

47. f ( x ) = ( x - 3) - 1

Solution 2

f ( x ) = ( x - 3) - 1 a = 1  up, vertex: (3, - 1) 2

0 = ( x - 3) - 1 2

1 = ( x - 3) 1 = x -3 31= x

x = 2, x = 4  (2, 0) , (4, 0) f (0) = 8  (0, 8) axis of symmetry: x = 3

(6, 8) on graph by symmetry

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964


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2

48. f ( x ) = ( x + 3) - 1

Solution 2

f ( x ) = ( x + 3) - 1 a = 1  up, vertex: (-3, - 1) 2

0 = ( x + 3) - 1 2

1 = ( x + 3) 1 = x +3 -3  1 = x

x = -4, x = -2  (-4, 0) , (-2, 0) f (0) = 8  (0, 8) axis of symmetry: x = -3

(-6, 8) on graph by symmetry

2

49. f ( x ) = 2 ( x + 1) - 2

Solution 2

f ( x ) = 2 ( x + 1) - 2 a = 2  up, vertex: (-1, - 2) 2

0 = 2 ( x + 1) - 2 2

2 = 2 ( x + 1) 2

1 = ( x + 1) 1 = x + 1 -1  1 = x

x = -2, x = 0  (-2, 0) , (0, 0) f (0) = 0  (0, 0) axis of symmetry: x = -1

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965


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

f (1) = 6  (1, 6)

(-3, 6) on graph by symmetry

50. f ( x ) = -

2 3 x - 2) ( 4

Solution 2

f ( x ) = - 43 ( x - 2)

a = - 43  down, vertex: (2, 0) 2

0 = - 43 ( x - 2) 2

0 = ( x - 2) 0 = x - 2

x = 2  (2, 0) f (0) = - 3  (0, - 3) axis of symmetry: x = 2

(4, - 3) on graph by symmetry

2

51. f ( x ) = - ( x + 4) + 1

Solution 2

f ( x ) = -( x + 4) + 1 a = -1  down, vertex: (-4, 1)

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966


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2

0 = - ( x + 4) + 1 2

( x + 4) = 1 x +4 = 1 x = -4  1 x = -5, x = -3  (-5, 0) , (-3, 0) f (0) = -15  (0, - 15) axis of symmetry: x = -4

(-8, - 15) on graph by symmetry

2

52. f ( x ) = - 3 ( x - 4 ) + 3

Solution 2

f ( x ) = -3 ( x - 4) + 3 a = -3  down, vertex: (4, 3) 2

0 = -3 ( x - 4) + 3 2

3 ( x - 4) = 3 2

( x - 4) = 1 x - 4 = 1 x =41 x = 3, x = 5  (3, 0) , (5, 0) f (0) = -45  (0, - 45) axis of symmetry: x = 4

(8, - 45) on graph by symmetry

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967


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2

53. f ( x ) = - 3 ( x - 2) + 6

Solution 2

f ( x ) = -3 ( x - 2) + 6 a = -3  down, vertex: (2, 6) 2

0 = -3 ( x - 2) + 6 2

3 ( x - 2) = 6 2

( x - 2) = 2 x -2 =  2 x = 2 2

(2 - 2, 0) , (2 + 2, 0) f (0) = -6  (0, - 6) axis of symmetry: x = 2

(4, - 6) on graph by symmetry

2

54. f ( x ) = 2 ( x - 3) - 4

Solution 2

f ( x ) = 2 ( x - 3) - 4 a = 2  up, vertex: (3, - 4)

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968


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2

0 = 2 ( x - 3) - 4 2

4 = 2 ( x - 3) 2

2 = ( x - 3)  2 = x -3 3 2 = x

(3 - 2, 0) , (3 + 2, 0) f (0) = 14  (0, 14)

axis of symmetry: x = 3

(6, 14) on graph by symmetry

55. f ( x ) =

2 1 x - 1) - 3 ( 3

Solution 2

f ( x ) = 31 ( x - 1) - 3

a = 31  up, vertex: (1, - 3) 2

0 = 31 ( x - 1) - 3 2

0 = ( x - 1) - 9 2

9 = ( x - 1) 3 = x - 1 x = 1 3

x = 4, x = -2  (4, 0) , (-2, 0) f (0) = - 83  (0, - 83 ) axis of symmetry: x = 1

(2, - 83 ) on graph by symmetry

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969


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

56. f ( x ) = -

2 1 x + 1) + 8 ( 2

Solution 2

f ( x ) = - 21 ( x + 1) + 8

a = - 21  down, vertex: (-1, 8) 2

0 = - 21 ( x + 1) + 8 1 2

2

( x + 1) = 8 2

( x + 1) = 16 x + 1 = 4 x = -1  4 x = 3, x = -5  (3, 0) , (-5, 0) f (0) = 152  (0, 152 ) axis of symmetry: x = -1

(-2, 152 ) on graph by symmetry

Graph each quadratic function given in general form. Identify the vertex, intercepts, and axis of symmetry. 57. f ( x ) = x 2 + 2 x

Solution f ( x ) = x 2 + 2 x; a = 1, b = 2, c = 0 x =-

b 2 == -1 2a 2 (1)

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970


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2

y = x 2 + 2 x = (-1) + 2 (-1) = -1 vertex: (-1, - 1) , a = 1  up 0 = x 2 + 2x 0 = x ( x + 2) x = 0 or x = -2  (0, 0) , (-2, 0) f (0) = 0  (0, 0) axis of symmetry: x = -1 f (1) = 3  (1, 3) on graph

(-3, 3) on graph by symmetry

58. f ( x ) = x 2 - 6 x

Solution

f ( x ) = x 2 - 6 x; a = 1, b = -6, c = 0 x =-

b -6 ==3 2a 2 (1) 2

y = x 2 - 6 x = (3) - 6 (3) = -9 vertex: (3, - 9) , a = 1  up 0 = x 2 - 6x 0 = x ( x - 6) x = 0 or x = 6  (0, 0) , (6, 0) f (0) = 0  (0, 0) axis of symmetry: x = 3 f (1) = -5  (1, - 5) on graph

(5, - 5) on graph by symmetry

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971


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

59. f ( x ) = x 2 - 6 x - 7

Solution f ( x ) = x 2 - 6 x - 7; a = 1, b = -6, c = -7 x =-

b -6 ==3 2a 2 (1) 2

y = x 2 - 6 x - 7 = (3) - 6 (3) - 7 = -16 vertex: (3, - 16) , a = 1  up

0 = x 2 - 6x - 7 0 = ( x + 1)( x - 7) x = -1 or x = 7  (-1, 0) , (7, 0) f (0) = -7  (0, - 7) axis of symmetry: x = 3

(6, - 7) on graph by symmetry

60. f ( x ) = x 2 - 4 x + 1

Solution f ( x ) = x 2 - 4 x + 1; a = 1, b = -4, c = 1 x =-

b -4 ==2 2a 2 (1)

y = x 2 - 4 x + 1 = 22 - 4 (2) + 1 = -3 vertex: (2, - 3) , a = 1  up

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972


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

0 = x2 - 4x + 1 x = 2  3 by quadratic formula

(2 - 3, 0) , (2 + 3, 0) f (0) = 0  (0, 1)

axis of symmetry: x = 2

(4, 1) on graph by symmetry

61. f ( x ) = -x 2 - 4 x + 1

Solution f ( x ) = -x 2 - 4 x + 1 a = -1, b = -4, c = 1 -4 b == -2 x =2a 2 (-1) 2

y = - (-2) - 4 (-2) + 1 = 5 vertex: (-2, 5) , a = -1  down

0 = -x 2 - 4 x + 1 x = -2  5 by quadratic formula

(-2 - 5, 0) , (-2 + 5, 0) f (0) = 0  (0, 1)

axis of symmetry: x = -2

(-4, 1) on graph by symmetry

62. f ( x ) = -x 2 - x + 6

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973


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

f ( x ) = -x 2 - x + 6 a = -1, b = -1, c = 6 x =-

1 b -1 ==2a 2 2 (-1) 2

æ 1ö æ 1ö 25 y = - çç- ÷÷÷ - çç- ÷÷÷ + 6 = çè 2 ÷ø çè 2 ÷ø 4 æ 1 25 ö÷ ÷÷ , a = -1  down vertex: ççç- , è 2 4 ø÷ 0 = -x 2 - x + 6 0 = x2 + x - 6 0 = ( x + 3)( x - 2) x = -3, x = 2  (-3, 0) , (2, 0) f (0) = 6  (0, 6) 1 2 h by symmetry 1, 6 on grap ( ) axis of symmetry: x = -

63. f ( x ) = 2 x 2 - 12 x + 10

Solution f ( x ) = 2 x 2 - 12 x + 10 a = 2, b = -12, c = 10 x =-

b -12 ==3 2a 2 (2) 2

y = 2 (3) - 12 (3) + 10 = -8 vertex: (3, - 8) , a = 2  up 0 = 2 x 2 - 12 x + 10 0 = 2 ( x - 1)( x - 5) x = 1 or x = 5  (1, 0) , (5, 0) f (0) = 10  (0, 10)

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974


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

axis of symmetry: x = 3

(6, 10) on graph by symmetry

64. f ( x ) = -3 x 2 - 3 x + 18

Solution f ( x ) = -3 x 2 - 3 x + 18 a = -3, b = -3, c = 18 x =-

1 b -3 ==2a 2 2 (-3) 2

æ 1ö æ 1ö 75 y = -3 çç- ÷÷÷ - 3 çç- ÷÷÷ + 18 = çè 2 ÷ø çè 2 ÷ø 4

æ 1 75 ö÷ ÷÷ , a = -3  down vertex: ççç- , è 2 4 ø÷ 0 = -3 x 2 - 3 x + 18 0 = -3 ( x + 3)( x - 2) x = -3 or x = 2  (-3, 0) or (2, 0) f (0) = 18  (0, 18) 1 2 1, 18 on graph by symmetry ( ) axis of symmetry: x = -

65. f ( x ) = -3 x 2 - 6 x - 9

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975


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

f ( x ) = -3 x 2 - 6 x - 9 a = -3, b = -6, c = -9 x =-

-6 b == -1 2a 2 (-3) 2

y = -3 (-1) - 6 (-1) - 9 = -6 vertex: (-1, - 6) , a = -3  down 0 = -3 x 2 - 6 x - 9 0 = -3 ( x 2 + 2 x + 3) impossible  no x-intercepts f (0) = -9  (0, - 9) axis of symmetry: x = -1

(-2, - 9) on graph by symmetry

66. f ( x ) = -4 x 2 - 4 x + 3

Solution f ( x ) = -4 x 2 - 4 x + 3 a = -4, b = -4, c = 3 b -4 1 x ===2a 2 2 (-4) 2

æ 1ö æ 1ö y = -4 çç- ÷÷÷ - 4 çç- ÷÷÷ + 3 = 4 çè 2 ÷ø çè 2 ÷ø æ 1 ö vertex: ççç- , 4÷÷÷ , a = -4  down è 2 ø÷ 0 = -4 x 2 - 4 x + 3 0 = 4x2 + 4x - 3 0 = (2 x + 3)(2 x - 1) æ 3 ö æ1 ö 3 1 x = - , x =  ççç- , 0÷÷÷ , ççç , 0÷÷÷ 2 2 è 2 ø÷ è 2 ø÷ f (0) = -9  (0, - 9)

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976


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

axis of symmetry: x = -1

(-2, - 9) on graph by symmetry

67. f ( x ) =

1 2 5 x - 2x 2 2

Solution

f ( x ) = 21 x 2 - 2x - 52 a = 21 , b = -2, c = - 52 x =-

b -2 ==2 2a 2 ( 21 ) 2

y = 21 (2) - 2 (2) - 52 = - 92

vertex: (2, - 92 ) , a = 21  up 0 = 21 x 2 - 2 x - 52 0 = x2 - 4x - 5 0 = ( x + 1)( x - 5) x = -1, x = 5  (-1, 0) , (5, 0) f (0) = - 52  (0, - 52 ) axis of symmetry: x = 2

(4, - 52 ) on graph by symmetry

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977


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

68. f ( x ) = -

1 2 x -x +4 2

Solution

f ( x ) = - 21 x 2 - x + 4 a = - 21 , b = -1, c = 4 x =-

-1 b == -1 2a 2 (- 21 ) 2

y = - 21 (-1) - 1(-1) + 4 = 92

vertex: (-1, 92 ) , a = - 21  down 0 = - 21 x 2 - x + 4 0 = x 2 + 2x - 8 0 = ( x + 4)( x - 2) x = -4, x = 2  (-4, 0) , (2, 0) f (0) = 4  (0, 4) axis of symmetry: x = -1

(-2, 4) on graph by symmetry

Fix It In exercises 69 and 70, identify the step the first error is made and fix it.

 

69. Write f x  3 x 2  12 x  11 in standard form.

Solution Step 5 was incorrect.

  Step 2: f  x   3  x  4 x  4  4   11 Step 3: f  x   3  x  4 x  4   3  4   11

Step 1: f  x   3 x 2  4 x  11 2

2

Step 4: f  x   3 x 2  4 x  4  12  11 Step 5: f  x   3  x  2   11 2

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978


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

70. Graph the parabola f  x   –2  x  1  10 by determining the direction it opens, finding 2

its vertex, finding its x-intercepts, finding its y-intercept, and then drawing the curve that passes through those points.

Solution Step 3 is incorrect.

Step 3: x-intercepts are 1  5, 0 and 1  5, 0

Applications 71. Minimum product Find a pair of numbers whose difference is 20 and whose product is minimized. What is the minimum product?

Solution Let x = first number and let y = second number. Then, x  y  20 y  x  20

Product = xy

 x  x  20   x 2  20 x Minimum product results from x 

b 20   10 2a 2

10  y  20  y  10 y  10 Therefore, the minimum product 2 numbers whose difference is 10 is 10  10  100 72. Minimum sum The sum of two positive numbers is 10. What is the smallest possible value of the sum of their squares?

Solution If the sum of 2 positive numbers is 10, then x  y  10, and y  10  x . The sum of their squares is represented by: S  x   x 2   10  x 

2

S  x   x 2  100  20 x  x 2 S  x   2 x 2  20 x  100  b 20  5 2a 4 y  10  5  5

Minimum sum results from x 

Therefore, the minimum sum of squares is 52  52  50

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979


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

73. Police investigations A police officer seals off the scene of an accident using a roll of yellow tape that is 300 feet long. What dimensions should be used to seal off the maximum rectangular area around the collision? Find the maximum area.

Solution Let x = the width of the region.

300 - 2x = 150 - x = the length. 2 Area = width ⋅ length

Then

y = x (150 - x ) y = -x 2 + 150 x a = -1, b = 150, c = 0 b 150 x === 75 2a 2 (-1) 150 - x = 150 - 75 = 75 y = 75 (150 - 75) = 5625

The dimensions are 75 ft by 75 ft, with an area of 5625 ft2. 74. Maximizing area A rectangular flower bed has a width of x feet and a perimeter of 100 feet. Find x such that the area of the rectangle is maximized.

Solution Let x = the width of the region. 100 - 2 x = 50 - x = the length. Then 2 Area = width ⋅ length y = x (50 - x ) y = -x 2 + 50 x a = -1, b = 50, c = 0 x =-

50 b == 25 ft 2a 2 (-1)

75. Maximizing land area Jaeden has 800 feet of fencing to enclose a rectangular plot of land that borders a river. If Jaeden doesn’t need a fence along the side of the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

Solution Let x = the width. Then 800 - 2x = the length.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

980


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Area = lw y = (800 - 2 x ) x y = -2 x 2 + 800 x a = -2, b = 800, c = 0

x =-

b 800 == 200 2a 2 (-2)

y = (800 - 400) 200 = 80,000 The maximum area of 80,000 ft2 has dimensions 200 ft by 400 ft. 76. Maximizing parking lot area A rectangular parking lot is being constructed for your college football stadium. If the parking lot is bordered on one side by a street and there are 750 yards of fencing available for the other three sides, find the length and width of the lot that will maximize the area. What is the largest area that can be enclosed?

Solution Let x = the width.

Then 750 - 2 x = the length. Area = lw y = (750 - 2 x ) x y = -2 x 2 + 750 x a = -2, b = 750, c = 0 b 750 == 187.5 x =2a 2 (-2) y = (750 - 375) 187.5 = 70, 312.5 The maximum area of 70,312.5 yd2 has dimensions 187.5 yd by 375 yd. 77. Maximizing storage area A farmer wants to partition a rectangular feed storage area in a corner of his barn, as shown in the illustration. The barn walls form two sides of the stall, and the farmer has 50 feet of partition for the remaining two sides. What dimensions will maximize the area?

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981


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution Let x = the width. Then 50 - x = the length. Area = lw y = (50 - x ) x y = -x 2 + 50 x a = -1, b = 50, c = 0 b 50 x === 25 2a 2 (-1) y = (50 - 25) 25 = 625

The maximum area occurs when the dimensions are 25 ft by 25 ft. 78. Maximizing grazing area A rancher wishes to enclose a rectangular partitioned corral with 1800 feet of fencing. (See the illustration.) What dimensions of the corral would enclose the largest possible area? Find the maximum area.

Solution Set up the variables:

Area = lw æ 1800 - 3 x ö÷ ÷÷ y = x ççç 2 è ø÷ y = - 32 x 2 + 900 x a = - 32 , b = 900, c = 0 x =-

b 900 == 300 2a 2 (- 32 ) 2

y = - 32 (300) + 900 (300) = 135,000 The maximum area occurs with dimension of 300 ft by 450 ft, for an area of 135,000 ft2

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982


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

79. Sheet metal fabrication A 24-inch-wide sheet of metal is to be bent into a rectangular trough with the cross section shown in the illustration. Find the dimensions that will maximize the amount of water the trough can hold. That is, find the dimensions that will maximize the cross-sectional area.

Solution Set up the variables:

Area = lw y = x (24 - 2 x ) y = 24 x - 2 x 2 y = -2 x 2 + 24 x a = -2, b = 24, c = 0 x =-

b 24 ==6 2a 2 (-2) 2

y = -2 x 2 + 24 x = -2 (6) + 24 (6) = 72 The maximum area occurs when the depth is 6 inches and the width is 12 inches. 80. Maximizing cross-sectional area A 90-foot-wide sheet of metal is to be bent to form a rectangular trough from which your animals will drink water. Find the dimensions that will maximize the amount of water the trough can hold. That is, find the crosssectional area of the trough.

Solution Set up the variables:

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

983


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Area = lw y = x (90 - 2 x ) y = 90 x - 2 x 2 y = -2 x 2 + 90 x a = -2, b = 90, c = 0 90 b x === 22.5 2a 2 (-2) y = -2 x 2 + 90 x 2

= -2 (22.5) + 90 (22.5) = 1012.5 The maximum area occurs when the depth is 22.5 ft and the width is 45 ft. 81. Architecture A parabolic arch has an equation of x2 + 20y – 400 = 0, where x is measured in feet. Find the maximum height of the arch.

Solution x 2 + 20 y - 400 = 0 20 y = -x 2 + 400 y =-

1 2 x + 20 20

1 , b = 0, c = 20 20 b 0 ==0 x =2a 2 (- 201 )

a=-

2 1 2 1 x + 20 = 0) + 20 = 20 ( 20 20 The maximum height is 20 feet.

y =-

82. Path of a guided missile A guided missile is propelled from the origin of a coordinate system with the x-axis along the ground and the y-axis vertical. Its path, or trajectory, is given by the equation y = 400x – 16x2. Find the object’s maximum height.

Solution y = 400 x - 16 x 2 y = -16 x 2 + 400 x a = -16, b = 400, c = 0

x =-

b 400 25 == 2a 2 2 (-16) 2

æ 25 ö æ 25 ö y = 400 x - 16 x 2 = 400 çç ÷÷÷ - 16 çç ÷÷÷ èç 2 ø÷ èç 2 ø÷ = 2500 The maximum height is 2500 units.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

984


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

83. Height of a basketball The path of a basketball thrown from the free throw line can be modeled by the quadratic function f (x) = –0.06x2 + 1.5x + 6, where x is the horizontal distance (in feet) from the free throw line and f(x) is the height (in feet) of the ball. Find the maximum height of the basketball.

Solution

f ( x ) = -0.06 x 2 + 1.5 x + 6  a = -0.06, b = 1.5, c = 6 x =-

b 1.5 == 12.5 2a 2 (-0.06) 2

f (12.5) = -0.06 (12.5) + 1.5 (12.5) + 6 = 15.375 The maximum height is about 15.4 ft. 84. Projectile motion Devin throws a ball up a hill that makes an angle of 458 with the horizontal. The ball lands 100 feet up the hill. Its trajectory is a parabola with equation y = –x2 + ax for some real number a. Find a.

Solution Since the triangle is a 45 –45 –90 triangle, we get the figure below:

Use the Pythagorean Theorem to find x: x 2 + x 2 = 1002 2 x 2 = 1002 1002 2 1002 100 x= = 2 2

x2 =

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985


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Use the positive value for x. æ 100 100 ö÷ 2 The point ççç , ÷÷ must be on the graph: y = –x + ax çè 2 2 ÷ø 2

æ 100 ö÷ æ 100 ö÷ ÷÷ + a çç ÷÷ = - ççç çç çè 2 ÷ø è 2 ø÷ 2 æ 100 2 ö÷ 100 2 1002 ç ÷÷ =+ a çç ççè 2 ÷ø÷ 2 2 100

(

)

100 2 = -1002 + 100 2 a

(

)

100 2 + 1002 = 100 2 a 100 2 + 1002 100 2

= a  a = 1+

100 2

 a = 1 + 50 2

85. Height of a football A football is thrown by a quarterback from the 10-yard line and caught by the wide receiver on the 50-yard line. The football’s path on this interval can be modeled by the quadratic function f ( x ) = - 201 x 2 + 3 x - 19, where x is the horizontal distance in yards from the goal line and f(x) is the height of the football in feet. Find the maximum height reached by the football.

Solution

f ( x ) = - 201 x 2 + 3 x - 19  a = - 201 , b = 3, c = -19 x =-

b 3 == 30 2a 2 (- 201 ) 2

y = - 201 x 2 + 3 x - 19 = - 201 (30) + 3 (30) - 19 = 26 The maximum height is about 26 ft. 86. Maximizing height A ball is thrown straight up from the top of a building 144 feet tall with an initial velocity of 64 feet per second. The height sstd (in feet) of the ball from the ground, at time t (in seconds), is given by s (t ) = 144 + 64t - 16t 2 . Find the maximum height attained by the ball.

Solution

s (t ) = 144 + 64t - 16t 2 = -16t 2 + 64t + 144  a = -16, b = 64, c = 144 t =-

b 64 ==2 2a 2 (-16) 2

s (2) = -16 (2) + 64 (2) + 144 = 208

The maximum height is 208 ft. 87. Maximum height of hiking trail Parabola, Lookout, and the Wall Loop Trail is a 3.8 mile moderately trafficked loop trail located near Vernon, British Columbia, Canada that

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986


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

features a lake. The height in feet of the first three miles can be modeled by the quadratic function

f  x   326.5 x 2  1002.5 x  1537 where x represents the mile marker number. At what mile marker does the trail reach maximum height and what is the maximum height? Round both answers to one decimal place.

Solution

f  x   326.5 x 2  1002.5 x  1537

The maximum height occurs at x 

b 1002.5   1.5 miles 2a 2  326.5 

f  1.5   2306.1 88. Path of a javelin A college student throws a javelin for his college team and its path is a parabola modelled by the quadratic function

f  x   0.007x 2  0.32x  1.5 where f(x) is the height of the javelin in meters and x is the horizontal distance in meters. a. Find the height of the javelin when x = 10 m b. Find the maximum height of the javelin. c. Determine the horizontal distance the javelin travels.

Solution

f  x   0.007x 2  0.32x  1.5

a.

f  10   0.007  10   0.32  10   1.5  4 m 2

b. Maximum height occurs at x 

b 0.32   22.857 2a 2  0.007 

f  22.857   52 m Therefore, the maximum height of the javelin is 5.2 m. c. The horizontal distance of the javelin can be found when the height of the javelin is 0, or hits the ground.

0  0.007x 2  0.32x  1.5 Solve using the quadratic formula. x

0.32 

 0.32  4  0.007  1.5  22.857 2  0.007  2

0.32  0.38  50 and  4.3 0.014

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987


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Since negative distance does not make sense, 50 m is the distance the javelin travels. 89. Zero-gravity adventure To allow passengers to experience zero gravity, a Boeing 727 flies in parabolic arcs to create a weightless environment. The aircraft flies level, then gradually increases to a 45 angle until reaching a specific height, then travels through a parabolic arc, and finally gradually decreases to a 45 angle. This is repeated several times.

If the first parabolic arc traveled is modeled by the function

s  t   0.0128t 2 + 0.832t  20.48 where t is the time in seconds and s(t) is the height in thousands of feet, determine the time the aircraft reaches maximum height. Round to the nearest tenth. What is that maximum height? Round to the nearest thousand.

Solution The plane will reach its maximum height at x 

b 0.832   32.5 seconds. 2a 2  0.0128

The maximum height will be s 32.5  34,000 ft 90. Flat-screen television sets A wholesaler of appliances finds that they can sell (1200 – x) flat-screen television sets each week when the price is x dollars. What price will maximize revenue?

Solution

Revenue  Price  # Sold y  x  1200  x  y  1200 x  x 2 y   x 2  1200 x a  1, b  1200, c  0 Find the vertex:

x

b 1200   600 2a 2  1

The maximum revenue occurs when the prices is $600. 91. Maximizing revenue A seller of contemporary desks finds that they can sell (820 – x) desks each month when the price is x dollars. What price will maximize revenue?

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988


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution Revenue  Price  # Sold

y  x  820  x  y  820 x  x 2

y   x 2  820 x a  1, b  820, c  0 Find the vertex: x

b 820   410 2a 2  1

The maximum revenue occurs when the price is $410. 92. Minimizing cost A company that produces and sells digital cameras has determined that the total weekly cost C(x), in dollars of producing x digital cameras is given by the function C(x) = 1.5x2 – 144x + 5856. Determine the production level that minimizes the weekly cost for producing the digital cameras and find that weekly minimum cost.

Solution

C  x   1.5 x 2  144 x  5856  a  1.5, b  144, c  5856 x

144 b   48 2a 2  1.5 

C  48   1.5  48   144  48   5856  2400 2

48 camera should be made, for a minimum cost of $2400.

93. Maximizing profit A company that produces and sells chandeliers has determined that the total monthly profit P(x) in dollars of producing and selling x chandeliers is given by the function P(x) = –1.5x2 + 153x + 7215. Determine the production level that maximizes the monthly profit, and find that maximum profit.

Solution

P  x   1.5 x 2  153 x  7215  a  1.5, b  153, c  7215 x

b 153   51 2a 2  1.5 

C  51  1.5  51  153  51  7215  11, 116.5 2

51 chandeliers should be made, for a maximum profit of $11,116.50.

94. Finding mass transit fares The Municipal Transit Authority serves 150,000 commuters daily when the fare is $1.80. Market research has determined that every penny decrease in the fare will result in 1000 new riders. What fare will maximize revenue?

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989


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution Let x  # of penny decreases.

Then Fare  180  x  cents 

# Riders  150,000  1000 x Revenue  Fare  # Riders y   180  x  150, 000  1000 x  y  27, 000, 000  30, 000 x  1000 x 2 a  1000, b  30, 000, c  27,000, 000 Find the vertex: b 30,000  15  x 2a 2  1000 

The maximum revenue occurs when the fare is decreased by 15 pennies, or when the fare is deceased to $1.65. 95. Selling concert tickets Tickets for a concert are cheaper when purchased in quantity. The first 100 tickets are priced at $10 each, but each additional block of 100 tickets purchased decreases the cost of each ticket by 50¢. How many blocks of tickets should be sold to maximize the revenue?

Solution Let x  # of tickets sold (over one).

Then Charge  10  0.50 x  cents  # Tickets  100  100 x Revenue  Charge  # Tickets

y   10  0.50 x  100  100 x 

y  1000  950 x  50 x 2 a  50, b  950, c  1000 Find the vertex:

x

b 950   9.5 2a 2  50 

The maximum revenue occurs when 9 or 10 additional blocks are sold, or when there are a total of 10 or 11 blocks sold. 96. Finding hotel rates A 300-room hotel is two-thirds filled when the nightly room rate is $90. Experience has shown that each $5 increase in cost results in 10 fewer occupied rooms. Find the nightly rate that will maximize income.

Solution Let x  # of $5 increases. Then Rate  90  5 x

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990


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# Rooms  200  10 x Revenue  Rate  # Rooms

y   90  5 x  200  10 x  y  18,000  100 x  50 x 2

a  50, b  100, c  18,000 Find the vertex:

x

b 100  1 2a 2  50 

The maximum revenue occurs when the room rate increases by 1 five-dollar increment, or when the rate is $95. 97. Finding Hilton rates A 500-room Hilton hotel is 80% filled when the nightly room rate is $160. Experience has shown that each $5 increase in the rate results in 10 fewer occupied rooms. Find the nightly rate that will maximize the nightly revenue.

Solution Let x  # of $5 increases. Then Rate  160  5 x

# Rooms  0.80  500   10 x  400  10 x Revenue  Rate  # Rooms

y   160  5 x  400  10 x  y  64,000  400 x  50 x 2

a  50, b  400, c  64, 000

Find the vertex:

x

b 400  1 2a 2  50 

The maximum revenue occurs when the room rate increases by 4 five-dollar increment, or when the rate is $180.

An object is tossed vertically upward from ground level. Its height s(t), in feet, at time t seconds is given by the position function s(t) = –16t2 + 80t. Use the position function for Exercises 98–100. 98. In how many seconds does the object reach its maximum height?

Solution

s  16t 2  18t a  16, b  80, c  0 Find the x-coord. of the vertex: b 80 5 xt     2.5 2a 2  16  2 The max. height occurs after 2.5 seconds.

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991


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

99. In how many seconds does the object return to the point from which it was thrown?

Solution Let s  0 s  16t 2  18t 0  16t 2  80t 0  16t  t  5 

t  0, or t  5: The object returns after 5 sec.

100. What is the maximum height reached by the object?

Solution s  16t 2  18t a  16, b  80, c  0 Find the y -coord. of the vertex. Note: The x-coord. was found in #83. y  s  16t 2  18t  16  2.5   80  2.5   100 2

The max. height is 100 ft. Use a graphing calculator to determine the coordinates of the vertex of each parabola. You will have to select appropriate viewing windows. 101. y = 2 x 2 + 9 x - 56 Solution

y = 2 x 2 + 9x - 56 Vertex: (-2.25, - 66.13)

102. y = 14 x -

x2 5

Solution

x2 5 Vertex: (35, 245) y = 14 x -

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992


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

103. y = ( x - 7)(5 x + 2) Solution

y = ( x - 7)(5 x + 2) Vertex: (3.3, - 68.5)

104. y = -x (0.2 + 0.1x ) Solution y = - x (0.2 + 0.1x ) Vertex: (- 1, 0.1)

Use a graphing calculator and quadratic regression to find the quadratic function that best fits the given set of data. 105. {(–1, 6), (0, –1), (1, –3), (2, –1.5), (3, 5), (4, 10)} Round to three decimal places. Solution

f ( x ) = 1.679 x 2 - 3.907 x - 0.229 106. {(–3, 4), (–1, 5), (0, 7), (2, 9), (5, 7), (6, 5)} Round to three decimal places. Solution

f ( x ) = -0.176 x 2 + 0.769 x + 7.219

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993


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

107. Alligators The length (in inches) and weight (in pounds) of 25 alligators are shown in the table. Find the quadratic function that best fits the data. Round a, b, and c to six decimal places. Use the regression function to estimate the weight of an alligator that is 130 inches long. Round the weight to the nearest pound. Length

Weight

Length

Weight

Length

Weight

94

130

72

38

90

106

74

51

128

366

89

84

147

640

85

84

68

39

58

28

82

80

76

42

86

80

86

83

114

197

94

110

88

70

90

102

63

33

72

61

78

57

86

90

74

54

69

36

61

44

Solution

f ( x ) = 0.086616 x 2 - 11.317553 x + 410.484123 2

f (130) = 0.086616 (130) - 11.317553 (130) + 410.484123 » 403 lb 108. Alligators Refer to Exercise 107. If an alligator weighs 125 pounds, what is its approximate length? Round to the nearest inch. Solution f ( x ) = 0.086616 x 2 - 11.317553 x + 410.484123 = 125 0.086616 x 2 - 11.317553 x + 285.484123 = 0

a = 0.086616, b = -11.317553, c = 285.484123 Use the quadratic formula. x =

-b + b2 - 4ac 2 (a)

» 97 inches

Discovery and Writing Exercises 109. What is a quadratic function? Solution Answers may vary. 110. Describe two ways of finding the vertex of a parabola given in general form. Solution Answers may vary.

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994


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

111. What is an axis of symmetry of a parabola? Solution Answers may vary. 112. Share the strategy you would use to solve a maximum or minimum application problem. Solution Answers may vary. 113. Find the dimensions of the largest rectangle that can be inscribed in the right triangle ABC shown in the illustration.

Solution

The equation of the line is y = - 43 x + 9. Thus the point ( x, y ) = ( x, - 43 x + 9) . Area = x (- 43 x + 9) y = - 43 x 2 + 9 x a = - 43 , b = 9, c = 0 Find the x-coord of the vertex: b 9 ==6 x =2a 2 (- 43 ) Thus, the dimensions are 6 by 4 21 units. 114. Point P lies in the first quadrant and on the line x + y = 1 in such a position that the area of triangle OPA is maximum. Find the coordinates of P. (See the illustration.)

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995


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

The point ( x , y ) = ( x , 1 - x ) . Area = 21 bh

y = 21 x (1 - x ) y = 21 x - 21 x 2

a = - 21 , b = 21 , c = 0 Find the x-coord of the vertex: x =-

1 b =- 2 = 1 2a 2 (- 21 ) 2

Thus, point P has coordinates ( 21 , 21 ) . 115. The sum of two numbers is 6, and the sum of the squares of those two numbers is as small as possible. What are the numbers? Solution Let x = one number Then 6 - x = the other number 2

Sum of squares = x 2 + (6 - x )

y = x 2 + 36 - 12 x + x 2 y = 2 x 2 - 12 x + 36

a = 2, b = -12, c = 36 Find the x-coord of the vertex: b -12 x ===3 2a 2 (2) Thus, the numbers are both 3.

116. What number most exceeds its square? Solution Let x = the number. Then x 2 = its square Amt. by which it = x - x 2 exceeds square y = x - x 2

a = -1, b = 1, c = 0 Find the x-coord of the vertex: b 1 1 x === 2a 2 2 (-1) The number 21 is the number which most exceeds its square.

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996


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 117. If the sum of a number and its square is a minimum, then the two numbers are - 21 and 41 . Solution True. 118. The graphs of some quadratic functions have no x-intercepts. Solution True. 119. The graphs of some quadratic functions have no y-intercepts. Solution False. The graph of a quadratic function always has exactly one y-intercept. 120. The graph of a quadratic function is never constant. Solution True. 121. If f ( x ) = a ( x - h) + k , and a > 0, then the range is (-¥, k ùúû . 2

Solution False. The range is (k, ∞). 122. If f ( x ) = a ( x - h) + k , and a < 0, then the graph of f (x) is increasing on (-¥, hùúû . 2

Solution True 123. If g ( x ) = ax 2 + bx + c, and a > 0, then the graph of the function is increasing on é b ö ê, ¥÷÷÷ . ê 2a ÷ø ë

Solution True 124. The axis of symmetry of the parabola f ( x ) = 444 x 2 - 888 x + 222 is x = –1. Solution False. The axis is x = 1.

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997


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

EXERCISES 4.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Solve 2 x 3 – 11x 2  5 x  0 by factoring. Solution

x 2 x 2 – 11x  5 x  0 x  2 x  1 x  5   0 x  0, x 

1 , x 5 2

2. Solve x 4 – 16 x 2  63  0 by factoring. Solution

 x  9 x  7   0  x  3 x  3  x  7   0 2

2

2

x  3 x  3 x 2  7 x 7 So, x   3,  7

3. Solve x 3 – x 2 – 25 x  25  0 by factoring. Solution x 2  x – 1 – 25  x  1  0

 x – 1  x – 25  0  x  1 x  5 x  5  0 2

x  1,  5 4. Solve 2 x 5 – 6 x 3  0 by factoring. Solution

2x 3 x 2  3  0

2x 3  0

 x  3  0

x 0

x 3

2

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998


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 5. The degree of the function f (x) = x4 – 3 is __________. Solution 4 6. Peaks and valleys on a polynomial graph are called __________ points. Solution turning 7. The roots of the polynomial equation f (x) = 0 are known as __________ of the polynomial function. Solution zeros 8. The zeros of a polynomial function appear as __________ on the graph of the polynomial function. Solution x-intercepts 9. If the degree of the polynomial is odd and the leading coefficient is positive, then the graph of the polynomial function __________ on the left and __________ on the right. Solution falls, rises 10. If the degree of the polynomial is odd and the leading coefficient is negative, then the graph of the polynomial function __________ on the left and __________ on the right. Solution rises, falls 11. If the degree of the polynomial is even and the leading coefficient is positive, then the graph of the polynomial function __________ on the left and __________ on the right. Solution rises, rises 12. If the degree of the polynomial is even and the leading coefficient is negative, then the graph of the polynomial function __________ on the left and __________ on the right. Solution falls, falls 13. If (x + 5)3 occurs as a factor of a polynomial function, then the __________ of the zero x = –5 is 3.

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999


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution multiplicity 14. The graph of a nth degree polynomial function can have at most __________ turning points. Solution n–1

 

 

15. If P(x) is a polynomial with real coefficients and P a  P b for a  b, then P(x) takes on all values between __________ in the interval a, b . Solution P(a) and P(b) 16. If P(x) has real coefficients and P(a) and P(b) have opposite signs, there is at least one

number r in a, b for which __________. Solution P(r) = 0 Practice Determine whether or not the functions are polynomial functions. For those that are, state the degree. 17. f  x  

1 5 x  5 x 3  3 x  10 2

Solution polynomial degree = 5 18. f  x   0.8 x 6  5 x 3  2 x  5 Solution polynomial degree = 6 19. f  x   11x 7  3 x 2  11x  1 Solution polynomial degree = 7 20. f  x   2 x 8  3 6 x 4  2 x 3  2 x  9 Solution polynomial degree = 8 21. f  x   x 4 

x 7

Solution not a polynomial

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1000


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 

22. f x  3 x 2  5 x 1  7 Solution not a polynomial 23. f  x   6 x 2  13 

1 x

Solution not a polynomial 24. f  x   3 7 x 3  9 x 2  x Solution not a polynomial Determine whether or not the graph of the functions shown are polynomial functions. 25.

Solution polynomial 26.

Solution polynomial 27.

Solution not a polynomial

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1001


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

28.

Solution not a polynomial Find the zeros of each polynomial function and state the multiplicity of each. State whether the graph touches the x-axis and turns or crosses the x-axis at each zero.

 

29. f x  4 x 2  25 Solution

f  x   4 x 2  25 4 x 2  25  0

 2x  5 2x  5  0 x   52 , multiplicity 1, crosses x  52 , multiplicity 1, crosses

 

30. f x  64  9 x 2 Solution

f  x   64  9 x 2 64  9 x 2  0

8  3x 8  3x   0 x   83 , multiplicity 1, crosses x  83 , multiplicity 1, crosses

 

31. f x  2 x 2  7 x  15 Solution

f  x   2 x 2  7 x  15 2 x 2  7 x  15  0

 2x  3 x  5  0 x  32 , multiplicity 1, crosses x  5, multiplicity 1, crosses

 

32. f x  6 x 2  x  2 Solution

f  x   6x 2  x  2

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1002


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

6x 2  x  2  0

 2x  1 3x  2  0 x   21 , multiplicity 1, crosses x  23 , multiplicity 1, crosses

 

33. g x  2 x 3  7 x 2  15 x Solution

g  x   2 x 3  7 x 2  15 x 2 x 3  7 x 2  15 x  0

x 2 x 2  7 x  15  0 x  2 x  3 x  5   0 x  0, multiplicity 1, crosses x   32 , multiplicity 1, crosses x  5, multiplicity 1, crosses

 

34. g x  x 3  8 x 2  16 x Solution

g  x   x 3  8 x 2  16 x x 3  8 x 2  16 x  0

x x 2  8 x  16  0 x  x  4  x  4   0 x  x  4  0 2

x  0, multiplicity 1, crosses x  4, multiplicity 2, touches

 

35. g x  x 3  6 x 2  4 x  24 Solution

g  x   x 3  6 x 2  4 x  24 x 3  6 x 2  4 x  24  0

x 2  x  6  4  x  6  0

 x  6  x  4   0  x  6 x  2 x  2  0 2

x  6, multiplicity 1, crosses x  2, multiplicity 1, crosses x  2, multiplicity 1, crosses

 

36. g x  x 3  2 x 2  9 x  18

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1003


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

g  x   x 3  2 x 2  9 x  18 x 3  2 x 2  9 x  18  0

x 2  x  2  9  x  2  0

 x  2  x  9  0  x  2 x  3 x  3  0 2

x  2, multiplicity 1, crosses x  3, multiplicity 1, crosses x  3, multiplicity 1, crosses

 

37. f x  x 4  2 x 3  3 x 2 Solution

f  x   x 4  2x 3  3x 2 x 4  2x 3  3x 2  0

x 2 x 2  2x  3  0 x  x  3  x  1  0 2

x  0, multiplicity 2, touches x  3, multiplicity 1, crosses x  1, multiplicity 1, crosses

 

38. f x  x 4  3 x 3  2 x 2 Solution

f  x   x 4  3x 3  2x 2 x 4  3x 3  2x 2  0

x 2 x 2  3x  2  0 x  x  1 x  2   0 2

x  0, multiplicity 2, touches x  1, multiplicity 1, crosses x  1, multiplicity 1, crosses

 

39. f x  x 4  15 x 2  44 Solution

f  x   x 4  15 x 2  44 x 4  15 x 2  44  0

 x  11 x  4  0 2

2

 x  11  x  11   x  2 x  2  0 © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1004


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x   11, multiplicity 1, crosses x  11, multiplicity 1, crosses x  2, multiplicity 1, crosses x  2, multiplicity 1, crosses

 

40. f x  x 4  19 x 2  48 Solution

f  x   x 4  19 x 2  48 x 4  19 x 2  48  0

 x  3 x  16  0 2

2

 x  3  x  3   x  4 x  4  0 x   3, multiplicity 1, crosses x  3, multiplicity 1, crosses x  4, multiplicity 1, crosses x  4, multiplicity 1, crosses 41. h  x   3 x 2  x  4   x  5  2

Solution h  x   3x 2  x  4   x  5 2

x  0, multiplicity 2, touches x  4, multiplicity 2, touches x  5, multiplicity 1, crosses

42. h  x   2 x  x  3   x  1 2

2

Solution h  x   2 x  x  3   x  1 2

2

x  0, multiplicity 1, crosses x  3, multiplicity 2, touches x  1, multiplicity 2, touches

43. h  x    2 x  5  x  3  x  1

2

Solution h  x    2 x  5  x  3  x  1

2

x  52 , multiplicity 1, crosses x  3, multiplicity 1, crosses x  1, multiplicity 2, touches

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1005


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

44. h  x    3 x  1  x  1 3

2

Solution

h  x    3x  1  x  1 3

2

x   31 , multiplicity 3, crosses x  1, multiplicity 2, crosses

In Exercises 45 and 46, use the graph of the polynomial functions f(x) shown.

a. Identify the zeros of f(x) and determine if their multiplicities are even or odd. b. Is the leading coefficient of f(x) positive or negative? c. Is the degree of f(x) even or odd? 45.

Solution a. –3 odd; –1 even; 2 odd; 4 odd b. negative c. odd

46.

Solution a. –4 odd; –2 odd; 0 even; 3 even b. positive. c. even

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1006


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

In Exercises 47 and 48, refer to the graph of the polynomial functions f(x) shown. Using a coefficient of 1 or –1, write a possible equation of the polynomial function in factored form with least degree.

47.

Solution

f  x   x 2  x  4

48.

Solution

f  x    x 2  x  2 x  3

Use the Leading Coefficient Test to determine the end behavior of each polynomial.

49. f  x   5 x 7  10 x 3  2 x Solution f  x   5 x 7  10 x 3  2 x

Degree  7  odd ; Lead Coef: pos. falls left, rises right

 

50. f x  4 x 9  7 x 2  5 x  12

Solution

f  x   4 x 9  7 x 2  5x  12

Degree  9  odd ; Lead Coef: pos. falls left, rises right

 

51. g x   21 x 5  3 x 4  2 x 2  4

Solution

g  x    21 x 5  3 x 4  2x 2  4

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1007


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Degree  5  odd ; Lead Coef: neg. rises left, falls right

 

52. g x  3 x 7  2 x 4  5 x  2

Solution

g  x   3x 7  2x 4  5 x  2

Degree  7  odd ; Lead Coef: neg. rises left, falls right

 

53. f x  7 x 4  2 x 2  1

Solution

f  x   7 x 4  2x 2  1

Degree  4 even ; Lead Coef: pos. rises left, rises right

 

54. f x  23 x 6  3 x 3  2 x

Solution

f  x   23 x 6  3x 3  2 x

Degree  6  even ; Lead Coef: pos. rises left, rises right

 

55. h x  3 x 4  5 x  1

Solution

h  x   3x 4  5 x  1

Degree  4  even ; Lead Coef: neg. falls left, falls right

 

56. h x   x 6  3 x 2  2

Solution

h  x    x 6  3x 2  2

Degree  4  even ; Lead Coef: neg. falls left, falls right Graph each polynomial function.

 

57. f x  x 3  9 x

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1008


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

f  x   x 3  9x x-int.

y-int.

f  0   03  9  0 

x 3  9x  0

y 0

x x2  9  0

0, 0

x  x  3  x  3   0 x  0, x  3, x  3

odd deg, pos coef  falls left, rises right Sign of f ( x ) = x 3 - 9x

Test point Graph of f(x)

+

+

(-¥, -3)

(-3, 0)

(0, 3)

(3, ¥)

–3

0

3

f (-4) = -28

f (-1) = 8

f (1) = -8

f (4) = 28

below axis

above axis

below axis

above axis

3

f (- x ) = (- x ) - 9 (- x ) = - x 3 + 9 x = -f ( x )  odd, symmetric about origin

 

58. f x  x 3  16 x

Solution

f  x   x 3  16 x x-int.

y-int.

x 3  16 x  0

x x 2  16  0 x  x  4  x  4   0

f  x   03  16  0  y 0

0, 0

x  0, x  4, x  4

0, 0  ,  4, 0  ,  4, 0 

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1009


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

odd deg, pos coef  falls left, rises right Sign of f ( x ) = x 3 - 16 x

Test point Graph of f(x)

+

+

(-¥, -4)

(-4, 0)

(0, 4)

(4, ¥)

–4

0

4

f (-5) = -45

f (-1) = 15

f (1) = -15

f (5) = 45

below axis

above axis

below axis

above axis

3

f (- x ) = (- x ) - 16 (- x ) = - x 3 + 16 x = -f ( x )  odd, symmetric about origin

 

59. f x   x 3  4 x 2

Solution

f  x    x3  4x2 x-int.

y-int.

x3  4x2  0

f  0    03  4  0 

 x2  x  4  0 x  0, x  4

0, 0 ,  4, 0

 

2

y 0

0, 0

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1010


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

odd deg, neg coef  rises left, falls right Sign of

+

(-¥, -4)

(-4, 0)

(0, ¥)

–4

0

f (-5) = 25

f (-1) = -3

f (1) = -5

above axis

below axis

below axis

f ( x ) = -x 3 - 4 x 2

Test point Graph of f(x) 3

2

f (- x ) = - (- x ) - 4 (- x ) = x 3 - 4 x 2  neither even nor odd, no symmetry

 

60. f x   x 3  2x

Solution

f  x    x 3  2x x-int.

y-int.

 x 3  2x  0

f  0    03  2  0 

 

x x2  2  0

y 0

0, 0 

x  0, x   2

0, 0  ,   2, 0  ,  2, 0 

odd deg, neg coef  rises left, falls right Sign of f ( x ) = -x 3 + 2 x

Test point Graph of f(x)

+

+

(-¥, - 2 )

(- 2, 0)

(0, 2 )

( 2, ¥)

- 2

0

2

f (-2) = 4

f (-1) = -1

f (1) = 1

f (2) = -4

above axis

below axis

above axis

below axis

3

f (- x ) = - (- x ) + 2 (- x ) = x 3 - 2 x = -f ( x )  odd, symmetric about origin

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1011


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 

61. f x  x 3  x 2

Solution

f  x   x3  x2 x-int.

y-int.

x3  x2  0

f  0   03  02

x 2  x  1  0

y 0

0, 0

x  0, x  1

0, 0  ,  1, 0 

odd deg, pos coef  falls left, rises right Sign of f (x) = x3 + x2

Test point Graph of f(x) 3

+

+

(-¥, -1)

(-1, 0)

(0, ¥)

–1

0

f (-2) = -4

f (- 21 ) = 81

f ( 1) = 2

above axis

below axis

above axis

2

f (- x ) = (- x ) + (- x ) = - x 3 + x 2  neither even nor odd, no symmetry

 

62. f x  x 3  x

Solution

f  x   x3  x x-int.

y-int.

x3  x  0

f  0   03  0

x x2  1  0

y 0

x  x  1 x  1  0

0, 0

x  0, x  1, x  1

0, 0  ,  1, 0  ,  1, 0  © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1012


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

odd deg, pos coef  falls left, rises right Sign of f (x) = x3 - x

Test point Graph of f(x)

+

+

(-¥, -1)

(-1, 0)

(0, 1)

(1, ¥)

–1

0

1

f (-2) = -6

f (- 21 ) = 83

f ( 21 ) = - 83

f (2) = 6

below axis

above axis

below axis

above axis

3

f (- x ) = (- x ) - (- x ) = - x 3 + x 2 = -f ( x )  odd, symmetric about origin

 

63. f x  x 3  9 x 2  18 x

Solution

f  x   x 3  9 x 2  18 x x-int.

y-int.

x 3  9 x 2  18 x  0

f  0   03  9  0   18  0 

x x 2  9 x  18  0 x  x  3  x  6   0

2

y 0

0, 0

x  0, x  3, x  6

0, 0 ,  3, 0 , 6, 0

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1013


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

odd deg, pos coef  falls left, rises right Sign of

+

+

(-¥, 0)

(0, 3)

(3, 6)

(6, ¥)

0

3

6

f ( x ) = x 3 - 9 x 2 + 18 x

Test point Graph of f(x) 3

f (-1) = -28

f (1) = 10

f (4) = -8

f (7) = 28

below axis

above axis

below axis

above axis

2

f (- x ) = (- x ) - 9 (- x ) + 18 (- x ) = - x 3 - 9 x 2 - 18 x  neither even nor odd, no symmetry

 

64. f x   x 3  9 x 2  18 x

Solution

f  x    x 3  9 x 2  18 x x-int.

y-int.

 x 3  9 x 2  18 x  0

f  0     0   9  0   18  0 

 x x 2  9 x  18  0 x  x  3  x  6   0

3

y 0

2

0, 0

x  0, x  3, x  6

0, 0 ,  3, 0 ,  6, 0

odd deg, neg coef  rises left, falls right Sign of f ( x ) = - x 3 - 9 x 2 - 18 x

Test point Graph of f(x) 3

+

+

(-¥, -6)

(-6, - 3)

(-3, 0)

(0, ¥)

–6

–3

0

f (-7) = 28

f (-4) = -8

f (-1) = 10

f (1) = -28

above axis

below axis

above axis

below axis

2

f (- x ) = - (- x ) - 9 (- x ) - 18 (- x ) = x 3 - 9 x 2 + 18 x  neither even nor odd, no symmetry

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1014


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 

65. f x  x 3  x 2  4 x  4

Solution

f  x   x3  x2  4x  4 x-int.

y-int.

x  x  4x  4  0 3

2

x 2  x  1  4  x  1  0

f 0  0  0   4 0  4 3

y 4

 x  1  x  4   0  x  1 x  2 x  2  0

2

0, 4

2

x  1, x  2, x  2

 1, 0  ,  2, 0  ,  2, 0 

odd deg, pos coef  falls left, rises right Sign of f (x) = x3 - x2 - 4x + 4

+

+

(-¥, -2)

(-2, 1)

(1, 2)

(2, ¥)

f (-3) = -20

f (0) = 4

f( )=-

below axis

above axis

below axis

–2 Test point Graph of f(x) 3

1 3 2

2 7 8

f (3) = 10 above axis

2

f (- x ) = (- x ) - (- x ) - 4 (- x ) + 4 = - x 3 - x 2 + 4 x + 4  neither even nor odd, no symmetry

 

66. f x  4 x 3  4 x 2  x  1

Solution

f  x   4x3  4x2  x  1 x-int.

y-int.

4x3  4x2  x  1  0

f 0  4 0  4 0   0  1

4 x  x  1   x  1  0 2

 x  1  4 x  1  0 2

x  1 or x  2

3

y 1

2

0, 1

1 4

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1015


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x  1, x  21 , x   21

 1, 0 ,  , 0 ,   , 0 1 2

1 2

odd deg, pos coef  falls left, rises right Sign of f (x) = 4x3 - 4x2 - x + 1

+

+

(-¥, - 21 )

(- 21 , 21 )

( 21 , 1)

(1, ¥)

- 21

Test point Graph of f(x) 3

1 2

1

f (-1) = -6

f (0) = 1

f( )=-

below axis

above axis

below axis

3 4

5 16

f (2) = 15 above axis

2

f (- x ) = 4 (- x ) - 4 (- x ) - (- x ) + 1 = -4 x 3 - 4 x 2 + x + 1  neither even nor odd, no symmetry

 

67. f x  x 4  2 x 2  1

Solution

f  x   x 4  2x 2  1 x-int. x 4  2x 2  1  0

 x  1 x  1  0 2

2

x2  1 x  1, x  1

y-int.

f 0  0  2 0  1 4

y 1

2

0, 1

 1, 0  ,  1, 0

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1016


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

even deg, pos coef  rises left, rises right Sign of f ( x ) = x 4 - 2x2 + 1

+

+

+

(-¥, -1)

(-1, 1)

(1, ¥)

–1 Test point Graph of f(x) 4

1

f (-2) = 9

f (0) = 1

f (2) = 9

above axis

above axis

above axis

2

f (-x ) = (-x ) - 2 (-x ) + 1 = x 4 - 2x 2 + 1 = f ( x )  even, symmetric about y-axis

 

68. f x  x 4  5 x 2  4

Solution

f  x   x 4  5x 2  4 x-int.

y-int.

x4  5x2  4  0

 x  1 x  4   0 2

2

x  1 or x  4 2

2

x  1, x  1, x  2, x  2

f 0  0  5 0  4 4

y 4

2

0, 4

 1, 0 ,  1, 0  ,  2, 0  ,  2, 0

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1017


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

even deg, pos coef  rises left, rises right Sign of

+

+

+

(-¥, -2)

(-2, - 1)

(-1, 1)

(1, 2)

(2, ¥)

f ( x ) = x 4 - 5x 2 + 4

–2 Test point Graph of f(x) 4

–1

1

2

f (-3) = 40

f (- 32 ) = - 35 16

f (0) = 4

f ( 32 ) = - 35 16

f (3) = 40

above axis

below axis

above axis

below axis

above axis

2

f (-x ) = (-x ) - 5 (-x ) + 4 = x 4 - 5x 2 + 4 = f ( x )  even, symmetric about y -axis

 

69. f x   x 4  5 x 2  4

Solution

f  x    x 4  5x 2  4

x-int.

y-int.

 x 4  5x 2  4  0

f 0   0  5 0  4 4

x4  5x2  4  0



y  4

0,  4 

x2  1 x2  4  0 x  1 or x  4 2

2

2

x  1, x  1, x  2, x  2

 1, 0 ,  1, 0  ,  2, 0  ,  2, 0

even deg, neg coef  falls left, falls right Sign of f ( x ) = -x 4 + 5 x 2 - 4

+

+

(-¥, -2)

(-2, - 1)

(-1, 1)

(1, 2)

(2, ¥)

–2 Test point Graph of f(x) 4

–1

1

2

f (-3) = 40

f (- ) =

35 16

f (0) = -4

f( )=

below axis

above axis

below axis

above axis

3 2

3 2

35 16

f (3) = -40 below axis

2

f (-x ) = -(-x ) + 5 (-x ) - 4 = -x 4 + 5x 2 - 4 = f ( x )  even, symmetric about y -axis

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1018


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 

70. f x   x 4  11x 2  18

Solution

f  x    x 4  11x 2  18 x-int.

y-int.

 x 4  11x 2  18  0

f  0    0   11  0   18 4

x  11x  18  0 4

2

y  18

 x  2 x  9  0 2

2

0,  18

x  2 or x  9 2

2

2

x  2, x   2, x  3, x  3

 2, 0 ,   2, 0 , 3, 0 ,  3, 0

even deg, neg coef  falls left, falls right Sign of

+

+

(-¥, -3)

(-3, - 2 )

(- 2, 2)

( 2, 3)

(3, ¥)

–3

- 2

2

3

f (-4) = -98

f (-2) = 10

f (0) = -18

f (2) = 10

f (4) = -98

below axis

above axis

below axis

above axis

below axis

f ( x ) = - x 4 + 11x 2 - 18

Test point Graph of f(x) 4

2

f (-x ) = -(-x ) + 11(-x ) - 18 = -x 4 + 11x 2 - 18 = f ( x )  even, symmetric about y -axis

 

71. f x   x 4  6 x 3  8 x 2

Solution

f  x    x 4  6 x 3  8x 2

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1019


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x-int.

y-int.

 x 4  6x 3  8x 2  0

f 0   0  6 0  8 0

4

 x2 x 2  6x  8  0

y 0

 x  x  2 x  4   0 2

3

2

0, 0

x  0, x  2, x  4

0, 0  ,  2, 0  ,  4, 0 

even deg, neg coef  falls left, falls right Sign of f ( x ) = -x 4 + 6 x 3 - 8 x 2

Test point Graph of f(x)

+

(-¥, 0)

(0, 2)

(2, 4)

(4, ¥)

0

2

4

f (-1) = -15

f (1) = -3

f (3) = 9

f (5) = -75

below axis

below axis

above axis

below axis

f (- x ) = - (- x ) + 6 (- x ) - 8 (- x 2 ) = - x 4 - 6 x 3 - 8 x 2 4

3

 neither even nor odd, no symmetry

 

72. f x   x 4  2 x 3  8 x 2

Solution

f  x    x 4  2x 3  8x 2 x-int.

y-int.

 x 4  2x 3  8x 2  0

f 0   0  2 0  8 0

 x 2 x 2  2x  8  0  x  x  2  x  4   0 2

4

y 0

3

2

0, 0

x  0, x  2, x  4

0, 0 ,  2, 0 ,  4, 0

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1020


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

even deg, neg coef  falls left, falls right Sign of f ( x ) = -x 4 + 2 x 3 + 8 x 2

Test point Graph of f(x)

+

+

(-¥, - 2)

(-2, 0)

(0, 4)

(4, ¥)

0

2

4

f (-3) = -63

f (-1) = 5

f ( 1) = 9

f (5) = -175

below axis

above axis

above axis

below axis

f (- x ) = - (- x ) + 2 (- x ) + 8 (- x 2 ) = - x 4 + 2 x 3 + 8 x 2 4

3

 neither even nor odd, no symmetry

73. f  x  

1 4 9 2 x  x 2 2

Solution 1 9 f  x   x4  x2 2 2 y-int.

x-int. 1 4 9 2 x  x 0 2 2 x 4  9x 2  0

x2 x2  9  0

4 2 1 9 0  0  2 2 y 0

f 0 

0, 0

x 2  x  3 x  3   0 x  0, x  3, x  3

0, 0 ,  3, 0 ,  3, 0

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1021


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

even deg, pos coef  rises left, rises right

Sign of f ( x) =

1 4 9 2 x - x 2 2

+

+

(-¥, - 3)

(-3, 0)

(0, 3)

(3, ¥)

0

3

f (-4) = 56

f (-1) = -4

f (1) = -4

f (4) = 56

above axis

below axis

below axis

above axis

–3 Test point Graph of f(x) 4

2

f (-x ) = 21 (-x ) - 92 (-x ) = 21 x 4 - 92 x 2 = f ( x )  even, symmetric about y-axis 1 4 x  8x 2 2

74. f  x   

Solution f x  

1 4 x  8x 2 2

x-int. 

1 4 x  8x 2  0 2  x 4  16 x 2  0

x

2

 x  16  0

y-int. f 0    y 0

2

4 2 1 0   8 0   2

0, 0

 x 2  x  4  x  4   0 x  0, x  4, x  4

0, 0 ,  4, 0 ,  4, 0

even deg, neg coef  falls left, falls right

Sign of 1 f ( x ) = - x 4 + 8x 2 2

+

+

(-¥, - 4)

(-4, 0)

(0, 4)

(4, ¥)

–4

0

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4

1022


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test point Graph of f(x) 4

f (-5) = -112.5

f (-1) = 7.5

f (1) = 7.5

f (4) = -112.5

below axis

above axis

above axis

below axis

2

f (-x ) = - 21 (-x ) + 8 (-x ) = - 21 x 4 + 8x 2 = f ( x )  even, symmetric about y-axis

 





75. f x  x x  3 x  2 x  1

Solution

f  x   x  x  3 x  2 x  1  x 4  4 x 3  x 2  6 x x-int.

y-int.

x  x  3 x  2 x  1  0

f 0  0

x  0, x  3, x  2, x  1

y 0

0, 0 ,  3, 0 ,  2, 0 ,  1, 0

0, 0

even deg, pos coef  rises left, rises right

Sign of f (x) =

+

+

+

(-¥, - 1)

(-1, 0)

(0, 2)

(2, 3)

(3, ¥)

–1

0

2

3

x ( x - 3)( x - 2)( x + 1)

Test point Graph of f(x) 4

3

f (-2) = 40

f (- 21 ) = - 35 16

f ( 1) = 4

f ( 52 ) = - 35 16

f (4 ) = 40

above axis

below axis

above axis

below axis

above axis

2

f (-x ) = (-x ) - 4 (-x ) + (-x ) + 6 (-x ) = x 4 + 4 x 3 + x 2 - 6x  neither even nor odd, no symmetry

 







76. f x   x  4 x  2 x  2 x  4

Solution

f  x     x  4  x  2 x  2 x  4    x 4  20 x 2  64

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1023


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x-int.

y-int.

  x  4  x  2  x  2  x  4   0

f  0   64

x  4, x  2, x  2, x  4

y  64

0,  64

 4, 0 ,  2, 0 ,  2, 0  ,  4, 0

even deg, pos coef  falls left, falls right Sign of

f (x) = 4

+

+

(-¥, - 4)

(-4, - 2)

(-2, 2)

(2, 4)

(4, ¥)

2

- x + 20 x - 64

Test point Graph of f(x) 4

–4

–2

2

f (-5) = -189

f (-3) = 35

f (0) = -64

f (3) = 35

f (5) = -189

below axis

above axis

below axis

above axis

below axis

4

2

f (-x ) = -(-x ) + 20 (-x ) - 64 = -x 4 + 20x 2 - 64 = f ( x )  even, symmetric about y-axis

 

77. f x  x 5  4 x 3

Solution

f  x   x5  4x3

x

x-int.

y-int.

x5  4x 3  0

f 0  0

3

 x  4  0 2

x  x  2  x  2   0 3

y 0

0, 0

x  0, x  2, x  2

0, 0 ,  2, 0 ,  2, 0

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1024


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

odd deg, pos coef  falls left, rises right Sign of

+

+

(-¥, - 2)

(-2, 0)

(0, 2)

(2, ¥)

f (x) = x5 - 4x3

Test point Graph of f(x) 5

–2

0

2

f (-3) = -135

f (-1) = 3

f (1) = -3

f (3) = 135

below axis

above axis

below axis

above axis

3

f (-x ) = (-x ) - 4 (-x ) = -x 5 + 4 x 3 = -f ( x )  odd, symmetric about origin

 

78. f x   x 5  8 x 3

Solution

f  x    x 5  8x 3 x-int.

y-int.

 x 5  8x 3  0

x

x3 x  8

3

 x  8  0 2

 x  8   0

f 0  0 y 0

0, 0

x  0, x  2 2, x  2 2

0, 0 ,  2 2, 0 ,  2 2, 0

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1025


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

odd deg, neg coef  rises left, falls right Sign of

+

+

(-¥, - 2 2)

(-2 2, 0)

(0, 2 2 )

(2 2, ¥)

-2 2

0

2 2

f (-3) = 27

f (-1) = -7

f ( 1) = 7

f (3) = -27

above axis

below axis

above axis

below axis

f ( x ) = -x 5 + 8 x 3

Test point Graph of f(x) 5

3

f (-x ) = -(-x ) + 8 (-x ) = x 5 = 8x 3 = -f ( x )  odd, symmetric about origin Use the Intermediate Value Theorem to show that each equation has at least one real zero between the specified numbers.

 

79. P x  2 x 2  x  3; –2 and –1

Solution

P  x   2x 2  x  3

P  2  3; P  1  2 Thus, there is a zero between  2 and  1.

 

80. P x  2 x 3  17 x 2  31x  20; –1 and 2

Solution

P  x   2 x 3  17 x 2  31x  20 P  1  36; P  2  126

Thus, there is a zero between  1 and 2.

 

81. P x  3 x 3  11x 2  14 x; 4 and 5

Solution

P  x   3 x 3  11x 2  14 x P  4   40; P  5  30

Thus, there is a zero between 4 and 5.

 

82. P x  2 x 3  3x 2  2 x  3; 1 and 2

Solution

P  x   2x 3  3x 2  2x  3 P  1  2; P  2  5

Thus, there is a zero between 1 and 2.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1026


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 

83. P x  x 4  8 x 2  15; 1 and 2

Solution

P  x   x 4  8x 2  15 P  1  8; P  2  1

Thus, there is a zero between 1 and 2.

 

84. P x  x 4  8 x 2  15; 2 and 3

Solution

P  x   x 4  8 x 2  15

P  2  1; P  3  24 Thus, there is a zero between 2 and 3.

 

85. P x  30 x 3  61x 2  39 x  10; 2 and 3

Solution

P  x   30 x 3  61x 2  39x  10 P  2  72; P  3  154

Thus, there is a zero between 2 and 3.

 

86. P x  30 x 3  61x 2  39 x  10; –1 and 0

Solution

P  x   30 x 3  61x 2  39x  10 P  1  42; P  0  10

Thus, there is a zero between 0 and 1.

 

87. P x  30 x 3  61x 2  39 x  10; 0 and 1

Solution

P  x   30 x 3  61x 2  39x  10 P  0  10; P  1  60

Thus, there is a zero between 0 and 1.

 

88. P x  5 x 3  9x 2  4 x  9; –1 and 0

Solution

P  x   5x 3  9x 2  4 x  9 P  1  1; P  0  9

Thus, there is a zero between  1 and 0.

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1027


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Fix It In Exercises 89 and 90, identify the step the first error is made and fix it.

 



89. Graph the polynomial function represented by the equation f x   x x  2 x  4 . State the end behavior, x-intercepts, y-intercept, and draw the graph of the polynomial function.

Solution Step 4 was incorrect. Step 4:

 

90. Graph the polynomial function f x  2x 4  12x 2 . State the end behavior, x-intercepts, y-intercepts, and draw the graph of the polynomial function.

Solution Step 2 was incorrect. Step 2: x-intercepts are  0, 0  ,

 6, 0 , and   6, 0

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1028


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Applications 91. Maximize volume An open box is to be constructed from a piece of cardboard 20 inches by 24 inches by cutting a square of length x from each corner and folding up the sides as shown in the figure.

a. Write a polynomial function V(x) that expresses the volume of the constructed box as a function of x. b. Use the theory learned about graphing polynomial functions and graph V(x). c. What is the domain of the function as it relates to the application problem? d. Use a graphing calculator to graph the function, and estimate the value of x that gives the maximum volume and then estimate the maximum volume. Round to one decimal place.

Solution a.

V  x   x  20  2 x  24  2 x   4 x 3  88 x 2  480 x

b. V  x   4 x 3  88 x 2  480 x  4 x x 2  22 x  120  4 x  x  10  x  12  x-int.

y-int.

4 x  x  10  x  12

f 0  0

x  0, x  10, x  12

y 0

0, 0 ,  10, 0 ,  12, 0

0, 0

odd deg, pos coef  falls left, rises right

Sign of f (x) = 4 x ( x - 10)( x - 12)

+

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

+

1029


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

(-¥, 0)

(0, 10) 0

Test point Graph of f(x)

(10, 12) 10

(12, ¥) 12

f (-1) = -572

f (1) = 396

f (11) = -44

f (13) = 156

below axis

above axis

below axis

above axis

c.

x ³ 0, 20 - 2 x ³ 0  x £ 10, 24 - 2 x ³ 0  x £ 12; Domain = éêë0, 10ùúû

d.

x » 3.6 in, V ( x ) » 774.2 in3

92. Maximize volume Repeat Exercise 91 using a piece of cardboard dimensions 30 inches by 36 inches.

Solution

V  x   x  30  2 x  36  2 x   4 x 3  132 x 2  1080 x

a.

b. V  x   4 x 3  132 x 2  1080 x  4 x x 2  33 x  270  4 x  x  15  x  18  x-int.

y-int.

4 x  x  15 x  18

f 0  0

x  0, x  15, x  18

y 0

0, 0 ,  15, 0 ,  18, 0

0, 0

odd deg, pos coef  falls left, rises right

Sign of f ( x) = 4 x ( x - 15)( x - 18)

+

+

(-¥, 0)

(0, 15)

(15, 18)

(18, ¥)

15

18

0 Test point Graph of f(x)

f (-1) = -1216

f (1) = 952

f (16) = -128

f (19) = 304

below axis

above axis

below axis

above axis

c.

x ³ 0, 30 - 2 x ³ 0  x £ 15, 36 - 2 x ³ 0  x £ 18; Domain = éêë0, 15ùúû

d.

x » 5.4 in, V ( x ) » 2612.8 in3

93. Maximize production If 270 apples trees are planted per acre, the production per tree is 840 pounds. For every tree, x, over 270 planted per acre, the production per tree decreases by (840 – 0.1x2) pounds per tree. a. Write a production function, P(x), for the number of pounds per acre as a function of x.

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1030


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

b. Use a graphing calculator to graph the function P(x) and determine the number of trees that should be planted per acre to produce the maximum number of pounds of apples per acre. Round to the nearest unit.

Solution a. # Trees = 270 + x; Prod/tree = 840 - 0.1x 2 P ( x ) = (# Trees)(Prod/tree) = (270 + x ) (840 - 0.1x 2 ) = - 0.1x 3 - 27 x 2 + 840 x + 226, 800

b.

x » 14 trees

94. Maximize volume for luggage Most airlines restrict the size of carry-on luggage to a total of 45 inches (length plus width plus height). The height of a piece of luggage is to be 4 inches less than the width. a. Write a function for the volume V as a function of x, the width of the luggage. b. Graph the function and determine the dimensions that will create a piece of luggage with the maximum volume. Round to the nearest inch.

Solution a.

Width = x; Height = x - 4; Length = 45 - x - ( x - 4) = 49 - 2 x V ( x ) = (Width)(Height )(Length) = x ( x - 4)(49 - 2 x ) = -2 x 3 + 57 x 2 - 196 x

b.

Width = x = 17 in; Height = x - 4 = 13 in; Length = 49 - 2 x = 15 in

95. Online DVD sales A Web-based company produces and sells DVDs. The monthly profit P(x), in hundreds of dollars, can be modeled by the polynomial function,

P  x   10 x 3  100 x 2  210 x, where 1  x  12 and x represents the month of the year

(x = 1 corresponds to January). a. What was the profit of the company in December? b. Identify the month(s) when the company’s profit was $0.

Solution a. b.

3

2

P (12) = 10 (12) - 100 (12) + 210 (12) = $5400

P (x) = 0 10 x 3 - 100 x 2 + 210 x = 0 10 x ( x 2 - 10 x + 21) = 0 10 x ( x - 3)( x - 7) = 0 x = 0 (not in domain) ; x = 3 (March) ; x = 7 (July )

96. Roller coaster A portion of a roller coaster’s tracks can be modeled by the polynomial

function f  x   0.0001 x 3  600 x 2  90, 000 x , where 0 ≤ x ≤ 400. f(x) represents the height of the roller coaster in feet and x represents the horizontal distance in feet. a. Find the height of the roller coaster when x = 100 yards.

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1031


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

b. Find the value(s) of x on the interval [0, 400] at which the height of the roller coaster is 0 yards.

Solution a.

3 2 é ù f (100) = 0.0001 ê(100) - 600 (100) + 90000 (100)ú = 0.0001 éëê4, 000, 000ùûú = 400 ft ë û

b.

f (x) = 0 0.0001( x 3 - 600 x 2 + 90000 x ) = 0 0.0001x ( x 2 - 600 x + 90000) = 0 0.0001x ( x - 300)( x - 300) = 0 x = 0; x = 300

97. Seattle temperature. The table shows the average low monthly temperature for six months (in degrees Fahrenheit) at the Space Needle in Seattle, Washington.

Month Avg. Temp

Jan 36

Mar 39

May 48

Jul 56

Sep 53

Nov 40

 

The polynomial function f x  0.127 x 3  1.720 x 2  3.321x  37.498 models the temperature with x = 1 corresponding to January. a. Using the model, what would be the average low temperature if you visit in February? Round to the nearest degree. b. Using the model, what would be the average low temperature if you visit in October? Round to the nearest degree.

Solution a.

f  2   0.127  2   1.720  2   3.321  2   37.498 3

2

 37 degrees

b.

f  10   0.127  10   1.720  10   3.321  10   37.498 3

2

 49 degrees

98. Walt Disney World temperature The table shows the average high monthly temperature for six months (in degrees Fahrenheit) at Walt Disney World, in Florida.

Month

Jan

Mar

May

Jul

Sep

Nov

Avg. Temp

72

77

87

92

90

79

The polynomial function f  x   0.088 x 3  1.012 x 2  0.263 x  70.496 models the temperature with x = 1 corresponding to January. a. Using the model, what would be the average high temperature if you visit in February? Round to the nearest degree. b. Using the model, what would be the average high temperature if you visit in October? Round to the nearest degree.

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1032


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution a.

f  2   0.088  2   1.012  2   0.263  2   70.496 3

2

 74 degrees

b.

f  10   0.088  10   1.012  10   0.263  10   70.496 3

2

 86 degrees

99. Containers A box has a length of 16 inches, a width of 10 inches, and a height of between 4 inches and 8 inches. Can it have a volume of 1000 in.3? Hint: Model the volume of the box with a function of h and use the Intermediate Value Theorem.

Solution

Let x  the height. Then the volume  V  x   lwh  16  10  x

V  x   16  10  4   640 in3 ; V  8   16  10  8   1280 in3

By the Intermediate Value Theorem, there is at least one value of x between 4 and 8 such that V(x) is between V(4) = 640 and V(8) = 1280, so there is a value of x with V(x) = 1000 in3. 100. Lollipops If a candy company makes spherical shaped lollipops with radii between 1 and 4 cm, use the Intermediate Value Theorem to determine whether a lollipop can be made with a volume of 200 cubic centimeters. Hint: The volume formula for a sphere is V  43  r 3 .

Solution

Let r  the radius. Then the volume  V  r   43  r 3

V  1  43   1  4.2cm3 ; V  4   43   4   268.1 cm3 3

3

By the Intermediate Value Theorem, there is at least one value of r between 1 and 4 such that V(r) is between V(1) ≈ 4.2 and V(4) ≈ 268.1, so there is a value of r with V(r) = 200 cm3.

Discovery and Writing 101. What is a zero of a polynomial function?

Solution Answers may vary. 102. Explain how to determine the zeros of a polynomial function.

Solution Answers may vary. 103. Describe how to determine the end behavior of a polynomial function.

Solution Answers may vary. 104. Describe an effective strategy to use for graphing a polynomial function.

Solution

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1033


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Answers may vary. 105. Is it possible for a polynomial function to have no x-intercept? Explain and give an example.

Solution Answers may vary. 106. Is it possible for a polynomial function to have no y-intercept? Explain and give an example.

Solution Answers may vary. 107. Explain why a polynomial function of odd degree must have at least one zero.

Solution Answers may vary. 108. What is the purpose of the Intermediate Value Theorem?

Solution Answers may vary. Use a graphing calculator to explore the properties of graphs of polynomial functions. Write a paragraph summarizing your observations. 109. Graph the function y = x2 + ax for several values of a. How does the graph change?

Solution Answers may vary. 110. Graph the function y = x3 + ax for several values of a. How does the graph change?

Solution Answers may vary. 111. Graph the function y = (x – a)(x – b) for several values of a and b. What is the relationship between the x-intercepts and the equation?

Solution Answers may vary. 112. Use the insight you gained in Exercise 99 to factor x3 – 3x2 – 4x + 12.

Solution Answers may vary. Critical Thinking Match each polynomial function with its graph shown below. 113. f(x) = (x – a)(x – b)2(x – c)

Solution b

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1034


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

114. f(x) = –(x – a)(x – b)(x – c)3

Solution a 115. g(x) = (x – a)2(x – b)

Solution d 116. g(x) = –(x – a)2(x – b)(x – c)

Solution c a.

b.

c.

d.

Determine if the statement is true or false. If the statement is false, then correct it and make it true. 117. Given the polynomial function f(x) = 100x100 + 50x50, x = 0 is a zero of multiplicity 50.

Solution True. 118. The polynomial function g(x) = (x – 100)100(x + 200)200 has at most 299 turning points.

Solution True.

EXERCISES 4.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Divide 412 by hand and identify the dividend, divisor, quotient, and remainder. 3

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1035


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution 1 137 ; dividend is 412; divisor is 3; quotient is 137; remainder is 1 3 2. Rewrite 3 x  x 3  5  7 x 2 in descending powers of x.

Solution

 x 3  7 x 2  3x  5

3. If x  7

 is one factor of a polynomial function, identify one zero.

Solution  7

4. If –8 is one zero of a polynomial function, identify one factor.

Solution

 x  8

5. Solve x 2  2 x  5  0 using the Quadratic Formula.

Solution

x

 2  4  15 2  1 2

2

2  16 2 2  4i  2  1  2i 

6. If

x 3  4 x 2  7 x  10  x 2  3 x  10, factor x 3  4 x 2  7 x  10 x1

Solution

 x  1  x  3x  10  x  1 x  5 x  2 2

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The variables in a polynomial have __________-number exponents.

Solution Whole

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1036


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

8. A zero of P(x) is any number c for which __________.

Solution P(r) = 0 9. The Remainder Theorem holds when c is __________ number.

Solution any 10. If P(x) is a polynomial function and P(x) is divided by __________, the remainder will be P(c).

Solution x–r 11. If P(x) is a polynomial function, then P(c) = 0 if and only if x – c is a __________ of P(x).

Solution factor 12. A shortcut method for dividing a polynomial by a binomial of the form x – c is called __________ division.

Solution synthetic Practice Use long division to perform each division 13.

4 x 3  2x 2  x  1 x1

Solution 4 x 2  2 x  1  x2 1 x  3 4 x 3  2x 2  x  1 4x3  4x2 2x 2  x 2x 2  2x x x

1 1 2

14.

2x 3 + 3x 2 - 5x + 1 x +3

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1037


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

2 x 2  3 x  4  x113 x  3 2x 3  3x 2  5x  2x  6 x 3

1

2

 3x 2  5x 3 x 2  9 x 4x 

1

4 x  12  11 15.

4

3

2

2 x + x + 2 x + 15 x - 5 x +2

Solution

2x 3  3x 2  8x 

1  x32

x  2 2 x 4  x 3  2 x 2  15 x  5 2x 4  4 x 3  3x 3  2x 2 3 x 3  6 x 2 8 x 2  15 x 8 x 2  16 x 

x 5

x 2 3

16.

4

3

2

x + 6x - 2x + x - 1 x-1

Solution

x 3  7 x 2  5 x  6  x5 1 x  1 x 4  6x 3  2x 2  x  1 x4  x3 7 x 3  2x 2 7 x3  7x2 5x 2 

x

5x  5x 2

6x  1 6x  6 5

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1038


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Find each value by substituting the given value of x into the polynomial and simplifying. Then find the value by performing long division and finding the remainder. 17. P ( x ) = 3 x 3 - 2 x 2 - 5 x - 7; P(2)

Solution

P  2  3  2  2  2  5  2  7 3

2

 3  8  2  4   5  2  7  24  8  10  7   1 3x 2  4 x 2  3

x  2 3x 3  2x 2  5x  7 3x 3  6x 2 4 x 2  5x 4 x 2  8x 3x  7 3x  6 1

18. P ( x ) = 5 x 3 + 4 x 2 + x - 1; P(–2)

Solution

P  2  5  2  4  2   2  1 3

2

 5  8  4  4   2  1

 40  16  2  1   27 5x 2  6x 2  x  2 5x 3  4 x 2  5 x  10 x 3

13 x  1

2

 6x 2 

x

 6 x  12 x 2

13 x  1 13 x  26  27

19. P ( x ) = 7 x 4 + 2 x 3 + 5 x 2 - 1; P(–1)

Solution

P  1  7  1  2  1  5  1  1 4

3

2

 7  1  2  1  5  1  1  7251 9

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1039


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

7 x 3  5 x 2  10 x  10 x  1 7 x 4  2x 3  5x 2  0x  1 7x4  7x3  5x 3  5x 2  5x 3  5x 2 10 x 2  0 x 10 x 2  10 x  10 x  1  10 x  10 9 20. P ( x ) = 2 x 4 - 2 x 3 + 5 x 2 - 1; P(2)

Solution

P  2  2  2  2  2  5  2  1 4

3

2

 2  16   2  8  5  4   1  32  16  20  1  35 2 x 3  2 x 2  9 x  18

x  2 2x 4  2x 3  5x 2  0 x  1 2x 4  4 x 3 2x 3  5x 2 2x 3  4 x 2 9x 2  0x 9 x 2  18 x 18 x  1 18 x  36 35 21. P ( x ) = 2 x 5 + x 4 - x 3 - 2 x + 3; P(1)

Solution

P  1  2  1   1   1  2  1  3 5

4

3

 2  1   1   1  2  3  2 1 12 3  3

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1040


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2x 4  3x 3  2x 2  2x x  1 2x 5  x 4  x 3  0 x 2  2x  3 2x5  2x 4 3x 4  x 3 3x 4  3x 3 2x 3  0x 2 2x 3  2x 2 2x 2  2x 2x 2  2x 0x  3 22. P ( x ) = 3x 5 + x 4 - 3 x 2 + 5 x + 7 ; P(–2)

Solution

P  2   3  2    2   3  2   5  2   7 5

4

2

 3  32    16   3  4   10  7

 96  16  12  10  7   95 3 x 4  5 x 3  10 x 2  23 x  x  2 3x 5  x 4  0x 3  3x 2  3x  6x 5

51 5x 

7

4

 5x4  0x 3  5 x 4  10 x 3 10 x 3  3 x 2 10 x 3  20 x 2  23 x 2  5 x  23 x 2  46 x 51x 

7

51x  102 95

Use the Remainder Theorem to find the remainder that occurs when

P ( x ) = 3 x 4 - 5 x 3 - 4 x 2 - 2 x + 1 is divided by each binomial. 23. x + 2

Solution 4

3

2

remainder = P (-2) = 3 (-2) + 5 (-2) - 4 (-2) - 2 (-2) + 1

= 3 (16) + 5 (-8) - 4 (4) - 2 (-2) + 1 = 48 - 40 - 16 + 4 + 1 = -3

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1041


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

24. x - 1

Solution 4

3

2

remainder = P (1) = 3 (1) + 5 (1) - 4 (1) - 2 (1) + 1

= 3 (1) + 5 (1) - 4 (1) - 2 (1) + 1 = 3 + 5 - 4 - 2 + 1 = 3 25. x - 2

Solution 4

3

2

remainder = P (2) = 3 (2) + 5 (2) - 4 (1) - 2 (1) + 1

= 3 (16) + 5 (8) - 4 (4) - 2 (2) + 1 = 48 + 40 - 16 - 4 + 1 = 69 26. x + 1

Solution 4

3

2

remainder = P (-1) = 3 (-1) + 5 (-1) - 4 (-1) - 2 (-1) + 1

= 3 (1) + 5 (-1) - 4 (1) - 2 (-1) + 1 = 3 - 5 - 4 + 2 + 1 = -3 27. x + 3

Solution 4

3

2

remainder = P (-3) = 3 (-3) + 5 (-3) - 4 (-3) - 2 (-3) + 1

= 3 (81) + 5 (-27) - 4 (9) - 2 (-3) + 1 = 243 - 135 - 36 + 6 + 1 = 79 28. x - 3

Solution 4

3

2

remainder = P (3) = 3 (3) + 5 (3) - 4 (3) - 2 (3) + 1

= 3 (81) + 5 (27) - 4 (9) - 2 (3) + 1 = 243 + 135 - 36 - 6 + 1 = 337 29. x - 4

Solution 4

3

2

remainder = P (4) = 3 (4) + 5 (4) - 4 (4) - 2 (4) + 1

= 3 (256) + 5 (64) - 4 (16) - 2 (4) + 1 = 768 + 320 - 64 - 8 + 1 = 1017 30. x + 4

Solution 4

3

2

remainder = P (-4) = 3 (-4) + 5 (-4) - 4 (-4) - 2 (-4) + 1

= 3 (256) + 5 (-64) - 4 (16) - 2 (-4) + 1 = 768 - 320 - 64 + 8 + 1 = 393 Use the Factor Theorem to determine whether each statement is true. If the statement is not true, so indicate. 31. x – 1 is a factor of P(x) = x7 – 1.

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1042


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

( x - 1) is a factor if P (1) = 0. P (1) = 17 - 1 = 0: true 32. x – 2 is a factor of P(x) = x3 – x2 + 2x – 8.

Solution

( x - 2) is a factor if P (2) = 0. P ( 1) = 23 - 22 + 2 (2) - 8 = 0: true 33. x – 1 is a factor of P(x) = 3x5 + 4x2 – 7.

Solution

( x - 1) is a factor if P (1) = 0. 5 2 P (1) = 3 (1) + 4 (1) - 7 = 0: true 34. x + 1 is a factor of P(x) = 3x5 + 4x2 – 7.

Solution

( x + 1) is a factor if P (-1) = 0. 5 2 P (-1) = 3 (-1) + 4 (-1) - 7 = -6: false 35. x + 3 is a factor of P(x) = 2x3 – 2x2 + 1.

Solution

( x + 3) is a factor if P (-3) = 0. P (-3) = 2 (-3) - 2 (-3) + 1 3

2

= -71: false 36. x – 3 is a factor of P(x) = 3x5 – 3x4 + 5x2 – 13x – 6.

Solution

( x - 3) is a factor if P (3) = 0. P (3) = 3 (3) - 3 (3) + 5 (3) - 13 (3) - 6 5

4

2

= 486: false 37. x – 1 is a factor of P(x) = x1984 – x1776 + x1492 – x1066.

Solution 1984

( x - 1) is a factor if P (1) = 0. P (1) = (1)

1776

- (1)

1492

+ (1)

1066

- ( 1)

= 0: true

38. x + 1 is a factor of P(x) = x1984 + x1776 – x1492 – x1066.

Solution 1984

( x + 1) is a factor if P (-1) = 0. P (-1) = (-1)

1776

+ (- 1)

1492

- (- 1)

1066

- (- 1)

= 0: true

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1043


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Use the Division Algorithm and synthetic division to express the polynomial function P(x) = 3x3 – 2x2 – 6x – 4 in the form (divisor) (quotient) + remainder for each divisor. 39. x – 1

Solution 1 3 -2 -6 -4 1 -5

3 3

1

- 5 -9

( x - 1)(3x + x - 5) - 9 2

40. x – 2

Solution 2 3 -2 -6 -4 6 3

4

8

4

2

0

( x - 2)(3x + 4 x + 2) + 0 2

41. x – 3

Solution 3 3 -2 -6 -4 9 3

7

21

45

15

41

( x - 3)(3x + 7 x + 15) + 41 2

42. x – 4

Solution 4 3 -2 - 6 12 3

-4

40 136

10 34 132

( x - 4)(3x 2 + 10 x + 34) + 132 43. x + 1

Solution -1 3 -2 -6 -4 -3

5

1

3 -5 -1

-3

( x + 1)(3 x - 5 x - 1) - 3 2

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1044


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

44. x + 2

Solution -2 3 - 2 -6 -6

-4

16 -20

3 -8

10 -24

( x + 2)(3x - 8 x + 10) - 24 2

45. x + 3

Solution -3 3 -2 -6 -9

-4

33 -81

3 -11

27 -85

( x + 3)(3 x - 11x + 27) - 85 2

46. x + 4

Solution -4 3 -2 -6

-4

-12

56 -200

3 -14

50 -204

( x + 4)(3x - 14 x + 50) - 204 2

Use synthetic division to perform each division. 47.

x3 + x2 + x - 3 x-1 Solution 1 1 1 1 -3 1 2

3

1 2 3

0

2

x + 2x + 3

48.

x 3 - x 2 - 5x - 6 x -2 Solution 2 1 -1 -5

2 -6

2 1

1

6

-3

0

2

x + x -3

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1045


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

49.

7 x 3 - 3x 2 - 5x + 1 x+1

Solution -1 7 -3 -5

-7

10 -5

7 -10

5 -4

7 x 2 - 10 x + 5 +

50.

1

-4 x+1

2x 3 + 4 x 2 - 3x + 8 x -3

Solution 3 2 4 -3

6

8

30 81

2 10 27 89 2 x 2 + 10 x + 27 +

51.

89 x -3

4 x 4 - 3x 3 - x + 5 x -3

Solution 3 4 -3

0 -1

12 27 4

5

81 240

9 27 80 245

4 x 3 + 9 x 2 + 27 x + 80 +

52.

245 x -3

x 4 + 5x 3 - 2x 2 + x - 1 x+1

Solution -1 1 5 -2

1

-1

- 1 -4 6 -7 1

4 -6 7 -8

x 3 + 4x 2 - 6x + 7 +

-8 x+1

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1046


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

53.

3 x 5 - 768 x x -4 Solution 4 3 0 3

0 -768

0

0

12 48 192

768

0

12 48 192

0

0

4

3

2

3 x + 12 x + 48 x + 192 x

54.

x5 - 4x2 + 4x + 4 x +3

Solution

-3 1

1

-4

0 0

4

4

- 3 9 -27 93

- 291

- 3 9 -31

-287

97

x 4 - 3 x 3 + 9 x 2 - 31x + 97 +

-287 x +3

Let P(x) = 5x3 + 2x2 – x + 1. Use synthetic division to find each value. 55. P(2)

Solution 2 5 2 -1

1

10 24 46 5 12 48 47

P (2) = 47

56. P(–2)

Solution -2 5 2 -1

1

-10 16 -30 5

- 8 15 -29

P (-2) = -29

57. P(–5)

Solution -5 5 2

-1

1

-25 115 -570 5 -23 114 -569 P (-5) = -569

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1047


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

58. P(3)

Solution 3 5 2 -1

15

1

51 150

5 17 50 151

P (3) = 151

59. P(i)

Solution i 5

-1

2

1

5i -5 + 2i -2 - 6i 2 + 5i -6 + 2i

5

- 1 - 6i

P (i ) = -1 - 6i 60. P(–i)

Solution i 5

-1

2

1

- 5i -5 - 2i -2 + 6i 2 - 5i -6 - 2i -1 + 6i

5

P (i ) = -1 + 6i Let P(x) = 2x4 – x2 + 2. Use synthetic division to find each value. æ 1ö 61. P çç ÷÷÷ çè 2 ÷ø

Solution 1 2 0 -1 2

0

1

- 41 - 81

2

1 2

1 - 21 - 41

2

15 8

P ( 21 ) = 158

æ 1ö 62. P ççç ÷÷÷ è 3 ø÷

Solution 1 2 0 -1 3 2 3

2

2 3

2 9

0

2

- 277 - 817

- 97 - 277

155 81

P ( 31 ) = 155 81

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1048


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

63. P(i)

Solution i 2 0 -1

0 2

2i -2 -3i 3 2 2i -3 -3i 5

P (i ) = 5

64. P(–i)

Solution -i 2 0

-1

0 2

-2i -2

3i 3

2 -2i -3 -3i 5 P (-i ) = 5 Let P(x) = x4 – 8x3 + 8x + 14x2 – 15. Write the terms of P(x) in descending powers of x and use synthetic division to find each value. 65. P(1)

Solution 1 1 -8 14

1 -7 1

-7

8 -15 7

15

7 15

0

P (1) = 0

66. P(0)

Solution 1 1 -8 14

0

0

8 -15 0

0

1 -8 14 8 -15

P (0) = -15

67. P(–3)

Solution -3 1 -8

14

8 -15

- 3 33 -141 399 1

- 11 47 -133 384 P (-3) = 384

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1049


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

68. P(–1)

Solution -1 1 -8

14

-1 1

8 -15

9 -23

15

- 9 23 -15

0

P (-1) = 0

69. P(–i)

Solution -i 1

1

-8

14

8

-15

-i

- 1 + 8i

8 - 13i

- 13 - 16i

-8-i

13 + 8i

16 - 13i

- 28 - 16i

-8

14

8

-15

i

- 1 - 8i

8 + 13i

- 13 + 16i

-8+ i

13 - 8i

16 + 13i

- 28 + 16i

P (-i ) = -28 - 16i

70. P(i)

Solution -i 1

1

P (i ) = -28 + 16i

Let P(x) = 8 – 8x2 + x5 – x3. Write the terms of P(x) in descending powers of x and use synthetic division to find each value. 71. P(i)

Solution i 1 0 -1

-8

0

8

i -1

- 2i

2 - 8i

8 + 2i

i -2 -8 - 2i

2 - 8i

16 + 2i

1

P (i ) = 16 + 2i

72. P(–i)

Solution -i 1 0 -1

-8

0

8

-i -1

2i

2 + 8i

8 - 2i

- i -2 -8 + 2i

2 + 8i

16 - 2i

1

P (-i ) = 16 - 2i

73. P(–2i)

Solution -2i 1 0

-1

-8

0

8

-2i -4

10i

20 + 16i

32 - 40i

1 -2i -5 -8 + 10i

20 + 16i

40 - 40i

P (-2i ) = 40 - 40i

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1050


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

74. P(2i)

Solution 2i 1 0 -1

-8

0

8

2i -4

- 10i

20 - 16i

32 + 40i

1 2i -5 -8 - 10i

20 - 16i 40 + 40i

P (2i ) = 40 + 40i

Use the Factor Theorem and synthetic division to determine whether the given polynomial is a factor of the polynomial function P(x). 75. P(x) = 3x3 – 13x2 – 10x + 56; x + 2

Solution

x + 2 is a factor if P (-2) = 0 -2 3 -13

-10

-6

56

38 -56

3 -19 -28

0 yes

76. P(x) = 2x3 + 3x2 – 32x + 15; x + 5

Solution

x + 5 is a factor if P (-5) = 0. -5 2

2

3 -32

15

-10

35 -15

-7

3

0

yes

77. P(x) = x4 – 3x3 + 4x2 – 2x + 4; x – 1

Solution

x - 1 is a factor if P (1) = 0. 1 1 -3

4 -2 4

1 -2 -2

1

2

2 0 0 4

no

78. P(x) = 2x4 – x3 – 2x2 + x + 1; x – 2

Solution

x - 2 is a factor if P (2) = 0. 2 2 -1 -2

2

1

1

4

6 8 18

3

4 9 19

no

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1051


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

79. P(x) = 3x5 – 22x3 + 15x2 + 3x + 9; x + 3

Solution

x + 3 is a factor if P (-3) = 0. -3 3

0 -22 -9

15 3

9

27 -15 0 -9

3 -9

5

0 3

0

yes

80. P(x) = x5 + 5x4 – 17x2 – 2x + 8; x + 4

Solution

x + 4 is a factor if P (-4) = 0. -4 1

1

0 -17 -2

5

8

-4 - 4

16

4 -8

1 -4

-1

2

0

yes

Determine whether the given number is a zero of the polynomial function P(x). 81. P(x) = –3x3 + 13x2 + 10x – 56; 4

Solution 4 is a zero if x - 4 is a factor. 4 -3 13 10 -56 -12 -3

4

56

1 14

0

yes

82. P(x) = –2x3 – 3x2 + 32x – 15; 3

Solution 3 is a zero if x - 3 is a factor. 3 - 2 -3 32 -15 -6 -27 - 2 -9

5

15 0

yes

83. P(x) = 4x4 + x3 + 20x2 – 4; –2

Solution -2 is a zero if x + 2 is a factor. -2 4 1 20 0 -4 -8

14 -68 136

4 -7 34 -68 132

no

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1052


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

84. P(x) = 2x4 – x3 – 64x2 – 2; –6

Solution -6 is a zero if x + 6 is a factor. -6 2 -1 -64 0 -2 -12

78 -84 504

2 -13

14 -84 502

no

85. P(x) = 4x5 – 2x4 + 6x3 + 5x2 – 6x + 1;

1 2

Solution 1 is a zero if x - 21 is a factor. 2 1 2

4 -2 6 5 -6 2 0 3

4 -1

0 6 8 -2

4

1

0

yes

86. P(x) = 6x5 + 7x4 – 3x3 + 6x2 + 13x – 5;

1 3

Solution 1 is a zero if x - 31 is a factor. 3 1 3

6 7 -3 6

6

13 -5

2

3 0

2

5

9

0 6 15

0

yes

A partial solution set is given for each polynomial equation. Find the complete solution set. 87. x3 + 3x2 – 13x – 15 = 0; {–1}

Solution x = –1 is a solution, so (x + 1) is a factor. Use synthetic division to divide by (x + 1). -1 1

1 3

3

-13 -15

-1

-2

15

2 -15

0

2

x + 3 x - 13 x - 15 = 0

( x + 1)( x 2 + 2 x - 15) = 0 ( x + 1)( x + 5)( x - 3) = 0 Solution set: {-1, - 5, 3}

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1053


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

88. x3 + 6x2 + 5x – 12 = 0; {1}

Solution x = 1 is a solution, so (x – 1) is a factor. Use synthetic division to divide by (x – 1). 5 -12

1 1 6 1 1

7

12

7 12

0

3

2

x + 6 x + 5 x - 12 = 0

( x - 1)( x 2 + 7 x + 12) = 0 ( x - 1)( x + 3)( x + 4) = 0 Solution set: {1, - 3, - 4} ïì 1 ïü 89. 2x3 + x2 – 18x – 9 = 0; ïí- ïý ïïî 2 ïïþ

Solution x = - 21 is a solution, so ( x + 21 ) is a factor. Use synthetic division to divide by ( x + 21 ) .

- 21 2

1 -18 -9

-1 2

0

9

0 -18

0

2 x 3 + x 2 - 18 x - 9 = 0

( x + 21 )(2 x 2 - 18) = 0 2 ( x + 21 )( x 2 - 9) = 0 2 ( x + 21 )( x + 3)( x - 3) = 0 Solution set: {- 21 , - 3, 3} ïì 1 ïü 90. 2x3 – 3x2 – 11x + 6 = 0; ïí ïý ïîï 2 ïþï

Solution x = 21 is a solution, so ( x - 21 ) is a factor. Use synthetic division to divide by ( x - 21 ) . 1 2

2 -3

-11 -6

1

- 1 -6

2 -2

- 12

0

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1054


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

2 x 3 - 3 x 2 - 11x + 6 = 0

( x - 21 )(2 x 2 - 2 x - 12) = 0 2 ( x - 21 )( x 2 - x - 6) = 0 2 ( x - 21 )( x + 2)( x - 3) = 0 Solution set: { 21 , - 2, 3} 91. x3 – 6x2 + 7x + 2 = 0; {2}

Solution x = 2 is a solution, so ( x - 2) is a factor. Use synthetic division to divide by ( x - 2) . 2 1 -6

7

- 8 -2

2 1 -4 3

2

-1

0

2

x - 6x + 7 x + 2 = 0

( x - 2)( x 2 - 4 x - 1) = 0 Use the quadratic formula to finish.

{

Solution set: 2, 2 + 5, 2 - 5

}

92. x3 + x2 – 8x – 6 = 0; {–3}

Solution x = –3 is a solution, so ( x + 3) is a factor. Use synthetic division to divide by ( x + 3) . -3 1

1 -8 -6 -3

6

6

1 -2 -2

0

3

2

x + x - 8x - 6 = 0

( x + 3)( x 2 - 2 x - 2) = 0 Use the quadratic formula to finish.

{

Solution set: -3, 1 + 3, 1 - 3

}

93. x3 – 3x2 + x + 57 = 0; {–3}

Solution x = –3 is a solution, so ( x + 3) is a factor. Use synthetic division to divide by ( x + 3) .

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1055


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

-3 1 -3

1

-3

18 -57

1 -6 3

57

19

0

2

x - 3 x + x + 57 = 0

( x + 3)( x 2 - 6 x + 19) = 0 Use the quadratic formula to finish.

{

Solution set: -3, 3  10 i

}

94. 2x3 – x2 + x – 2 = 0; {1}

Solution x = 1 is a solution, so ( x - 1) is a factor. Use synthetic division to divide by ( x - 1) . 1 2 -1

2

1 -2

2

1

2

1

2

0

2x 3 - x 2 + x - 2 = 0

( x - 1)(2 x 2 + x + 2) = 0 Use the quadratic formula to finish. ïì 1 15 ïüï Solution set: ïí1, -  iý ïï 4 4 ïï îï þï

95. x4 – 2x3 – 2x2 + 6x – 3 = 0; {1, 1}

Solution x = 1 is a solution, so ( x - 1) is a factor. Use synthetic division to divide by ( x - 1) . 1 1 -2

-2

1 1 4

-1 3

6 -3

- 1 -3

3

-3

0

3 2

x - 2x - 2x + 6x - 3 = 0

( x - 1)( x 3 - x 2 - 3x + 3) = 0 Use the fact that x = 1 is a double root and divide the depressed polynomial (x – 1):

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1056


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

1 1 -1

-3

0 -3

1 1

3

-3

0

0

( x - 1)( x - x - 3x + 3) = 0 ( x - 1)( x - 1)( x 2 - 3) = 0 3

2

{

Solution set: 1, 1, 3, - 3

}

96. x5 + 4x4 + 4x3 – x2 – 4x – 4 = 0; {1, –2, –2}

Solution x = –2 is a solution, so ( x + 2) is a factor. Use synthetic division to divide by ( x + 2) . -2 1

-1 -4 -4

4

4

-2

-4

0

2

4

2

0

- 1 -2

0

4

3

1 5

2

x + 4x + 4x - x - 4x - 4 = 0

( x + 2)( x 4 + 2 x 3 - x - 2) = 0 Use the fact that x = –2 is a double root and divide the depressed polynomial (x + 2): -2 1

2

0 -1 -2

-2

0

0

2

0

0 -1

0

1

( x + 2)( x + 2x - x - 2) = 0 ( x + 2)( x + 2)( x 3 - 1) = 0 4

3

Use the fact that x = 1 is a double root and divide the depressed polynomial (x – 1):

1 1

1

0 0 -1

-1

1

0

1

1

0

( x + 2)( x + 2)( x 3 - 1) = 0 ( x + 2)( x + 2)( x - 1)( x 2 + x + 1) = 0 ìï 1 3 üïï Solution set: ïí-2, - 2, 1, -  iý ïï 2 2 ïï ïî ïþ

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1057


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

97. x4 – 5x3 + 7x2 – 5x + 6 = 0; {2, 3}

Solution x = 2 is a solution, so ( x - 2) is a factor. Use synthetic division to divide by ( x - 2) . 2 1 -5

7

2

-6

1

-3

4

3

-5

6

2 -6 -3

1

0

2

x - 5x + 7 x - 5x + 6 = 0

( x - 2)( x 3 - 3x 2 + x - 3) = 0 x = 3 is a root, so (x – 3) is a factor. Use synthetic division to divide by (x – 3).

3 1 -3

1 -3

3 1

0

0

3

1

0

( x - 2)( x - 3x + x - 3) = 0 ( x - 2)( x - 3)( x 2 + 1) = 0 3

2

x 2 + 1 = 0  x 2 = -1  x = i Solution set: {2, 3,  i } 98. x4 + 2x3 – 3x2 – 4x + 4 = 0; {1, –2}

Solution x = 1 is a solution, so ( x - 1) is a factor. Use synthetic division to divide by ( x - 1) . 1 1 2 -3 1

0 3

4

0 -4

3

1 3 4

-4

-4

0

2

x + 2x - 3x - 4 x + 4 = 0

( x - 1)( x 3 + 3x 2 - 4) = 0 x = –2 is a root, so (x + 2) is a factor. Use synthetic division to divide by (x + 2).

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1058


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

-2 1

3

0 -4

-2

-2

4

1

-2

0

1

( x - 1)( x + 3x 2 - 4) = 0 ( x - 1)( x + 2)( x 2 + x - 2) = 0 ( x - 1)( x + 2)( x + 2)( x - 1) = 0 Solution set: {1, - 2, 1, - 2} 3

Find a polynomial function P(x) with the given zeros. There is no unique answer for P(x). 99. 4, 5

Solution

( x - 4)( x - 5) = x - 9x + 20 2

100. –3, 5

Solution

( x + 3)( x - 5) = x - 2x - 15 2

101. 1, 1, 1

Solution

( x - 1)( x - 1)( x - 1) = ( x 2 - 2 x + 1)( x - 1) = x 3 - 3 x 2 + 3 x - 1 102. 1, 0, –1

Solution

( x - 1)( x - 0)( x + 1) = ( x 2 - x ) ( x + 1) = x 3 - x 103. 2, 4, 5

Solution

( x - 2)( x - 4)( x - 5) = ( x 2 - 6 x + 8) ( x - 5) = x 3 - 11x 2 + 38 x - 40 104. 7, 6, 3

Solution

( x - 7 )( x - 6)( x - 3) = ( x 2 - 13 x + 42) ( x - 3) = x 3 - 16 x 2 + 81x - 126 105. 1, –1,

2, - 2

Solution

( x - 1)( x + 1)( x - 2 )( x + 2 ) = ( x 2 - 1)( x 2 - 2) = x 4 - 3x 2 + 2

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1059


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

106. 0, 0, 0,

3, - 3

Solution

( x - 0)( x - 0)( x - 0)( x - 3 )( x + 3 ) = x 3 ( x 2 - 3) = x 5 - 3x 3

107.

2 , i, –i

Solution

( x - 2)( x - i )( x + i ) = ( x - 2)( x - i ) = ( x - 2)( x + 1) = x - 2x + x - 2 2

2

2

3

2

108. i, i, 2

Solution

( x - i )( x - i )( x - 2) = ( x 2 - 2ix + i 2 )( x - 2) = ( x 2 - 2ix - 1) ( x - 2) = x 3 - 2 x 2 - 2ix 2 + 4ix - x + 2 = x 3 - (2 + 2i ) x 2 - (1 - 4i ) x + 2 109. 0, 1 + i, 1 – i

Solution ( x - 0) éëê x - (1 + i )ùûú éëê x - (1 - i )ùûú = x éëê x 2 - (1 - i ) x - (1 + i ) x + (1 + i )(1 - i )ùûú = x éê x 2 - x + ix - x - ix + 1 - i 2 ùú ë û = x éê x 2 - 2 x + 2ùú = x 3 - 2 x 2 + 2 x ë û 110. i, 2 + i, 2 – i

Solution ( x - i ) éêë x - (2 + i )ùúû éêë x - (2 - i )ùúû = ( x - i ) éêë x 2 - (2 - i ) x - (2 + i ) x + (2 + i )(2 - i )ùúû = ( x - i ) éê x 2 - 2 x + ix - 2 x - ix + 4 - i 2 ùú ë û 2 é ù = ( x - i ) ê x - 4 x + 5ú ë û = x 3 - 4 x 2 + 5 x - ix 2 + 4ix - 5i

= x 3 - (4 + i ) x 2 + (5 + 4i ) x - 5i Fix It In exercises 111 and 112, identify the error made and fix it. 111. Use synthetic division to perform the division. x 3  5x2  3x  6 x3

Solution The divisor is x – 3, so 3 should be written in the box.

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1060


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

-5

3 1

3 -6

3 -6 -9 1 - 2 - 3 -15 x2  2x  3 

15 x 3

112. Use synthetic division to perform the division. 4 x 3  2x 2  4 x2

Solution A zero must be written as a coefficient for x in the dividend.

-2 4

-2

0

-8

20 -40

4

4 -10 20 -36 4 x 2 - 10 x + 20 -

36 x +2

Discovery and Writing 113. State the Division Algorithm and explain how it can be used to verify the results of long division.

Solution Answers may vary. 114. State the Remainder Theorem and describe why it is used in algebra.

Solution Answers may vary. 115. State the Factor Theorem and describe why it is used in algebra.

Solution Answers may vary. 116. Describe the steps used to perform synthetic division.

Solution Answers may vary. 117. If 0 is a zero of P ( x ) = an x n + an–1 x n–1 +  + a1 x + a0 , find a0.

Solution

P (0) = 0 n

n–1

an (0) + an–1 (0)

+  + a1 (0) + a0 = 0

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1061


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

0 + 0 +  + 0 + a0 = 0 a0 = 0 118. If 0 occurs twice as a zero of P ( x ) = an x n + an-1 x n-1 +  + a1 x + a0 , find a1.

Solution From #117, a0 = 0. Then, P ( x ) = an x n + an- 1 x n- 1 +  + a1 x = x (an x n- 1 + an- 1 x n-2 +  + a2 x + a1 ) .

If 0 is a double root, then it is a root of the depressed polynomial, so n- 1

an (0)

n- 2

+ an- 1 (0)

+  + a2 (0) + a1 = 0 = a1 = 0 (as in # 117) .

119. If P (2) = 0 and P (-2) = 0, explain why x2 – 4 is a factor of P(x).

Solution P(2) = 0  (x – 2) is a factor. P(–2) = 0  (x + 2) is a factor. The product of two factors will also be a factor, so (x – 2)(x + 2) = x2 – 4 is a factor of the polynomial P(x). 120. If P ( x ) = x 4 - 3 x 3 + kx 2 + 4 x - 1 and P(2) = 11, find k.

Solution P (2) = 11 3

2

24 - 3 (2) + k (2) + 4 (2) - 1 = 11 16 - 3 (8) + 4k + 8 - 1 = 11 4k - 1 = 11 4k = 12 k=3

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 121. Synthetic division can be used to perform the division

55 x 44 + 44 x 33 + 33 x 22 + 11 . x - 66

Solution True. 122. Synthetic division can be used to perform the division

55 x 44 + 44 x 33 + 33 x 22 + 11 . x 2 - 66

Solution False. The divisor must be linear.

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1062


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

123. –1 is a zero of P(x) = x13,579 + x2468.

Solution P(–1) = 0. True. 124. (x – 1) is a factor of P(x) = x13,579 – x2468.

Solution P(1) = 0. True. 125. If P(x) = x444 + x44 + x4, then P(i) = 3i.

Solution False. P(i) = 3. 126. i is a zero of P(x) = 222x2 + 222.

Solution P(i) = 0. True. 127. If (135x – 246) is a factor of P(x), then x =

135 is a zero of P(x). 246

Solution False. 246 is a zero. 135 128. If 0 is a zero of P(x), then the constant term of P(x) is 0.

Solution True. (see #117)

EXERCISES 4.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

 

Identify the number of real zeros the linear function f x  3 x  5 has.

Solution 1

 

2. Identify the number of complex zeros the quadratic function f x  x 2  16 has.

Solution 2 3. Identify the conjugate of 4  9i .

Solution 4  9i

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1063


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

4. Identify the conjugate of 5  7i .

Solution 5  7i

 

 

5. Find f  x if f x  5 x 4  2 x 3  4 x 2  6 x  1.

Solution f x   5 x   2 x   4 x   6 x   1 4

3

2

 5x 4  2x 3  4 x 2  6x  1

 

6. Identify the number of sign changes of P x  6 x 5  3 x 4  2 x 3  4 x 2  11x  1.

Solution 4 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If P(x) is a polynomial function with positive degree, then P(x) has at least one __________.

Solution zero 8. The statement in Exercise 1 is called the __________.

Solution Fundamental Theorem of Algebra 9. The __________ of a + bi is a – bi.

Solution conjugate 10. The polynomial 6x4 + 5x3 – 2x2 – 3 has __________ variations in sign.

Solution 2 11. The polynomial (–x)3 – (–x)2 – 4 has __________ variations in sign.

Solution (–x)3 – (–x)2 – 4 = –x3 – x2 – 4 0 variations

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1064


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

12. The polynomial function P(x) = 7x4 + 5x3 – 2x + 1 can have at most __________ positive zeros.

Solution 2 variations  at most 2 positive roots 13. The polynomial function P ( x ) = 7 x 4 + 5 x 3 – 2 x + 1 can have at most __________ negative zeros.

Solution 4

3

7 x 4 + 5 x 3 – 2 x + 1  7 (- x ) + 5 (- x ) - 2 (- x ) + 1  7 x 4 – 5 x 3 + 2 x + 1  2 variations  at most 2 negative roots

14. Complex zeros occur in complex __________ pairs. (Assume that the equation has real coefficients.)

Solution conjugate 15. If no number less than d can be a zero of P(x) = 0, then d is called a(n) __________.

Solution lower bound 16. If no number greater than c can be a zero of P(x) = 0, then c is called a(n) __________.

Solution upper bound Practice Determine how many zeros each polynomial function has. 17. P ( x ) = x 10 - 1

Solution

P ( x ) = x 10 - 1  10 zeros 18. P ( x ) = x 40 - 1

Solution

P ( x ) = x 40 - 1  40 zeros 19. P ( x ) = 3 x 4 – 4 x 2 – 2 x + 7

Solution

P ( x ) = 3 x 4 – 4 x 2 – 2 x + 7  4 zeros

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1065


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

20. P ( x ) = –32 x 111 – x 5 – 1

Solution

P ( x ) = –32 x 111 – x 5 – 1  111 zeros 21. One zero of P(x) = x(3x4 – 2) – 12x is 0. How many other zeros are there?

Solution

P ( x ) = x (3 x 4 – 2) – 12 x

= 3 x 5 - 14 x 5 total zeros  4 other zeros 22. Two zeros of P(x) = 3x2(x7 – 14x + 3) are 0. How many other zeros are there?

Solution

P ( x ) = 3 x 2 ( x 7 – 14 x + 3)

= 3 x 9 - 42 x 3 + 9 x 2 9 total zeros  7 other zeros Determine how many linear factors and zeros each polynomial function has. 23. P(x) = x4 – 81

Solution P ( x ) = x 4 – 81 4 linear factors, 4 zeros

24. P(x) = x40 + x39

Solution P ( x ) = x 40 + x 39 40 linear factors, 40 zeros

25. P(x) = 4x5 + 8x3

Solution P ( x ) = 4 x 5 + 8x 3 5 linear factors, 5 zeros

26. P(x) = x3 + 144x

Solution P ( x ) = x 3 + 144 x 3 linear factors, 3 zeros

Write a quadratic function with real coefficients and the given zero. 27. 2i

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1066


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution If 2i is a zero, then –2i is a zero also:

P ( x ) = ( x - 2i )( x + 2i )

= x 2 - 4i 2 = x 2 + 4 28. –3i

Solution If –3i is a zero, then 3i is a zero also:

P ( x ) = ( x + 3i )( x - 3i )

= x 2 - 9i 2 = x 2 + 9 29. 3 – i

Solution If 3 – i is a zero, then 3 + i is a zero also:

P ( x ) = éê x - (3 - i )ùú éê x - (3 + i )ùú = x 2 - (3 + i ) x - (3 - i ) x + (3 - i )(3 + i ) ûë ë û

= x 2 - 3 x - ix - 3 x + ix + 9 - i 2 = x 2 - 6 x + 10 30. 4 + 2i

Solution If 4 + 2i is a zero, then 4 – 2i is a zero also:

P ( x ) = éê x - (4 + 2i )ùú éê x - (4 - 2i )ùú = x 2 - (4 - 2i ) x - (4 + 2i ) x + (4 + 2i )(4 - 2i ) ë ûë û

= x 2 - 4 x + 2ix - 4 x - 2ix + 16 - 4i 2 = x 2 - 8 x + 20 Write a third-degree polynomial function with real coefficients and the given zeros. 31. 3, –i

Solution If –i is a zero, then i is a zero also: P ( x ) = ( x - 3)( x + i )( x - i ) = ( x - 3)( x 2 - i 2 ) = ( x - 3)( x 2 + 1) = x 3 - 3x 2 + x - 3

32. 1, i

Solution If i is a zero, then –i is a zero also:

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1067


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

P ( x ) = ( x - 1)( x - i )( x + i ) = ( x - 1)( x 2 - i 2 ) = ( x - 1)( x 2 + 1) = x3 - x2 + x - 1

33. 2, 2 + i

Solution If 2 + i is a zero, then 2 – i is a zero also: P ( x ) = ( x - 2) éê x - (2 + i )ùú éê x - (2 - i )ùú ë ûë û 2 = ( x - 2) éê x - (2 - i ) x - (2 + i ) x + (2 + i )(2 - i )ùú ë û = ( x - 2) éê x 2 - 2 x + ix - 2 x - ix + 4 - i 2 ùú ë û = ( x - 2) éê x 2 - 4 x + 5úù = x 3 - 6 x 2 + 13 x - 10 ë û 34. –2, 3 – i

Solution If 3 – i is a zero, then 3 + i is a zero also: P ( x ) = ( x + 2) éê x - (3 - i )ùú éê x - (3 + i )ùú ë ûë û 2 é = ( x + 2) ê x - (3 + i ) x - (3 - i ) x + (3 - i )(3 + i )ùú ë û = ( x + 2) éê x 2 - 3 x - ix - 3 x + ix + 9 - i 2 ùú ë û = ( x + 2) éê x 2 - 6 x + 10ùú = x 3 - 4 x 2 - 2 x + 20 ë û Write a fourth-degree polynomial function with real coefficients and the given zeros. 35. 3, 2, i

Solution If i is a zero, then –i is a zero also:

P ( x ) = ( x - 3)( x - 2)( x - i )( x + i )

= ( x 2 - 5 x + 6)( x 2 - i 2 ) = ( x 2 - 5 x + 6)( x 2 + 1) = x 4 - 5 x 3 + 7 x 2 - 5 x + 6 36. 1, 2, 1 + i

Solution If 1 + i is a zero, then 1 – i is a zero also: P ( x ) = ( x - 1)( x - 2) éê x - (1 + i )ùú éê x - (1 - i )ùú ë ûë û 2 2 = ( x - 3 x + 2) éê x - (1 - i ) x - (1 + i ) x + (1 + i )(1 - i )ùú ë û = ( x 2 - 3 x + 2) éê x 2 - x + ix - x - ix + 1 - i 2 ùú ë û = ( x 2 - 3 x + 2) éê x 2 - 2 x + 2ùú = x 4 - 5 x 3 + 10 x 2 - 10 x + 4 ë û

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1068


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

37. i, 1 – i

Solution If i and 1 – i are zero, then –i and 1 + i are zero also: P ( x ) = ( x - i )( x + i ) éê x - (1 - i )ùú éê x - (1 + i )ùú ë ûë û 2 2 é 2 = ( x - i ) ê x - (1 + i ) x - (1 - i ) x + (1 - i )(1 - i )ùú ë û = ( x 2 + 1) éê x 2 - x - ix - x + ix + 1 - i 2 ùú ë û = ( x 2 + 1) éê x 2 - 2 x + 2ùú = x 4 - 2 x 3 + 3 x 2 - 2 x + 2 ë û 38. i, 2 – i

Solution If i and 2 – i are zero, then –i and 2 + i are zero also: P ( x ) = ( x - i )( x + i ) éê x - (2 - i )ùú éê x - (2 + i )ùú ë ûë û 2 2 é 2 = ( x - i ) ê x - (2 + i ) x - (2 - i ) x + (2 - i )(2 + i )ùú ë û = ( x 2 + 1) éê x 2 - 2 x - ix - 2 x + ix + 4 - i 2 ùú ë û = ( x 2 + 1) éê x 2 - 4 x + 5ùú = x 4 - 4 x 3 + 6 x 2 - 4 x + 5 ë û Write a quadratic function with the given repeated zero. 39. 2i

Solution If 2i is a double zero, then there are two factors of (x – 2i). [The problem does not specify real coefficients, so we do not include –2i as a zero.]

P ( x ) = ( x - 2i )( x - 2i )

= x 2 - 4ix + 4i 2 = x 2 - 4ix - 4 40. –2i

Solution If –2i is a double zero, then there are two factors of (x + 2i). [The problem does not specify real coefficients, so we do not include 2i as a zero.]

P ( x ) = ( x + 2i )( x + 2i )

= x 2 + 4ix + 4i 2 = x 2 + 4ix - 4 Use Descartes’ Rule of Signs to find the number of possible positive, negative, and nonreal zeros of each function. 41. P(x) = 3x3 + 5x2 – 4x + 3

Solution

P ( x ) = 3x 3 + 5x 2 – 4 x + 3 2 sign variations  2 or 0 positive zeros

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1069


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

3

2

P (-x ) = 3 (-x ) + 5 (-x ) – 4 (-x ) + 3

= -3 x 3 + 5 x 2 + 4 x + 3 1 sign variation  1 negative zeros # pos 2 0

# neg 1 1

# nonreal 0 2

42. P(x) = 3x3 – 5x2 – 4x – 3

Solution

P ( x ) = 3x 3 - 5x 2 – 4 x - 3 1 sign variations  1 positive zero 3

2

P (-x ) = 3 (-x ) - 5 (-x ) – 4 (-x ) - 3

= -3 x 3 - 5 x 2 + 4 x - 3 2 sign variations  2 or 0 negative zeros # pos 1 1

# neg 2 0

# nonreal 0 2

43. P(x) = 2x3 + 7x2 + 5x + 5

Solution P ( x ) = 2x 3 + 7 x 2 + 5x + 5 0 sign variations  0 positive zeros 3

2

P (-x ) = 2 (-x ) + 7 (-x ) + 5 (-x ) + 3

= -2 x 3 + 7 x 2 - 5 x + 3 3 sign variations  3 or 1 negative zeros # pos 0 0

# neg 3 1

# nonreal 0 2

44. P(x) = –2x3 – 7x2 – 5x – 4

Solution P ( x ) = –2 x 3 – 7 x 2 – 5 x – 4 0 sign variations  0 positive zeros 3

2

P (-x ) = -2 (-x ) - 7 (-x ) - 5 (-x ) - 4 = 2x 3 - 7 x 2 + 5x - 4 3 sign variations  3 or 1 negative zeros

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1070


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 0 0

# neg 3 1

# nonreal 0 2

45. P(x) = 8x4 – 5

Solution P ( x ) = 8x 4 – 5 0 sign variations  0 positive zeros 4

P (-x ) = 8 (-x ) + 5 = 8x 4 + 5 0 sign variations  0 negative zeros

# pos 0

# neg 0

# nonreal 4

46. P(x) = –3x3 + 5

Solution P ( x ) = –3 x 3 + 5 1 sign variation  1 positive zero 3

P (-x ) = -3 (-x ) + 5 = 3x 3 + 5 0 sign variations  0 negative zeros # pos 1

# neg 0

# nonreal 2

47. P(x) = x4 + 8x2 – 5x – 10

Solution P ( x ) = x 4 + 8 x 2 – 5 x – 10: 1 sign variation  1 positive zero 4

2

P (-x ) = (-x ) + 8 (-x ) - 5 (-x ) - 10 = x 4 + 8 x 2 + 5 x - 10: 1 sign variation  1 negative zero # pos 1

# neg 1

# nonreal 2

48. P(x) = 5x7 + 3x6 – 2x5 + 3x4 + 9x3 + x2 + 1

Solution P ( x ) = 5 x 7 + 3 x 6 – 2 x 5 + 3 x 4 + 9 x 3 + x 2 + 1: 2 sign variation  2 or 0 positive zeros 7

6

5

4

3

2

P (-x ) = 5 (-x ) + 3 (-x ) – 2 (-x ) + 3 (-x ) + 9 (-x ) + (-x ) + 1 = -5 x 7 + 3 x 6 + 2 x 5 + 3 x 4 - 9 x 3 + x 2 + 1: 3 sign variation  3 or 1 negative zeros

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1071


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 2 2 0 0

# neg 3 1 3 1

# nonreal 2 4 4 6

49. P(x) = –x10 – x8 – x6 – x4 – x2 – 1

Solution P ( x ) = –x 10 – x 8 – x 6 – x 4 – x 2 – 1: 0 sign variation  0 positive zeros 10

8

6

4

2

P (-x ) = – (-x ) – (-x ) – (-x ) – (-x ) – (-x ) – 1 = -x 10 – x 8 – x 6 – x 4 – x 2 – 1: 0 sign variation  0 negative zeros # pos 0

# neg 0

# nonreal 10

50. P(x) = x10 + x8 + x6 + x4 + x2 + 1

Solution P ( x ) = x 10 + x 8 + x 6 + x 4 + x 2 + 1: 0 sign variation  0positive zeros 10

8

6

4

2

P (-x ) = (-x ) + (-x ) + (-x ) + (-x ) + (-x ) + 1 = x 10 + x 8 + x 6 + x 4 + x 2 + 1: 0 sign variation  0 negative zeros # pos 0

# neg 0

# nonreal 10

51. P(x) = x9 + x7 + x5 + x3 + x (Is 0 a zero?)

Solution P ( x ) = x 9 + x 7 + x 5 + x 3 + x = x ( x 8 + x 6 + x 4 + x 2 + 1) : 0 sign variation  0 positive zeros 8 6 4 2 é ù P (- x ) = (- x ) ê(- x ) + (- x ) + (- x ) + (- x ) + 1ú ë û 8 6 4 2 é ù = - x ê x + x + x + x + 1ú : 0 sign variation  0 negative zeros ë û

# pos 0

# neg 0

#zero 1

# nonreal 8

52. P(x) = –x9 – x7 – x5 – x3 – x (Is 0 a zero?)

Solution P ( x ) = - x 9 - x 7 - x 5 - x 3 - x = - x ( x 8 + x 6 + x 4 + x 2 + 1) : 0 sign var  0positive zeros 8 6 4 2 é ù P (-x ) = - (- x ) ê(- x ) + (-x ) + (- x ) + (-x ) + 1ú ë û 8 6 4 2 é ù = x ê x + x + x + x + 1ú : 0 sign variation  0 negative zeros ë û

# pos 0

# neg 0

#zero 1

# nonreal 8

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1072


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

53. P(x) = –2x4 – 3x2 + 2x + 3

Solution P ( x ) = -2 x 4 - 3 x 2 + 2 x + 3: 1 sign variation  1 positive zero 4

2

P (-x ) = -2 (-x ) - 3 (-x ) + 2 (-x ) + 3 = -2 x 4 - 3 x 2 - 2 x + 3: 1 sign variation  1 negative zero # pos 1

# neg 1

# nonreal 2

54. P(x) = –7x5 – 6x4 + 3x3 – 2x2 + 7x – 4

Solution

P ( x ) = -7 x 5 - 6 x 4 + 3 x 3 - 2 x 2 + 7 x - 4: 4 sign variation  4 or 2 or 0 positive zeros 5

4

3

2

P (-x ) = -7 (-x ) - 6 (-x ) + 3 (-x ) - 2 (-x ) + 7 (-x ) - 4 = 7 x 5 - 6 x 4 - 3 x 3 - 2 x 2 - 7 x - 4: 1 sign variation  1 negative zero # pos 4 2 0

# neg 1 1 1

# nonreal 0 2 4

Find integer bounds for the zeros of each function. Answers can vary. 55. P(x) = x2 – 2x – 4

Solution P ( x ) = x 2 - 2x - 4 4 1 -2 -4

-2 1 -2 -4

4

8

-2

8

2

4

1 -4

4

1

Upper bound: 4

Lower bound: - 2

56. P(x) = 9x2 – 6x – 1

Solution P ( x ) = 9x 2 - 6x - 1 1 9 - 6 -1

9

-1 9

-6 -1

9

3

-9

15

3

2

1 -15

14

Upper bound: 1

Lower bound: - 1

57. P(x) = 18x2 – 6x – 1

Solution

P ( x ) = 18 x 2 - 6 x - 1

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1073


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

1 18 -6 -1 18 12 18

12

11

Upper bound: 1 -1 18 -6 -1 -18

24

18 -24 23 Lower bound: - 1 58. P(x) = 2x2 – 10x – 9

Solution

P ( x ) = 2 x 2 - 10 x - 9 6 2 -10 -9

2

12

12

2

3

Upper bound: 6

-1 2

2

-10 -9 -2

12

- 12

3

Lower bound: - 1 59. P(x) = 6x3 – 13x2 – 110x

Solution P ( x ) = 6 x 3 - 13 x 2 - 110 x 6 6 -13 -110

6

0

36

138 168

23

28 168

Upper bound: 6

-4 6

2

-13 -110

0

- 24

148 -152

- 37

38 -152

Lower bound: - 4 60. P(x) = 12x3 + 20x2 – x – 6

Solution

P ( x ) = 12 x 3 + 20 x 2 - x - 6

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1074


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

1 12 20 -1 -6 12 32

31

12 32 31

25

Upper bound: 1 20 -1 -6 -2 12 -24

8 -14

12 - 4 7 -20 Lower bound: - 2 61. P(x) = x5 + x4 – 8x3 – 8x2 + 15x + 15

Solution

P ( x ) = x 5 + x 4 - 8 x 3 - 8 x 2 + 15 x + 15 1 -8 -8

15

15

3

12

12

12

81

1 4

4

4 27 96

3 1

Upper bound: 3

-4 1

1 -8

-4

-8

15

12 -16

1 -3

15

96 -444

4 -24 111

- 429

Lower bound: - 4 62. P(x) = 3x4 – 5x3 – 9x2 + 15x

Solution P ( x ) = 3 x 4 – 5 x 3 – 9 x 2 + 15 x 3 3 -5 -9

15

0

9

12

9

72

4

3 24

72

3

Upper bound: 3

-2 3 -5 -9 -6 3 -11

15

0

22 -26 22 13 -11

22

Lower bound: - 2 63. P(x) = 3x5 – 11x4 – 2x3 + 38x2 – 21x – 15

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1075


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution P ( x ) = 3 x 5 - 11x 4 - 2 x 3 + 38 x 2 - 21x - 15 4 3 -11 -2 38

-21 -15

12

4

8

184 652

1

2 46

163 637

3

Upper bound: 4 -2 3 -11 -2 -6

38 -21

34 -64

1 -17 32

-15

52 -62

- 26

31 -77

Lower bound: - 2 64. P(x) = 3x6 – 4x5 – 21x4 + 4x3 + 8x2 + 8x + 32

Solution

P ( x ) = 3 x 6 - 4 x 5 - 21x 4 + 4 x 3 + 8 x 2 + 8 x + 32 4 3 -4 -21

3

4

8

12

32 44

192

8

11

200 808 3264

48

8

32

800 3234

Upper bound: 4 -3 3 -4 -21

4

8

8

32

-9

39 -54

150 -474

1398

3 -13

18 -50

158 -466

1430

Lower bound: - 3

Fix It In Exercises 65 and 66, identify the step the first error is made and fix it. 65. Use Descartes Rule of signs to determine the number of possible positive real zeros of

P  x   3 x 2  5 x 4  3 x  1  x 3 . To do so, write the polynomial in descending powers of

the variable, determine the number of variations of sign of P(x), and then state your answer.

Solution Step 3 was incorrect. Step 3: Number of possible positive real zeros is 3 or 1. 66. Use Descartes Rule of signs to determine the number of possible negative real zeros of

P  x   2 x  3 x 6  5 x 4  3 x 3  1  x 5  x 2 . To do so, write the polynomial in descending

powers of the variable,

Solution Step 2 was incorrect.

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1076


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 

Step 1: P x  3 x 6  x 5  5 x 4  3 x 3  x 2  2 x  1

 

Step 2: P  x  3x 6  x 5  5 x 4  3 x 3  x 2  2x  1

 

Step 3: Number of variations in signs of P  x is 4. Step 4: Number of possible positive real zeros is 4, 2, or 0.

Discovery and Writing 67. Explain why the Fundamental Theorem of Algebra guarantees that every polynomial function of positive degree has at least one complex zero.

Solution Answers may vary. 68. Explain why the Fundamental Theorem of Algebra and the Factor Theorem guarantee that an nth-degree polynomial function has n zeros.

Solution Answers may vary. 69. State the Conjugate Pairs Theorem and explain why it is important in algebra.

Solution Answers may vary. 70. What is Decartes’ Rule of Signs? Explain how to apply the rule to a polynomial function.

Solution Answers may vary. 71. Prove that any odd-degree polynomial function with real coefficients must have at least one real zero.

Solution The number of nonreal zeros must occur in conjugate pairs, so the number of nonreal zeros will always be even. Since a polynomial of odd degree has an odd number of zeros, at least one zero must not be nonreal. Thus, at least one zero of such a polynomial will be real. 72. If a, b, c, and d are positive numbers, prove that P(x) = ax4 + bx2 + cx – d has exactly two nonreal zeros.

Solution According to Descartes' Rule of Signs, the polynomial will have 1 positive zero and 1 negative zero. Since the polynomial has a total of 4 zeros, the other 2 zeros must be nonreal (and conjugates).

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1077


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 73. The Fundamental Theorem of Algebra states that every polynomial function of positive degree has at least one imaginary zero.

Solution False. The theorem states that every polynomial function has at least one complex zero. 74. If a polynomial function is of degree n, then, counting multiple zeros separately, the function has n zeros.

Solution True 75. If 55 – 77i is a zero of the polynomial function P(x), then 55 + 77i is also a zero of P(x).

Solution False. You need to know that the coefficients are real numbers to use the Conjugate Pairs Theorem. 76. If 22 + 44i is a zero of the polynomial function P(x) with real coefficients, then 22 – 44i is also a zero of P(x).

Solution True. 77. The polynomial function P(x) = x123 + x456 + x789 has exactly 123 zeros.

Solution False. It will have 789 zeros. 78. The polynomial function P(x) = x4 – x2 + 4x7 – 3x3 – 1 has three variations in sign.

Solution False. It has 1 variation in sign.

EXERCISES 4.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Given the function P(x) in factored form. Identify all zeros and classify the zeros as real, rational, irrational, or nonreal numbers. State all that apply.







P  x    x  2 2 x  1 x  2 x  2 x   2  2i  x   2  2i 

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1078


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution 1 : real and rational; 2 irrational; 2  2i : nonreal; 2  2i : nonreal

2: real and rational;

2: real and irrational;  2: real and

2. List all of the positive and negative factors of 36.

Solution  1,  2,  3,  4,  6,  9,  12,  18,  36 factors of p

3. If p = –3 and q = 5, write the factors of p divided by the factors of q, factors of q , and simplify each fraction.

Solution 3 3 1 1    3,  ,   1,  1 5 1 5 4. Is

 

1 a zero of P x  24 x 3  26 x 2  9 x  1? 2

Solution 3

2

 1  1  1  1 P    24    26    9    1  0 so, yes. 2 2 2 2 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 5. The rational zeros of the function P(x) = 3x3 + 4x – 7 will have the form qp , where p is a factor of _________ and q is a factor of 3.

Solution –7 6. The rational zeros of the function P(x) = 5x3 + 3x2 – 4 will have the form qp , where p is a factor of –4 and q is a factor of __________.

Solution 5 7. Consider the synthetic division of P(x) = 5x3 – 7x2 – 3x – 63 by x – 3.

3

5 -7 -3 -63

5

15

24

63

8

21

0

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1079


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Since the remainder is 0, 3 is a __________ of the function.

Solution zero 8. In Exercise 7, the depressed equation is __________.

Solution 5x2 + 8x + 21 = 0 Practice Use the Rational Zero Theorem to list all possible rational zeros of the polynomial function. 9. P(x) = x3 + 10x2 + 5x – 12

Solution num:  1,  2,  3,  4,  6,  12; den:  1 possible zeros:  1,  2,  3,  4,  6,  12 10. P(x) = –x3 + 3x2 – 4x – 8

Solution num:  1,  2,  4,  8; den:  1 possible zeros:  1,  2,  4,  8 11. P(x) = 2x4 – x3 + 10x2 + 5x – 6

Solution num:  1,  2,  3,  6; den:  1,  2 possible zeros:  1,  2,  3,  6,  21 ,  32 12. P(x) = 3x4 – x3 + 7x2 – 5x – 8

Solution num:  1,  2,  4,  8; den:  1,  3 possible zeros:  1,  2,  4,  8,  31 ,  23 ,  43 ,  83 13. P(x) = 4x5 – x4 – x3 + x2 + 5x – 10

Solution num:  1,  2,  5,  10; den:  1,  2,  4 possible zeros:  1,  2,  5,  10,  21 ,  52 ,  41 ,  45 14. P(x) = 6x4 – 2x3 + x2 – x + 3

Solution num:  1,  3; den:  1,  2,  3,  6 possible zeros:  1,  3,  21 ,  23 ,  31 ,  61 Find all rational zeros of each polynomial function. 15. P(x) = x3 – 5x2 – x + 5

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1080


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

P ( x ) = x 3 - 5x 2 - x + 5 Possible rational zeros:  1,  5 Descartes’ Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

Test x = -1: -1 1 -5 -1 -1 1 -6

5

6 -5 5

0

P ( x ) = x - 5x - x + 5 3

2

= ( x + 1)( x 2 - 6 x + 5) = ( x + 1)( x - 5)( x - 1) zeros: {-1, 5, 1}

16. P(x) = x3 + 7x2 – x – 7

Solution

P (x) = x3 + 7x2 - x - 7 Possible rational zeros:  1,  7 Descartes’ Rule of Signs: # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = 1: 1 1 7 -1 -7 1 1

8

8

7

7

0

P (x) = x + 7x2 - x - 7 3

= ( x - 1)( x 2 + 8 x + 7) = ( x - 1)( x + 1)( x + 7) zeros: {1, - 1, - 7}

17. P(x) = x3 – 2x2 – x + 2

Solution

P ( x ) = x 3 - 2x 2 - x + 2 Possible rational zeros:  1,  2

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1081


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Descartes’ Rule of Signs: # neg 1 1

# pos 2 0

# nonreal 0 2

Test x = -1: -1 1 -2 -1 -1 1 -3

2

3 -2 2

0

P ( x ) = x - 2x - x + 2 3

2

= ( x + 1)( x 2 - 3 x + 2) = ( x + 1)( x - 1)( x - 2) zeros: {-1, 1, 2}

18. P(x) = x3 + x2 – 4x – 4

Solution

P (x) = x3 + x2 - 4x - 4 Possible rational zeros:  1,  2,  4 Descartes’ Rule of Signs: # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = -1: -1 1

1 -4 -4 -1

1

0

4

0 -4

0

P (x) = x + x - 4x - 4 3

2

= ( x + 1)( x 2 - 4) = ( x + 1)( x + 2)( x - 2) zeros: {-1, - 2, 2}

19. P(x) = x3 – x2 – 4x + 4

Solution

P (x) = x3 - x2 - 4x + 4 Possible rational zeros:  1,  2,  4 Descartes’ Rule of Signs:

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1082


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 2 0

# neg 1 1

# nonreal 0 2

Test x = 1: 1 1 -1 -4

0 -4

1 1

4

0 -4

0

P (x) = x - x - 4x + 4 3

2

= ( x - 1)( x 2 - 4) = ( x - 1)( x + 2)( x - 2) zeros: {1, - 2, 2}

20. P(x) = x3 + 2x2 – x – 2

Solution

P ( x ) = x 3 + 2x 2 - x - 2 Possible rational zeros:  1,  2 Descartes' Rule of Signs: # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = 1: 1 1 2 -1 -2

1

1

3

2

3

2

0

P ( x ) = x + 2x 2 - x - 2 3

= ( x - 1)( x 2 + 3 x + 2) = ( x - 1)( x + 1)( x + 2) zeros: {1, - 1, - 2}

21. P(x) = x3 – 2x2 – 9x + 18

Solution

P ( x ) = x 3 - 2 x 2 - 9 x + 18 Possible rational zeros:  1,  2,  3,  6,  9,  18 Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

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1083


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 2: 2 1 -2 -9

18

0 -18

2

0 -9

1

0

P ( x ) = x - 2 x - 9 x + 18 3

2

= ( x - 2)( x 2 - 9) = ( x - 2)( x + 3)( x - 3) zeros: {2, - 3, 3}

22. P(x) = x3 + 3x2 – 4x – 12

Solution

P ( x ) = x 3 + 3 x 2 - 4 x - 12 Possible rational zeros:  1,  2,  3,  4,  6,  12 Descartes' Rule of Signs: # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = 2: 2 1 3 -4 -12 2 1

10

5

12

6

0

P ( x ) = x + 3 x - 4 x - 12 3

2

= ( x - 2)( x 2 + 5 x + 6) = ( x - 2)( x + 3)( x + 2) zeros: {2, - 3, - 2}

23. P(x) = 2x3 – x2 – 2x + 1

Solution Possible rational zeros:  1,  21 Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

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1084


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 1: 1 2 -1 -2

1 -1

2 2

1

1

-1

0

P ( x ) = 2x - x - 2x + 1 3

2

= ( x - 1)(2 x 2 + x - 1) = ( x - 1)(2 x - 1)( x + 1) zeros: {1, 21 , - 1}

24. P(x) = 3x3 + x2 – 3x – 1

Solution

3x 3 + x 2 - 3x - 1 = 0 Possible rational zeros:  1,  31 Descartes' Rule of Signs: # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = 1: 1 3

1 -3 -1 3

4

1

3 4

1

0

P ( x ) = 3x + x 2 - 3x - 1 3

= ( x - 1)(3 x 2 + 4 x + 1) = ( x - 1)(3 x + 1)( x + 1) zeros: {1, - 31 , - 1}

25. P(x) = 3x3 + 5x2 + x – 1

Solution

3x 3 + 5x 2 + x - 1 = 0 Possible rational zeros:  1,  31 Descartes' Rule of Signs: # pos 1 1

# neg 2 0

# nonreal 0 2

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1085


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = -1: -1 3

1 -1

5 -3

3

-2

1

-1

2

0

P ( x ) = 3x + 5x + x - 1 3

2

= ( x + 1)(3 x 2 + 2 x - 1) = ( x + 1)(3 x - 1)( x + 1) zeros: {-1, 31 , - 1}

26. P(x) = 2x3 – 3x2 + 1

Solution Possible rational zeros:  1,  21 Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

Test x = 1: 1 2 -3 2 2 -1

0

1

- 1 -1 -1

0

P ( x ) = 2x - 3x + 1 3

2

= ( x - 1)(2 x 2 - x - 1) = ( x - 1)(2 x + 1)( x - 1) zeros: {1, - 21 , 1}

27. P(x) = 30x3 – 47x2 – 9x + 18

Solution Possible rational zeros:  1, ,  2,  3,  6,  9,  18,  21 ,  32 ,  92 ,  31 ,  23 ,  51 ,  52 ,  53 ,  65 ,  95 ,  185 ,  61 ,  101 ,  103 ,  109 ,  151 ,  152 ,  301

Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

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1086


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 23 : 2 3

30 -47

-9

18

- 18 -18

20 30 -27

- 27

0

P ( x ) = 30 x - 47 x - 9 x + 18 2

3

= ( x - 23 )(30 x 2 - 27 x - 27) = 3 ( x - 23 )(10 x 2 - 9 x - 9) = (3 x - 2)(2 x - 3)(5 x + 3) zeros: { 23 , 32 , - 53 } 28. P(x) = 20x3 – 53x2 – 27x + 18 Solution Possible rational zeros: 1, ,  2,  3,  6, 9,  18,  21 ,  32 ,  92 ,  41 ,  43 ,  94 ,

 51 ,  52 ,  53 ,  65 ,  95 ,  185 ,  101 ,  103 , 3 9  109 ,  201 ,  20 ,  20

Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

Test x = 3: 3 20 -53 -27

21 -18

60 20

18

-6

7

0

P ( x ) = 20 x - 53 x - 27 x + 18 3

2

= ( x - 3)(20 x 2 + 7 x - 6) = ( x - 3)(4 x 2 + 3) (5 x - 2)

zeros: {3, - 43 , - 52 } 29. P(x) = 15x3 – 61x2 – 2x + 24 Solution Possible rational zeros:

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1087


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 1,  2,  3,  6,  6,  8,  12,  24,  31 ,  23 ,  43 ,  83 ,  51 ,  52 ,  53 ,  45 , ,  65 ,  85 ,  125 ,  24 5  151 ,  152 ,  154 ,  158

Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

Test x = 4: 4 15 -61 -2

24

60 -4 -24 - 1 -6

15

0

P ( x ) = 15 x - 61x - 2 x + 24 3

2

= ( x - 4)(15 x 2 - x - 6) = ( x - 4)(3 x - 2)(5 x + 3) zeros: {4, 23 , - 53 }

30. P(x) = 20x3 – 44x2 + 9x + 18 Solution Possible rational zeros:  1,  2,  3,  6,

 9,  18,  21 ,  23 ,  92 ,  41 ,  43 ,  49 ,  51 ,  52 ,  53 ,  65 ,  95 ,  185 ,  101 ,  103 ,  109 ,  101 , 3 9  20 ,  20

Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

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1088


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 32 : 3 2

20 -44

9

18

30 -21 -18 20 -14

- 12

0

P ( x ) = 20 x - 44 x + 9 x + 18 2

3

= ( x - 32 )(20 x 2 - 14 x - 12) = 2 ( x - 32 )(10 x 2 - 7 x - 6)

= (2 x - 3)(2 x + 1)(5 x - 6) zeros: { 32 , - 21 , 65 }

31. P(x) = 24x3 – 82x2 + 89x – 30 Solution Possible rational zeros: 1,  2,  3,  5,

6,  10,  15, 30,  21 ,  23 ,  52 ,  152 ,  31 ,  23 ,  53 ,  103 ,  41 ,  43 ,  45 ,  154 ,  61 ,  65 ,  81 ,  83 ,  85 ,  158 ,  121 , 5  125 ,  241 ,  24

Descartes' Rule of Signs: # pos 3 1

# neg 0 0

# nonreal 0 2

Test x = 32 : 3 2

24 -82

89 -30

36 -69 24 -46

30

20

0

P ( x ) = 24 x - 82 x + 89 x - 30 3

2

= ( x - 32 )(24 x 2 - 46 x + 20) = 2 ( x - 32 )(12 x 2 - 23 x + 10) = (2 x - 3)(4 x - 5)(3 x - 2)

zeros: { 32 , 45 , 23 }

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1089


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

32. P(x) = x4 – 10x3 + 35x2 – 50x + 24 Solution Possible rational zeros:  1,  2,  3,  4,  6,  8,  12,

24 Descartes' Rule of Signs: # pos 4 2 0

# neg 0 0 0

# nonreal 0 2 4

Test x = 1: 1 1 -10 35 -50 1 -9

26 -24

1 - 9 26 -24 Test x = 2: 2 1 -9

24

0

26 -24

2 -14 1 -7

24

12

0

P ( x ) = x - 10 x + 35 x 2 - 50 x + 24 4

3

= ( x - 1)( x 3 - 9 x 2 + 26 x - 24) = ( x - 1)( x - 2)( x 2 - 7 x + 12) = ( x - 1)( x - 2)( x - 3)( x - 4) zeros: {1, 2, 3, 4} 33. P(x) = x4 + 4x3 + 6x2 + 4x + 1 Solution Possible rational zeros:  1

Descartes' Rule of Signs: # pos 0 0 0

# neg 4 2 0

# nonreal 0 2 4

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1090


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = -1: -1 1 4 6

4

1

-1 -3 -3 -1 1 3 3 Test x = -1:

1

-1 1

1

3

3

0

-1 -2 -1 1

2

1

0

P ( x ) = x + 4x + 6x2 + 4x + 1 4

3

= ( x + 1)( x 3 + 3 x 2 + 3 x + 1) = ( x + 1)( x + 1)( x 2 + 2 x + 1) = ( x + 1)( x + 1)( x + 1)( x + 1) zeros: {-1, - 1, - 1, - 1} 34. P(x) = x4 – 8x3 + 14x2 + 8x – 15 Solution Possible rational zeros:  1,  3,  5,  15

Descartes' Rule of Signs: # pos 3 1

# neg 1 1

# nonreal 0 2

Test x = -1: -1 1 -8 14

8 -15

-1

9 -23

15

1 -9

23 -15

0

Test x = 1: 1 1 -9 23 -15 1 -8 1 -8

15

15 0

P ( x ) = x 4 - 8 x 3 + 14 x 2 + 8 x - 15 = ( x + 1)( x 3 - 9 x 2 + 23 x - 15) = ( x + 1)( x - 1)( x 2 - 8 x + 15) = ( x + 1)( x - 1)( x - 3)( x - 5) zeros: {-1, 1, 3, 5}

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1091


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

35. P(x) = 4x4 – 8x3 – x2 + 8x – 3

Solution Possible rational zeros:  1,  3,  21 ,  32 ,  41 ,  43

Descartes' Rule of Signs: # neg 1 1

# pos 3 1

# nonreal 0 2

Test x = -1: -1 4 - 8 - 1 -4

8 -3

12 -11

4 -12 11 Test x = 1: 1 4 -12

-3

3 0

11 -3

4 -8 1 -8

3

3

0

P ( x ) = 4 x - 8x - x 2 + 8x - 3 4

3

= ( x + 1)(4 x 3 - 12 x 2 + 11x + 3) = ( x + 1)( x - 1)(4 x 2 - 8 x + 3) = ( x + 1)( x - 1)(2 x - 3)(2 x - 1) zeros: {-1, 1, 32 , 21 } 36. P(x) = 3x4 – 14x3 + 11x2 + 16x – 12

Solution Possible rational zeros: 1,  2,  3, 4,  6,  12,

 31 ,  23 ,  43 Descartes' Rule of Signs: # neg 1 1

# nonreal 0 2

Test x = -1: -1 3 -14 11

16 -12

# pos 3 1

-3 3 -17

17 -28

12

- 12

0

28

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1092


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 2: 2 3 -17 28 -12 6 -22 3

- 11

12

6

0

P ( x ) = 3 x - 14 x + 11x 2 + 16 x - 12 4

3

= ( x + 1)(3 x 3 - 17 x 2 + 28 x - 12) = ( x + 1)( x - 2)(3 x 2 - 11x + 6) = ( x + 1)( x - 2)(3 x - 2)( x - 3) zeros: {-1, 2, 23 , 3} 37. P(x) = 12x4 + 20x3 – 41x2 + 20x – 3

Solution Possible rational zeros: 1,  3,  21 ,  32 ,  31 ,  41 ,  43 ,  61 ,  121

Descartes' Rule of Signs: # pos 3 1

# neg 1 1

# nonreal 0 2

Test x = -3: -3 12

20

-41

-36 12 -16

20 -3

48 -21

3

-1

0

7

Test x = 21 : 1 2

12 -16

7 -1

6 -5 12 -10

2

1 0

P ( x ) = 12 x 4 + 20 x 3 - 41x 2 + 20 x - 3 = ( x + 3)(12 x 3 - 16 x 2 + 7 x - 1) = ( x + 3)( x - 21 )(12 x 2 - 10 x + 1) = 2 ( x + 3)( x - 21 )(6 x 2 - 5 x + 1)

= ( x + 3)(2 x - 1)(3 x - 1)(2 x - 1) zeros: {-3, 21 , 31 , 21 }

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1093


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

38. P(x) = x5 + 3x4 – 5x3 – 15x2 + 4x + 12

Solution Possible rational zeros: 1,  2, 3,  4,

6,  12 Descartes' Rule of Signs: # pos 2 2 0 0

# neg 3 1 3 1

# nonreal 0 2 2 4

Test x = -1: -1 1 3 -5 -15

4

-1

-2

7

2

-7

- 8 12

1

Test x = 1: 1 1 2 -7 1

3

-8

12

8 -12 0

12

- 4 -12

1 3 -4 -12 Test x = 2:

0

2 1 3 -4 -12 2 1

10

5

12

6

0

P ( x ) = x + 3 x - 5 x 3 - 15 x 2 + 4 x + 12 5

4

= ( x + 1)( x 4 + 2 x 3 - 7 x 2 - 8 x + 12) = ( x + 1)( x - 1)( x 3 + 3 x 2 - 4 x - 12) = ( x + 1)( x - 1)( x - 2) ( x 2 + 5 x + 6) = ( x + 1)( x - 1)( x - 2)( x + 2)( x + 3) zeros: {-1, 1, 2, - 2, - 3}

39. P(x) = x5 – 3x4 – 5x3 + 15x2 + 4x – 12

Solution Possible rational zeros:  1,  2,

 3,  4,  6,  12 Descartes' Rule of Signs:

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1094


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 3 3 1 1

# neg 2 0 2 0

Test x = 1: 1 1 -3 -5 -1

# nonreal 0 2 2 4 15

4 -12

- 2 -7

8

12

-7

8

12

0

Test x = -1: -1 1 -2 -7

8

12

1 -2

-1

4 -12

3

1 -3 -4 Test x = 2:

12

2 1 -3 -4

12

0

2 -2 -12 1

- 1 -6

0

P ( x ) = x - 3 x - 5 x 3 + 15 x 2 + 4 x - 12 5

4

= ( x - 1)( x 4 - 2 x 3 - 7 x 2 + 8 x + 12) = ( x - 1)( x + 1)( x 3 - 3 x 2 - 4 x + 12) = ( x - 1)( x + 1)( x - 2)( x 2 - x - 6) = ( x - 1)( x + 1)( x - 2)( x + 2)( x - 3) zeros: {1, - 1, 2, - 2, 3}

40. P(x) = 6x5 – 7x4 – 48x3 + 81x2 – 4x – 12

Solution Possible rational zeros: 1,  2,  3, 4,  6,  12,  21 ,  32 ,  31 ,  23 ,  43 ,  61

Descartes' Rule of Signs: # pos 3 3 1 1

# neg 2 0 2 0

# nonreal 0 2 2 4

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1095


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 2: 2 6 -7 -48

81 -4 -12

10 -76

12

5 - 38

6

Test x = -3: 5 -38 -3 6 -18

10

12

5

6

0

5

6

39 -3 -6

6 -13 Test x = 2:

1

2 6 -13

1

2

0

2

12 -2 -2 -1

6

-1

0

P ( x ) = 6 x - 7 x - 48 x 3 + 81x 2 - 4 x - 12 5

4

= ( x - 2)(6 x 4 + 5 x 3 - 38 x 2 + 5 x + 6) = ( x - 2)( x + 3)(6 x 3 - 13 x 2 + x + 2) = ( x - 2)( x + 3)( x - 2)(6 x 2 - x - 1) = ( x - 2)( x + 3)( x - 2)(2 x - 1)(3 x + 1) zeros: {2, - 3, 2, 21 , - 31 }

41. P(x) = x7 – 12x5 + 48x3 – 64x

Solution First, factor out the common factor of x: x 7 - 12 x 5 + 48 x 3 - 64 x = x ( x 6 - 12 x 4 + 48 x 2 - 64)

Possible rational zeros:  1,  2,  4,  8,  16,  32,  64 Descartes' Rule of Signs: # pos 3 3 1 1

# neg 3 1 3 1

Test x = 2: 2 1 0 -12 2 1

#zero 1 1 1 1 0

48

# nonreal 0 2 2 4 0 -64

4 -16 -32 32

2 -8

- 16

16 32

64 0

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1096


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 2: 2 1 2 -8 -16

1

16

32

0 -32 -32

2

8

4

0 -16 -16

0

Test x = 2: 0 -16 -16

2 1 4

1

2

12

24

16

6

12

8

0

Test x = -2: -2 1

6

12

8

- 2 -8 - 8 1

4

4

0

P ( x ) = x 7 - 12 x 5 + 48 x 3 - 64 x = x ( x 6 - 12 x 4 + 48 x 2 - 64) = x ( x - 2)( x 5 + 2 x 2 - 8 x 3 - 16 x 2 + 16 x + 32) = x ( x - 2)( x - 2)( x 4 + 4 x 3 - 16 x - 16) = x ( x - 2)( x - 2)( x - 2)( x 3 + 6 x 2 + 12 x + 8) = x ( x - 2)( x - 2)( x - 2)( x + 2)( x 2 + 4 x + 4) = x ( x - 2)( x - 2)( x - 2)( x + 2)( x + 2)( x + 2)  zeros: {0, 2, 2, 2, - 2, - 2, - 2} 42. P(x) = x7 + 7x6 + 21x5 + 35x4 + 35x3 + 21x2 + 7x + 1

Solution Possible rational zeros:

1 Descartes' Rule of Signs: # pos 0 0 0 0

# neg 7 5 3 1

Test x = -1: -1 1 7 21

1

# nonreal 0 2 4 6 35

35

21

7

1

-1

- 6 -15 -20 -15 -6 -1

6

15

20

15

6

1

0

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1097


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = -1: 15 -1 1 6

20

15

6

1

-1

- 5 -10 -10 -5 -1

5

10

1

10

5

1

0

Test x = -1:

-1 1

5

10

-1

- 4 -6 -4 -1

1

4

10

6

5

4

1

1

0

Test x = -1:

-1 1

4

6

4

1

- 1 - 3 - 3 -1 1

3

3

1

0

Test x = -1:

-1 1

3

3

1

-1 -2 -1 1

2

1

0

P ( x ) = x 7 + 7 x 6 + 21x 5 + 35 x 4 + 35 x 3 + 21x 2 + 7 x + 1 = ( x + 1)( x 6 + 6 x 5 + 15 x 4 + 20 x 3 + 15 x 2 + 6 x + 1) = ( x + 1)( x + 1)( x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 5 x + 1) = ( x + 1)( x + 1)( x + 1)( x 4 + 4 x 3 + 6 x 2 + 4 x + 1) = ( x + 1)( x + 1)( x + 1)( x + 1)( x 3 + 3 x 2 + 3 x + 1) = ( x + 1)( x + 1)( x + 1)( x + 1)( x + 1)( x 2 + 2 x + 1) = ( x + 1)( x + 1)( x + 1)( x + 1)( x + 1)( x + 1)( x + 1) Solution set = {-1, - 1, - 1, - 1, - 1, - 1, - 1}

Find all zeros of each polynomial function. 43. P(x) = x3 – 3x2 – 2x + 6

Solution Possible rational zeros:  1,  2,  3,  6 Descartes' Rule of Signs: # pos 2 0

# neg 1 1

# nonreal 0 2

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1098


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 3: 3 1 - 3 -2

6

0 -6

3

0 -2

1

0

P ( x ) = x - 3x - 2x + 6 3

2

= ( x - 3)( x 2 - 2)

x - 3 = 0 or x 2 - 2 = 0 x=3

x= 2

{

zeros: 3, - 2,

2

}

44. P(x) = x3 + 3x2 – 3x – 9

Solution Possible rational zeros:  1,  3,  9 Descartes' Rule of Signs: # neg 2 0

# pos 1 1

# nonreal 0 2

Test x = -3:

-3 1

3 -3 -9

-3

0

9

0 -3

1

0

P ( x ) = x + 3x - 3x - 9 3

2

= ( x + 3)( x 2 - 3) x +3 = 0

or x 2 - 3 = 0

x = -3

{

zeros: 3, - 3,

x= 3

3

}

45. P(x) = 2x3 – x2 + 2x – 1

Solution Possible rational zeros:  1,  21 Descartes' Rule of Signs: # pos 3 1

# neg 0 0

# nonreal 0 2

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1099


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 21 : 1 2

2 -1 2 -1

2

1 0

1

0 2

0

P ( x ) = 2x - x 2 + 2x - 1 3

= ( x - 21 )(2 x 2 + 2) x - 21 = 0 or 2 x 2 + 2 = 0 x = 21

x =  -1 x = i

zeros: { , - i , i } 1 2

46. P(x) = 3x3 + x2 + 3x + 1

Solution Possible rational zeros:  1,  31 Descartes' Rule of Signs: # pos 0 0

# neg 3 1

# nonreal 0 2

Test x = - 31 : - 31 3

1 3

1

-1 0 -1 3

0 3

0

P ( x ) = 3x + x + 3x + 1 3

2

= ( x + 31 )(3 x 2 + 3) or 3 x 2 + 3 = 0

x + 31 = 0 x = - 31

x =  -1 x = i

zeros: {- , - i , i } 1 3

 

47. P x  3 x 3  2 x 2  12 x  8

Solution Possible rational zeros:  83 ,  81 ,  43 ,  41 ,  31 ,  1,  23 ,  2 Descartes' Rule of Signs:

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1100


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 3 1

# neg 0 0

# nonreal 0 2

Test x = 23 : 2 3

3 -2 12 -8 2 3

0

8

0 12

0

P ( x ) = 3 x - 2 x + 12 x - 8 3

2

= (3 x - 2)( x 2 + 4) 3 x - 2 = 0 or x 2 + 4 = 0 x = 32

x =  -4 x =  2i

zeros: { , 2i , - 2i } 2 3

48. P(x) = x4 - 2x3 – 8x2 + 8x + 16

Solution Possible rational zeros:  1,  2,  4,  8,  16 Descartes' Rule of Signs: # neg 2 0 2 0

# pos 2 2 0 0

Test x = 2: 2 1 -2 -8 2

8

16

0 -16 -16

0 -8

1

# nonreal 0 2 2 4

-8

0

Test x = -2: 0 -8 -8 -2 1

-2 1

4

8

- 2 -4

0

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1101


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

P ( x ) = x 4 - 2 x 3 - 8 x 2 + 8 x + 16 = ( x - 2)( x 3 - 8 x - 8) = ( x - 2)( x + 2)( x 2 - 2 x - 4) Use the quadratic formula.

{

zeros: 2, - 2, 1  5

}

49. P(x) = x4 – 2x3 – 2x2 + 2x + 1

Solution Possible rational zeros:

1 Descartes' Rule of Signs: # pos 2 2 0 0

# neg 2 0 2 0

# nonreal 0 2 2 4

Test x = 1: 1 1 -2 -2

1

-1 -3 - 1

1 1 -1

2

- 3 -1

0

Test x = -1: -1 1 -1 -3 -1 -1 1

2

1

- 2 -1

0

P ( x ) = x - 2x - 2x 2 + 2x + 1 4

3

= ( x - 1)( x 3 - x 2 - 3 x - 1) = ( x - 1)( x + 1)( x 2 - 2 x - 1) Use the quadratic formula.

{

zeros: 1, - 1, 1  2

}

 

50. P x  36 x 4  x 2  2 x  1

Solution Possible rational zeros:  1,  21 ,  31 ,  41 ,  61 ,  91 ,  121 ,  181 ,  361 Descartes' Rule of Signs:

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1102


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 3 1

# neg 1 1

# nonreal 0 2

Test x = 31 : 1 3

0 -1 2 -1

36

12 36 12 Test x = -1 2

4

1

1

3

3

0

:

-1 2

0 -1

36

-18 36 -18

2 -1

9 -4

1

-2

0

8

P ( x ) = 36 x 4 - x 2 + 2 x - 1 æ 1 öæ 1ö = çç x - ÷÷÷ çç x + ÷÷÷ 6 (6 x 2 - x + 1) ÷ çè ç 3 øè 2 ÷ø 1 1 6 x 2 - x + 1 = 0 or x - = 0 or x + = 0 3 2

x=

1

2

(-1) - 4 (6)(1) 2 (6)

1  -23 12 -1 1  23i 1 x= or x = or x = 12 3 2 1 -1 1 23 1 23 + i, i zeros: , , 3 2 12 12 12 12 x=

51. P(x) = 2x4 – 4x3 – 2x2 + 4x – 4

Solution Possible rational zeros:  1,  2,  4,  21 Descartes' Rule of Signs: # neg 1 1

# pos 3 1

# nonreal 0 2

Test x = 1: 1 2 -4

2 4 -4

2 -2 0 2 -2

0

4

4 0

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1103


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = -1: -1 2 -2 0

4

-2 4 -4 2 -4 4

0

P ( x ) = 2x - 4 x 3 + 2x 2 + 4 x - 4 4

= ( x - 1)(2 x 3 - 2 x 2 + 4) = ( x - 1)( x + 1)(2 x 2 - 4 x + 4) Use the quadratic formula. zeros: {1, - 1, 1  i }

52. P(x) = 2x4 + x3 + 17x2 + 9x – 9

Solution Possible rational zeros:  1,  3,  9,  21 ,  32 ,  92 Descartes' Rule of Signs: # pos 1 1

# neg 3 1

# nonreal 0 2

Test x = -1: 1 17 -1 2 -2

9 -9

1 -18

2 -1 18 Test x = 21 : 1 2

-9

9 0

2 -1 18 -9 1 2

0

0 18

9 0

P ( x ) = 2 x + x + 17 x 2 + 9 x - 9 4

3

= ( x + 1)(2 x 3 - x 2 + 18 x - 9) = ( x + 1)( x - 21 )(2 x 2 + 18) 2 x 2 + 18 = 0  x =  -9 zeros: {-1, 21 ,  3i }

 

53. P x  x 4  3 x 3  7 x 2  9x  30

Solution Possible rational zeros:  1,  30,  15,  10,  3,  6,  5,  2

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1104


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Descartes' Rule of Signs: # neg 1 1

# pos 3 1

# nonreal 0 2

Test x = 2: 2 1 3 -7

9 -30

2

10

6

30

1 5

3

15

0

Test x = -5: 3 -7 -5 1 -5

9 -30

10 -15

1 -2

30

-6

3

0

P ( x ) = x + 3 x - 7 x + 9 x - 30 4

3

2

= ( x - 2)( x + 5)( x 2 + 3) x 2 + 3 = 0 or x - 2 = 0 or x + 5 = 0 x 2 = -3 x =  i 3 or x = 2 or x = -5

{

zeros: 2, - 5, i 3, - i 3

}

 

54. P x  2 x 4  x 3  2 x 2  4 x  40

Solution Possible rational zeros  1,  21 ,  2,  4,  5,  52 ,  8,  10,  20,  40,  1,  30,  15,  10,  3,  6,  5,  2 Descartes' Rule of Signs: # pos 1 1

# neg 1 3

# nonreal 2 0

Test x = -2: -2 2

-1 - 2

-4 -40

-4

10 -16

40

2 -5

8 -20

0

Test x = 52 : 5 2

2 -1 -2 -4 -40

2

5

10

20

40

4

8

16

0

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1105


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

P ( x ) = 2 x 4 - x 3 - 2 x 2 - 4 x - 40 æ 5ö = ( x + 2)ççç x - ÷÷÷ (2 x 2 + 8) 2 ø÷ è 5 2 x 2 + 8 = 0 or x - = 0 or x + 2 = 0 2 2 x 2 = -8 x =  2i ìï ïü 5 zeros: íï-2, , 2i , - 2i ýï ïïî ï 2 þï

 

55. P x  12 x 4  x 3  42 x 2  4 x  24

Solution Possible rational zeros

{1, 21 , 31 , 41 , 61 , 121 , 2, 23 , 3, 23 , 43 , 4, 43 , 6, 8, 83 , 12, 24} Descartes' Rule of Signs: # pos 1 1

# neg 1 3

# nonreal 2 0

Test x = 23 : 12

4 -24

1

42

8

6 32

24

12 9 48 36 Test x = - 43 :

0

2 3

- 43 12

1 42 -9

4 -24

6 -36

24

12 -8 48 -32

0

P ( x ) = 12 x 4 + x 3 + 42 x 2 + 4 x - 24 æ 2 öæ 3ö = çç x - ÷÷÷ çç x + ÷÷÷ (12 x 2 + 48) ÷ çè ç 3 øè 4 ø÷ 2 3 12 x 2 = -48 or x - = 0 or x + = 0 3 4 2 -3 2 x = -4 or x = or x = 3 4 x =  2i ïì 2 ïü 3 zeros: ïí , - , 2i , - 2i ïý ïïî 3 ïïþ 4

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1106


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

56. P(x) = x5 – 3x4 + 28x3 – 76x2 + 75x – 25

Solution Possible rational zeros:  1,  5,  25 Descartes' Rule of Signs: # pos 5 3 1

# neg 0 0 0

Test x = 1: 1 1 -3 28

-76

1 -2 2 -2

# nonreal 0 2 4 75 -25

26 -50

26 -50

25

Test x = 1: 1 1 -2 26 -50

25

1 -1 1

0

25 -25

- 1 25 -25

Test x = 1: 1 1 -1 25

-25

1

0

25

0 25

0

1

25

0

P ( x ) = x 5 - 3 x 4 + 28 x 3 - 76 x 2 + 75 x - 25 = ( x - 1)( x 4 - 2 x 3 + 26 x 2 - 50 x + 25) = ( x - 1)( x - 1)( x 3 - x 2 + 25 x - 25) = ( x - 1)( x - 1)( x - 1)( x 2 + 25) x 2 + 25 = 0  x =  -25 zeros: {1, 1, 1,  5i } 57. P(x) = x5 – 3x4 – 2x3 – 14x2 – 15x – 5

Solution Possible rational zeros:  1,  5 Descartes' Rule of Signs: # pos 1 1 1

# neg 4 2 0

# nonreal 0 2 4

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1107


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = -1: -1 1

3 -2

-14 -15 -5

- 1 -2 1

4

10

5

2 -4 -10

-5

0

Test x = -1: -1 1 2 -4 -10 -5 -1

-1

5

5

1

-5

-5

0

1

Test x = -1: -1 1 1 -5 -5 -1 1

0

5

-5

0

0

P ( x ) = x + 3 x - 2 x 3 - 14 x 2 - 15 x - 5 5

4

= ( x + 1)( x 4 + 2 x 3 - 4 x 2 - 10 x - 5) = ( x + 1)( x + 1)( x 3 + x 2 - 5 x - 5) = ( x + 1)( x + 1)( x + 1)( x 2 - 25)

x2 - 5 = 0  x =  5

{

zeros: -1, - 1, - 1,  5

}

58. P(x) = 2x5 – 3x4 + 6x3 – 9x2 – 8x + 12

Solution Possible rational zeros:  1,  2,  3,  4,  6,

 12,  21 ,  32 Descartes' Rule of Signs: # pos 4 2 0

# neg 1 1 1

# nonreal 0 2 4

Test x = 1: -1 2 -3

6

-9

2 -1

5

2 -1

5

-8

12

- 4 -12

- 4 -12

0

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1108


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = -1: -1 2 - 1 5

-4 -12

-2 3

-8

12

2 -3 8

- 12

0

Test x = : 3 2

3 2

2 -3

8

-12

3

0

12

0

8

0

2

P ( x ) = 2 x 5 - 3 x 4 + 6 x 3 - 9 x 2 - 8 x + 12 = ( x - 1)(2 x 4 - x 3 + 5 x 2 - 4 x - 12) = ( x - 1)( x + 1)(2 x 3 - 3 x 2 + 8 x - 12) = ( x - 1)( x + 1)( x - 32 )(2 x 2 + 8) 2 x 2 + 8 = 0  x =  -4

{

zeros: 1, - 1, 32 ,  2i

}

59. P(x) = 3x5 – x4 + 36x3 – 12x2 – 192x + 64

Solution Possible rational zeros:  1,  2,  4,  8,  16,  32,  64,  31 ,  23 ,  43 ,  83 ,  163 ,  32 ,  64 3 3 Descartes' Rule of Signs: # pos 4 2 0

# neg 1 1 1

Test x = 2: 2 3 -1 36

-12 -192

64

10

92

160 -64

5 46

80

- 32

6 3

# nonreal 0 2 4

0

Test x = -2: -2 3

5 46 -6

3

80 -32

2 -96

- 1 48

- 16

32 0

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1109


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 31 : 1 3

3 -1

48

-16

1

0

16

3

0

48

0

P ( x ) = 3 x - x + 36 x 3 - 12 x 2 - 192 x + 64 4

5

= ( x - 2)(3 x 4 + 5 x 3 + 46 x 2 + 80 x - 32) = ( x - 2)( x + 2)(3 x 3 - x 2 + 48 x - 16) = ( x - 2)( x + 2)( x - 31 )(3 x 2 + 48) 3 x 2 + 48 = 0  x =  -16 zeros: {2, - 2, 31 ,  4i }

 

60. P x  4 x 5  12 x 4  15 x 3  45 x 2  4 x  12

Solution Possible rational zeros:

{1, 21 , 41 , 2, 3, 32 , 43 , 4, 6, 12} Descartes' Rule of Signs: # pos 4 2 0

# neg 1 1 1

# nonreal 0 2 4

Test x = 3: 3 4 -12 15 -45 12

0

45

4 0 15 Test x = 21 :

0

1 2

4

-12

-4

0 -12 -4

15 -45

2 -5

12

0

-4

5 -20 -12

4 - 10 10 -40 -24 Test x = -21 : -1 2

4

4

12

-12

15 -45

-2

7

- 14

22

0

-4

-11

12

28 -12

- 56

24

0

P ( x ) = 4 x - 12 x + 15 x - 45 x - 4 x + 12 5

4

3

2

æ 1 öæ 1ö = ( x - 3)ççç x - ÷÷÷ ççç x + ÷÷÷ (4 x 2 + 16) 2 ÷ø è 2 ÷ø è

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1110


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

1 1 = 0 or x + = 0 2 2 1 1 2 x = -4 or x = 3 or x = or x = 2 2 x = 2i ìï 1 üï 1 zeros: ïí3, , - , 2i , - 2i ïý ïîï 2 ïþï 2 4 x 2 = -16 or x - 3 = 0 or x -

In Exercises 61–64, 1 + i is a zero of each polynomial function. Find the other zeros. 61. P(x) = x3 – 5x2 + 8x – 6

Solution If (1 + i ) is a zero, then so is (1 - i ) , and x - (1 + i ) and x - (1 - i ) are factors. Then éê x - (1 + i )ùú éê x - (1 - i )ùú = x 2 - 2 x + 2 is a factor. Divide it out: ë ûë û x- 3 x 2 - 2x + 2 x 3 - 5x 2 + 8x - 6 x 3 - 2x 2 + 2x - 3x 2 + 6x - 6 - 3x 2 + 6x - 6 0

P ( x ) = x - 5x + 8x - 6 3

2

= ( x 2 - 2 x + 2) ( x - 3) zeros: {1 + i , 1 - i , 3} 62. P(x) = x3 – 2x + 4

Solution If (1 + i ) is a zero, then so is (1 - i ) , and x - (1 + i ) and x - (1 - i ) are factors. Then éê x - (1 + i )ùú éê x - (1 - i )ùú = x 2 - 2 x + 2 is a factor. Divide it out: ë ûë û x +2

x 2 - 2x + 2 x 3 + 0x 2 - 2x + 4 x 3 - 2x 2 + 2x 2x 2 - 4 x + 4 2x 2 - 4 x + 4 0 P ( x ) = x 3 - 2x + 4 = ( x 2 - 2 x + 2) ( x + 2) zeros: {1 + i , 1 - i , - 2}

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1111


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

63. P(x) = x4 – 2x3 – 7x2 + 18x – 18

Solution If (1 + i ) is a zero, then so is (1 - i ) , and x - (1 + i ) and x - (1 - i ) are factors. Then éê x - (1 + i )ùú éê x - (1 - i )ùú = x 2 - 2 x + 2 is a factor. Divide it out: ë ûë û x2 - 9 x 2 - 2 x + 2 x 4 - 2 x 3 - 7 x 2 + 18 x - 18 x 4 - 2x 3 + 2x 2 - 9 x 2 + 18 x - 18 -9 x 2 + 18 x - 18 0

P ( x ) = x 4 - 2 x 3 - 7 x 2 + 18 x - 18 = ( x 2 - 2 x + 2)( x 2 - 9) = ( x 2 - 2 x + 2) ( x + 3)( x - 3) zeros: {1 + i , 1 - i , - 3, 3} 64. P(x) = x4 – 2x3 – 2x2 + 8x – 8

Solution If (1 + i ) is a zero, then so is (1 - i ) , and x - (1 + i ) and x - (1 - i ) are factors. Then éê x - (1 + i )ùú éê x - (1 - i )ùú = x 2 - 2 x + 2 is a factor. Divide it out: ë ûë û x2 - 4 x 2 - 2x + 2 x 4 - 2x 3 - 2x 2 + 8x - 8 x 4 - 2x 3 + 2x 2 - 4 x 2 + 8x - 8 -4 x 2 + 8 x - 8 0 P ( x ) = x 4 - 2x 3 - 2x 2 + 8x - 8 = ( x 2 - 2 x + 2)( x 2 - 4) = ( x 2 - 2 x + 2) ( x + 2)( x - 2) zeros: {1 + i , 1 - i , - 2, 2}

Solve each equation. 65. x 3 -

4 2 13 x x -2 = 0 3 3

Solution Possible rational zeros:  1,  2,  3,  6,  31 ,  23

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1112


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Descartes' Rule of Signs: # neg 2 0

# pos 1 1

# nonreal 0 2

Test x = -1 : -1 3 -4 -13 -6 -3

7

6

3 -7

-6

0

4 2 13 x x -2 = 0 3 3 3 x 3 - 4 x 2 - 13 x - 6 = 0 x3 -

( x + 1)(3x 2 - 7 x - 6) = 0 ( x + 1)(3x 2 + 2)( x - 3) = 0 solutions: {-1, - 23 , 3} 66. x 3 -

19 2 1 x + x+1=0 6 6

Solution Possible rational zeros:  1,  2,  3,  6,  31 ,  32 ,  31 ,  23 ,  61

Descartes' Rule of Signs: # neg 1 1

# pos 2 0

Test x = 3: 3 6 -19 1

# nonreal 0 2 6

18 -3 -6 6

- 1 -2

0

19 2 1 x + x +1=0 6 6 6 x 3 - 19 x 2 + x + 6 = 0

x3 -

( x - 3)(6 x 2 - x - 2) = 0 ( x - 3)(3x - 2)(2x + 1) = 0 solutions: {3, 23 , - 21 }

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1113


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

67. x–5 – 8x–4 + 25x–3 – 38x–2 + 28x–1 – 8 = 0

Solution

x –5 - 8 x -4 + 25 x -3 - 38 x -2 + 28 x -1 - 8 = 0 x 5 ( x –5 - 8 x -4 + 25 x -3 - 38 x -2 + 28 x -1 - 8) = x 5 (0) 1 - 8 x + 25 x 2 - 38 x 3 + 28 x 4 - 8 x 5 = 0 Possible rational zeros:  1,  21 ,  41 ,  81 Descartes' Rule of Signs: # neg 0 0 0

# pos 5 3 1

# nonreal 0 2 4

Test x = 1: 1 8 -28 38 -25 8 -20 8 -20

18

18 -7 -7

Test x = 1: 1 8 -20 18 -7 8 -12

8 -12

-8

0

1

0

6 -1

4 -4 8

1

1

6 -1

8 -12 6 -1 1 Test x = 2 : 1 2

8 -1

2

1 0

8 x 5 - 28 x 4 + 38 x 3 - 25 x 2 + 8 x - 1 = 0

( x - 1)(8 x 4 - 20 x 3 + 18 x 2 - 7 x + 1) = 0 ( x - 1)( x - 1)(8 x 3 - 12 x 2 + 6 x - 1) = 0 ( x - 1)( x - 1)( x - 21 )(8 x 2 - 8 x + 2) = 0 ( x - 1)( x - 1)( x - 21 )(4 x - 2)(2x - 1) = 0 solutions: {1, 1, 21 , 21 , 21 } 68. 1 – x–1 – x–2 – 2x–3 = 0

Solution

1 – x –1 – x –2 – 2 x –3 = 0 x 3 (1 – x –1 – x –2 – 2 x –3 ) = x 3 (0)  x 3 - x 2 - x - 2 = 0

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1114


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Possible rational zeros:  1,  2 Descartes' Rule of Signs: # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = 2: 2 1 -1 -1 -2

1

2

2

2

1

1

0

x3 - x2 - x - 2 = 0

( x - 2)( x 2 + x + 1) = 0 Use the quadratic formula on the second factor

{

solutions: 2, - 21 + 23 i, - 21 - 23 i

}

Fix It In Exercises 69 and 70, identify the step the first error is made and fix it. To do so, identify p and q, use the Rational Zeros Theorem and write the possible rational zeros, find one rational zero, and then solve the depressed equation and state all zeros.

 

69. Find all zeros of P x  x 3  7 x 2  25 x  39.

Solution Step 4 was incorrect. Step 4: The zeros are 3, 2 + 3i , 2 - 3i

 

70. Find all zeros P x  16 x 4  8 x 3  17 x 2  8 x  1.

Solution Step 4 was incorrect. Step 4: The zeros are

1 1 , , i, - i 4 4

Applications 71. Parallel resistance If three resistors with resistances of R1, R2, and R3 are wired in parallel, their combined resistance R is given by the following formula. The design of a voltmeter requires that the resistance R2 be 10 ohms greater than the resistance R1, that the resistance R3 be 50 ohms greater than R1, and that their combined resistance be 6 ohms. Find the value of each resistance. 1 1 1 1    R R1 R2 R3

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1115


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution Let x  R1 . Then x + 10 = R2 and x + 50 = R3. 1 1 1 1    R R1 R2 R3 1 1 1 1    6 x x  10 x  50 6 x  x  10  x  50   61  6 x  x  10  x1  x 110  x 150 

x  x  10  x  50   6  x  10  x  50   6 x  x  50   6 x  x  10  x 3  60 x 2  500 x  6 x 2  360 x  3000  6 x 2  300 x  6 x 2  60 x

x 3  42 x 2  220 x  3000  0

 x  10  x  52x  300  0 2

10 1 42 -220 -3000 10

520

3000

1 52

300

0

Use the quadratic formula on the second factor. The two solutions from that factor are negative. The only solution that makes sense is x = 10. The resistances are 10, 20 and 60 ohms. 72. Fabricating sheet metal The open tray shown in the illustration is to be manufactured from a 12-by-14-inch rectangular sheet of metal by cutting squares from each corner and folding up the sides. If the volume of the tray is to be 160 cubic inches and x is to be an integer, what size squares should be cut from each corner?

Solution



The volume is 12  2 x 14  2 x x  4 x 3  52 x 2  168 x.

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1116


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

4 x 3  52 x 2  168 x  160 4 x 3  52 x 2  168 x  160  0

4 x 3  13 x 2  42 x  40  0 x  13 x  42 x  40  0 3

2

 x  2  x  11x  20  0  x  2 x  2 x  9  0 2

The only solution that makes sense is x  2. 2 inch by 2 inch squares should be cut from the corners. Test x = 2: 2 1 -13 42 -40 2 -22 - 11

1

40

20

0

73. FedEx box The length of a FedEx 25-kg box is 7 inches more than its height. The width of the box is 4 inches more than its height. If the volume of the box is 4420 cubic inches, find the height of the box.

Solution Let x = the height. Then x + 7 = the length, and x + 4 = the width.



The volume is x x  7 x  4  x 3  11x 2  28 x. x 3  11x 2  28 x  4420 x  11x 2  28 x  4420  0 3

 x  13  x  24 x  340  0 2

The only real solution is x  13. The height is 13 inches. Test x = 13: 13 1 11 28 -4420 13 312

4420

1 24 340

0

74. Dr. Pepper can A Dr. Pepper aluminum can is approximately the shape of a cylinder. If the height of the can is 9 cm more than its radius and the volume of the can is approximately 108π cubic cm, find the radius of the can. The formula for the volume of a cylinder is V = πr2h

Solution Let x = the radius. Then x + 9 = the height.

The volume is  r 2 h   x 2  x  9    x 3  9 x 2 .

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1117


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

 x 3  9 x 2  108 x  9 x  108 3

2

x  9 x 2  108  0 3

 x  3  x  12 x  36   0  x  3 x  6  x  6   0 2

The only solution that makes sense is x  3. The radius is 3 cm. Test x = 3: 3 1 9 0 -108 3 36

108

1 12 36

0

75. Hilly terrain We are interested in the nature of some hilly terrain. Computer simulation has told us that for a cross section from west to east, the height h(x), in feet above sea level is related to the horizontal distance x (in miles) from a fixed point by the function. h(x) = –x4 + 5x3 + 91x2 – 545x + 550, x  [0, 9]. At what distances from the fixed point is the height 100 feet above sea level?

Solution Possible rational zeros:  1,  2,  3,

 5,  6,  9,  10,  15,  18,  25,  30,  45,  50,  75,  90,  150,  225,  450 Descartes' Rule of Signs: # pos 3 1

# neg 1 1

# nonreal 0 2

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1118


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 1: 1 1 -5 -91 1

545 -450

- 4 -95

1 -4 -95

450

Test x = 5: 4 1 -4 -95 5 1

450 0

450

5 -450

- 1 -90

0

The only solutions between 0 and 9 are 1, 5, and 9 miles. -x 4 + 5 x 3 + 91x 2 - 545 x + 550 = 100 x 4 - 5 x 3 - 91x 2 + 545 x - 450 = 0

( x - 1)( x 3 - 4 x 2 - 95 x + 450) = 0 ( x - 1)( x - 5)( x 2 + x - 90) = 0 ( x - 1)( x - 5)( x - 9)( x + 10) = 0 Solution: {1, 5, 9, - 10} 76. Velocity of a hot-air balloon A hot-air balloon is tethered to the ground and only moves up and down. You and a friend take a ride on the ballon for approximately 25 minutes. On this particular ride the velocity of the balloon, v(t) in feet per minute, as a function of time, t in minutes, is represented by the function v(t) = –t3 + 34t2 – 320t + 850 At what times is the velocity of the balloon 50 feet per minute?

Solution Possible rational zeros: 1,  2,  4,  5, 8,  10,  16,

20,  25,  32, 40,  50,  80, 100,  160,  200, 400,  800 Descartes' Rule of Signs: # pos 3 1

# neg 0 0

# nonreal 0 2

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1119


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test t = 4 : 4 1 -34 4 1

- 30

3

2

320 -800 - 120

800

200

0

-t + 34t - 320t + 850 = 50 t 3 - 34t 2 + 320t - 800 = 0

(t - 4)(t 2 - 30t + 200) = 0 (t - 4)(t - 10)(t - 20) = 0 Solution: {4, 10, 20} After 4, 10, and 20 minutes. 77. Value of stock The approximate value V(t) in dollars of a share of your stock over the past ten years can be approximated by the function

V  t   t 3  15t 2  54t  40 where t represents the year after it was purchased. Determine the year(s) when the approximate value per share was $80.

Solution

V  t   t 3  15t 2  54t  40 80  t 3  15t 2  54t  40 0  t 3  15t 2  54t  40 Possible rational zeros:  1,  2,  4,  5,  8,  10 Test x  1 1 1 15

54 40

1 14 1 14

40

40 0

t 3  15t 2  54t  40   t  1 t 2  14t  40

  t  1 t  4  t  10 

t  1  0 t  4  0 t  10  0 t  1, 4, 10 years

78. Train tracks The approximate height f(x) in yards of a train traveling over a 3-mile terrain can be approximated by the function

f  x   10 x 3  60 x 2  110 x  40 where x represents mile marker from the starting point. Determine the mile marker(s) when the height of the train is 100 yards.

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1120


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

f  x   10 x 3  60 x 2  110 x  40 100  10 x 3  60 x 2  110 x  40 0  10 x 3  60 x 2  110 x  60 Test x  2 2 10 60 110 60 20 80 60 10 40

30

0

0   x  2  10 x  40 x  30 2

0   x  2  10 x  4 x  3 2

0  10  x  2  x  3  x  1 x 20 x 30 x 10 x  mile makers 1, 2 and 3

79. Local maximum and minimum In calculus, to determine the local maximum and

 

minimum values of F x  3 x 4  16 x 3  6 x 2  72 x  24 the zeros of the polynomial

 

function f x  12 x 3  48 x 2  12 x  72 must be determined. Use the Rational Zeros Theorem and synthetic division to identity the zeros of f(x). Then use Desmos to graph F(x). Do the local maximum and minimum values occur at the zeros you determined?

Solution

f  x    x  3  12  x 2  x  2

 12  x  3  x  2  x  1

x 3 0 x 2 0 x 10 x3

x2

x  1

Yes, a minimum at x = –1, a maximum at x = 2, and a minimum at x = 3 80. Local maximum and minimum In calculus, to determine the local maximum and

 

minimum values of F x  x 4  4 x 3  26 x 2  60 x  100 the zeros of the polynomial

 

function f x  4 x 3  12 x 2  52 x  60 must be determined. Use the Rational Zeros Theorem and synthetic division to identity the zeros of f(x). Then use Desmos to graph F(x). Do the local maximum and minimum values occur at the zeros you determined?

Solution

  4  x  5  x  2x  3

f  x    x  5 4 x 2  8 x  12 2

 4  x  5  x  3  x  1

x 5 0 x 3 0 x 10 x  5

x 3

x  1

Yes, a minimum at x = –5, a maximum at x = –1, and a minimum at x = 3

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1121


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Discovery and Writing 81. State the Rational Zero Theorem and explain its purpose in algebra.

Solution Answers may vary. 82. Describe a strategy that can be used to determine all zeros of a polynomial function.

Solution Answers may vary. 83. If n is an even integer and c is a positive constant, show that P(x) = xn + c has no real zeros.

Solution Answers may vary. 84. If n is an even positive integer and c is a positive constant, show that P(x) = xn – c has two real zeros.

Solution Answers may vary. 85. Precalculus A rectangle is inscribed in the parabola y = 16 – x2, as shown in the illustration. Find the point (x, y) if the area of the rectangle is 42 square units.

Solution A coordinate of a point on the curve has coordinates ( x , 16 – x 2 ) . Thus, the area of the rectangle is A = 2 x (16 – x 2 ) = 32 x - 2 x 3 . 32 x - 2 x 3 = 42 -2 x 3 + 32 x - 42 = 0 x 3 - 16 x + 21 = 0

( x - 3)( x 2 + 3 x - 7) = 0 Using the quadratic formula on the second factor

-3  37 » 1.54 or - 4.54. 2 The only solutions that make sense are x = 3 or x = 1.54. yields the solutions x =

Points: (3, 7) or (1.54, 13.63)

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1122


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 3: 3 1 0 -16

9 -21

3 1

21

-7

3

0

86. Precalculus One corner of the rectangle shown is at the origin, and the opposite corner (x, y) lies in the first quadrant on the curve y = x3 – 2x2. Find the point (x, y) if the area of the rectangle is 27 square units.

Solution A coordinate of a point on the Curve has coordinates ( x , x 3 - 2 x 2 ) . Thus, the area of the rectangle is A = x ( x 3 - 2 x 2 ) = x 4 - 2 x 3 .

x 4 - 2 x 3 = 27 x 4 - 2 x 3 - 27 = 0

( x - 3)( x + x + 3x + 9) = 0 3

2

Since the coefficients of the second factor are all positive, Descartes' Rule of Signs indicates that there are no more positive solutions. Thus, x = 3: (3, 9) . Test x = 3: 3 1 -2 0 0 -27

1

3

3 9

27

1

3 9

0

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 87. 31 is a possible rational zero of P(x) = 3x3 – 5x4 – 2x2 – x + 1

Solution False. The possible rational zeros are  1 and  51 . 88. Every zero of a polynomial function has a corresponding x-intercept.

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1123


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution False. Only the real zeros of a polynomial function have corresponding x-intercepts. 89. If we have identified one zero of a third-degree polynomial function, we can always find the remaining two zeros by factoring, using the Square Root Property or the Quadratic Formula.

Solution True. 90. If every coefficient of a polynomial function is positive, then the function has no positive real zeros.

Solution True. 91. A polynomial function with real coefficients and degree 3 will always have at least one rational zero.

Solution False. It must have at least one real zero, but that zero need not be rational. 92. A polynomial function can have possible rational zeros but no actual rational zeros.

Solution True.

EXERCISES 4.6 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

State one number that x cannot equal. 3 f x  x 4

Solution 4 2. State two numbers that x cannot equal. 5x f x  2 x  25

Solution 5 or –5 3. Simplify the rational expression.

x2 2x 2  3x  2

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1124


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

x2

 2x  1 x  2

1 2x  1

4. Simplify the rational expression.

Solution

9  x2 x  2 x  15 2

9  x2 x  2 x  15 2

 3  x  3  x   3  x  x  5 x  3 x  5

5. Use the graph to identify the vertical line and horizontal line the graph approaches but never touches.

Solution x  3; y  2 x2  x  2 remainder using long division and write answer in the from quotient  . x 2 divisor

6. Divide

Solution x3 x  2 x2  x  2

 x 2  2x

3x  2

  3x  6 4 4 x 3 x 2

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1125


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. When a graph approaches a vertical line but never touches it, we call the line an __________.

Solution asymptote 8. A rational function is a function with a polynomial numerator and a __________ polynomial denominator.

Solution nonzero 9. To find a __________ asymptote of a rational function in simplest form, set the denominator polynomial equal to 0 and solve the equation.

Solution vertical 10. To find the __________ of a rational function, let x = 0 and solve for y or find f (0).

Solution y-intercept 11. To find the __________ of a rational function, set the numerator equal to 0 and solve the equation.

Solution x-intercept 12. In the function f ( x ) = QP (( xx )) , if the degree of P(x) is less than the degree of Q(x), the horizontal a symptote is __________.

Solution y=0 13. In the function f ( x ) = QP (( xx )) , if the degree of P(x) and Q(x) are the __________, the horizontal a symptote is y=

the leading coefficient of the numerator . the leading coefficient of the denominator

Solution same 14. In a rational function, if the degree of the numerator is 1 greater than the degree of the denominator, the graph will have a __________.

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1126


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution slant asymptote 15. A graph can cross a ___________ asymptote but can never cross a asymptote.

Solution horizontal or slant; vertical x2 - 4

16. The graph of f ( x ) = x +2 , will have a point __________.

Solution missing Find the equations of the vertical and horizontal asymptotes of each graph. Find the domain and range. 17.

Solution vertical: x = 2, horizontal: y = 1 domain: (-¥, 2) È (2, ¥) range: (-¥, 1) È (1, ¥)

18.

Solution vertical: x = -2, x = 2, horizontal: y = 0 domain: (-¥, - 2) È (-2, 2) È (2, ¥) range: (-¥, ¥)

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1127


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Practice The time t it takes to travel 600 miles is a function of the mean rate of speed r:

t = f ( r ) = 600 r Find t for the given values of r. 19. 30 mph

Solution

t = f (30) = 600 = 20 hr 30 20. 40 mph

Solution

t = f (40) = 600 = 15 hr 40 21. 50 mph

Solution

t = f (50) = 600 = 12 hr 50 22. 60 mph

Solution

t = f (60) = 600 = 10 hr 60 Suppose the cost (in dollars) of removing p% of the pollution in a river is given by the function C = f ( p) =

50, 000 p 100 - p

(0 £ p < 100)

Find the cost of removing each percent of pollution. 23. 10%

Solution c = f (10) =

50, 000 (10) 100 - 10

» $5555.56

24. 30%

Solution c = f (30) =

50, 000 (30) 100 - 30

» $21, 428.57

25. 50%

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1128


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution c = f (50) =

50, 000 (50) 100 - 50

» $50, 000.00

26. 80%

Solution c = f (80) =

50, 000 (80) 100 - 80

» $200, 000.00

Find the domain of each rational function. Do not graph the function. 27. f ( x ) =

x2 x -2

Solution x2 ; den = 0  x = 2 x -2 domain = (-¥, 2)  (2, ¥) f (x) =

28. f ( x ) =

x 3 - 3x 2 + 1 x +3

Solution x 3 - 3x 2 + 1 ; den = 0  x = -3 x +3 domain = (-¥, - 3)  (-3, ¥) f (x) =

29. f ( x ) =

2x 2 + 7 x - 2 x 2 - 25

Solution

f (x) =

2x 2 + 7 x - 2 2x 2 + 7 x - 2 = x 2 - 25 ( x + 5)( x - 5)

den = 0  x = -5, x = 5 domain = (-¥, - 5)  (-5, 5)  (5, ¥)

30. f ( x ) =

5x 2 + 1 x2 + 5

Solution f (x) =

5x 2 + 1

x2 + 5 den = 0  never true domain = (-¥, ¥)

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1129


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

31. f ( x ) =

x-1 x3 - x

Solution

f (x) =

x-1 x-1 = ; den = 0  x = 0, x = -1, x = 1 3 x -x x ( x + 1)( x - 1)

domain = (-¥, - 1) È (-1, 0) È (0, 1) È (1, ¥) 32. f ( x ) =

x +2 2

2x - 9x + 9

Solution

f ( x) =

x +2

=

x +2

(2x - 3)( x - 3) domain = (-¥, ) È ( , 3) È (3, ¥) 2

2x - 9x + 9 3 2

33. f ( x ) =

; den = 0  x =

3 , x=3 2

3 2

3x 2 + 5 x2 + 1

Solution

3x 2 + 5 ; den = 0  never true x2 + 1 domain = (-¥, ¥) f ( x) =

34. f ( x ) =

7x2 - x + 2 x4 + 4

Solution

7x2 - x + 2 ; den = 0  never true x4 + 4 domain = (-¥, ¥) f (x) =

Find the vertical asymptotes, if any, of each rational function. Do not graph the function. 35. f ( x ) =

x x -3

Solution

x ; den = 0  x = 3 x -3 vertical : x = 3

f ( x) =

36. f ( x ) =

2x 2x + 5

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1130


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

2x ; den = 0  x = - 52 2x + 5 vertical : x = - 52

f ( x) =

37. f ( x ) =

x +2 x2 - 1

Solution

f (x) =

x +2 x2 - 1

=

x +2

( x + 1)( x - 1)

den = 0  x = -1, x = 1 vertical : x = -1, x = 1 38. f ( x ) =

x -4 x 2 - 16

Solution

f (x) =

x -4 2

x - 16

=

x -4

=

1

( x + 4)( x - 4) ( x + 4)

den = 0  x = -4 vertical : x = -4 39. f ( x ) =

1 x2 - x - 6

Solution

f ( x) =

1 2

x - x -6

=

1

( x + 2)( x - 3)

den = 0  x = -2, x = 3 vertical : x = -2, x = 3 40. f ( x ) =

x+2 2x 2 - 6x - 8

Solution

f (x) =

x +2 2

2x - 6x - 8

= =

x -2

2 ( x - 3 x - 4) 2

( x + 2) 2 ( x + 1)( x - 4)

den = 0  x = -1, x = 4 vertical : x = -1, x = 4

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1131


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

41. f ( x ) =

x2 x2 + 5

Solution

x2 ; den = 0  never true x2 + 5 vertical : none

f (x) =

42. f ( x ) =

x 3 - 3x 2 + 1 2x 2 + 3

Solution

x 3 - 3x 2 + 1 ; den = 0  never true 2x 2 + 3 vertical : none

f (x) =

Find the horizontal asymptotes, if any, of each rational function. Do not graph the function. 43. f ( x ) =

2x - 1 x

Solution 2x - 1 f (x) = ; deg (num) = deg (den) x 2 horizontal : y = , or y = 2 1 44. f ( x ) =

x2 + 1 3x 2 - 5

Solution

x2 + 1 ; deg (num) = deg (den) 3x 2 - 5 1 horizontal : y = 3 f (x) =

45. f ( x ) =

x2 + x - 2 2x 2 - 4

Solution

x2 + x - 2 ; deg (num) = deg (den) 2x 2 - 4 1 horizontal : y = 2 f (x) =

46. f ( x ) =

5x 2 + 1 5 - x2

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1132


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

5x 2 + 1 ; deg (num) = deg (den) 5 - x2 5 horizontal : y = , or y = -5 -1 f (x) =

47. f ( x ) =

x+1 x - 4x 3

Solution

x+1

f (x) =

; deg (num) < deg (den) x - 4x horizontal : y = 0 48. f ( x ) =

3

x 2 x 2 - x + 11

Solution x

f (x) =

2

2 x - x + 11 horizontal : y = 0

49. f ( x ) =

; deg (num) < deg (den)

x2 x -2

Solution

x2 ; deg (num) > deg (den) x -2 horizontal : none f (x) =

50. f ( x ) =

x4 + 1 x -3

Solution x4 + 1 ; deg (num) > deg (den) x -3 horizontal : none f (x) =

Find the slant asymptote, if any, of each rational function. Do not graph the function. 51. f ( x ) =

x 2 - 5x - 6 x -2

Solution

x 2 - 5x - 6 -12 = x -3+ x -2 x -2 slant : y = x - 3

f (x) =

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1133


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

52. f ( x ) =

x 2 - 2 x + 11 x+3

Solution 26 x 2 - 2 x + 11 = x -5+ x +3 x +3 slant : y = x - 5

f (x) =

53. f ( x ) =

2x 2 - 5x + 1 x -4

Solution

2x 2 - 5x + 1 13 = 2x + 3 + x -4 x -4 slant : y = 2 x + 3

f (x) =

54. f ( x ) =

5x 3 + 1 x +5

Solution 5x3 + 1 ; deg (num) = 3 and x +5 deg (den) = 1; slant: none f (x) =

55. f ( x ) =

x 3 + 2x 2 - x - 1 x2 - 1

Solution f (x) =

x 3 + 2x 2 - x - 1 x2 - 1

= x +2+

1

x2 - 1 slant : y = x + 2

56. f ( x ) =

-x 3 + 3 x 2 - x + 1 x2 + 1

Solution f (x) =

-x 3 + 3 x 2 - x + 1 x2 + 1

= -x + 3 +

-2

x2 + 1 slant : y = -x + 3

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1134


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Graph each rational function. Check your work with a graphing calculator. 57. y =

1 x -2

Solution 1 y= x -2 Vert: x = 2; Horiz: y = 0 Slant: none; x-intercepts: none y -intercepts: (0, - 21 ) ; Symmetry: none

58. y =

3 x+3

Solution 3 y= x +3 Vert : x = -3; Horiz: y = 0 Slant: none; x-intercepts: none y -intercepts: (0, 1) ; Symmetry: none

59. y =

x x-1

Solution x y= x-1 Vert : x = 1; Horiz: y = 11 = 1

Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: none

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1135


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

60. y =

x x+2

Solution x y= x +2 Vert : x = -2; Horiz: y = 11 = 1

Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: none

61. f ( x ) =

x+1 x +2

Solution x+1 f (x) = x +2 Vert : x = -2; Horiz: y = 11 = 1

Slant: none; x-intercepts: (-1, 0) y -intercepts: (0, 21 ) ; Symmetry: none

62. f ( x ) =

x-1 x -2

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1136


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution x-1 f (x) = x -2 Vert : x = 2; Horiz: y = 11 = 1

Slant: none; x-intercepts: ( 1, 0) y -intercepts: (0, 21 ) ; Symmetry: none

63. f ( x ) =

2x - 1 x-1

Solution 2x - 1 f (x) = x-1 Vert : x = 1; Horiz: y = 21 = 2

Slant: none; x-intercepts: ( 21 , 0)

y -intercepts: (0, 1) ; Symmetry: none

64. f ( x ) =

3x + 2 x2 - 4

Solution 3x + 2 3x + 2 f ( x) = 2 = x - 4 ( x + 2)( x - 2) Vert : x = -2, x = 2; Horiz: y = 0 Slant: none; x-intercepts: (- 23 , 0)

y -intercepts: (0, - 21 ) ; Symmetry: none

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1137


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

65. g ( x ) =

x2 - 9 x2 - 4

Solution g (x) =

x 2 - 9 ( x + 3)( x - 3) = x 2 - 4 ( x + 2)( x - 2)

Vert : x = -2, x = 2; Horiz: y = 11 = 1

Slant: none; x-intercepts: (-3, 0) , (3, 0) y -intercepts: (0, 94 ) ; Symmetry: y -axis

66. g ( x ) =

x2 - 4 x2 - 9

Solution g (x) =

x 2 - 4 ( x + 2)( x - 2) = x 2 - 9 ( x + 3)( x - 3)

Vert : x = -3, x = 3; Horiz: y = 11 = 1

Slant: none; x-intercepts: (-2, 0) , (2, 0) y -intercepts: (0, 49 ) ; Symmetry: y -axis

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1138


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

67. g ( x ) =

x2 - x - 2 x2 - 4x + 3

Solution g (x) =

( x + 1)( x - 2) x2 - x - 2 = 2 x - 4 x + 3 ( x - 3)( x - 1)

Vert : x = 3, x = 1; Horiz: y = 11 = 1

Slant: none; x-intercepts: (-1, 0) , (2, 0) y -intercepts: (0, - 23 ) ; Symmetry: none

68. g ( x ) =

x 2 + 7 x + 12 x 2 - 7 x + 12

Solution g (x) =

x 2 + 7 x + 12 ( x + 3)( x + 4) = x 2 - 7 x + 12 ( x - 3)( x - 4)

Vert : x = 3, x = 4 ; Horiz: y = 11 = 1

Slant: none; x-intercepts: (-3, 0) , (-4, 0) y -intercepts: (0, 1) ; Symmetry: none

Because of the differences in scale, 3 different views of the graph are needed to see all of the characteristic parts of the graph:

69. y =

x 2 + 2x - 3 x 3 - 4x

Solution

y=

x 2 + 2x - 3 x3 - 4x

=

( x - 1)( x + 3) x ( x + 2)( x - 2)

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1139


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Vert : x = 0, x = -2, x = 2; Horiz: y = 0 Slant: none; x-intercepts: (1, 0) , (-3, 0) y -intercepts: none; Symmetry: none

70. y =

3x 2 - 4 x + 1 2x 3 + 3x 2 + x

Solution y=

(3 x - 1)( x - 1) 3x 2 - 4 x + 1 = 2x 3 + 3x 2 + x x (2 x + 1)( x + 1)

Vert : x = 0, x = - 21 , x = -1; Horiz: y = 0

Slant: none; x-intercepts: ( 31 , 0) , ( 1, 0) y -intercepts: none; Symmetry: none

Because of the differences in scale, 2 different views of the graph are needed to see all of the characteristic parts of the graph:

71. y =

x2 - 9 x2

Solution

y=

x2 - 9 2

=

( x + 3)( x - 3)

x x2 Vert : x = 0; Horiz: y = 11 = 1

Slant: none; x-intercepts: (3, 0) , (-3, 0) y -intercepts: none; Symmetry: y -axis

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1140


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

72. y =

3 x 2 - 12 x2

Solution 3 x 2 - 12 3 ( x + 2)( x - 2) = x2 x2 Vert : x = 0; Horiz: y = 31 = 3 y=

Slant: none; x-intercepts: (2, 0) , (-2, 0) y -intercepts: none; Symmetry: y -axis

73. f ( x ) =

x 2

( x + 3)

Solution f (x) =

x 2

( x + 3)

Vert : x = -3; Horiz: y = 0 Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: none

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1141


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

74. f ( x ) =

x 2

( x - 1)

Solution f (x) =

x 2

( x - 1)

Vert : x = 1; Horiz: y = 0 Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: none

75. f ( x ) =

x+1 x ( x - 2) 2

Solution f (x) =

x+1 x ( x - 2) 2

Vert : x = 0, x = 2; Horiz: y = 0 Slant: none; x-intercepts: (-1, 0) y -intercepts: none; Symmetry: none

76. f ( x ) =

x-1 2

x 2 ( x + 2)

Solution

f ( x) =

x-1 2

x ( x + 2) 2

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1142


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Vert : x = 0, x = -2; Horiz: y = 0 Slant: none; x-intercepts: (1, 0) y -intercepts: none; Symmetry: none

77. y =

x 2

x +1

Solution x y= 2 x +1 Vert : none; Horiz: y = 0 Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: origin

78. y =

x-1 x2 + 2

Solution x-1 y= 2 x +2 Vert : none; Horiz: y = 0 Slant: none; x-intercepts: (1, 0)

y -intercepts: (0, - 21 ) ; Symmetry: none

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1143


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

79. y =

3x 2 x2 + 1

Solution 3x 2 y= 2 x +1 Vert : none; Horiz: y = 31 = 3

Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: y -axis

80. y =

x2 - 9 2x 2 + 1

Solution y=

x2 - 9 2

=

( x + 3)( x - 3)

2x + 1 2x 2 + 1 Vert : none; Horiz: y = 21

Slant: none; x-intercepts: (3, 0) , (-3, 0) y -intercepts: (0, - 9) ; Symmetry: y -axis

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1144


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

81. h ( x ) =

x2 - 2x - 8 x-1

Solution x 2 - 2 x - 8 ( x + 2)( x - 4) = x-1 x-1 Vert : x = 1; Horiz: Slant: y = x - 1

h(x) =

x-intercepts: (4, 0) , (-2, 0) y -intercepts: (0, 8) ; Symmetry: none

82. h ( x ) =

x2 + x - 6 x +2

Solution x 2 + x - 6 ( x + 3)( x - 2) = x +2 x +2 -4 = x - 1+ x +2 Vert : x = -2; Horiz: none; Slant: y = x - 1 h(x) =

x-intercepts: (2, 0) , (-3, 0) y -intercepts: (0, - 3) ; Symmetry: none

83. f ( x ) =

x 3 + x 2 + 6x x2 - 1

Solution f (x) =

x ( x 2 + x + 6) x 3 + x2 + 6x = x2 - 1 ( x + 1)( x - 1) = x + 1+

7x + 1 x2 - 1

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1145


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Vert : x = -1, x = 1; Horiz: none Slant: y = x + 1; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: none

84. f ( x ) =

x 3 - 2x 2 + x x2 - 4

Solution 2

x ( x - 1) x 3 - 2x 2 + x f (x) = = 2 x -4 ( x + 2)( x - 2) = x -2+

5x - 8

x2 - 4 Vert : x = -2, x = 2; Horiz: none Slant: y = x - 2; x-intercepts: (0, 0) , (1, 0) y -intercepts: (0, 0) ; Symmetry: none

Graph each rational function. Note that the numerator and denominator of the fraction share a common factor. 85. f ( x ) =

x2 x

Solution x2 f (x) = = x (if x ¹ 0) x

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1146


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

86. f ( x ) =

x2 - 1 x-1

Solution x 2 - 1 ( x + 1)( x - 1) = = x+1 x-1 x-1 (if x ¹ 1) f (x) =

87. f ( x ) =

x3 + x x

Solution x ( x + 1) x3 + x f (x) = = = x2 + 1 x x (if x ¹ 0) 2

88. f ( x ) =

x3 - x2 x-1

Solution x ( x - 1) x3 - x2 = = x2 x-1 x-1 (if x ¹ 1) f (x) =

2

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1147


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

89. f ( x ) =

x 2 - 2x - 1 x-1

Solution f (x) =

90. f ( x ) =

x 2 - 2 x - 1 ( x - 1)( x - 1) = x-1 x-1 = x - 1 (if x ¹ 1)

2x 2 + 3x - 2 x +2

Solution f (x) =

91. f ( x ) =

2 x 2 + 3 x - 2 (2 x - 1)( x + 2) = x +2 x +2 = 2 x - 1 (if x ¹ -2)

x3 - 1 x-1

Solution f (x) =

2 x 3 - 1 ( x - 1)( x + x + 1) = x-1 x-1 = x 2 + x + 1 (if x ¹ 1)

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1148


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

92. f ( x ) =

x2 - x x2

Solution f (x) =

x ( x - 1) x2 - x = 2 x x2 x-1 = (if x ¹ 0) x

Fix It In Exercises 93 and 94, identify the step where the first error is made and fix it. Be sure and identify the vertical asymptote(s), horizontal asymptote(s), complete a table of values, and then draw the graph of the function. 93. Graph f  x  

x1 . x2  1

Solution Step 1 was incorrect. Step 1: The vertical asymptote is x = 1. 94. Graph f  x  

4x2 . x2  1

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1149


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution Step 3 was incorrect. Step 3: f  2  

16 5

Applications 95. Cell phone plan A cell phone provider offers a new phone for $500 with a monthly plan of $50. a. Write a rational function C ( x ) that represents the average cost of the phone per month x. b. What is the decrease in the average cost per month from the fifth month to the tenth month. Round to the nearest dollar.

Solution 500  50 x x

a.

C( x ) 

b.

C(10)  C(5) 

500  500 500  250   100  150  50, so a decrease of $50 10 5

A national hiking club wants to publish a directory of its members. Some investigation shows that the cost of typesetting and photography will be $700, and the cost of printing each directory will be $3.25. 96. a. Find a function that gives the total cost C of printing x directories. b. Find the total cost of printing 500 directories. c. Find a function that gives the mean cost per directory C of printing x directories. d. Find the mean cost per directory if 500 directories are printed. e. Find the mean cost per directory if 1000 directories are printed. f.

Find the mean cost per directory if 2000 directories are printed.

Solution a.

c  x   3.25x  700

b.

c  500  3.25  500  700  $2325

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1150


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

3.25 x  700 x

c.

c x 

d.

c  500  

e.

c  10 0 0  

f.

c  2000  

3.25  500   700 500

 $4.65

3.25  1000   700 1000

 $3.95

3.25  2000   700 2000

 $3.60

An electric company charges $10 per month plus 20¢ for each kilowatt-hour (kwh) of electricity used. 97. a. Find a function that gives the total cost C of x kwh of electricity. b. Find the total cost for using 775 kwh. c. Find a function that gives the mean cost per kwh, C, when using x kwh. d. Find the mean cost per kwh when 775 kwh are used. Round to the nearest hundredth. e. Find the mean cost per kwh when 3200 kwh are used. Round to the nearest hundredth.

Solution a.

c  x   0.20x  10

b.

c  775  0.20  775  10  165

c.

c x 

d.

c  775  

e.

c  3200  

0.20 x  10 x

0.20  775   10 775

 $0.21

0.20  3200   10 3200

 $0.20

98. Utility costs An overseas electric company charges $8.50 per month plus 9.5¢ for each kilowatt-hour (kwh) of electricity used. a. Find a linear function that gives the total cost C of x kwh of electricity. b. Find a rational function that gives the average cost per kwh when using x kwh. c. Find the average cost per kwh when 850 kwh are used.

Solution a.

c  x   0.095x  8.50

b.

c x 

0.095 x  8.50 x

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1151


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

c.

c x 

0.095  850   8.50

850  $0.105  10.5¢

99. Boyle’s law For a given mass at a constant temperature, the product of the pressure P in atmospheres (atm) and volume V in milliliters (ml) is a constant k. That is, PV = k or P 

k V

a. Write the rational function that models Boyle’s law if k = 1200. b. What is the pressure that corresponds to a volume of 30 ml? c. Graph the rational function for P > 0.

Solution a. b.

1200 V 1200 P  30    40 atm: 30

P V  

c.



100. Drug concentration The function C t  t102 t 4 describes the concentration of a drug taken orally in micrograms per milliliter (mcg/ml) in the blood stream in terms of time t in minutes. a. What is the concentration after 20 minutes? Round to the nearest tenth. b. At what time is the concentration of the drug in the bloodstream 2.5 mcg/ml? c. Identify the horizontal asymptote of the graph of the concentration function. In the context of the problem what does it represent?

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1152


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution a.

C  20  

200  0.5 mcg /ml 404

10t t2  4 2.5 t 2  4  10t 2.5 

b.

2.5t  10  10t  0 2

2.5 t 2  4t  4  0 2.5  t  2  t  2   0

t  2, 2, but time must be positive, so 2 minutes c. C = 0; As time passes by, the amount of drug in the blood stream approaches 0 mcg/ml. 101. Speed of airplane Matthew flies in still air to Glacier National Park which is 800 km away. On the return flight a tail wind increases his speed by 40 km/h. a. Using the fact that time equals distance divided by rate, write a rational function that represents the total travel time in terms of rate r. b. If the total travel time was 5 hours, what was the speed of the plane. Round to the nearest tenth.

Solution 1, 600r  32, 800 r 2  40r

a.

t r  

b.

t  5  301.4 km/h

102. Scheduling work crews The following rational function gives the number of days it would take two construction crews, working together, to frame a house that crew 1 (working alone) could complete in x days and crew 2 (working alone) could complete in (x + 3) days. f x 

x 2  3x 2x  3

a. If crew 1 could frame a certain house in 21 days, how long would it take both crews working together? b. If crew 2 could frame a certain house in 25 days, how long would it take both crews working together?

Solution a. Let x  21:

f  21 

212  3  21 2  21  3

 11.2 days

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1153


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

b. Let x  3  25, so x  22:

f  22 

222  3  22 2  22  3

 11.7 days

Discovery and Writing 103. What is a rational function?

Solution Answers may vary. 104. If you are given the equation of a rational function, explain how you determine the vertical and horizontal asymptotes, if any.

Solution Answers may vary. 105. How do you know when a rational function has a slant asymptote? If one exists, explain how to determine its equation.

Solution Answers may vary. 106. Describe a strategy that can be used to graph a rational function.

Solution Answers may vary. In Exercises 107–110, a, b, c, and d are nonzero constants. 107. Show that y = 0 is a horizontal asymptote of the graph of f  x  

ax  b . cx 2  d

Solution

ax  b ax b a b  2  2 2 x x2 x x x y  2    d cx  d cx 2  d cx 2 d c 2  2 2 2 x x x x 00  0. Thus the horizontal asymptote is y  0. As x approaches  , y  c0 ax  b

108. Show that y  ac x is a slant asymptote of the graph of f  x  

ax 3  b . cx 2  d

Solution

ax 3  b ax 3 b b  2 ax  2 2 2 ax  b x x y  2   x2 x  d cx  d cx 2  d cx d c 2  2 x x2 x2 x 3

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1154


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

As x approaches  , y 

ax  0 a a  x. Thus the slant asymptote is y  x. c0 c c

109. Show that y  ac is a horizontal asymptote of the graph of f  x  

ax 2  b . cx 2  d

Solution

ax 2  b ax 2 b b  2 a 2 2 2 ax  b x x x x y  2    d cx  d cx 2  d cx 2 d c 2  2 2 2 x x x x a0 a a As x approaches  , y   . Thus the horizontal asymptote is y  . c0 c c 2

 

3

110. Graph the rational function f x  x x 1 and explain why the curve is said to have a parabolic asymptote.

Solution

x3 1 1  x2  x 1 x2  0 x x x    x2 . . As x approaches  , y  y  1 1 x x x 3

The dotted graph to the left is the graph of the equation y  x 2 . x3  1 . x Notice that for x-coordinates more than 2 units away from x  0, the two graphs are

The solid graph is the graph of the equation y 

very similar. Thus y  x 2 is called a parabolic asymptote of the function f  x  

x3  1 . x

Use a graphing calculator to perform each experiment. Write a brief paragraph describing your findings. 111. Investigate the positioning of the vertical asymptotes of a rational function by graphing

f  x   x x k for several values of k. What do you observe?

Solution Answers may vary.

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1155


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

112. Investigate the positioning of the vertical asymptotes of a rational function by graphing

f  x   x 2x k for k = 4, 1, –1, and 0. What do you observe?

Solution Answers may vary.

 

2

113. Find the range of the rational function f x  xkx2  1 for several values of k. What do you observe?

Solution Answers may vary. 114. Investigate the positioning of the x-intercepts of a rational function by graphing

f  x   x x k for k = 1, –1, and 0. What do you observe? 2

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 115. A rational function can have two horizontal asymptotes.

Solution True. 116. All rational functions have vertical asymptotes.

Solution False. Some rational functions have vertical asymptotes. 117. The graph of the rational function f  x  

x 7 has two vertical asymptotes, x = –7 x 2  49

and x = 7.

Solution False. The function has two vertical asymptotes, x   7 and x  7. 118. The graph of the rational function f  x  

x 100  100 has a horizontal asymptote at y = 0. x 101  101

Solution True. 119. The graph of the rational function f  x  

x 101  101 x 100  100

Solution True.

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1156


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

120. The graph of a rational function will never cross a vertical asymptote.

Solution True. 121. The graph of a rational function will never cross a horizontal asymptote.

Solution False. A rational function can cross a horizontal asymptote. 122. A rational function can have two slant asymptotes.

Solution False. A rational function can have at most one slant asymptote.

CHAPTER REVIEW SOLUTIONS Determine whether the graph of each quadratic function opens upward or downward. State whether a maximum or minimum point occurs at the vertex of the parabola. 1.

f x 

1 2 x 4 2

Solution 1 f  x   x2  4 2 a  21  a  0 upward, minimum

2.

f  x    4  x  1  5 2

Solution f  x   4  x  1  5 2

a  4  a  0 downward, maximum

Find the vertex of each parabola. 3.

f  x   2  x  1  6 2

Solution

f  x   2  x  1  6 2

Vertex :  1, 6 

4.

f  x   2  x  4   5 2

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1157


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

f  x   2  x  4   5 2

Vertex :  4,  5 

5.

f  x   x 2  6x  4 Solution

f  x   x 2  6 x  4; a  1, b  6, c  4 x

b 6   3 2a 2  1

y  x 2  6 x  4    3   6  3   4 2

 13

Vertex :  3,  13 

6.

f  x   4 x 2  4 x  9 Solution

f  x   4 x 2  4 x  9; a  4, b  4, c  9 x

b 4 1   2a 2 2  4 

y  4 x 2  4 x  9  4  21   4  21   9 2

Vertex :  ,  8 

 8

1 2

Graph each quadratic function and label the vertex on the graph. 7.

f  x    x  2  3 2

Solution

f  x    x  2  3 2

a  1  up, vertex:  2,  3  0   x  2  3 2

3   x  2

2

 3  x 2

2 3  x

2  3, 0 , 2  3, 0

f  0   1   0, 1

f  1  2   1,  2 

 3,  2 on graph by symmetry

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1158


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

8.

f  x     x  4  4 2

Solution

f  x     x  4  4 2

a  1  down, vertex:  4, 4  0    x  4  4 2

 x  4  4 2

x  4  2 x 42

x  2 or x  6   2, 0  ,  6, 0 

f  0   12   0,  12  f  3   3   3, 3 

5, 3 on graph by symmetry

9.

y  x2  x

Solution

f  x   x 2  x; a  1, b  1, c  0 x

1 b 1   2a 2  1 2

y  x 2  x   21   21   41 2

vertex:  21 ,  41  , a  1  up

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1159


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

0  x2  x

0  x  x  1

x  0 or x  1   0, 0  ,  1, 0 

f  0   0   0, 0 

f  2   2   2, 2  on graph

 1, 2 on graph by symmetry

10. y  x  x 2

Solution

f  x   x  x 2 ; a  1, b  1, c  0 x

b 1 1   2a 2  1 2

y  x  x 2  21   21   41 2

vertex:  21 , 41  , a  1  down 0  x  x2

0  x 1  x 

x  0 or x  1   0, 0  ,  1, 0 

f  0   0   0, 0 

f  2   2   2,  2  on graph

 1,  2 on graph by symmetry

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1160


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

11.

y  x 2  3x  4

Solution y  x 2  3 x  4; a  1, b  3, c  4 3 b 3 x   2a 2  1 2 y  x 2  3 x  4   32   3  32   4 2

  25 4

vertex:  32 ,  25 , a  1  up 4  0  x 2  3x  4

0   x  1 x  4 

x  1 or x  4   1, 0  ,  4, 0 

f  0   4   0,  4 

f  2  6   2,  6  on graph

 1,  6 on graph by symmetry

12. y  3 x 2  8 x  3

Solution y  3 x 2  8 x  3; a  3, b  8, c  3 4 b 8 x   2a 2  3 3 y  3 x 2  8 x  3  3  43   8  43   3 2

  25 3

vertex:  43 ,  25 , a  3  up 3  0  3x 2  8x  3

0   3 x  1 x  3 

x   31 or x  3    31 , 0  ,  3, 0 

f  0   3   0,  3 

f  1  8   1,  8  on graph

 ,  8 on graph by symmetry 5 3

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1161


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

13. Architecture A parabolic arch has an equation of 3x2 + y – 300 = 0. Find the maximum height of the arch.

Solution

3 x 2  y  300  0 y  3 x 2  300 a  3, b  0, c  300 b 0 x  0 2a 2  3  y  3  0   300  300 2

The maximum height is 300 units.

14. Puzzle problem The sum of two numbers is 1, and their product is as large as possible. Find the numbers.

Solution Let the numbers be x and 1  x .

Product  x  1  x 

y  x  x 2 : a  1, b  1, c  0 vertex: x  

1 1 b   2a 2 2  1

Both numbers are

1 . 2

15. Maximizing area A rancher wishes to enclose a rectangular corral with 1400 feet of fencing. What dimensions of the corral will maximize the area? Find the maximum area.

Solution Let x  the width of the region. 1400  2 x  700  x  the length. Then 2 Area  width  length y  x  700  x 

y   x 2  700 x a  1, b  700, c  0

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1162


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x

b 700   350 2a 2  1

700  x  700  350  350

y  350  700  350   122, 500

The dimensions are 350 ft by 350 ft, with an area of 122,500 ft 2 .

16. Digital cameras A company that produces and sells digital cameras has determined that the total weekly cost C of producing x digital cameras is given by the function C(x) = 1.5x2 – 150x + 4850. Determine the production level that minimizes the weekly cost for producing the digital cameras and find that weekly minimum cost.

Solution

C  x   1.5 x 2  150 x  4850 a  1.5, b  1.50, c  4850 b 150 x   50 2a 2  1.5

C  50  1.5  50  150  50  4850  1100 2

50 cameras should be made, for a minimum cost of $1100. Find the zeros of each polynomial function and state the multiplicity of each. State whether the graph touches the x-axis and turns or crosses the x-axis at each zero. 17.

g(x) = x3 – 6x2 + 9x

Solution

g  x   x 3  6x 2  9x x 3  6x 2  9x  0

x x 2  6x  9  0 x  x  3  0 2

x  0, multiplicity 1, crosses x  3, multiplicity 2, touches 18. g(x) = x3 + 7x2 – 4x – 28

Solution

g  x   x 3  7 x 2  4 x  28 x 3  7 x 2  4 x  28  0

x 2  x  7   4  x  7  x  7  x 2  4  0

 x  7  x  2 x  2  0

x  7, multiplicity 1, crosses x  2, multiplicity 1, crosses x  2, multiplicity 1, crosses

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1163


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

19. f(x) = x4 – 4x3 + 3x2

Solution

f  x   x 4  4 x 3  3x 2 x 4  4 x 3  3x 2  0

x2 x2  4x  3  0 x  x  1 x  3  0 2

x  0, multiplicity 2, touches x  1, multiplicity 1, crosses x  3, multiplicity 1, crosses 20. f(x) = x4 – 10x2 + 24

Solution

f  x   x 4  10 x 2  24 x 4  10 x 2  24  0

 x  6 x  4   0 2

2

 x  6  x  6   x  2 x  2  0 x   6 , multiplicity 1, crosses x  6, multiplicity 1, crosses x  2, multiplicity 1, crosses x  2, multiplicity 1, crosses

Use the Leading Coefficient Test to determine the end behavior of each polynomial function. 21.

f (x) =

2x5 + 9x 3 - 7 x

Solution

f ( x ) = 2x 5 + 9x 3 - 7 x Degree = 5 (odd); Lead Coef: pos. falls left, rises right 22. g ( x ) = - 21 x 7 + 5 x 4 + 6 x 2 - 7

Solution g ( x ) = - 21 x 7 + 5 x 4 + 6 x 2 - 7 Degree = 7 (odd); Lead Coef: pos. rises left, falls right

23. f ( x ) = 7 x 6 - 5 x 2 + 4

Solution

f ( x ) = 7 x 6 - 5x 2 + 4

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1164


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Degree = 4 (even); Lead Coef: pos. rises left, rises right

24. h ( x ) = -2 x 4 - 3 x - 8

Solution f ( x ) = -2 x 4 - 3 x - 8 Degree = 4 (even); Lead Coef: pos. falls left, falls right

Graph each polynomial function. 25. y = x 3 - x

Solution

f  x   x3  x x-int.

y-int.

x3  x  0

f  0   03  0

x x2  1  0

y 0

x  x  1 x  1  0

0, 0

x  0, x  1, x  1

odd deg, pos coef  falls left, rises right Sign of f (x) = x3 - x

+

-

+

(-¥, - 1)

(-1, 0)

(0, 1)

(1, ¥)

–1 Test point Graph of f(x)

0

1

f (-2) = -6

f (- 21 ) = 83

f ( 21 ) = - 83

f (2) = 6

below axis

above axis

below axis

above axis

3

f (- x ) = (- x ) - (- x ) = - x 3 + x = -f ( x )  odd, symmetric about origin

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1165


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

26. y = x 3 - x 2

Solution

f  x   x3  x2 x-int.

y-int.

x3  x2  0

f  0   03  02

x 2  x  1  0

y 0

0, 0

x  0, x  1

odd deg, pos coef  falls left, rises right Sign of f (x) = x3 - x2

+

(-¥, 0)

(0, 1)

(1, ¥)

0 Test point Graph of f(x) 3

1

f (-1) = -2

f ( 21 ) = - 81

f (2) = 4

below axis

below axis

above axis

2

f (- x ) = (- x ) - (- x ) = - x 3 - x 2  neither even nor odd, no symmetry

27. f ( x ) = -x 3 - 7 x 2 - 10 x

Solution

f ( x ) = -x 3 - 7 x 2 - 10 x x-int.

y-int.

 x 3  7 x 2  10 x  0

f  0     0   7  0   10  0 

 x x 2  7 x  10  0  x  x  2  x  5   0

x  0, x  2, x  5

3

y 0

2

0, 0

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1166


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

odd deg, neg coef  rises left, falls right Sign of f ( x ) = -x 3 - 7 x 2 - 10 x

Test point Graph of f(x) 3

+

+

(-¥, - 5)

(-5, - 2)

(-2, 0)

(0, ¥)

–5

–2

0

f (-6) = 24

f (-3) = -6

f (-1) = 4

f (1) = -18

above axis

below axis

above axis

below axis

2

f (- x ) = - (- x ) - 7 (- x ) - 10 (- x ) = x 3 - 7 x 2 + 10 x  neither even nor odd, no symmetry

28. f ( x ) = -x 4 + 18 x 2 - 32

Solution

f ( x ) = -x 4 + 18 x 2 - 32 y-int.

x-int.

 x 2

 x 4  18 x 2  32  0

f  0  32

 x  18 x  32  0

y  32

    x  2 x  16   0 4

2

2

2

 x  2   x  4 x  4  0

0,  32

x   2, x  2, x  4, x  4

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1167


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

even deg, neg coef  falls left, falls right Sign of f ( x ) =

+

+

(-¥, - 4)

(-4, - 2)

(- 2, 2)

( 2, 4)

(4, ¥)

–4

- 2

- x 4 + 18 x 2 - 32

Test point Graph of f(x)

2

4

f (-5) = -207

f (-2) = 24

f (0) = -32

f (2) = 24

f (5) = -207

below axis

above axis

below axis

above axis

below axis

4

2

f (- x ) = - (- x ) + 18 (- x ) - 32 = - x 4 + 18 x 2 - 32 = f ( x )  even, symmetric about y -axis

Use the Intermediate Value Theorem and show that each polynomial function has a zero between the two given numbers. 29. f(x) = 5x3 + 37x2 + 59x + 18; –1 and 0

Solution

f ( x ) = 5 x 3 + 37 x 2 + 59x + 18 f (-1) = -9; f (0) = 18 Thus, there is a zero between - 1 and 0. 30. f(x) = 6x3 – x2 – 10x – 3; 1 and 2

Solution

f ( x ) = 6 x 3 - x 2 - 10 x - 3 f (1) = -8; f (2) = 21 Thus, there is a zero between 1 and 2. Let P(x) = 4x4 + 2x3 – 3x2 – 2. Find the remainder when P(x) is divided by each binomial. 31.

x–1

Solution 4

3

2

P (1) = 4 (1) + 2 (1) - 3 (1) - 2 = 1 The remainder is 1.

32. x – 2

Solution 4

3

2

P (2) = 4 (2) + 2 (2) - 3 (2) - 2 = 66 The remainder is 66.

33. x + 3

Solution 4

3

2

P (- 3) = 4 (- 3) + 2 (- 3) - 3 (- 3) - 2 = 241  The remainder is 241.

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1168


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

34. x + 2

Solution 4

3

2

P (- 2) = 4 (- 2) + 2 (- 2) - 3 (- 2) - 2 = 34  The remainder is 34 .

Use the Factor Theorem to determine whether each statement is true. 35. x – 2 is a factor of x3 + 4x2 – 2x + 4.

Solution 3

2

P (2) = (2) + 4 (2) - 2 (2) + 4 = 24  The remainder is 24.  not a factor

36. x + 3 is a factor of 2x4 + 10x3 + 4x2 + 7x + 21.

Solution 4

3

2

P (- 3) = 2 (- 3) + 10 (- 3) + 4 (- 3) + 7 (- 3) + 21 = - 72  The re mainder is - 72.

37. x – 5 is a factor of x5 – 3125.

Solution 5

P (5) = (5) - 3125 = 0 The remainder is 0.  factor

38. x – 6 is a factor of x5 – 6x4 – 4x + 24.

Solution 5

4

P (6) = (6) - 6 (6) - 4 (6) + 24 = 0 The remainder is 0.  factor

Use synthetic division to divide the polynomial by the given polynomial. 39. 3x4 + 2x2 + 3x + 7; x – 3

Solution 3 3 0 2

3

7

9 27 87

270

3 9 29 90

277

3 x 3 + 9 x 2 + 29 x + 90 +

277 x -3

40. 2x4 – 3x2 + 3x – 1; x – 2

Solution 2 2 0 -3

3

-1

4

8 10

26

2 4

5 13

25

2 x 3 + 4 x 2 + 5 x + 13 +

25 x -2

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1169


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

41. 5x5 – 4x4 + 3x3 – 2x2 + x – 1; x + 2

Solution

-2 5

-4

-2

3

1

-10 28 -62

5 -14

31 -64

-1

128 -258 129 -259

5 x 4 - 14 x 3 + 31x 2 - 64 x + 129 +

-259 x+2

42. 4x5 + 2x4 – x3 + 3x2 + 2x + 1; x + 1

Solution -1 4 2 -1

-4

3

2 -1

4 -2

1

2

1

-2 0

2

0 1

4 x 4 - 2x 3 + x 2 + 2x +

1 x+1

Let P(x) = 5x3 + 2x2 – x + 1. Use synthetic division to find each value. 43. P(3)

Solution 3 5 2 -1

15

1

51 150

5 17 50 151

P (3) = 151

44. P(–3)

Solution -3 5 2 -1

1

-15 39 -114 5 -13 38 -113 P (-3) = -113 æ 1ö 45. P ççç ÷÷÷ è 2 ÷ø

Solution 1 5 2 -1 2

1

5 2

9 4

5 8

9 2

5 4

13 8

5

P ( 21 ) = 138

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1170


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

46. P(i)

Solution i 5 2

5i

-1

1

- 5 + 2i -2 - 6i

5 2 + 5i -6 + 2i

- 1 - 6i

P (i ) = -1 - 6i A partial Solution set is given for each polynomial equation. Find the complete solution set. 47. 2x3 – 3x2 – 11x + 6 = 0; {3}

Solution x = 3 is a solution, so (x – 3) is a factor. Use synthetic division to divide by (x – 3). 3 2 -3 -11

6

9 -6

6

3 -2

2 3

0

2

2 x - 3 x - 11x + 6 = 0

( x - 3)(2 x 2 + 3x - 2) = 0 ( x - 3)(2 x - 1)( x + 2) = 0 Solution set: {3, 21 , - 2} 48. x4 + 4x3 – x2 – 20x – 20 = 0; {–2, –2}

Solution x = –2 is a solution, so (x + 2) is a factor. Use synthetic division to divide by (x + 2). -2 1

1 4

4

-1 -20 -20

-2 -4

10

20

2 -5

-10

0

3

2

x + 4 x - x - 20 x - 20 = 0

( x + 2)( x 3 + 2 x 2 - 5x - 10) = 0 Use the fact that x = –2 is a double root and divide the depressed polynomial by (x + 2):

-2 1

2 -5 -10 -2

1

0

10

0 -5

0

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1171


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

( x + 2)( x 3 + 2 x 2 - 5 x - 10) = 0 ( x + 2)( x + 2)( x 2 - 5) = 0

{

Solution set: -2, - 2,

5, - 5

}

Find the polynomial function of lowest degree with integer coefficients and the given zeros. 49. –1, 2, and

3 2

Solution

2 ( x + 1)( x - 2)( x - 32 ) = 2 ( x 2 - x - 2) ( x - 32 ) = 2 ( x 3 - 52 x 2 - 21 x + 3) = 2x 3 - 5x 2 - x + 6

50. 1, –3, and

1 2

Solution

2 ( x - 1)( x + 3)( x - 21 ) = 2 ( x 2 + 2 x - 3) ( x - 21 ) = 2 ( x 3 + 32 x 2 - 4 x + 32 ) = 2x 3 + 3x 2 - 8x + 3

51.

2, –5, i, and –i

Solution

( x - 2)( x + 5)( x - i )( x + i ) = ( x 2 + 3 x - 10)( x 2 - i 2 ) = ( x 3 + 3 x - 10)( x 2 + 1) = x 4 + 3 x 3 - 9 x 2 + 3 x - 10

52. –3, 2, i, and –i

Solution

( x + 3)( x - 2)( x - i )( x + i ) = ( x 2 + x - 6)( x 2 - i 2 ) = ( x 2 + x - 6)( x 2 + 1) = x 4 + x 3 - 5x 2 + x - 6

How many zeros does each function have? 53. P(x) = 3x6 – 4x5 + 3x + 2

Solution 3x 6 - 4 x 5 + 3x + 2 = 0 6 zeros

54. P(x) = 2x6 – 5x4 + 5x3 – 4x2 + x – 12

Solution 2 x 6 - 5 x 4 + 5 x 3 - 4 x 2 + x - 12 = 0 6 zeros

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1172


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

55. P(x) = 3x65 – 4x50 + 3x17 + 2x

Solution 3 x 65 - 4 x 50 + 3 x 17 + 2 x = 0 65 zeros

56. P(x) = x1984 – 12

Solution x 1984 - 12 = 10 1984 zeros

Determine how many linear factors and zeros each polynomial function has. 57. P(x) = x4 – 16

Solution

P ( x ) = x 4 - 16 4 linear factors, 4 zeros 58. P(x) = x40 + x30

Solution

P ( x ) = x 40 + x 30 40 linear factors, 40 zeros 59. P(x) = 4x5 + 2x3

Solution

P ( x ) = 4 x 5 + 2x 3 5 linear factors, 5 zeros 60. P(x) = x3 – 64x

Solution

P ( x ) = x 3 - 64 x 3 linear factors, 3 zeros Find another zero of a polynomial function with real coefficients if the given quantity is one zero. 61. 2 + i

Solution 2 - i is also a zero. 62. –i

Solution -i = 0 - i , so 0 + i = i is also a zero.

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1173


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Write a third-degree polynomial function with real coefficients and the given zeros. 63. 4, –i

Solution If - i is a zero, then i is a zero also:

( x - 4)( x + i )( x - i ) = 0 ( x - 4) ( x 2 - i 2 ) = 0 ( x - 4)( x 2 + 1) = 0 x3 - 4x2 + x - 4 = 0 64. –5, i

Solution If i is a zero, then - i is a zero also:

( x + 5)( x - i )( x + i ) = 0 ( x + 5)( x 2 - i 2 ) = 0 ( x + 5)( x 2 + 1) = 0 x 3 + 5x 2 + x + 5 = 0 Find the number of possible positive, negative, and nonreal zeros for each polynomial function. 65. P(x) = 3x4 + 2x3 – 4x + 2

Solution P ( x ) = 3 x 4 + 2 x 3 - 4 x + 2: 2 sign variations  2 or 0 positive zeros 4

3

P (-x ) = 3 (-x ) + 2 (-x ) - 4 (-x ) + 2 = 3 x 4 + 2 x 3 - 4 x + 2: 2 sign variations  2 or 0 negative zeros

# pos 2 2 0 0

# neg 2 0 2 0

# nonreal 0 2 2 4

66. P(x) = 2x4 – 3x3 + 5x2 + x – 5

Solution

P ( x ) = 2 x 4 - 3 x 3 + 5 x 2 + x - 5: 3 sign variations  3 or 1 positive zeros 4

3

2

P (-x ) = 2 (-x ) - 3 (-x ) + 5 (-x ) + (-x ) - 5 = 2x 4 + 3x 3 + 5x 2 - x - 5: 1 sign variation  1 negative zero

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1174


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 3 1

# neg 1 1

# nonreal 0 2

67. P(x) = 4x5 + 3x4 + 2x3 + x2 + x – 7

Solution P ( x ) = 4 x 5 + 3 x 4 + 2 x 3 + x 2 + x - 7 : 1 sign variation  1 positive zero 5

4

3

2

P (-x ) = 4 (-x ) + 3 (-x ) + 2 (-x ) + (-x ) + (-x ) - 7 = -4 x 5 + 3 x 4 - 2 x 3 + x 2 - x - 7: 4 sign variations  4 or 2 or 0 negative zeros

# pos 1 1 1

# neg 4 2 0

# nonreal 0 2 4

68. P(x) = 3x7 – 4x5 + 3x3 + x – 4

Solution P ( x ) = 3 x 7 - 4 x 5 + 3 x 3 + x - 4: 3 sign variations  3 or 1 positive zeros 7

5

3

P (-x ) = 3 (-x ) - 4 (-x ) + 3 (-x ) + (-x ) - 4 = -3 x 7 + 4 x 5 - 3 x 3 - x - 4: 2 sign variations  2 or 0 negative zeros

# pos 3 3 1 1

# neg 2 0 2 0

# nonreal 2 4 4 6

69. P(x) = x4 + x2 + 24,567

Solution P ( x ) = x 4 + x 2 + 24, 567 : 0 sign variations  0 positive zeros 4

2

P (-x ) = (-x ) + (-x ) + 24, 567 = x 4 + x 2 + 24, 567 : 0 sign variations  0 negative zeros

# pos 0

# neg 0

# nonreal 4

70. P(x) = –x7 – 5

Solution P ( x ) = -x 7 - 5: 0 sign variations  0 positive zeros 7

P (-x ) = - (-x ) - 5 = x 7 - 5 : 1 sign variation  1 negative zero

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1175


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

# pos 0

# neg 1

# nonreal 6

Find integer bounds for the zeros of each function. Answers can vary. 71.

P(x) = 5x3 – 4x2 – 2x + 4

Solution P ( x ) = 5x 3 - 4 x 2 - 2x + 4 2 5 -4 -2 10

4

12 20

5 6 10 24 Upper bound: 2 -1 5 -4 -2 4 -5

9 -7

5 -9

7 -3

Lower bound: - 1 72. P(x) = x4 + 3x3 – 5x2 – 9x + 1

Solution P ( x ) = x 4 + 3x 3 - 5x 2 - 9x + 1 2 1 3 -5 -9 2

10

1

10 2

1 5 5 1 3 Upper bound: 2 -5 1 3 -5 -9

1

-5

10 -25 170

1 -2

5 -34 171

Lower bound: - 5

Use the Rational Zero Theorem to list all possible rational zeros of the polynomial function. 73. P(x) = 2x4 + x3 – 3x2 – 5x – 6

Solution num:  1,  2,  3,  6; den:  1,  2 possible zeros:  1,  2,  3,  6,  21 ,  32 74. P(x) = 4x5 – 2x4 + 3x3 – 5x – 10

Solution num:  1,  2,  5,  10; den:  1,  2,  4 possible zeros:  1,  2,  5,  10,  21 ,  52 ,  41 ,  45

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1176


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Find all rational zeros of each polynomial function. 75. P(x) = x3 – 10x2 + 29x – 20

Solution

P ( x ) = x 3 - 10 x 2 + 29 x - 20 Possible rational zeros  1,  2,  4,  5,  10,  20 Descartes' Rule of Signs # pos 3 1

# neg 0 0

# nonreal 0 2

Test x = 1: 1 1 -10

29 -20 -9

1 1

-9

20

20

0

P ( x ) = x - 10 x + 29 x - 20 3

2

= ( x - 1)( x 2 - 9 x + 20) = ( x - 1)( x - 5)( x - 4)

Solution set: {1, 5, 4}

76. P(x) = x3 – 8x2 – x + 8

Solution

P ( x ) = x 3 - 8x 2 - x + 8 Possible rational zeros  1,  2,  4,  8 Descartes' Rule of Signs # pos 2 0

# neg 1 1

# nonreal 0 2

Test x = 1: 1 1 -8

-1

8

- 7 -8

1 1 -7

-8

0

P ( x ) = x - 8x - x + 8 3

2

= ( x - 1)( x 2 - 7 x - 8) = ( x - 1)( x - 8)( x + 1)

Solution set: {1, 8, - 1}

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1177


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

77. P(x) = 2x3 + 17x2 + 41x + 30

Solution Possible rational zeros  1,  2,  3,  5,  6,  10,  15,  30,  21 ,  32 ,  52 ,  152 Descartes' Rule of Signs # pos 0 0

# neg 3 1

# nonreal 0 2

Test x = -2: -2 2

17 -4

2

41

30

- 26 -30

13

15

0

P ( x ) = 2 x + 17 x + 41x + 30 3

2

= ( x + 2) (2 x 2 + 13 x + 15) = ( x + 2)(2 x + 3)( x + 5)

Solution set: {-2, - 32 , - 5}

78. P(x) = 3x3 + 2x2 + 2x – 1

Solution Possible rational zeros  1,  31 Descartes' Rule of Signs # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = 31 : 1 3

3 2 2 -1 1

1

1

3 3

3

0

P ( x ) = 3 x 3 + 2 x 2 + 2 x - 10 = ( x - 31 ) (3 x 2 + 3 x + 3) 3 x 2 + 3 x + 3 does not factor rationally. Rational solutions: { 31 }

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1178


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

79. P(x) = 4x4 – 25x2 + 36

Solution Possible rational zeros  1,  2,  3,  4,  6,  9,  12,  18,  36,  21 ,  32 ,  92 ,  41 ,  43 ,  94

Descartes' Rule of Signs # pos 2 2 0 0

# neg 2 0 2 0

# nonreal 0 2 2 4

Test x = 2: 2 4 0 -25

36

16 -18 -36

8 4 8

0

- 9 -18

0

Test x = -2:

-2 4

8 -9 -18

-8 4

0

18

0 -9

0

P ( x ) = 4 x - 25 x + 36 4

2

= ( x - 2)(4 x 3 + 8 x 2 - 9 x - 18) = ( x - 2)( x + 2)(4 x 2 - 9) = ( x - 2)( x + 2)(2 x + 3)(2 x - 3) Solution set: {2, - 2, - 32 , 32 } 80. P(x) = 2x4 – 11x3 – 6x2 + 64x + 32

Solution Possible rational zeros  1,  2,  4,  8,  16,  32,  21 Descartes' Rule of Signs # pos 2 2 0 0

# neg 2 0 2 0

# nonreal 0 2 2 4

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1179


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Test x = 4: 4 2 -11 -6

64

32

- 12 -72 -32

8

2 -3 - 18 Test x = 4:

-8

0

4 2 -3 -18 -8 8 2

20

5

8

2

0

P ( x ) = 2 x - 11x - 6 x 2 + 64 x + 32 4

3

= ( x - 4)(2 x 3 - 3 x 2 - 18 x - 8) = ( x - 4)( x - 4)(2 x 2 + 5 x + 2) = ( x - 4)( x - 4)(2 x + 1)( x + 2) Solution set: {4, 4, - 21 , - 2} Find all zeros of each function. 81. P(x) = 3x3 – x2 + 48x – 16

Solution Possible rational zeros  1,  2,  4,  8,  16,  31 ,  32 ,  43  83 ,  163

Descartes' Rule of Signs # pos 3 1

# neg 0 0

# nonreal 0 2

Test x = 31 : 1 3

3 -1

48 -16

1 3

0

16

0 48

0

P ( x ) = 3 x - x + 48 x - 16 3

2

= ( x - 31 )(3 x 2 + 48)

x = 31

or

x =  -16 x =  4i

Solution set: { 31 , - 4i , 4i }

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1180


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

82. P(x) = x4 – 2x3 – 9x2 + 8x + 20

Solution Possible rational zeros

 1,  2,  4,  5  10,  20 Descartes' Rule of Signs # pos 2 2 0 0

# neg 2 0 2 0

# nonreal 0 2 2 4

Test x = 2: 2 1 -2

-9

8

0 -18 -20

2

0 -9

1

20

- 10

0

Test x = -2: 0 -9 -10 -2 1

-2

4

10

1 - 2 -5

0

P ( x ) = x - 2 x - 9 x 2 + 8 x + 20 4

3

= ( x - 2)( x 3 - 9 x - 10) = ( x - 2)( x - 2)( x 2 - 2 x - 5) Use the quadratic formula.

{

Solution set: 2, - 2, 1  6

}

Find the domain of each rational function. 83. f ( x ) =

3x 2 + x - 2 x 2 - 25

Solution

f ( x) =

3x 2 + x - 2 2

x - 25

=

(3x - 2)( x + 1) ( x + 5)( x - 5)

den = 0  x = -5 or x = 5 domain = (-¥, - 5) È (-5, 5) È (5, ¥) 84. f ( x ) =

2x 2 + 1 x2 + 7

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1181


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

2x 2 + 1 x2 + 7 den = 0  never true f (x) =

domain = (-¥, ¥) Find the vertical asymptotes, if any, of each rational function. 85. f ( x ) =

x +5 x2 - 1

Solution

f (x) =

x +5 2

x -1

=

x +5

( x + 1)( x - 1)

den = 0  x = -1 or x = 1 vertical: x = -1 or x = 1 86. f ( x ) =

x -7 x 2 - 49

Solution

f ( x) =

x -7 x -7 = 2 x - 49 ( x + 7)( x - 7)

1 x +7 den = 0  x = -7 vertical: x = -7 =

87. f ( x ) =

x 2

x + x -6

Solution

f (x) =

x x2 + x - 6

=

x

( x + 3)( x - 2)

den = 0  x = -3 or x = 2 vertical: x = -3 or x = 2 88. f ( x ) =

5x + 2 2x 2 - 6x - 8

Solution

f (x) =

5x + 2 5x + 2 = 2 x - 6 x - 8 (2 x + 2)( x - 4) 2

den = 0  x = -1 or x = 4 vertical: x = -1 or x = 4

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1182


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Find the horizontal asymptotes, if any, of each rational function. 89. f ( x ) =

2x 2 + x - 2 4x2 - 4

Solution

2x 2 + x - 2 ; deg(num) = deg(den) 4x2 - 4 2 1 horizontal: y = , or y = 4 2 f (x) =

90. f ( x ) =

5x 2 + 4 4 - x2

Solution

5x 2 + 4 ; deg(num) = deg(den) 4 - x2 5 horizontal: y = , or y = -5 -1 f (x) =

91.

f (x) =

x+1 x3 - 4x

Solution

x+1 ; deg(num) < deg(den) x3 - 4x horizontal: y = 0 f ( x) =

92. f ( x ) =

x3 2 x 2 - x + 11

Solution

x3 ; deg(num) > deg(den) 2x 2 - 6x - 8 horizontal: none f (x) =

Find the slant asymptote, if any, for each rational function. 93. f ( x ) =

2x 2 - 5x + 1 x -4

Solution

2x 2 - 5x + 1 13 = 2x + 3 + x -4 x -4 slant: y = 2 x + 3

f (x) =

94. f ( x ) =

5x 3 + 1 x +5

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1183


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution

5x 3 + 1 ; deg(num) = 3, x +5 deg(den) = 1  slant: none f ( x) =

Graph each rational function. 95. f ( x ) =

2x x -4

Solution 2x f (x) = x -4 vertical : x = 4; horizontal: y = 21 = 2 Slant: none; x-intercepts: (0, 0)

y -intercepts: (0, 0) ; Symmetry: none

96. f ( x ) =

-4 x x +4

Solution -4 x f (x) = x +4 Vert : x = -4; Horiz: y = -14 = -4 Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: none

97. f ( x ) =

x 2

( x - 1)

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1184


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Solution f (x) =

x 2

( x - 1)

Vert : x = 1; Horiz: y = 0 Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: none

2

98. f ( x ) =

( x - 1) x

Solution 2

f (x) =

( x - 1)

: Vert : x = 0 x Horiz: none; Slant: y = x - 2 x-intercepts: (1, 0) y -intercepts: none; Symmetry: none

99. f ( x ) =

x2 - x - 2 x2 + x - 2

Solution y=

x 2 - x - 2 ( x + 1)( x - 2) = x 2 + x - 2 ( x + 2)( x - 1)

Vert : x = -2, x = 1; Horiz: y = 1

Slant: none; x-intercepts: (-1, 0) , (2, 0) y -intercepts: (0, 1) ; Symmetry: none

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1185


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

100. f ( x ) =

x3 + x x2 - 4

Solution x ( x 2 + 1) x3 + x = y= 2 x - 4 ( x + 2)( x - 2) Vert : x = -2, x = 2; Slant: y = x; x-int: (0, 0) y -int: (0, 0) ; Symmetry: none

CHAPTER TEST SOLUTIONS Find the vertex of each parabola. 1.

y = 3(x – 7)2 – 3

Solution 2

y = 3 ( x - 7 ) - 3; Vertex: (7, - 3)

2. f(x) = 3x2 – 24x + 38

Solution y = 3 x 2 - 24 x + 38 a = 3, b = -24, c = 38 -24 b vertex: x = ==4 2a 2 (3) 2

y = 3 x 2 - 24 x + 38 = 3 (4) - 24 (4) + 38 = - 10

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1186


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Graph the function. 3. f(x) = (x – 3)2 + 1

Solution 2

y = ( x - 3) + 1; Shift y = x 2 U 1, R 3.

Assume that an object tossed vertically upward reaches a height of h feet after t seconds, where h = 100t – 16t2. 4. In how many seconds does the object reach its maximum height?

Solution h = 100t - 16t 2 ; a = -16, b = 100, c = 0 x =-

100 25 b == seconds 2a 8 2 (-16)

5. What is that maximum height?

Solution

h = 100t - 16t 2 ; a = -16, b = 100, c = 0 From# 4, x = 285 . 2

y = 100 ( 25 - 16 ( 25 = 6245 feet 8 ) 8 ) 6. Suspension bridges The cable of a suspension bridge is in the shape of the parabola x2 – 2500y + 25,000 = 0 in the coordinate system shown in the illustration. (Distances are in feet.) How far above the roadway is the cable’s lowest point?

Solution The roadway is at y = 0, so the distance to the lowest point will be the y-coord. of the vertex

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1187


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x 2 - 2500 y + 25, 000 = 0 x 2 + 25, 000 = 2500 y 1 2500

y = c-

x 2 + 10 = y

02 b2 = 10 = 10 4a 4 ( 25100 )

The lowest point is 10 ft above.

Graph each function. 7. f(x) = x4 – x2

Solution

y  x4  x2 f x   x   x  4

2

 x 4  x 2  f  x   even x-int.

y-int.

x4  x2  0

y  04  02 y 0

x2 x2  1  0

0, 0

x  0, x  1 2

2

x  0, x  1, x  1

0, 0  ,  1, 0  ,  1, 0 

8. f(x) = x5 – x3

Solution y  x5  x3 f x   x   x  5

  x5    x 

3

3

  x 5  x 3  f  x   odd

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1188


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

x-int.

y-int.

x x 0

y  05  03 y 0

5

3

x3 x2  1  0

0, 0

x  0, x  1 3

2

x  0, x  1, x  1

0, 0  ,  1, 0  ,  1, 0 

9. Is x = –2 a zero of P(x) = x2 + 5x + 6?

Solution P  2    2   5  2   6  4  10  6  0;  2 is a zero of P  x  . 2

Use long division and the Remainder Theorem to find each value. 10. P(x) = x5 + 2; P(–2)

Solution x 4  2 x 3  4 x 2  8 x  16 x  2 x5  0x 4  0x 3  0x 2  0x  2 x5  2x 4  2x 4  0x 3  2x 4  4 x 3

4x3  0x2 4 x 3  8x 2  8x 2  0x  8 x 2  16 x

16 x  2 16 x  32 30

11. Use the Factor Theorem to determine whether x – 3 is a factor of 2x4 – 10x3 + 4x2 + 7x + 21.

Solution P  3   2  3   10  3   4  3   7  3   21 4

3

2

 30  remainder  30 x  3 is not a factor of the polynomial.

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1189


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Use synthetic division to express P(x) = 2x3 – 3x2 – 4x – 1 in the form (divisor)(quotient) + remainder for the divisor. 12. x – 2

Solution 2 2 -3 -4

2 -4

4 2

-1

- 2 -5

1

( x - 2)(2x + x - 2) - 5 2

Use synthetic division to perform each division. 13.

2 x 2 - 7 x - 15 x -5

Solution

5 2 -7 -15 10

15

3

0

2 14.

2x + 3

3x 3 + 7 x 2 + 2x x+2

Solution -2 3 7

2 0

-6

-2 0

1

0

3

0

3x 2 + x

Let P(x) = 3x3 – 2x2 + 4. Use synthetic division to find each value. æ 1ö 15. P ççç- ÷÷÷ è 3 ø÷

Solution

- 31 3 -2

3

0

4

-1

1 - 31

-3

1

P (- 31 ) = 113

11 3

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1190


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

16. P(i)

Solution i 3 -2

3i

0

4

- 3 - 2i 2 - 3i

3 -2 + 3i -3 - 2i 6 - 3i P (i ) = 6 - 3i Find a polynomial function with the given zeros. 17. 5, –1, 0

Solution

 x  5 x  1 x  0    x  4 x  5 x 2

 x 3  4 x 2  5x

18. i, –i,

3, - 3

Solution

 x  i  x  i   x  3  x  3    x  i  x  3   x  1 x  3  x  2x  3 2

2

2

2

2

4

2

19. How many linear factors and zeros does P(x) = 3x3 + 2x2 – 4x + 1 have?

Solution 3 linear factors, 3 zeros 20. If 3 – 2i is a zero of P(x) where P(x) has real number coefficients, find another zero.

Solution 3  2i must also be a zero. Use Descartes’ Rule of Signs to find the number of possible positive, negative, and nonreal zeros of the polynomial function. 21. P(x) = 3x5 – 2x4 + 2x2 – x – 3

Solution P ( x ) = 3 x 5 - 2 x 4 + 2 x 2 - x - 3: 3 sign variations  3 or 1 positive zeros 5

4

2

P (-x ) = 3 (-x ) - 2 (-x ) + 2 (-x ) - (-x ) - 3 = -3 x 5 - 2 x 4 + 2 x 2 + x - 4: 2 sign variations  2 or 0 negative zeros # pos 3 3 1 1

# neg 2 0 2 0

# nonreal 0 2 2 4

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1191


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Find integer bounds for the zeros of each polynomial function. Answers can vary. 22. P(x) = x5 – x4 – 5x3 + 5x2 + 4x – 5

Solution

P ( x ) = x 5 - x 4 - 5x 3 + 5x 2 + 4 x - 5 3 1 -1 -5 5 3

4 -5

6 3 24

-3 1

84

1 2 1 8 28 79 Upper bound: 3

1

-1

-5

-3

12 -21 48 -156

-4

7 -16 52 -161

5

4

-5

Lower bound: - 3

23. Use the Rational Zero Theorem to list all possible rational zeros of P(x) = 5x3 + 4x2 + 3x + 2.

Solution num:  1,  2; den:  1,  5; possible zeros:  1,  2,  51 ,  52 24. Find all zeros of the polynomial function. P(x) = 2x3 + 3x2 – 11x – 6.

Solution Possible rational zeros

 1,  2,  3,  6,  31 ,  23 Descartes' Rule of Signs # pos 1 1

# neg 2 0

# nonreal 0 2

Test x = 2: 2 2 3 -11 -6

2 3

4

14

6

7

3

0

2

2 x + 3 x - 11x - 6 = 0

( x - 2)(2 x 2 + 7 x + 3) = 0 ( x - 2)(2 x + 1)( x + 3) = 0 Solution set: {2, - 21 , - 3} 25. Find all zeros of the function P(x) = x4 + x3 + 3x2 + 9x – 54.

Solution Possible rational zeros

 1,  2,  3,  6,  9,  18,  27,  54

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1192


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

Descartes' Rule of Signs # neg 3 1

# pos 1 1

# nonreal 0 2

Test x = 2: 2 1 1 3 9 -54 2 6 18

54

1 3 9 27 Test x = -3: -3 1 3 9

0 27

-3 0 -27 1

0 9

0

x 4 + x 3 + 3 x 2 + 9 x - 54 = 0

( x - 2)( x 3 + 3x 2 + 9x + 27) = 0 ( x - 2)( x + 3)( x 2 + 9) = 0 x 2 + 9 = 0  x 2 = -9  x =  3i Solution set: {2, - 3,  3i } 26. Does the polynomial P(x) = 3x3 + 2x2 – 4x + 4 have a zero between the values x = 1 and x = 2?

Solution

P (1) = 5, P (2) = 28; The Intermediate Value Theorem does not guarantee a zero between 1 and 2.

Find all asymptotes of the graph of each rational function. Do not graph the function. 27. f ( x ) =

x-1 x2 - 9

Solution

y=

x-1 2

x -9

=

x-1

( x + 3)( x - 3)

Vert: x = -3, x = 3; Horiz: y = 0 28. f ( x ) =

x 2 - 5 x - 14 x -3

Solution

y=

x 2 - 5 x - 14 ( x - 7)( x + 2) = x -3 ( x - 3)

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1193


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

-20 x -3 vert: x = 3; Slant: y = x - 2 = x -2+

Graph the rational function. 29. f ( x ) =

x2 x -9 2

Solution f (x) =

x2 x2 = x 2 - 9 ( x + 3)( x - 3)

Vert : x = -3, x = 3; Horiz: y = 1 Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: y -axis

Graph the rational function. The numerator and denominator share a common factor. 30. f ( x ) =

x x -x 2

Solution x y= 2 x +1 Vert : none; Horiz: y = 0 Slant: none; x-intercepts: (0, 0) y -intercepts: (0, 0) ; Symmetry: origin

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1194


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

GROUP ACTIVITY SOLUTIONS Polynomial Function Trending What is Polynomial Function Trending? Polynomial function trending describes a pattern in data that is curved or breaks from a straight linear trend. It often occurs in a large set of data that contains many fluctuations. As more data becomes available, the trends often become less linear, and a polynomial trend takes its place. Many statistical packages now regularly include polynomial function trend lines as part of their analysis.

Real-World Example of Polynomial Trending Data Polynomial function trending would be apparent on the graph that shows the relationship between the value of stock and the time the stock is sold. As expected the value of the investment fluctuates over the years, sometimes being worth more than you paid, other times, less.

Group Activity This year, you decide to invest some money in the stock market. As expected the value will fluctuate over time. Based on past performance of the stock, polynomial trending can be used to predict the approximate value of your investment in future years. The polynomial trend function

V  t   t 4  12t 3  44t 2  48t 0, 8 models the approximate value of your stock over the next eight years, with V(t) representing the gain or loss in hundreds of dollars, and t representing the year, with this year being represented by t = 0. a. Determine the years when the gain or loss on your stock will be zero. b. Sketch the graph of the polynomial function over the given interval. c. Confirm the graph of your polynomial function using Desmos. d. In what years was your investment worth less than you paid? e. If you decide to sell your stock in 8 years, describe what your financial situation will be like.

Solution

0  t 4  12t 3  44t 2  48t Possible rational zeros:  1,  2,  3,  4,  6,  8,  12,  16,  24,  48

Test t  6 6 1 12 44 48 6 36 48 1

6

8

0

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1195


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 4: Polynomial and Rational Functions

a.

V  t    t  6  t 2  6t  8

  t  6  t  4  t  2 

t 2  0 t2

t 4  0 t 6  0 t4

t 6

Stock will be zero at years 2, 4, and 6 b. and c.

d. years 0–2 and years 4–6, as the function values are negative on these values e.

V  8  $38, 400

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1196


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER 5: EXPONENTIAL AND LOGARITHMIC F UNCTIONS

TABLE OF CONTENTS End of Section Exercise Solutions ................................................................................. 1197 Exercises 5.1 ............................................................................................................................ 1197 Exercises 5.2 .......................................................................................................................... 1236 Exercises 5.3 .......................................................................................................................... 1250 Exercises 5.4 .......................................................................................................................... 1286 Exercises 5.5 .......................................................................................................................... 1296 Exercises 5.6 .......................................................................................................................... 1320 Chapter Review Solutions.............................................................................................. 1359 Chapter Test Solutions .................................................................................................. 1384 Cumulative Review Solutions ........................................................................................ 1390 Group Activity Solutions ................................................................................................. 1401

END OF SECTION EXERCISE SOLUTIONS EXERCISES 5.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Simplify. a. 63 ∙ 67 b. (63)7 Solution a. 610 b. 621

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1197


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

2. Evaluate. a. 34 b. 30 c. 3–4 Solution a. 81 b. 1 c.

1 81

3. Evaluate. 3

a

 1   5  1   5

0

b.

3

c.

 1   5

Solution a.

1 25

b. 1 c. 125 4. Fill in the blanks. To draw the graph of g(x) = (x – 3)2 + 5, translate the graph of f(x) = x2 ___ units to the right and 5 units ______. Solution 3 units right and 5 units upward 5. Use transformations to graph f(x) = –(x + 2)2 – 2. Solution

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1198


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

6. Which of the following functions are exponential functions? a. f(x) = 4x b. f(x) = (x – 4)2 c.

f ( x)  f ( x) 

4 x 4

d. f(x) = x3 + 4 e. f(x) = 4x f.

f (x)  x  4

g.

f ( x)  4 x

h.

f ( x)  3 x  4

i.

 1 f ( x)    4

x

Solution e and i are exponential functions Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If b > 0 and b

1, f(x) = bx represents an __________ function.

Solution exponential 8. If f(x) = bx represents an increasing function, then b > ______. Solution 1 9. In interval notation, the domain of the exponential function f(x) = bx is ______. Solution

 ,  

10. The number b is called the __________ of the exponential function f(x) = bx. Solution base

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1199


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

11. The range of the exponential function f(x) = bx is __________. Solution

0,  

12. The graphs of all exponential functions f(x) = bx have the same _____-intercept, the point ______. Solution y, (0, 1) 13. If b > 0 and b ≠ 1, the graph of f(x) = bx approaches the x-axis, which is called a horizontal __________ of the curve. Solution asymptote 14. If f(x) = bx represents a decreasing function, then _____ < b < _____. Solution 0, 1 15. If b > 1, then f(x) = bx defines a (an) __________ function. Solution increasing 16. The graph of an exponential function f(x) = bx always passes through the points (0, 1) and __________. Solution (1, b) 17. To two decimal places, the value of e is __________. Solution 2.72 18. The continuous compound interest formula is A = __________. Solution Pert 19. Since e > 1, the base-e exponential function is a (an) __________ function. Solution increasing 20. The graph of the exponential function f(x) = ex passes through the points (0, 1) and __________. Solution (1, e)

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1200


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Practice Use a calculator to approximate each expression. Round your answer to four decimal places. 21. 25.7 Solution 51.9842 22. 3–1.6 Solution 0.1724 2

23. 6 5

Solution –0.4884 3

24. 7 2 Solution –18.5203 25. 4 3 Solution 4 3  11.0357

26. 5 2 Solution 5 2  9.7385

27. 7 Solution 7   451.8079 28. 3   Solution 3    0.0317 Use properties of exponents to simplify each expression. 29. e5 Solution 148.4132

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1201


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

30. e–3 Solution 0.0498 31. –e–4.2 Solution –0.0150 32. e 5 Solution –9.3565 33. 5 2 5 2 Solution

 

5 2 5 2  5 2  2  52 2  52

 

34. 5 2

2

 25 2

2

Solution

5   5 2

2

 

35. a 8

2 2

 52  25

8 2

 a 16  a4

2

Solution

a   a 8

2

36. a 12 a 3 Solution a 12 a 3  a 12  3  a2 3  3  a3 3

Evaluate each exponential function at the given values and simplify. 37. f(x) = 9x a. f(2) b. f(0) c. f(–2) Solution a. f(2) = 92 = 81 b. f(0) = 90 = 1 c.

f  2   9 2 

1 81

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1202


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

38. f(x) = 6x a. f(3) b. f(0) c. f(–3) Solution a. f(3) = 63 = 216 b. f(0) = 60 = 1 c.

f  3  93 

 1 39. f  x     3

1 216

x

a. f(3) b. f(0) c. f(–3) Solution 3

a.

 1 1 f  3     3 27  

b.

 1 f 0     1 3

c.

 1 f  3     3

0

 1 40. f  x      5

3

 33  27

x

a. f(4) b. f(0) c. f(–4) Solution 4

a.

 1 1 f 4       625 5

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1203


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions 0

b.

 1 f  0       1 5

c.

 1 f  4      5

4

   5   625 4

41. f(x) = ex a. f(5) b. f(0) c. f(–5) Solution a. f(5) = e5 b. f(0) = e0 = 1 c.

f ( 5)  e5 

1 e5

42. f(x) = e2x a. f(3) b. f(0) c. f(–3) Solution a. f(x) = e2(3) = e6 b. f(x) = e2(0) = e0 = 1 c.

2 3

e

 e6 

1 e6

43. f(x) = 8–x a. f(2) b. f(0) c. f(–2) Solution a.

f  2  8

2

2

 1 1    64 8

b. f(0) = 80 = 1 c. f(–2) = 82 = 64

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1204


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

44. f(x) = 7–x a. f(3) b. f(0) c. f(–3) Solution a.

f  3   7 3 

1 1  3 343 7

b. f(0) = 70 = 1 c. f(–3) = 73 = 343 x

 1 45. f  x      3 2 a. f(1) b. f(0) c. f(–1) Solution 1

a.

 1 1 6 7 f  1     3    2 2 2 2  

b.

 1 f  0     3  1  3  4 2

c.

 1 f  1     3  2  3  5 2

0

1

x

 1 46. f  x      7 4 a. f(1) b. f(0) c. f(–1) Solution 1

a.

 1 1 28 27  f  1     7   4 4 4 4

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1205


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions 0

b.

 1 f  0     7  1  7  6 4

c.

 1 f  1     7  4  7  3 4

1

47. f(x) = 11x – 1 a. f(2) b. f(0) c. f(–2) Solution a. f(2) = 112 – 1 = 11 01

b.

f (0)  11

c.

f ( 2)  11

 111 

2  1

1 11

 113 

1 1331

48. f(x) = –2x + 3 a. f(3) b. f(0) c. f(–3) Solution a. f(3) = –23 + 3 = –26 = –64 b. f(0) = –20 + 3 = –8 c.

f (3)  233  20  1

49. f(x) = 3ex – 2 a. f(2) b. f(0) c. f(–2) Solution a. f(2) = 3e2 – 2 b. f(0) = 3e0 – 2 = 3 – 2 = 1 c.

f ( 2)  3e2  2 

3 2 e2

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1206


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

50. f(x) = ex – 7 a. f(7) b. f(0) c. f(–7) Solution a.

f 7   e

77

 e0  1

b.

f  0  e

07

 e7 

c.

f  7   e

7  7

1 e7

 e14 

1 e14

51. f(x) = 2x – 1 + 1 a. f(1) b. f(0) c. f(–1) Solution a.

f  1  2

b.

f  0  2

c.

f  1  2

1 1

 1  20  1  2

01

 1  21  1 

1  1

1 3 1 2 2

 1  22  1 

1 5 1 4 4

52. f(x) = 4x + 1 – 1 a. f(1) b. f(0) c. f(–1) Solution a.

f  1  4

b.

f 0  4

c.

f  1  4

1 1

 1  42  1  15

01

1413

1  1

 1  40  1  0

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1207


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

The graph of an exponential function is shown. Identify its domain, range, equation of its horizontal asymptote, and whether it is increasing or decreasing on its domain. 53.

Solution

domain ,  ; range 4,  ; horizontal asymptote: y  4; increasing on its domain 54.

Solution

domain ,  ; range 6,  ; horizontal asymptote: y  6; increasing on its domain 55.

Solution

domain ,  ; range 4,  ; horizontal asymptote: y  4; decreasing on its domain 56.

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1208


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

domain ,  ; range 3,  ; horizontal asymptote: y  3; decreasing on its domain 57.

Solution

domain ,  ; range , 1 ; horizontal asymptote: y  1; decreasing on its domain 58.

Solution

domain ,  ; range , 2 ; horizontal asymptote: y  2; decreasing on its domain 59.

Solution

domain ,  ; range , 2 ; horizontal asymptote: y  2; increasing on its domain 60.

Solution

domain ,  ; range , 9 ; horizontal asymptote: y  9; increasing on its domain

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1209


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Graph each exponential function.

 

61. f x  3x Solution

f  x   3x

point :  0, 1 ,  1, 3 

 

62. f x  5x Solution

f  x   5x

point :  0, 1 ,  1, 5 

 1 63. f  x     5

x

Solution

 1 f x    5

x

 1 point :  0, 1 ,  1,   5

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1210


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 1 64. f  x     3

x

Solution x

 1 f x    3  1 point :  0, 1 ,  1,   3

3 65. f  x     4

x

Solution x

3 f x    4  3 point :  0, 1 ,  1,   4

4 66. f  x     3

x

Solution x

4 f x    3  4 point :  0, 1 ,  1,   3

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1211


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

67. f  x    1.5 

x

Solution

f  x    1.5

x

point :  0, 1 ,  1, 1.5

68. f  x    0.3

x

Solution

f  x    0.3 

x

point :  0, 1 ,  1, 0.3 

 

69. f x  3 x Solution

f  x   3 x

 1 point :  0, 1 ,  1,   3

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1212


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

70. f(x) = 4-x Solution

71. f(x) = –6x Solution

 

72. f x  5 x Solution

f  x   5 x

point :  0,  1 ,  1,  5 

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1213


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 1 73. f  x      5

x

Solution x

 1 f x    5  1 point :  0,  1 ,  1,   5 

 1 74. f  x     3

x

Solution x

 1 f x    3 point :  0, 1 ,  1, 3

Determine whether the graph could represent an exponential function of the form f(x) = bx. 75.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote. YES

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1214


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

76.

Solution The graph does not pass through (0, 1). No 77.

Solution The graph does not pass through (0, 1). No 78.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote. YES Find the value of b, if any, that would cause the graph of f(x) = bx to look like the graph indicated. 79.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote, so it could be an  1 1 exponential function. It passes through the point  1,    1, b  . b  2 2  

80.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote, so it could be an exponential function. It passes through the point (1, 7)  (1, b). b  7

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1215


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

81.

Solution The graph does not pass through (0, 1). It is not an exponential function. 82.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote, so it could be an exponential function. It passes through the point (1, 3)  (1, b). b  3 83.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote, so it could be an exponential function. It passes through the point (1, 2)  (1, b). b  2 84.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote, so it could be an exponential function.

y  bx 1  b 1 3 1 1  1 1    b 3 3b

 

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1216


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

85.

Solution The graph passes through (0, 1) and has the x-axis as an asymptote, so it could be an exponential function. y  bx e2  b2 eb

86.

Solution The graph does not pass through (0, 1). It is not exponential function. Graph each function by using transformations.

 

87. f x  3x  1 Solution

f  x   3x  1

Shift y  3 x D1.

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1217


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

88. f x  2x  3 Solution

f  x   2x  3

Shift y  2x U3.

 

89. f x  2x  1 Solution

f  x   2x  1

Shift y  2x U1.

 

90. f x  4 x  4 Solution

f  x   4x  4

Shift y  4 x D4.

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1218


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

91. f x  3x  1 Solution

f  x   3x  1

Shift y  3x R1.

 

92. f x  2x  3 Solution

f  x   2x  3

Shift y  2x L3.

 

93. f x  3x  1 Solution

f  x   3x  1

Shift y  3x L1.

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1219


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

94. f x  2x 3 Solution

f  x   2x  3

Shift y  2x R3.

 

95. f x  ex  4 Solution

f  x   ex  4

Shift y  e x D4.

 

96. f x  ex  2 Solution

f  x   ex  2

Shift y  e x U2.

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1220


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

97. f x  e x 2 Solution

f  x   e x 2

Shift y  e x R2.

 

98. f x  e x  3 Solution

f  x   ex 3

Shift y  e x L3.

 

99. f x  2x  1  2 Solution

f  x   2x  1  2

Shift y  2 x L1, D2.

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1221


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

100. f x  3x  1  2 Solution

f  x   3x  1  2

Shift y  3 x R1, U2.

x 2 101. y  3  1

Solution f ( x )  3x 2  1 Shift y  3 x R2, U1.

x 2

102. y  3

1

Solution f ( x )  3x  2  1 Shift y  3 x L2, D1.

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1222


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

103. f x  3x  1 Solution

f  x   3 x  1

Reflect y  3x about x, Shift U1

 

104. f x  2x  3 Solution

f  x   2x  3

Reflect y  2x about x, Shift D3

 

105. f x  2 x  3 Solution

f  x   2 x  3

Reflect y  2x about y , Shift D3

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1223


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

106. f x  4 x  4 Solution

f  x   4 x  4

Reflect y  4 x about y , Shift U4

 

107. f x  ex  2 Solution

f  x   e x  2

Reflect y  e x about x, Shift U2

 

108. f x  e x  3 Solution

f  x   e x  3

Reflect y  e x about y , Shift U3

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1224


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

109. f(x) = ex – 1 + 3 Solution

110. f(x) = ex + 2 – 4 Solution

Use a graphing calculator to graph each function.

 

111. f  x   5 2 x Solution

 

y  5 2x

 

112. f  x   2 5 x Solution

 

y  2 5x

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1225


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

113. f x  3 x Solution

y  3 x

 

114. f x  2 x Solution

y  2 x

 

115. f x  2e x Solution

y  2ex

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1226


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

116. f x  3e x Solution

y  3e x

 

117. f x  5e0.5 x Solution

y  5e0.5x

 

118. f x  3e2 x Solution

y  3e2x

Fix It In exercises 119 and 120, identify the step the first error is made and fix it. 3 119. Given the exponential function f(x) = 4–x + 2 determine f   . 2

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1227


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution Step 5 was incorrect. 3 3 Step 1: f    4 2  2 2

3 1 Step 2: f    3  2 2   42

3 Step 3: f    2

1

  4

3

2

3 1 Step 4: f     2 2 8  3  17 Step 5: f    2 8

120. Use the graph of f(x) = 3x to graph g(x) = –3x + 5 –2. First graph f(x) = 3x. Then apply the following sequence of transformations: reflect the graph, translate the graph horizontally, and translate the graph vertically. Solution Step 3 was incorrect: Step 1:

Step 2:

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1228


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Step 3:

Step 4:

Applications In Exercises 121–124, assume that there are no deposits or withdrawals. 121. Compound interest An initial deposit of $10,000 earns 8% interest, compounded quarterly. How much will be in the account in 10 years? Solution  r A  p1   n 

nt

 0.08   10000  1   4    $22, 080.40

4  10 

122. Compound interest An initial deposit of $1000 earns 9% interest, compounded monthly. How much will be in the account in 4 21 years?

Solution  r A  p1   n 

nt

 0.09   1000  1   12    $1497.04

12 4.5 

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1229


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

123. Comparing interest rates How much more interest could $500 earn in 5 years, compounded semiannually (two times a year), if the annual interest rate were 5 21 % instead of 5%?

Solution 5% interest:  r A  P1  n 

Difference  655.83  640.04

5 21 % interest:

nt

 0.05   500  1   2    $640.04

2 5 

 r A  P1  n 

 $15.79 more

nt

 0.055   500  1   2    $655.83

2 5 

124. Comparing savings plans Which institution in the ads provides the better investment?

Solution Assume 1 year for each account: 5.25% interest: 5.35% interest:  r A  P 1   n  

nt

 0.0525   P 1   12    P 1.0538

12 1

 r A  P 1   n  

The 5.25% rate compounded monthly provides a better return.

nt

 0.0535   P 1   1    P  1.0535 

1 1

125. Compound interest If $1 had been invested on July 4, 1776, at 5% interest, compounded annually, what would it be worth on July 4, 2076?

Solution  r A  P 1  n 

nt

1 300 

 0.05   1 1   1    $2, 273, 996.13

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1230


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

126. 360/365 method Some financial institutions pay daily interest, compounded by the 360/365 method, using the following formula.

 r  A  A0  1   360  

365t

t is in years

Using this method, what will an initial investment of $1000 be worth in 5 years, assuming a 7% annual interest rate?

Solution  r  A  A0  1   360  

365t

 0.07   1000  1   360    $1425.93

365 5 

127. Carrying charges A college student takes advantage of the ad shown and buys a bedroom set for $1100. They plan to pay the $1100 plus interest when the income tax refund comes in 8 months. At that time, what will they need to pay?

Solution  r A  P 1   n 

nt

 0.0175   1100  1   1    $1263.77

1 8 

128. Credit card interest A bank credit card charges interest at the rate of 21% per year, compounded monthly. If a senior in college charges $1500 to pay for college expenses and intends to pay it in one year, what will they have to pay?

Solution  r A  P 1  n 

nt

 0.21   1500  1   12    $1847.16

12 1

129. Continuous compound interest An initial investment of $5000 earns 8.2% interest, compounded continuously. What will the investment be worth in 12 years?

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1231


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution A  Pert  5000e0.082( 12)  $13, 375.68

130. Continuous compound interest An initial investment of $2000 earns 8% interest, compounded continuously. What will the investment be worth in 15 years?

Solution 0.08 15 

A  Pert  2000e

 $6640.23 131. Comparison of compounding methods An initial deposit of $5000 grows at an annual rate of 8.5% for 5 years. Compare the final balances resulting from continuous compounding and annual compounding.

Solution Continuous: 0.085 5 

A  Pe  5000e  $7647.95 rt

Annually:  r A  P 1  n 

nt

 0.085   5000  1   1  

1 5 

 $7518.28

132. Comparison of compounding methods An initial deposit of $30,000 grows at an annual rate of 8% for 20 years. Compare the final balances resulting from continuous compounding and annual compounding.

Solution Continuous: 0.08 20 

A  Pe  30, 000e  $148, 590.97 rt

Annually:  r A  P1  n 

nt

 0.08   30, 000  1   1  

1 20 

 $139, 828.71

133. Frequency of compounding $10,000 is invested in each of two accounts, both paying 6% annual interest. In the first account, interest compounds quarterly, and in the second account, interest compounds daily. Find the difference between the accounts after 20 years. Solution Quarterly: Daily: Difference nt nt 33, 197.90  32, 906.63   r r A  P 1  A  P 1   $291.27 n n    0.06   10000  1   4    $32, 906.63

4  20 

 0.06   10000  1   365    $33, 197.90

365 20 

134. Determining an initial deposit An account now contains $11,180 and has been accumulating interest at a 7% annual rate, compounded continuously, for 7 years. Find the initial deposit.

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1232


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution A  Pert 0.07  7 

11, 180  Pe 11, 180

P 0.07  7  e $6849.16  P

135. Saving for college In 20 years, a parent wants to accumulate $40,000 to pay for his daughter’s college expenses. If he can get 6% interest, compounded quarterly, how much must he invest now to achieve his goal?

Solution

 r A  P 1  n 

nt

 0.06  40,000  P  1   4   40,000 P 4 20  1  0.064   

4  20 

$12, 155.61  P 136. Saving for college In Exercise 135, how much should the parent invest to achieve his goal if he can get 6% interest, compounded continuously?

Solution A  Pert 0.06 20 

40, 000  Pe 40, 000

P 0.06 20  e $12, 047.77  P 137. Population of a city The population P(t) of a small city can be approximated by the exponential function, P(t) = 1200e0.2t, where t represents time in years. What will be the population of the city in 12 years? Round to a whole number.

Solution

P  t   1200e0.2t

P  12   1200e

0.2  12 

 13, 228

138. Amount of drug present Typically the amount of a drug A(t), in mg, present in the bloodstream t hours after being intravenously administered can be approximated by the exponential function, A(t) = –1000e–0.3t + 1250. How much of the drug is present in the bloodstream after 14 hours? Round to a whole number.

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1233


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

A  t   1000e0.3t  1250

A  14   1000e

0.3 14 

 1250

 1235 mg

Discovery and Writing 139. What is an exponential function? Give three examples.

Solution Answers may vary. 140. What strategy would you use to graph an exponential function?

Solution Answers may vary. 141. Define e and describe the natural exponential function.

Solution Answers may vary. 142. Explain compound interest.

Solution Answers may vary. 143. Financial planning To have $P available in n years, $A can be invested now in an account paying interest at an annual rate r, compounded annually. Show that A  P 1  r 

n

Solution  r A  P 1   n  

P  A1  r  P

nt

n

A

1  r  P 1  r   A n

n

144. If 2t+4 = k2t, find k.

Solution

2t  4  k 2t 2t  24  2t  k 24  k 16  k

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1234


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

145. If 53t = kt, find k.

Solution

53t  k t

5   k 3

t

t

125t  k t 125  k 146. a. If et+3 = ket, find k. b. If e3t = kt, find k.

Solution a. et  3  ket et  e3  et  k e3  k

b.

e 3t  k t

e   k 3

t

t

e3  k

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 147. The domain of the exponential function f(x) = 7x is all real numbers.

Solution True.

148. The range of the exponential function is f(x) = 7x is 7,  .

Solution

False. The range is 0,  . 149. The graph of f(x) = –7–x has a y-intercept at (0, –7).

Solution False. The y-intercept is (0, –1).

 1 150. The graph of f  x      7

x

has a horizontal asymptote at y = –7.

Solution False. The horizontal asymptote is y = 0. 151. The graph of f(x) = 7x – 7 + 7.7 has a horizontal asymptote at y = 7.7.

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1235


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution True.

 1 152. The graphs of f(x) = 7 and g  x     7 x

x

are identical.

Solution True. 153. The graphs of f(x) = 7x and g(x) = 7–x intersect at the point (0, 7).

Solution False. They intersect at (0, 1). 154. To obtain the graph of g(x) = ex + 7 – 7, we can use the graph of f(x) = ex and shift it 7 units to the left and 7 units down.

Solution True.

EXERCISES 5.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

If the half-life of Cobalt-60 is 5.27 years, then in 5.27 years, how much of it has decomposed?

Solution half t

2. If A  t   3  2  1600 , calculate A(800) and round to two decimal places. 

Solution 800

1

A  800   3  2  1600  3  2  2  2.12 

3. If I(x) = 12(0.6)x, determine I(7) and round to two decimal places.

Solution I  7   12  0.6   0.34 7

4. If the birth rate for a specific country is 18 per thousand and the annual death rate for that country is 6 per thousand, what is the annual growth rate of population written as a decimal?

Solution

18 6 12    0.012 1000 1000 1000

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1236


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

5. If P(t) = 200,000,000e0.011t, calculate P(40) and round to the nearest whole number.

Solution P  40   200, 000, 000e

0.011 40 

 310, 541, 444

1,000,000 , calculate P(5) and round to the nearest whole number. 1  999e0.3t



6. If P t 

Solution

p 5 

1,000,000 1  999e

0.3 5

 4466

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The Malthusian model assumes a constant __________ rate and a constant __________ rate.

Solution birth, death 8. The Malthusian prediction is pessimistic, because a __________ grows exponentially, but food supplies grow __________.

Solution population, linearly Applications Use a calculator to help solve each problem. 9. Tritium decay Tritium, a radioactive isotope of hydrogen, has a half-life of 12.4 years. Of an initial sample of 50 grams, how much will remain after 100 years?

Solution

A  A0 2t /h

100/  12.4 

 50  2  0.1868 grams 10. Chernobyl In April 1986, the world’s worst nuclear power disaster occurred at Chernobyl in the former USSR. An explosion released about 1000 kilograms of radioactive cesium-137 (137Cs) into the atmosphere. If the half-life of 137Cs is 30.17 years, how much will remain in the atmosphere in 100 years?

Solution

A  A0 2t /h

100/  30.17 

 1000  2  101 kg

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1237


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

11. Chernobyl Refer to Exercise 10. How much 137Cs will remain in 200 years?

Solution

A  A0 2t /h

200/  30.17 

 1000  2  10 kg

12. Carbon-14 decay The half-life of radioactive carbon-14 is 5700 years. How much of an initial sample will remain after 3000 years?

Solution

A  A0 2t /h  A0  23000/5700  A0  0.694 

About 69.4% will remain. 13. Plutonium decay One of the isotopes of plutonium, 237Pu, decays with a half-life of 40 days. How much of an initial sample will remain after 60 days?

Solution

A  A0 2t /h  A0  260/40  A0  0.354   About 35.4% will remain.

14. Comparing radioactive decay One isotope of holmium, 162Ho, has a half-life of 22 minutes. The half-life of a second isotope, 164Ho, is 37 minutes. Starting with a sample containing equal amounts, find the ratio of the amounts of 162Ho to 164Ho after one hour. Solution 162

164

Ho:

Ho:

A  A0 2t /h  A0  260/22  A0  0.151

About 15.1% will remain.

amt. of 162Ho amt. of 164Ho

0.151  0.465 0.325

A  A0 2t /h  A0  260/22

 A0  0.325  About 32.5% will remain.

15. Drug absorption in smokers The biological half-life of the asthma medication theophylline is 4.5 hours for smokers. Find the amount of the drug retained in a smoker’s system 12 hours after a dose of 1 unit is taken.

Solution

A  A0 2t /h

12/  4.5

 1 2  0.1575 unit 16. Drug absorption in nonsmokers For a nonsmoker, the biological half-life of theophylline is 8 hours. Find the amount of the drug retained in a nonsmoker’s system 12 hours after taking a one-unit dose.

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1238


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution A  A0 2t /h  1  212/8  0.3536 unit

17. Oceanography The intensity I of light (in lumens) at a distance x meters below the surface is given by I = I0kx, where I0 is the intensity at the surface and k depends on the clarity of the water. At one location in the Arctic Ocean, I0 = 8 lumens and k = 0.5. Find the intensity at a depth of 2 meters.

Solution I  I0 k x

 8  0.5

2

 2 lumens

18. Oceanography At one location in the Atlantic Ocean, I0 = 14 lumens and k = 0.7. Find the intensity of light at a depth of 12 meters. (See Exercise 17.)

Solution I  I0 k x

 14  0.7 

12

 0.194 lumen

19. Oceanography At a depth of 3 meters at one location in the Pacific Ocean, the intensity I of light is 1 lumen and k = 0.5. Find the intensity I0 of light at the surface.

Solution I  I0 k x

1  I0  0.5 

1

0.5

3

 I0

3

8 lumens  I0

20. Oceanography At a depth of 2 meters at one location off the coast of Belize, the intensity I of light is 2 lumens and k = 0.2. Find the intensity I0 of light at the surface.

Solution I  I0 k x

2  I0  0.2  1

0.2

2

2

 I0

50 lumens  I0

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1239


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

21. Bluegill population A Wisconsin lake is stocked with 10,000 bluegill. The population is expected to grow exponentially according to the model P(t) = P02t/2. How many bluegill will be in the lake in 5 years?

Solution P  p0 2t /2  10, 000  25/2  56, 570 fish

22. Community growth The population of Eagle River is growing exponentially according to the model P(t) = 375(1.3)t, where t is measured in years from the present date. Find the population in 3 years.

Solution

P  375  1.3 

t

 375  1.3

3

 824 people 23. Newton’s law of cooling Some hot water, initially at 100°C, is placed in a room with a temperature of 40°C. The temperature T of the water after t hours is given by T(t) = 40 + 60(0.75)t. Find the temperature in 3 21 hours.

Solution

T  40  60  0.75

t

 40  60  0.75

3.5

 61.9C 24. Bacterial cultures A colony of 6 million bacteria is growing in a culture medium. The population P after t hours is given by the formula P(t) = (6 106)(2.3)t. Find the population after 4 hours.

Solution

    6  10   2.3 

P  6  106  2.3  6

t

4

 167, 904, 600

25. Population growth The growth of a town’s population is modeled by P(t) = 173e0.03t. How large will the population be when t = 20?

Solution

P  173e0.03t 0.03 20

 173e  315

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1240


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

26. Population decline The decline of a city’s population is modeled by P(t) = 1.2 106e–0.008t. How large will the population be when t = 30?

Solution P  1.2  106 e0.008t  1.2  106 e

0.008 30 

 9.44  105 27. Epidemics The spread of hoof and mouth disease through a herd of cattle can be modeled by the formula P(t) = P0e0.27t, where P is the size of the infected population, P0 is the infected population size at t = 0, and t is in days. If a rancher does not act quickly to treat two cases, how many cattle will have the disease in one week?

Solution

P  P0e0.27t

0.27  7 

 2e  13 cases 28. Alcohol absorption Typically, the percent of alcohol absorbed into the bloodstream after drinking two glasses of wine is given by the following formula. Find the percent of alcohol absorbed into the blood after 21 hour.

P  t   0.3 1  e 0.05t

Solution

P  0.3 1  e0.05t

 0.3 1  e

0.05 30 

 where t is in minutes.

 0.233  23.3%

29. World population growth The population of the Earth is approximately 6 billion people and is growing at an annual rate of 1.9%. Assuming a Malthusian growth model, find the world population in 30 years.

Solution

P  P0ekt

0.019 30 

 6e  10.6 billion 30. World population growth See Exercise 29. Assuming a Malthusian growth model, find the world population in 40 years.

Solution

P  P0ekt

0.019 40

 6e  12.8 billion 31. World population growth See Exercise 29. By what factor will the current population of the Earth increase in 50 years?

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1241


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution P  P0ekt

0.019 50 

 6e

 15.5 billion 15.5  a factor of 2.6 6 32. World population growth See Exercise 29. By what factor will the current population of the Earth increase in 100 years?

Solution P  P0ekt

0.019 100 

 6e

 40.1 billion 40.1  a factor of 6.7 6 33. Drug absorption The percent P of the drug triazolam (a drug for treating insomnia) remaining in a person’s bloodstream after t hours is given by P(t) = e–0.3t. What percent will remain in the bloodstream after 24 hours?

Solution

P  e0.3t e

0.3 24 

 0.0007  0.07% 34. Medicine The concentration x of a certain drug in an organ after t minutes is given by y(t) = 0.08(1 – e–0.1t). Find the concentration of the drug in 21 hour.

Solution

x  0.08 1  e0.1t

 0.08 1  e

0.1 30 

 0.076

35. Medicine Refer to Exercise 34. Find the initial concentration of the drug (Hint: when t = 0).

Solution

Let t  0:

x  0.08 1  e0.1t

 0.08 1  e

 0.08 1  e

0.1 0 

0

  0.08  1  1  0

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1242


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

36. Spreading the news Suppose the function

N  t   P 1  e 0.1t

is used to model the length of time t (in hours) it takes for N people living in a town with population P to hear a news flash. How many people in a town of 50,000 will hear the news between 1 and 2 hours after it happened?

Solution

N  P 1  e0.1t

 50, 000 1  e  4758

0.1 1

N  50, 000 1  e

0.1 2

 9063 9063  4758  4305 37. Spreading the news How many people in the town described in Problem 30 will not have heard the news after 10 hours?

Solution

N  P 1  e0.1t

 50,000 1  e

0.1 10 

 31,606 50,000  31,606  18, 394 38. Epidemics Refer to Example 5. How many people will be infected with HIV in 5 years?

Solution

P 

1, 200,000

1   1200  1 e0.4t 1, 200, 000

1   1200  1 e

0.4  5 

 7350 people 39. Epidemics Refer to Example 5. How many people will be infected with HIV in 8 years?

Solution 1, 200,000 P 1   1200  1 e0.4t

1, 200,000

1   1200  1 e

0.4  8 

 24, 060 people

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1243


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

40. Epidemics In a city with a population of 450,000, there are currently 1000 cases of hepatitis. If the spread of the disease is projected by the following logistic function, how many people will contract the hepatitis virus after 6 years?

P t  

450,000

1   450  1 e0.2t

Solution 450,000 P 1   450  1 e0.2t

450,000

1   450  1 e

0.2 6 

 3303 people 41. Epidemics In a city with a population of 55,000, there are currently 100 cases of the avian bird flu. If the spread of the disease is projected by the following formula, how many people will contract the bird flu after 2 years?

P t  

55,000

1   550  1 e0.8t

Solution

P 

55,000

1   550  1 e0.8t 55, 000

1   550  1 e

0.8 2

 492 people 42. Life expectancy The life expectancy l of white females can be estimated by using the function l(x) = 78.5(1.001)x, where x is the current age. Find the life expectancy of a white female who is currently 50 years old. Give the answer to the nearest tenth.

Solution

I  78.5  1.001

x

 78.5  1.001

50

 82.5 years 43. Oceanography The width w (in millimeters) of successive growth spirals of the sea shell Catapulus voluto, shown in the illustration, is given by the function w(n) = 1.54e0.503n, where n is the spiral number. To the nearest tenth of a millimeter, find the width of the fifth spiral.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1244


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

w  1.54e0.503n 0.503 5

 1.54e  19.0 mm

44. Skydiving Before the parachute opens, the velocity v (in meters per second) of a skydiver is given by v(t) = 50(1 – e–0.2t). Find the initial velocity.

Solution

v  50 1  e0.2t

 50 1  e

 50 1  e

0.2 0 

0

 0 meters/second 45. Skydiving Refer to Exercise 44 and find the velocity after 20 seconds.

Solution

v  50 1  e0.2t

 50 1  e

0.2 20 

 49 meters/second

46. Free-falling objects After t seconds, a certain falling object has a velocity v given by v(t) = 50(1 – e–0.3t). Which is falling faster after 2 seconds, this object or the skydiver in Exercise 44?

Solution

v  50 1  e0.2t

 50 1  e

0.2 2 

 16.5 meters/second

v  50 1  e0.3t

 50 1  e

0.3 2

This object will be falling faster.

 22.6 meters/second

47. Population growth In 2009, the male population of a country was about 133 million, and the female population was about 139 million. Assuming a Malthusian growth model with a 1% annual growth rate, how many more females than males will there be in 20 years?

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1245


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution Males

P  P0e

Females P  P0ekt

kt

 133e0.01t

 139e0.01t

0.01 20

0.01 20

 133e  139e  162.4 million  169.8 million There will be about 7 million more females. 48. Population growth See Exercise 47. How many more females than males will there be in 50 years?

Solution Males

P  P0e

Females P  P0ekt

kt

 133e0.01t 0.01 50

 139e0.01t 0.0150 

 133e  139e  219.3 million  229.2 million There will be about 10 million more females. Use a graphing calculator to solve each problem. 49. In Example 4, suppose that better farming methods change the formula for food growth to f(x) = 31x + 2000. How long will the food supply be adequate?

Solution Find where these graphs meet:

y  1000e0.02t , y  31x  2000

It will take about 72.2 years. 50. In Example 4, suppose that a birth control program changed the formula for population growth to P(t) = 1000e0.01t. How long will the food supply be adequate?

Solution Find where these graphs meet:

y  1000e0.01t , y  30.62x  2000

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1246


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

It will take about 215 years.

Discovery and Writing 51. Exponential regression A population of Escherichia coli bacteria doubles every 20 minutes. Construct a table that shows the growth of a single E. coli bacterium for a 2-hour period. Then use a graphing calculator to plot the data and determine an exponential regression equation to model this growth.

Solution

y  1.035264924x 52. Refer to Exercise 51. At what point will the population reach 200 cells?

Solution Graph y  1.035264924 and y  200 and find the intersection point. x

It will happen after about 150 minutes, or after about 2 hours and 30 minutes.

ex  e x from x = –2 to x = 2. The 2 graph will look like a parabola, but it is not. The graph, called a catenary, is important in the design of power distribution networks, because it represents the shape of a uniform flexible cable whose ends are suspended from the same height. The function is called the hyperbolic cosine function. The hyperbolic cosine function was used in the design and construction of the Gateway Arch in St. Louis, Missouri.

 

53. Graph the function defined by the equation f x 

Solution

y

ex  e x 2

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1247


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

54. Graph the function defined by the equation f x 

ex  e x from x = –2 to x = 2. The 2

function is called the hyperbolic sine function.

Solution

y

ex  e x 2

55. Graph the following logistic function, first discussed in Example 5. Use WINDOW settings of [0, 20] for x and [0, 1,500,000] for y.

P t  

1, 200,000

1   1199 e0.4t

Solution

P

1, 200,000 1  1199e0.4t

56. Use the TRACE capabilities of your graphing calculator to explore the logistic function of Example 5 and Exercise 55. As time passes, what value does P approach? How many years does it take for 20% of the population to become infected? For 80%?

Solution 20%: about 14.26 years 80%: about 21.19 years

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1248


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

57. The value of e can be calculated to any degree of accuracy by adding the first several terms of the following list. The more terms that are added, the closer the sum will be to e. Add the first six numbers in the following list. To how many decimal places is the sum accurate?

1, 1,

1 1 1 1 , , , , 2 23 234 2345

Solution 1 1 1 1 1 1     2.716; 2 23 234 2345 e  2.718: accurate to 2 places 58. Mixture problem The tank in the illustration initially contains 20 gallons of pure water. A brine solution containing 0.5 pounds of salt per gallon is pumped into the tank, and the well-stirred mixture leaves at the same rate. The amount A of salt in the tank after

t minutes is given by A  t   10 1  e 0.03t .

a. b. c. d.

Graph this function. What is A when t = 0? Explain why that value is expected. What is A after 2 minutes? After 10 minutes? What value does A approach after a long time (as t becomes large)? Explain why this is the value you would expect.

Solution a.

b.

A  10 1  e

0.03 0 

 10  1  1  0

At the beginning, it is pure water, so there is no salt. © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1249


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

c.

A  10 1  e

0.03 2 

 0.58 lb

A  10 1  e

0.03 10 

 2.59 lb d. A will approach 10 lb after a long time. This makes sense because 0.5(20) = 10.

EXERCISES 5.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Simplify each expression. 1

a.

492

b.

10003

c.

813

1

1

Solution 1

a.

49 2  49  7

b.

1000 3  3 1000  10

c.

814  4 81  3

1

1

2. Simplify each expression. a. 8–1 b. 9–2 c. 10–3

Solution a.

81 

1 8

b.

92 

1 1  2 81 9

c.

103 

1 1  3 1000 10

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1250


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

3. Fill in the boxes. a.

2  128

b.

3  243

c.

10  10,000

Solution a. 7 b. 5 c. 4 4. Fill in the boxes. a.

  1 1    4 64  

b.

  1   1 4

c.

  1    64 4

Solution a. 3 b. 0 c. –3 5. Simplify the expression. 2

a.

83

b.

 32

c.

 1 2   4

2 5

3

Solution 2

 8  2  4 2

a.

83 

b.

 32   32    2  4

c.

3  1 2  1   1 1          8 4 2  4

3

2 5

3

2

2

5

2

3

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1251


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

6. The domain of the logarithmic function f(x) = log(4 – x) consists of all x for which 4 – x > 0. Solve the linear inequality.

Solution 4x 4  x  4 x 4

The domain is , 4

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The equation y = logbx is equivalent to __________.

Solution x  by 8. The domain of the logarithmic function y = logbx is the interval __________.

Solution

0,  

9. The __________ of the logarithmic function y = logbx is the interval ,  .

Solution range logb x

10. b

= _____.

Solution x 11. Because the exponential function is one-to-one, it has an __________ function.

Solution inverse 12. The inverse of an exponential function is called a __________ function.

Solution logarithmic 13. logbx is the __________ to which b is raised to get x.

Solution exponent

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1252


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

14. The y-axis is an __________ of a graph of f(x) = logbx.

Solution asymptote 15. The graph of f(x) = logbx passes through the points ______ and ______.

Solution

 b, 1 ,  1, 0

16. log1010x = ______.

Solution x 17. ln x means __________.

Solution

loge x 18. The domain of the function f(x) = ln x is the interval __________.

Solution

0,  

19. The range of the function f(x) = ln x is the interval __________.

Solution

 ,  

20. The graph of f(x) = ln x has the __________ as an asymptote

Solution y-axis 21. In the expression log x, the base is understood to be ______.

Solution 10 22. In the expression ln x, the base is understood to be ______.

Solution e Practice Write each equation in logarithmic form. 23. 82  64

Solution 82  64 log 8 64  2

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1253


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

24. 103  1000

Solution 103  1000 log 10 1000  3

25. 42 

1 16

Solution 42  log 4

1 16

1  2 16

26. 34 

1 81

Solution 34  log 3

 1 27.   2

1 81

1  4 81

5

 32

Solution 5

 1    32 2 log 1/2 32  5  1 28.   3

3

 27

Solution 3

 1    27 3 log 1/3 27  3 29. x y  z

Solution xy  z log x z  y

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1254


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions n 30. m  p

Solution mn  p log m p  n

Write each equation in exponential form. 31. log3 81  4

Solution log 3 81  4 34  81

32. log7 7  1

Solution log 7 7  1 71  7

33. log 1/2

1 3 8

Solution 1 log 1/2  3 8 3  1 1    2 8   34. log 1/5 1  0

Solution log 1/5 1  0 0

 1   1 5 35. log 4

1  3 64

Solution 1 log 4  3 64 1 43  64

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1255


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

1  2 36

36. log 6

Solution log 6

1  2 36 1 6 2  36

37. log   1

Solution log    1

1   38. log 7

1  2 49

Solution 1 log 7  2 49 1 7 2  49 Evaluate each logarithmic expression without using a calculator. 39. log7 343

Solution 3 40. log2 1024

Solution 10 41. log 12

1 12

Solution –1 42. log 6

1 36

Solution –2

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1256


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

43. log 7 7

Solution

1 2 44. log 13 13

Solution

1 2 45. log 121 11

Solution

1 2 46. log 144 12

Solution

1 2 47. log 5 3 5

Solution

1 3 48. log 11 3 11

Solution

1 3 49. log8 2

Solution

1 3 50. log125 5

Solution

1 3

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1257


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

51. log6 1

Solution 0 52. log18 1

Solution 0 53. log 2 3

16 81

Solution 4 54. log 1 64 8

Solution –2 Solve each logarithmic equation for x. 55. log2 8  x

Solution log 2 8  x 2x  8 x3

56. log3 9  x

Solution log 3 9  x 3x  8 x2

57. log 4

1 x 64

Solution 1 x log 4 64 1 4x  64 x  3

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1258


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

58. log6 216  x

Solution log 6 216  x 6 x  216 x3

59. log 1/2

1 x 8

Solution 1 log 1/2  x 8 x  1 1    2 8   x3 60. log 1/3

1 x 81

Solution 1 log 1/3 x 81 x  1 1    3 81   x4 61. log9 3  x

Solution log 9 3  x 9x  3 1 x 2

62. log125 5  x

Solution log 125 5  x 125 x  5 1 x 3

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1259


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

63. log1/2 8  x

Solution log 1/2 8  x x

 1   8 2 x  3 64. log 1/2 16  x

Solution log 1/2 16  x x

 1    16 2 x  4 65. log8 x  2

Solution log 8 x  2 82  x 64  x

66. log7 x  0

Solution log 7 x  0 70  x 1 x

67. log7 x  1

Solution log 7 x  1 71  x 7x

68. log2 x  8

Solution log 2 x  8 28  x 256  x

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1260


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

1 2

69. log25 x 

Solution 1 2 251/2  x 5x

log 25 x 

70. log4 x 

1 2

Solution 1 2 41/2  x

log 4 x 

2x

71. log5 x  2

Solution

log 5 x  2 52  x 1 x 52 1 x 25 72. log3 x  4

Solution

log 3 x  4 34  x 1 x 34 1 x 81 73. log 36 x  

1 2

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1261


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution log 36 x   361/2  x 1 x 361/2 1 x 6

74. log 27 x  

1 2

1 3

Solution log 27 x   27 1/3  x 1 x 27 1/3 1 x 3

1 3

75. log x 53  3

Solution log x 53  3

x 3  53 x 5

76. log x 5  1

Solution log x 5  1 x1  5 x 5

77. log x

9 2 4

Solution

9 2 4 9 x2  4 3 x 2

log x

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1262


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

78. log x

3 1  3 2

Solution log x

3 1  3 2

3 3 2  3 2 1/2  x   3    3 1 x  9 3 x 1/2 

79. log x

1  3 64

Solution 1 log x  3 64 1 x 3  64 1 1  3 3 x 4 x4 80. log x

1  2 100

Solution log x

81. log x

1  2 100 1 x 2  100 1 1  2 2 x 10 x  10

9  2 4

Solution 9 log x  2 4 9 x 2  4

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1263


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 x    94  1

2

1

4 9 2 x 3

x2 

82. log x

3 1  3 2

Solution 3 1  3 2 3 x 1/2  3 2  3 2 1/2  x   3   

log x

2

 3  x     3 9 x 3 3 83. 2log 5  x 2

Solution From the definition: log 2 5

5

log 3 4

x

2 84. 3

Solution From the definition: log 3 4

4

log 4 6

6

3

85. x

Solution From the definition:

x

log 4 6

6 x 4

86. x

log 3 8

8

Solution From the definition:

x

log 3 8

8 x 3

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1264


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Use a calculator to find each value to four decimal places. 87. log 3.25

Solution log 3.25  0.5119 88. log 0.57

Solution log 0.57  0.2441 89. log 0.00467

Solution log 0.00467  2.3307 90. log 375.876

Solution log 375.876  2.5750 91. ln 45.7

Solution ln 45.7  3.8221

92. ln 0.005

Solution ln 0.005   5.2983

93. ln

2 3

Solution

ln

94. ln

2  0.4055 3

12 7

Solution

ln

12  0.5390 7

95. ln 35.15

Solution ln 35.15  3.5596

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1265


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

96. ln 0.675

Solution ln 0.675   0.3930

97. ln 7.896

Solution ln 7.896  2.0664

98. ln 0.00465

Solution ln 0.00465   5.3709

99. log In 1.7

Solution

log In 1.7   0.2752

100. ln log 9.8

Solution

ln log 9.8  0.0088

101. ln log 0.1

Solution

ln log 0.1 : undefined

102. log ln 0.01

Solution

log ln 0.01 : undefined

Use a calculator to find y to four decimal places, if possible. 103. log y  1.4023

Solution log y  1.4023

y  25.2522 104. log y  0.926

Solution log y  0.926

y  8.4333

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1266


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

105. log y  3.71

Solution log y  3.71

y  1.9498  104 106. log y  log 

Solution log y  log 

y  3.1416 107. ln y  1.4023

Solution ln y  1.4023

y  4.0645 108. ln y  2.6490

Solution ln y  2.6490

y  14.1399 109. ln y  4.24

Solution ln y  4.24

y  69.4079 110. ln y  0.926

Solution ln y  0.926

y  2.5244 111. ln y   3.71

Solution ln y  3.71

y  0.0245 112. ln y   0.28

Solution ln y  0.28

y  0.7558

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1267


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

113. log y  In 8

Solution log y  In 8

y  120.0719 114. ln y  log 7

Solution ln y  log 7

y  2.3282 Find each value without using a calculator. 115. log 10,000

Solution log 10, 000  x 10x  10, 000 x4 log 10, 000  4 116. log 1, 000, 000

Solution log 1, 000, 000  x 10 x  1, 000, 000 x 6 log 1, 000, 000  6 117. log 0.001

Solution log 0.001  x 10 x  0.001 x  3 log 0.001  3 118. log

1 100, 000

Solution 1 x log 100, 000 1 100, 000 x  5

10 x 

log

1  5 100, 000

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1268


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

119. eln 7

Solution From the definition:

eln 7  7 120. eln 9

Solution From the definition:

eln 9  9

 

121. ln e4

Solution From the definition:

 

ln e4  4

 

122. ln e6

Solution From the definition:

 

ln e6  6 Find the value of b, if any, that would cause the graph of f(x) = logbx to look like the graph shown. 123.

Solution The graph passes through the point

 b, 1   2, 1 .  b  2

124.

Solution The graph passes through the point

 b, 1   2 , 1 .  b  21 © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1269


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

125.

Solution y  log b x by  x 1 2 1 1  1 1 b   2 b2 b1 

  126.

Solution

y  log b x by  x b1  2

 b    2 1

1

b

1

1 2

Find the domain of each logarithmic function. Write the answer in interval notation. 127. f(x) = log2(x – 3)

Solution x 3 0

x3

 3,   128. f(x) = 2log3(x – 5)

Solution x 5  0

x 5

5,  

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1270


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

129. f(x) = 1 + log5(4 – x)

Solution 4x 0  x  4 x4

 , 4 

130. f(x) = –6 + log6(2x)

Solution 2x  0

x 0

0,   131. f(x) = ln(x – 1)

Solution x10

x 0

0,  

132. f(x) = 5 + ln(2 – x)

Solution 2 x  0  x  2 x2

 , 2

Evaluate the logarithmic functions at each of the given x-values. 133. f(x) = 2 + log2 x a. f(4) b. f(1) c.

 1 f  4

Solution a.

f  4   2  log 2  4   2  2  4

b.

f  1  2  log 2  1  2  0  2

c.

 1  1 f    2  log 2    2   2   0 4 4

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1271


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

134. f(x) = –2 + log3 x a. f(27) b. f(1) c.

 1 f  9

Solution a.

f  27   2  log 3  27   2  3  1

b.

f  1  2  log 3  1  2  0  2

c.

 1  1 f    2  log 3    2   2   4 9 9

135. f(x) = log2 (x – 1) a. f(2) b. f(33) c.

5 f  4

Solution a.

f  2  log 2  2  1  log 2  1  0

b.

f  33  log 2  33  1  log 2  32  5

c.

5 5   1 f    log 2   1   log 2    2 4 4  4

136. f(x) = log3 (3 – x) a. f(–6) b. f(2) c.

8 f  3

Solution

a.

f  6   log 3 3   6   log 3  9   2

b.

f  2  log 3  3  2  log 3  1  0

c.

8   1 8 f    log 3  3    log 3    0 3 3     3

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1272


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

For each logarithmic function graph shown, sate the domain, range, and equation of the vertical asymptote. 137.

Solution

domain: 4,  ; range: ,  ; x  4 138.

Solution

domain: 1,  ; range: ,  ; x  1 139.

Solution

domain: 2,  ; range: ,  ; x  2 140.

Solution

domain: 0,  ; range: ,  ; x  0

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1273


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Graph each function. 141. f ( x)  log3 x

Solution f ( x )  log 3 x

points:  1, 0  ,  3, 1

 

142. f x  log 4 x

Solution

f  x   log 4 x

points:  1, 0  ,  4, 1

 

143. f x  log 1/3 x

Solution

f  x   log 1/3 x

1  points:  1, 0  ,  , 1  3 

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1274


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

144. f x  log 1/4 x

Solution

f  x   log 1/4 x

1  points:  1, 0  ,  , 1  4 

 

145. f x   log 5 x

Solution

f  x    log 5 x

Reflect y  log 5 x about x

 

146. f x   log 2 x

Solution

f  x    log 2 x

Reflect y  log 2 x about x

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1275


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

147. f x  2  log 2 x

Solution

f  x   2  log 2 x

Shift y  log 2 x U2.

 

148. f x  3  log 3 x

Solution

f  x   3  log 3 x

Shift y  log 3 x D3.

 

149. f x  log 3 x  2

Solution

f  x   log 3  x  2 

Shift y  log 3 x L2.

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1276


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

150. f x  log 2 x  1

Solution

f  x   log 2  x  1

Shift y  log 2 x R1.

 

151. f x  3  log 3 x  1

Solution

f  x   3  log 3  x  1

Shift y  log 3 x U3, L1.

 

152. f x  3  log 3 x  1

Solution

f  x   3  log 3  x  1

Shift y  log 3 x D3, L1.

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1277


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

153. f x  3  ln x

Solution

f  x   3  ln x

Shift y  ln x D3.

 

154. f x  2  ln x

Solution

f  x   2  ln x

Shift y  ln x U2.

 

155. f x  ln x  4

Solution

f  x   ln  x  4 

Shift y  ln x R4.

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1278


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

156. f x  ln x  1

Solution

f  x   ln  x  1

Shift y  ln x L1.

 

157. f x  1  ln x

Solution

f  x   1  ln x

Reflect y  ln x about x , shift U1.

 

158. f x  2  ln x

Solution

f  x   2  ln x

Reflect y  ln x about x, shift U2.

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1279


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

159. f(x) = 2 + ln(x + 3)

Solution

160. f(x) = –1 + ln(x – 2)

Solution

Use a graphing calculator to graph each function.

 

 

161. f x  log 3 x

Solution

f  x   log  3 x 

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1280


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

x 162. f  x   log   3

Solution x f  x   log   3

 

 

163. f x  log  x

Solution

f  x   log   x 

 

164. f x   log x

Solution

f  x   log x

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1281


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

1  165. f  x   ln  x  2 

Solution 1  f  x   ln  x  2 

 

166. f x  ln x 2

Solution

f  x   ln x 2

 

 

167. f x  ln  x

Solution

f  x   ln   x 

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1282


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

 

168. f x  ln 3 x

Solution

f  x   ln  3 x 

Fix It In exercises 169 and 170, identify the step the first error is made and fix it. 169. Solve the logarithmic equation log x

4  2 for x. To begin, write the equation in 25

exponential form.

Solution Step 5 was incorrect. Step 1: x 2 

Step 2:

 25

1   2 25 x

Step 3: 25  4x 2 2 Step 4: x 

Step 5: x 

5 4

5 2

170. Graph f(x) = log(x + 6) and state it’s domain, range, and vertical asymptote.

Solution Step 4 was incorrect. Step 4: vertical asymptote is x  6

Discovery and Writing 171. Describe how to convert an equation in logarithmic form to an equation in exponential form.

Solution Answers may vary.

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1283


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

172. Describe how to convert an equation in exponential form to an equation in logarithmic form.

Solution Answers may vary. 173. Explain how you would evaluate the logarithmic expression log2256.

Solution Answers may vary. 174. What are the differences between common logarithms and natural logarithms?

Solution Answers may vary. 175. How do you find the domain of a logarithmic function?

Solution Answers may vary. 176. What strategy would you use to graph f(x) = log5x?

Solution Answers may vary. 177. Consider the following graphs. Which is larger, a or b, and why?

Solution Answers may vary. 178. Consider the following graphs. Which is larger, a or b, and why?

Solution Answers may vary.

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1284


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

179. Choose two numbers and add their common logarithms. Then find the common logarithm of the product of those two numbers. What do you observe? Does it work for three numbers?

Solution Answers may vary. 180. Choose two numbers and subtract their common logarithms. Then find the common logarithm of the quotient of those two numbers. What do you observe?

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 181. log97,214 1  0

Solution True. 182. If loga b  88, then logb a  88.

Solution False. log b a 

1 . 88

183. log313 0 is undefined.

Solution True.

184. log 10 1, 000,000  6

Solution False. It is undefined. 185. log 2

1  10 1024

Solution False. log 2

1  10. 1024

186. log 12 12369  369

Solution True.

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1285


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions log 99 77

187. 99

 77

Solution True. 188. log 5 19 5  19

Solution False. log 5 19 5 

1 . 19

189. log 2468 ln e2468  1

Solution True.

190. The domain of f  x   log 7 x 2  4 is  0,   .

Solution

 

False. The domain is ,  2  2,  .

EXERCISES 5.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Evaluate the logarithmic expression 20log xy for x = 30 and y = 0.5. Round your answer to the nearest whole number.

Solution  30  20log    20log  60   36  0.5 

2. Evaluate the logarithmic expression log xy for x = 4500 and y = 0.06. Round your answer to the nearest tenth.

Solution  4500  log    4.9  0.06 

3. Evaluate the logarithmic expression  x1 ln 1  zy

 for x = 0.02, y = 60, and z = 70. Round

your answer to the nearest whole number.

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1286


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution  1    1 60    ln  1    50 ln    97 70   0.02   7

4. Evaluate the logarithmic functions f  t  

ln 2 t

at t = 0.03. Round your answer to the

nearest whole number.

Solution f  0.03  

ln 2 0.03

 23

Vocabulary and concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 5. dB gain = __________

Solution E 20 log O EI 6. The intensity of an earthquake is measured by the formula R = __________.

Solution

log

A P

7. The formula for charging batteries is __________.

Solution t

1  C ln  1   k  M

8. If a population grows exponentially at a rate r, the time it will take for the population to double is given by the formula t = __________.

Solution

ln 2 r 9. The formula for isothermal expansion is __________.

Solution V  E  RT ln  f   Vi 

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1287


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

10. The logarithm of a negative number is __________.

Solution undefined Applications Use a calculator to solve each problem. 11. Gain of an amplifier An amplifier produces an output of 17 volts when the input signal is 0.03 volt. Find the decibel voltage gain. Round to the nearest decibel.

Solution

dB gain  20 log

E0

EI 17  20 log 0.03  55 dB

12. Transmission lines A 4.9-volt input to a long transmission line decreases to 4.7 volts at the other end. Find the decibel voltage loss. Round to two decimals.

Solution

dB gain  20 log

E0

EI 4.7  20 log 4.9  0.36 dB gain  0.36 dB loss

13. Gain of an amplifier Find the dB gain of an amplifier whose input voltage is 0.71 volt and whose output voltage is 20 volts. Round to the nearest decibel.

Solution

dB gain  20 log

E0

EI 20  20 log 0.71  29 dB

14. Gain of an amplifier Find the dB gain of an amplifier whose output voltage is 2.8 volts and whose input voltage is 0.05 volt.

Solution

dB gain  20 log

E0

EI 2.8  20 log 0.05  35 dB

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1288


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

15. dB gain Find the dB gain of the amplifier shown below. Round to one decimal.

Solution

dB gain  20 log

E0

EI 30  20 log 0.1  49.5 dB

16. dB gain Find the dB gain of the amplifier shown below. Round to one decimal.

Solution

dB gain  20 log

E0

EI 80  20 log 0.12  56.5 dB

17. Earthquakes An earthquake has an amplitude of 5000 micrometers and a period of 0.2 second. Find its measure on the Richter scale. Round to one decimal.

Solution A P 5000  log 0.2  4.4

R  log

18. Earthquakes An earthquake has an amplitude of 8000 micrometers and a period of 0.008 second. Find its measure on the Richter scale. Round to the nearest whole number.

Solution A P 8000  log 0.008 6

R  log

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1289


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

19. Earthquakes An earthquake with a period of 41 second has an amplitude of 2500 micrometers. Find its measure on the Richter scale. Round to the nearest whole number.

Solution R  log  log

A P 2500 1 4

4

20. Earthquakes An earthquake has a period of 21 second and an amplitude of 5 cm. Find its measure on the Richter scale. (Hint: 1 cm = 10,000 micrometers) Round to the nearest whole number.

Solution R  log  log

A P 50, 000 1 2

5

21. Earthquakes An earthquake measuring between 3.5 and 5.4 on the Richter scale is often felt but rarely causes damage. Suppose an earthquake in Northern California has an amplitude of 6000 micrometers and a period of 0.3 second. Is it likely to cause damage?

Solution A P 6000  log 0.3  4.3  no damage

R  log

22. Earthquakes An earthquake measuring between 7 and 7.9 on the Richter scale is a major earthquake and can cause serious damage over larger areas. Suppose an earthquake in Chile has an amplitude of 198.5 cm and a period of 0.1 second. Would it cause serious damage over large areas? (Hint: 1 cm = 10,000 micrometers)

Solution A P 1985000  log 0.1  7.3  damage

R  log

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1290


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

23. Battery charge If k = 0.116, how long will it take a battery to reach a 90% charge? Assume that the battery was fully discharged when it began charging. Round to one decimal place.

Solution

C 1  t   ln  1   k  M  1 0.9M  t ln  1   M  0.116  1 t ln  1  0.9  19.8 minutes 0.116 24. Battery charge If k = 0.201, how long will it take a battery to reach a 40% charge? Assume that the battery was fully discharged when it began charging. Round to one decimal place.

Solution

C 1  ln  1   k  M  1 0.4M  t ln  1   0.201  M  1 t ln  1  0.4   2.5 minutes 0.201

t

25. Population growth A town’s population grows at the rate of 12% per year. If this growth rate remains constant, how long will it take the population to double? Round to one decimal place.

Solution ln 2 ln 2 t   5.8 years 0.12 r 26. Fish population growth One thousand bass were stocked in Catfish Lake in Eagle River, Wisconsin, a lake with no bass population. If the population of bass is expected to grow at a rate of 25% per year, how long will it take the population to double? Round to one decimal place.

Solution ln 2 ln 2 t    2.8 years 0.25 r 27. Population growth A population growing at an annual rate r will triple in a time t given by the formula t =

ln 3 r

. How long will it take the population of the town in Exercise 25

to triple? Round to one decimal place.

Solution ln 3 ln 3 t   9.2 years 0.12 r

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1291


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

28. Fish population growth How long would it take the fish population in Exercise 26 to triple? Round to one decimal place.

Solution ln 3 ln 3 t   4.4 years r 0.25 29. Isothermal expansion One mole of gas expands isothermically to triple its volume. If the gas temperature is 400 K, what energy is absorbed? Round to the nearest joule.

Solution V  E  RT ln  f   Vi   3V  E   8.314  400  ln  i   Vi  E   8.314  400  ln  3  E  3654 joules

30. Isothermal expansion One mole of gas expands isothermically to double its volume. If the gas temperature is 300K, what energy is absorbed? Round to the nearest joule.

Solution V  E  RT ln  f   Vi   2V  E   8.314  300  ln  i   Vi  E   8.314  300  ln  2  E  1729 joules

If an investment is growing continuously for t years, its annual growth rate r is given by the following formula, where P is the current value and P0 is the amount originally invested. 1 P r  ln t P0

31. Investing A company grew continuously from 2013 to 2023. A $10,000 investment in the stock in 2013 would be worth $100,000 in 2023. Find the company’s average annual growth rate during this period. Round to the nearest percent.

Solution 1 P r  ln t P0

1 100,000 ln 10 10,000  0.23  about 23% per year 

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1292


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

32. Investing A company has grown continuously from 1988 to 2023. A $10,000 investment in the stock in 1988 would be worth $2,500,000 in 2023. Find the company’s average annual growth rate during this period. Round to the nearest percent.

Solution 1 P r  ln t P0

1 2, 500, 000 ln 35 10,000  0.16  16% per year 

33. Depreciation In business, equipment is often depreciated using the double decliningbalance method. In this method, a piece of equipment with a life expectancy of N years, costing $C, will depreciate to a value of $V in n years, where n is given by the following formula.

n

log V  log C  2 log  1   N  

If a computer that cost $37,000 has a life expectancy of 5 years and has depreciated to a value of $8000, how old is it?

Solution log V  log C n log  1  N2 

log 8000  log 37,000 log  1  23 

 3 year old 34. Depreciation A word processor worth $470 when new had a life expectancy of 12 years. If it is now worth $189, how old is it? (See Exercise 33.) Round to the nearest year.

Solution log V  log C n log  1  N2 

log 189  log 470 log  1  122 

 5 year old 35. Annuities If $P is invested at the end of each year in an annuity earning interest at an annual rate r, the amount in the account will be $A after n years, where  Ar  log   1 P  n log  1  r 

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1293


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

If $1000 is invested each year in an annuity earning 12% annual interest, when will the account be worth $20,000? Round to one decimal place.

Solution n

log  ArP  1 log  1  r 

log  

20000  0.12  1000

 1 

log  1  0.12 

 10.8 years 36. Annuities If $5000 is invested each year in an annuity earning 8% annual interest, when will the account be worth $50,000? (See Exercise 35.) Round to one decimal place.

Solution n

log  ArP  1 log  1  r 

log  

50000 0.08 5000

 1 

log  1  0.08

 7.6 years 37. Breakdown voltage The coaxial power cable shown has a central wire with radius R1 = 0.25 centimeter. It is insulated from a surrounding shield with inside radius R2 = 2 centimeters. The maximum voltage the cable can withstand is called the breakdown voltage V of the insulation. V is given by the formula V  ER1 ln

R2 R1

where E is the dielectric strength of the insulation. If E = 400,000 volts/centimeter, find V. Round to the nearest volt.

Solution V  ER1 ln

R2 R1

 2  V   400, 000  0.25  ln    0.25  V  208, 000 V

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1294


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

38. Breakdown voltage In Exercise 37, if the inside diameter of the shield were doubled, what voltage could the cable withstand? Round to the nearest volt.

Solution V  ER1 ln

R2 R1

 4  V   400, 000  0.25  ln    0.25  V  277, 000 V

39. Suppose you graph the function f(x) = ln x on a coordinate grid with a unit distance of 1 centimeter on the x- and y-axes. How far out must you go on the x-axis so that f(x) = 12? Give your result to the nearest mile.

Solution

f  x   12  ln x  12  x  162755  You would go out 162, 755 cm. 162,755 cm 

162, 755 cm 1

1 in 1 ft 1 mi    1 mile 2.54 cm 12 in 5280 ft

40. Suppose you graph the function f(x) = log x on a coordinate grid with a unit distance of 1 centimeter on the x- and y-axes. How far out must you go on the x-axis so that f(x) = 12? Give the result to the nearest mile. Why is this result so much larger than the result in Exercise 39?

Solution

f  x   12  log x  12  x  1012 cm  You would go out 1012 cm. 1012 cm 

1012 cm

1 in

1 ft

1 mi

 6, 214,000 mile 1 2.54 cm 12 in 5280 ft It is begger because In has a base of e  2.72 while log has a base of 10.

Discovery and Writing 41. Describe how to determine the intensity of an earthquake.

Solution Answers may vary. 42. Explain how to determine the doubling time of a population.

Solution Answers may vary. 43. One form of the logistic function is given by the following function. Explain how you would find the y-intercept of its graph.

f x 

1 1  e2 x

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1295


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution Set x  0 1 y  2 0  1e 1  1  e0 1    1 1 2

 1 y -intercept:  0,   2

44. Graph the function f(x) = ln |x|. Explain why the graph looks the way it does.

Solution Explanations may vary.

EXERCISES 5.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Simplify each expression. a.

xm  xn

b.

xm xn

c.

x  m

n

Solution a.

x m n

b.

x m n

c.

x mn

2. Simplify. a.

x 3  x 5  x 2

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1296


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

x4  x5 x2

b.

Solution

 3  5   2 

a.

x 3  x 5  x 2  x

b.

x4  x5 x9  2  x7 x2 x

y  y  3

5

3. Simplify.

 x 4 

1 x4

4

Solution

y  y   y y 3

5

4

20

12

 y8

4. Simplify each. What do you observe about the two answers? a.

log 3  27  3

b.

log3 27  log3 3

Solution a. 4 b. 4

They are equal.

5. Simplify each expression. What do you observe about the two answers? a.

 125  log 5    25 

b.

log5 125  log5 25

Solution a. 1 b. 1

They are equal.

6. Simplify each expression. What do you observe about the two answers? a.

log 2 43

b.

3log2 4

Solution a. 6 b. 6

They are equal.

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1297


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7.

logb 1  _____ Solution 0

8.

logb b  _____ Solution 1

9.

logb MN  logb _____  log b _____ Solution M, N log b x

10. b

 _____

Solution x 11. If logb x  logb y , then _____ = _____.

Solution x, y 12. log b

M  log b M _____ logb N N

Solution – 13. log b x p  p  log b _____

Solution x 14. log b bx  _____

Solution x

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1298


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

15. log b A  B _____ logb A  logb B

Solution ≠ 16. logb A  logb B _____ logb AB

Solution = Simplify each expression. 17. log4 1  _____

Solution

log4 1  0 18. log4 4  _____

Solution

log4 4  1 19. log 4 47  _____

Solution

log 4 47  7log 4 4  7 20. 4log 4 8  _____

Solution log 4 8

8

log 5 10

 _____

4

21. 5

Solution log 5 10

5

 10

22. log5 52  _____

Solution

log 5 52  2log 5 5  2 23. log5 5  _____

Solution

log5 5  1

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1299


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

24. log5 1  _____

Solution

log5 1  0 Practice Use a calculator to verify each equation. 25. log   3.7  2.9    log 3.7  log 2.9

Solution Answers may vary. 26. ln

9.3  ln 9.3  ln 2.1 2.1

Solution Answers may vary. 27. ln  3.7   3 ln 3.7 3

Solution Answers may vary. 28. log 14.1 

1 log 14.1 2

Solution Answers may vary. 29. log 3.2 

ln 3.2 ln 10

Solution Answers may vary. 30. ln 9.7 

log 9.7 log e

Solution Answers may vary. Assume that x, y, and z, are positive numbers. Use the properties of logarithms to write each expression in terms of the logarithms of x, y, and z. 31. log2 2xy

Solution

log2 2xy  log2 2  log2 x  log2 y

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1300


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

32. log2 3xz

Solution

log2 3xz  log2 3  log2 x  log2 z 33. log 3

2x y

Solution 2x  log 3 2 x  log 3 y log 3 y  log 3 2  log 3 x  log 3 y 34. log 3

x yz

Solution x log 3  log 3 x  log 3 yz yz  log 3 x  log 3 y  log 3 z   log 3 x  log 3 y  log 3 z

35. log 4 x 2 y 3

Solution log 4 x 2 y 3  log 4 x 2  log 4 y 3  2log 4 x  3log 4 y

36. log 4 x 3 y 2 z

Solution log 4 x 3 y 2 z  log 4 x 3  log 4 y 2  log 4 z  3log 4 x  2log 4 y  log 4 z

37. log 5  xy 

1/3

Solution log 5  xy 

1/3

1 log 5 xy 3 1  log 5 x  log 5 y  3 

38. log 5 x 1/2 y 3

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1301


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution log 5 x 1/2 y 3  log 5 x 1/2  log 5 y 3 1  log 5 x  3log 5 y 2

39. log 6 x z

Solution log 6 x z  log 6 xz 1/2  log 6 x  log 6 z 1/2  log 6 x  40. log 6

1 log 6 z 2

xy

Solution

log 6 xy  log 6  xy 

1/2

1 log 6 xy 2 1  log 6 x  log 6 y  2 

41. log 10

3

x

3

yz

Solution

log 10

3

x

3

yz

 log 10 3 x  log 10 3 yz  log 10 x 1/3  log 10  yz 

1/3

1 1 log 10 x  log 10 yz 3 3

1 1 log 10 x  log 10 y  log 10 z  3 3 1 1 1  log 10 x  log 10 y  log 10 z 3 3 3 

42. log 10 4

x3 y 2 z4

Solution log 10

4

 x3 y 2  x3 y 2  log 10   z 4  z4  

1/4

x3 y 2 1 1 log 10  log 10 x 3 y 2  log 10 z 4 4 4 4 z

 

1 log 10 x 3  log 10 y 2  log 10 z 4 4 1   3log 10 x  2log 10 y  4log 10 z  4 1 1  log 10 x  log 10 y  log 10 z 4 2 

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1302


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

43. ln x 7 y 8

Solution ln x 7 y 8  ln x 7  ln y 8  7 ln x  8 ln y

44. ln

4x y

Solution 4x  ln 4 x  ln y ln y  ln 4  ln x  ln y 45. ln

x y 4z

Solution x ln 4  ln x  ln y 4 z y z

 ln x  ln y 4  ln z

 ln x   4 ln y  ln z   ln x  4 ln y  ln z 46. ln x y

Solution ln x y  ln xy 1/2  ln x  ln y 1/2  ln x 

1 ln y 2

Assume that x, y, and z, are positive numbers. Use the properties of logarithms to write each expression as the logarithm of one quantity.

47. log 7 x  1  log 7 x

Solution

log 7  x  1  log7 x  log7

x1 x

48. log 7 x  log 7 x  2  log 7 8

Solution log 7 x  log 7  x  2   log 7 8  log 7 x  x  2   log 7 8  log 7

x  x  2 8

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1303


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

49. 2log 8 x 

1 log 8 y 3

Solution

2log 8 x 

1 log 8 y  log 8 x 2  log 8 y 1/3  log 8 x 2 y 1/3  log 8 x 2 3 y 3

50. 2log8 x  3log8 y  log8 z

Solution

2log8 x  3log8 y  log 8 z  log 8 x 2  log 8 y 3  log8 z  log8 x 2 y 3 z  log 8

51. 3log9 x  2log9 y 

z x y3 2

1 log9 z 2

Solution 3log 9 x  2log 9 y 

z 1 log 9 z  log 9 x 3  log 9 y 2  log 9 z 1/2  log 9 x 3 y 2 z  log 9 3 2 2 x y

52. 3 log 9 x  1  2 log 9 x  2  log 9 x

Solution 3 log 9  x  1  2 log 9  x  2   log 9 x  log 9  x  1  log 9  x  2  3

2

 log 9 x  log 9

x  x  1

 x  2

3

2

x  y  53. log 10   x   log 10   y  z  z 

Solution

x z  zx  x  x x  y  x  xz x log 10   z   log 10   y   log 10 yz  log 10  log 10  log 10 y z z y yz y  y  z zy     z

54. log 10 xy  y 2  log 10  xz  yz   log 10 z

Solution

log 10 xy  y

2

  log  10

 xy  y  z  log  x  y  yz  log y xz  yz  log z  log 

2

10

10

xz  yz

10

z x  y

10

55. ln x  ln x  5  ln 9

Solution ln x  ln  x  5   ln 9  ln x  x  5   ln 9  ln

x  x  5 9

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1304


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

56. 5 ln x 

1 ln y 5

Solution

5 ln x 

1 ln y  ln x 5  ln y 1/5  ln x 5 5 y 5

57.  6 ln x  2 ln y  ln z

Solution

6 ln x  2 ln y  ln z  ln y 6  ln y 2  ln z  ln x 6 y 2 z  ln

58. 2 ln x  3 ln y 

z x y2 6

1 ln z 3

Solution 2 ln x  3 ln y 

3 z 1 ln z  ln x 2  ln y 3  ln z 1/3  ln x  2 y 3 z 1/3  ln 2 3 3 x y

Determine whether each statement is true or false. 59. logb ab  logb a  1

Solution

logb ab  logb a  logb b  logb a  1 60. log b

TRUE

1   log b a a

Solution

log b

1  log b a1   log b a a

TRUE

61. logb 0  1

Solution

logb 0 is undefined.

FALSE

62. logb 2  log2 b

Solution

logb 2  log2 b FALSE

(except for b = 2)

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1305


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

63. log b x  y  log b x  log b y

Solution

log b  x  y   log b x  log b y , so log b  x  y   log b x  log b y

TRUE unless x  y  x  y 



64. log b xy  log b x log b y

Solution

log b  x  y   log b x  log b y , so log b  xy   log b x log b y  FALSE

65. If loga b  c, then logb a  c.

Solution If log a b  c, then log b a  c FALSE

66. If log a b  c, then log b a 

1 . c

Solution

If log a b  c, then log b a  log a b  c  ac  b

a  c

1/ c

1 c

 b1/c

a  b1/c  log b a  TRUE

1 c

67. log 7 77  7

Solution log 7 77  7  7 7  7 7 TRUE

68. 7log 7 7  7

Solution 7

log 7 7

7 TRUE

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1306


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

69. log b  x   log b x

Solution  log b x  log b x 1  log b FALSE

1 x

70. If log b a  c, then log b a p  pc.

Solution log b a p  p log b a  pc TRUE

71.

log b A log b B

 log b A  log b B

Solution  A log b    log b A  log b B, so B log b A  log b A  log b B log b B FALSE 72. log b  A  B  

Solution

log b  A  B  

log b A log b B

log b  A log b  B 

FALSE 73. log b

1   log b 5 5

Solution 1 log b  log b 51   log b 5 5 TRUE 74. 3 log b 3 a  log b a

Solution 1 3 log b 3 a  3log b a 1/3  3  log b a 3  log b a TRUE

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1307


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

75.

1 log b a3  log b a 3 Solution 1 1 log b a3   log b a  log b a 3 3 TRUE

76. log4/3 y   log3/4 y

Solution Let log 4/3 y  c.

Then  43   y . c

    y    y  log c

1

3 4

3 4

c

3/4

y  c.

Thus, log 4/3 y  c   log 3/4 y .  TRUE

77. logb y  log1/b y  0

Solution Let log 1/ b y  c.

Then  b1   y . c

    y 1

c

 b

c

b 1

 y  log b y  c.

log 1/ b y  log b y  c   c   0. TRUE

log 10 3

78. log 10 103  3 10

Solution log 10 103  3

  3  3  9 log 10  3  10  log 10 3

3 10

log 10 3

3

10

FALSE



79. ln xy  ln x ln y

Solution

ln  xy   ln x  ln y , so ln  xy   ln x ln y  FALSE

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1308


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

80.

ln A ln B

 ln A  ln B

Solution

A , so B ln A

ln A  ln B  ln ln A  ln B 

ln B

FALSE 81.

1 ln a5  ln a 5 Solution 1 1 ln a5   ln a  ln a 5 5 TRUE

82. ln y  ln

1 y

Solution 1 ln  ln y 1    1 ln y  ln y y TRUE Given that log10 4 0.6021, log10 7 0.8451, and log10 9 0.9542, use these values and the properties of logarithms to approximate each value. Do not use a calculator. 83. log10 28

Solution

log 10 28  log 10  4  7   log 10 4  log 10 7  0.6021  0.8451  1.4472

84. log 10

7 4

Solution 7 log 10  log 10 7  log 10 4 4  0.8451  0.6021  0.2430 85. log 10 2.25

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1309


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution 9 log 10 2.25  log 10   4  log 10 9  log 10 4  0.9542  0.6021  0.3521

86. log10 36

Solution

log 10 36  log 10  4  9   log 10 4  log 10 9  0.6021  0.9542  1.5563

87. log 10

63 4

Solution 63 log 10  log 10 63  log 10 4 4  log 10  7  9  log 10 4

 log 10 7  log 10 9  log 10 4  0.8451  0.9542  0.6021  1.1972 88. log 10

4 63

Solution 4 log 10  log 10 4  log 10 63 63  log 10 4  log 10  7  9

 log 10 4  log 10 7  log 10 9  0.6021  0.8451  0.9542  1.1972 89. log 10 252

Solution

log 10 252  log 10  4.63

 log 10  4  7  9  log 10 4  log 10 7  log 10 9  0.6021  0.8451  0.9542  2.4014

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1310


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

90. log 10 49

Solution

log 10 49  log 10 72  2log 10 7

 2  0.8451  1.6902

91. log 10 112

Solution

log 10 112  log 10  4  28 

 log 10  4  7  4   log 10 4  log 10 7  log 10 4  0.6021  0.8451  0.6021  2.0493

92. log 10 324

Solution

log 10 324  log 10  4  81

 log 10  4  9  9  log 10 4  log 10 9  log 10 9  0.6021  0.9542  0.9542  2.5105

93. log 10

144 49

Solution  144  log 10    log 10 144  log 10 49  49 

 log 10  4  4  9   log 10  7  7   log 10 4  log 10 4  log 10 9  log 10 7  log 10 7  0.6021  0.6021  0.9542  0.8451  0.8451  0.4682

94. log 10

324 63

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1311


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution  324  log 10    log 10 324  log 10 63  63 

 log 10  4  9  9   log 10  7  9   log 10 4  log 10 9  log 10 9  log 10 7  log 10 9  0.6021  0.9542  0.9542  0.8451  0.9542  0.7112

Use a calculator and the Change-of-Base Formula to find each logarithm. Round to four decimal places. 95. log3 7

Solution log 10 7 log 3 7   1.7712 log 10 3 96. log7 3

Solution log 10 3 log 7 3   0.5646 log 10 7 97. log 3

Solution log 10 3  0.9597 log  3  log 10  98. log3 

Solution log 10   1.0420 log 3   log 10 3 99. log3 8

Solution log 10 8 log 3 8   1.8928 log 10 3 100. log5 10

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1312


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution log e 10

log 5 10 

log e 5

 1.4307

101. log 2 5

Solution log 2 5 

log e 5 log e 2

 2.3219

102. log e

Solution log e e  0.8736 log  e  log e  Fix It In exercises 103 and 104, identify the step the first error is made and fix it. 103. Use the properties of logarithms to write log 5

x y z

as sum or difference of logarithms.

Expand as much as possible.

Solution Step 4 was incorrect. 1

 x  y 2 Step 1: log 5    z  Step 2:

1 xy log5 2 z

Step 3:

1 log  x  y   log5  z    2 5

Step 4:

1 1 log5  x  y   log5  z  2 2

104. Write 5ln x – 2 ln 3 

1 ln y as a single logarithm. 3

Solution Step 3 was incorrect. 1

Step 1: ln x 5  ln 32  ln y 3 Step 2: ln x 5  ln 9  ln 3 y

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1313


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

x5  ln 3 y 9

Step 3: ln

 x5   Step 4: ln   93 y   

Applications 105. pH of water slide The water in the Abyss, a water slide at the Atlantis Resort in the Bahamas, has a hydrogen ion concentration of 6.3 10–8 gram-ions per liter. Find the pH. Round to two decimal places.

Solution

pH   log H+ 

  log 6.3  108

 7.20 106. pH of swimming pool The ideal pH for a swimming pool is 7.2, the same pH as our eyes. The swimming pool at the local YMCA has a hydrogen ion concentration of 1.6 10–7 gram-ions per liter. Find the pH of the pool. Round to two decimal places. Is this ideal?

Solution

pH   log H+ 

  log 1.6  107

 6.80  not ideal 107. pH of a solution Find the pH of a solution with a hydrogen ion concentration of 1.7 10–5 gram-ions per liter. Round to two decimal places.

Solution

pH   log H+ 

  log 1.7  105

 4.77 108. pH of calcium hydroxide Find the hydrogen ion concentration of a saturated solution of calcium hydroxide whose pH is 13.2.

Solution pH   log H+  13.2   log H+  13.2  log H+  6.3  1014  H+ 

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1314


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

109. pH of apples The pH of apples can range from 2.9 to 3.3. Find the range in the hydrogen ion concentration.

Solution pH   log H+ 

pH   log H+ 

2.9   log H+ 

3.3   log H+ 

2.9  log H+ 

 3.3   log H+ 

1.26  103  H+ 

5.01  104  H+ 

The hydrogen ion concentration can range from 5.01

10–4 to 1.26

10–3.

110. pH of sour pickles The hydrogen ion concentration of sour pickles is 6.31 the pH. Round to one decimal place.

10–4. Find

Solution

pH   log H+ 

  log 6.31  104

 3.2 111. dB gain An amplifier produces a 40-watt output with a 21 -watt input. Find the dB gain. Round to the nearest decibel.

Solution P dB gain  10log O PI 40  10log 1 2

 19 dB 112. dB loss Losses in a long telephone line reduce a 12-watt input signal to an output of 3 watts. Find the dB gain. (Because it is a loss, the “gain” will be negative.) Round to the nearest decibel.

Solution dB gain  10log

PO

PI 3  10log 12  6 dB

113. Weber–Fechner Law What increase in intensity is necessary to quadruple the loudness?

Solution L  k ln I

4L  4k ln I  k  4ln I  k ln I 4 The original intensity must be raised to the fourth power.

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1315


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

114. Weber–Fechner Law What decrease in intensity is necessary to make a sound half as loud?

Solution L  k ln I 1 1 1 L  k ln I  k  ln I  k ln I 1/2 2 2 2 The original intensity must be raised to the one-half power (take the square root). 115. Isothermal expansion If a certain amount E of energy is added to one mole of a gas, it expands from an initial volume of 1 liter to a final volume V without changing its temperature according to the formula E  8300 ln V

Find the volume if twice that energy is added to the gas.

Solution E  8300 ln V

2E  2  8300 ln V  8300 ln V 2 The original volume is squared. 116. Richter scale By what factor must the amplitude of an earthquake change to increase its severity by 1 point on the Richter scale? Assume that the period remains constant. The Richter scale is given by

R  log

A P

where A is the amplitude and P the period of the tremor.

Solution R  log

A  A A A 10 A ; R  1  log  1  log  log 10  log   10   log P P P P P 

The amplitude must be multiplied by a factor of 10.

Discovery and Writing 117. Explain the Product Rule of logarithms and give an example.

Solution Answers may vary. 118. Explain the Quotient Rule of logarithms and give an example.

Solution Answers may vary. 119. Explain the Power Rule of logarithms and give an example.

Solution Answers may vary. 120. Explain the Change-of-Base Formula and give an example. © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1316


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution Answers may vary. 4log3 2

121. Simplify: 3

1 log 25 5

 52

Solution 4log3 2

3

1 log 25 5

 52

log3 24

3

log 251/2

5 5

 24  251/2  16  5  21

122. Find the value of a – b:

5log x 

1 1 5 log y  log x  log y  log x a y b 3 2 6

Solution 1 1 5 log y  log x  log y  log x 5  log y 1/3  log x 1/2  log y 5/6 3 2 6  log x 5 y 1/3 x 1/2 y 5/6  log x 9/2 y 1/2

5log x 

a  92 , b   21  a  b  92    21   102  5 123. Prove Property 6 of logarithms:

log b

M  log b M  log b N N

Solution Let log b M  x and log b N  y . Then b x  M and b y  N.

M b2   bx  y . N by M So log b  x  y , or N M log b  log b M  log b N. N 124. Show that  logb x  log1/b x.

Solution

 log b x  Q log b x 1  Q

bQ  x  1

b    x  Q

1

1

1

b Q  x

b   x 1

Q

  x 1 b

Q

log 1/ b x  Q Thus,  log b x  Q  log 1/ b x.

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1317


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

125. Show that e x ln a  a x .

Solution x ln a

e

ln ax

e

 ax

126. Show that eln x  x.

Solution

ln x  M  eM  x , by definition. ln x

Thus, e

 eM  x .

 

127. Show that ln e x  x .

Solution

 

ln e x  xln e  x  1  x

128. If logb 3x  1  logb x, find b.

Solution log b 3 x  1  log b x log b 3  log b x  1  log b x log b 3  1 b1  3  b  3 129. Explain why ln(log 0.9) is undefined.

Solution

log 0.9   0, so ln log 0.9  is undefined.

130. Explain why logb(ln 1) is undefined.

Solution

ln 1  0, so log b ln 1 is undefined.

In Exercises 131–132, A and B both are negative. Thus, AB and BA are positive, and log AB and log BA are defined. 131. Is it still true that log AB = log A + log B? Explain.

Solution Answers may vary. 132. Is it still true that log BA  log A  log B ? Explain.

Solution Answers may vary.

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1318


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Critical Thinking In Exercises 133–140, match the logarithmic expression on the left with an equivalent logarithmic expression on the right. Assume x, y, z, and w are positive numbers.

133. log500 

w 100 x 200 z 300 y 400

a.

134. log 500 

w 100 x 200 y 400 z 300

b.

100 log 500 w  200 log 500 x  300 log 500 z  400 log 500 y 100 log 500 w  200 log 500 x  300 log 500 z  400 log 500 y

135. log 200 x 100 y

c.

1 1 log 200 x  log 200 y 100 100

136. log 200 100 xy

d.

log 200 x 

137. log 300 x 100 y 200

e.

200 log300 x  100 log300 y

138. log 300 x 200 y 100

f.

100 log300 x  200 log300 y

139.  log 300 x 100 y 200

g.

log 300 x 100  log 300 y 200

h.

log 300

140.  log 300

1

x

100

y

200

1

x

100

1 log 200 y 100

 log 300

1

y

200

Solution 133. log 500

w 100 x 200 z 300  log 500 w 100 x 200 z 300  log 500 y 400 400 y 100

200

300

500

400

500

500

 100 log 500 w  200 log 500 x  300 log 500 z  400 log 500 y  134. log 500

. b

 log 500

    w   log  x   log  z   log  y 

w 100 x 200  log 500 w 100 x 200  log 500 y 400 z 300 y 400 z 300 100

200

400

500

500

300

500

 100 log 500 w  200 log 500 x  400 log 500 y  300 log 500 z 

135. log 200 x 100 y  log 200 xy 1/100  log 200 x  log 200 y 1/100  log 200 x 

. a

 log 500

    w   log  x   log  y   log  z 

1 log 200 y  d. 100

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1319


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

136. log 200 100 xy  log 200  xy 

1/100

1 1 1 log 200  xy   log 200 x  log 200 y  c. 100 100 100

  log 1

 1 1  1 1  100  200   log 300 100  log 300 200  y  x y x

  1 140.  log 300 100 200  log 300  100 200   log 300 x 100 y 200  log 300 x 100  log 300 y 200  x y x y  1

. g

1

300

. h

139.  log 300 x 100 y 200  log 300 x 100 y 200

. e

138. log 300 x 200 y 100  log 300 x 200  log 300 y 100  200 log 300 x  100 log 300 y 

. f

137. log 300 x 100 y 200  log 300 x 100  log 300 y 200  100 log 300 x  200 log 300 y 

EXERCISES 5.6 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Solve each equation. a. 4(x + 2) = –x – 7 b. 6x2 = 13x + 5

Solution a.

4  x  2   x  7 4x  8  x  7 5 x  15 x  3 6 x 2  13 x  5

b.

6 x 2  13 x  5  0

 3x  1 2x  5  0 3x  1  0 1 x 3

2x  5  0 5 x 2

2. Solve each equation. a.

42 6 x2

b.

2x 15 x x 2 2

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1320


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution 42 a. 6 x2 42  6  x  2  42  6 x  12 30  6 x 5x

b.

2x 15 x x 2 2  2x 15  2  x  2  2  x  2  x   x 2 2 

4 x  2 x  x  2   15  x  2  4 x  2 x 2  4 x  15 x  30 0   2 x  3  x  10 

2x  3  0

x  10  0

3 2

x  10

x

3. Fill in the boxes. a.

5? 

1 125

b.

e? 

1 e8

Solution a. −3 b. −8 4. a. Use the Power Rule to rewrite: log10 12x b. Write as a single logarithm: log3 x + log3 (x + 5) c. Write as a single logarithm: ln x – ln (x + 5)

Solution a.

x log10 12

b.

log 3 x 2  5 x

c.

 x  ln    x  5

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1321


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

5. Simplify: a. ln e b. ln(e2x – 7) c. eln 4x

Solution a. 1 b.

2x  7

c.

4x

6. Evaluate: a. log10 0 b. log10 (–6) c. ln 0 d. ln (–7)

Solution a. undefined b. undefined c. undefined d. undefined

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. An equation with a variable in its exponent is called a(n) __________ equation. Solution exponential 8. An equation with a logarithmic expression that contains a variable is a(n) __________ equation.

Solution logarithmic 9. The formula for carbon dating is A = __________.

Solution

A0 2t /h 10. The formula for population growth is P = __________.

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1322


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

P0ekt Practice Solve each exponential equation using like bases. 11. 23 x  2  16 x

Solution

23 x  2  16x

 

23 x  2  24

x

23 x  2  24 x 3x  2  4 x 2x 12. 32 x  2  27 x  12

Solution 32 x  2  27 x  12

2  5

x 2

 27 x  12

25 x  10  27 x  12 5 x  10  7 x  12 2  2 x 1  x

13. 27 x  1  32 x  1

Solution

27 x  1  32 x  1

3  3

x 1

 32 x  1

33 x  3  32 x  1 3x  3  2x  1 x  2 14. 3 x  1  92 x

Solution

3x  1  92 x

 

3x  1  32

2x

3x  1  34 x x  1  4x 1  3 x  31  x

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1323


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

15. 54 x  1  25 x  2

Solution

54 x  1  25 x 2

 

54 x  1  52

 x 2

54 x  1  52 x 4 4 x  1  2 x  4 6 x  5 x

5 6

16. 52 x  1  125 x

Solution

52 x  1  125x

 

52 x  1  53

x

52 x  1  53 x 2x  1  3x 1 x 17. 4 x  2  8 x

Solution

4 x  2  8x

2  2

x 2

 

 23

x

22 x 4  23 x 2x  4  3x 4  x 18. 16 x  1  82 x  1

Solution

16 x  1  82 x  1

2  4

x 1

 

 23

2x 1

24 x  4  26 x  3 4 x  4  6x  3 1  2x 1 x 2 19. 812 x  27 2 x  5

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1324


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

812 x  272 x 5

3   3  2x

4

3

2 x 5

38 x  36 x  15 8 x  6 x  15 2 x  15 15 2

x

20. 625 x  9  125 x  12

Solution

625x 9  125x  12

5  4

x 9

 

 53

x  12

54 x 36  53 x 36 4 x  36  3 x  36 x 0 2

21. 2 x  2 x  8

Solution 2

2x  2 x  8 2

2x 2 x  23 x 2  2x  3 x 2  2x  3  0

 x  3 x  1  0

x  3  0 or x  1  0 x3

x  1

2

22. 5 x  3 x  625

Solution 2

5x  3 x  625 2

5x  3 x  54 x 2  3x  4 x 2  3x  4  0

 x  4 x  1  0

x  4  0 or x  1  0 x  4 2

x1 2

23. 36 x  216 x  3

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1325


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution 2

2

36 x  216 x  3

6   6  x2

2

3

2

x2 3

2

62 x  63 x 9 2x 2  3x 2  9 9  x2 3  x 2

24. 25 x 5 x  31254 x

Solution 2

25x 5 x  31254 x

5  2

 

x2 5 x

 55

4x

2

52 x  10 x  520 x 2 x 2  10 x  20 x 2 x 2  30 x  0

2 x  x  15   0 x  0 or x  15 2

25. 7 x  3 x 

1 49

Solution

1 49 2 7 x  3 x  7 2 2

7 x 3 x 

x 2  3 x  2 x 2  3x  2  0

 x  2 x  1  0 x20

x10

or

x  2 2

26. 3 x  4 x 

x  1

1 81

Solution

1 81 3 x 2 4 x  34 2

3x  4 x 

x 2  4 x  4

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1326


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

x2  4x  4  0

 x  2 x  2  0 x20

or

x20

x  2

x  2

27. e  x  6  e x

Solution e x  6  e x x  6  x 6  2x 3x

28. e2 x  1  e3 x  11

Solution

e2 x  1  e3 x  11 2 x  1  3 x  11 12  x 2

29. e x  1  e24

Solution 2

e x  1  e24 x 2  1  24 x 2  25  0

 x  5 x  5  0 x  5  0 or x  5 2

30. e x  7 x 

x 5 0 x 5

1 e12

Solution 2

ex 7 x  e

x2  7 x

1

e12  e12

x  7 x  12 2

x  7 x  12  0 2

 x  3 x  4   0 x  3  0 or x  3

x4 0 x  4

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1327


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solve each exponential equation and express the answer in terms of common or natural logarithms. Then use a calculator to approximate each solution, rounding to four decimal places. 31. 4x = 5 Solution 4x  5 ln 4 x  ln 5 x ln 4  ln 5 ln 5  1.1610 x ln 4

32. 7x = 12 Solution 7 x  12 ln 7 x  ln 12 x ln 7  ln 12 ln 12  1.2770 x ln 7

33. 13x – 1 = 2 Solution

13x  1  2 ln 13x  1  ln 2

 x  1 ln 13  ln 2 x1 x

ln 2 ln 13 ln 2 ln 13

 1  1.2770

34. 5x + 1 = 3 Solution

5x  1  3 ln 5x  1  ln 3

 x  1 ln 5  ln 3 x1 x

ln 3 ln 5 ln 3 ln 5

 1  0.3174

35. 2x + 1 = 3x

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1328


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution 2x  1  3x ln 2x  1  ln 3x

 x  1 ln 2  x ln 3 x ln 2  x ln 3  ln2

x ln2  x ln3   ln2 x

ln2 ln2  ln3

 1.7095

36. 5 x  3  32 x Solution

5x  3  32 x ln 5x  3  ln 32 x

 x  3 ln5  2x ln3 x ln 5  3 ln 5  2 x ln3 x ln 5  2 x ln 3  3 ln5 x ln5  2ln3   3 ln 5 x

3 ln 5 ln5  2 ln3

 8.2144

37. 2 x  3 x Solution 2x  3x ln 2 x  ln 3 x x ln 2  x ln 3 x ln 2  x ln 3  0

x ln 2  ln 3   0 x 0

38. 32 x  4 x Solution 32 x  4 x ln 32 x  ln 4 x 2 x ln 3  x ln 4 2 x ln 3  x ln 4  0

x  2 ln 3  ln 4   0 x 0

39. 42 x 3  7 x  1

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1329


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

42 x  3  7 x  1 ln 42 x  3  ln 7 x  1

 2x  3 ln 4   x  1 ln 7 2 x ln 4  3 ln 4  x ln 7  ln 7 2 x ln 4  x ln 7  3 ln 4  ln 7 x  2 ln 4  ln 7   3 ln 4  ln 7 x

3 ln 4  ln 7 2 ln 4  ln 7

 7.3847

40. 52 x  1  6 x  1 Solution

52 x  1  6 x  1 ln 52 x  1  ln 6 x  1

 2 x  1 ln 5   x  1 ln 6 2 x ln 5  ln 5  x ln 6  ln 6 2 x ln 5  x ln 6  ln 5  ln 6 x  2 ln 5  ln 6   ln 5  ln 6 x

ln 5  ln 6 2 ln 5  ln 6

 2.3833

2

41. 7 x  10 Solution 2

7 x  10 2

ln 7 x  ln 10 x 2ln 7  ln 10 x2 

ln 10 ln 7

x

ln 10 ln 7

 1.0878

2

42. 8x  11 Solution 2

8x  11 2

ln 8x  ln 11 x 2ln 8  ln 11 x2 

ln 11 ln 8

x

ln 11 ln 8

  1.0738

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1330


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions 2

43. 8 x  9 x Solution 2

8x  9x 2

ln 8x  ln 9x x 2ln 8  x ln 9 x 2ln 8  x ln 9  0

x  x ln 8  ln 9   0 x 0

x ln 8  ln 9  0 x ln 8  ln 9 x

ln 9 ln 8

 1.1610

2

44. 5 x  25 x

Solution 2

5x  25 x 2

ln 5 x  ln 25 x x 2ln 5  5 x ln 2 x 2ln 5  5 x ln 2  0

x  x ln 4  5 ln 2   0

x 0

x ln 5  5 ln 2  0 x ln 5  5 ln 2 x

5 ln 2 ln 5

 2.1534

45. e x  10

Solution e x  10 ln e x  ln 10 x  ln 10  2.306

46. 8e x  16

Solution

8e x  16 ex  2 ln e x  ln 2 x  ln 2  0.6931 47. 4e 2 x  24

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1331


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

4e2 x  24 e2 x  6 ln e2 x  ln 6 2 x  ln 6 x

1 ln 6 2

48. 2e5 x  18

Solution 2e5 x  18 e5 x  9 ln e5 x  ln 9 5x ln e  ln 9 5 x  ln 9 1 x  ln 9  0.4394 5

49. e 2 x  4  16

Solution e2 x  4  16 e2 x  12 ln e2 x  ln 12 2 x ln e  ln 12 2 x  ln 12 1 x  ln 12  1.2425 2

50. 7e 3 x  4  17

Solution

7e3 x  4  17 7e3 x  21 e3 x  3 ln e3 x  ln 3 3 x ln e  ln 3 3 x  ln 3 1 x  ln 3 3 51. e x  3  8  14

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1332


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution e x  3  8  14 ex 3  6 ln e x  3  ln 6

 x  3 ln e  ln 6

x  3  ln 6 x  3  ln 6  1.2082

52. e2 x  1  4  1

Solution

e2 x  1  4  1 e2 x  1  5 ln e2 x  1  ln 5

 2x  1 ln e  ln 5 2 x  1  ln 5 2 x  1  ln 5 1  ln 5 x  1.3047 2

Solve each equation. If an answer is not exact, give the answer in decimal form. Round to four decimal places.

t n i H

t n i H

53. 4 x  2  4 x  15

: 4 x +2  4 x 4 2.

Solution 4 x  2  4 x  15 4 x 42  4 x  15 16  4 x  4 x  15 15  4 x  15 4x  1 x 0

54. 3 x  3  3 x  84

: 3 x +3  3 x 33.

Solution 3 x  3  3 x  84 3 x 33  3 x  84 27  3 x  3 x  84 28  3 x  84 3x  3 x1

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1333


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

55. 2 3 x  62 x

Solution

  log 2  3    log 6   2 3x  62 x x

2x

log 2  log 3x  2 x log 6 log 2  x log 3  2 x log 6 x log 3  2 x log 6   log 2

x log 3  2log 6    log 2

 log 2 log 3  2log 6 x  0.2789 x

56. 2 3 x  1  3 2 x  1 Solution

 log 2  3    log 3  2       2 3x  1  3 2x  1 x 1

x 1

log 2  log 3x  1  log 3  log 2x  1

log 2   x  1 log 3  log 3   x  1 log 2 log 2  x log 3  log 3  log 3  x log 2  log 2 x log 3  x log 2  2log 2

x log 3  log 2  2log 2 2log 2 log 3  log 2 x  3.4190 x

t n i H

 

57. 22 x  10 2 x  16  0

: Let y  2 x .

Solution

 

22 x  10 2 x  16  0 y  10 y  16  0 2

 y  2 y  8  0

y  2  0 or y  8  0 y 2

y 8

2 2

2x  8

x1

x3

x

t n i H

 

58. 32 x  10 3 x  9  0

: Let y  3 x .

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1334


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

 

32 x  10 3x  9  0 y  10 y  9  0 2

 y  1 y  9  0

y  1  0 or y  9  0 y 1

y 9

3 1

3x  9

x 0

x2

x

t n i H

59. 22 x  1  2 x  1

: 2a  b  2 a 2 b .

Solution

22 x  1  2x  1 22 x 21  2x  1  0

  2 2   1 2  1  0 2 2   1  0 or 2  1  0 2 1 2  2   1 2 2x  2x  1  0 x

x

x

x

x

x

x 0

2x   21 impossible

t n i H

 

60. 32 x  1  10 3 x  3  0

: 3 a  b  3a 3 b .

Solution

  3 3  10  3   3  0 3  3   10  3   3  0 3 3   1 3  3  0 3  3   1  0 or 3  3  0 3 3   1 3 3 32 x  1  10 3 x  3  0 2x

1

x

2x

x

x

x

x

x

x

x

3 x  31

x1

x  1

Solve each logarithmic equation. Obtain decimal approximations for answers containing e. Round to four decimal places. 2 61. log x  2

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1335


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution log x 2  2 x 2  102 x 2  100 x   100   10 3 62. log x  3

Solution

log x 3  3 x 3  103 x 3  1000 x  3 1000  10 63. log

4x  1 0 2x  9

Solution 4x  1 log 0 2x  9 x  1 100  2x  9 4x  1 1 2x  9 2x  9  4x  1 8  2x 4x

64. log

5x  2

2  x  7

0

Solution log

5x  2

2  x  7

0

100 

5x  2

2  x  7

5x  2 2 x  14 2 x  14  5 x  2 3 x  12 1

x4

65. ln x  6

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1336


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution ln x  6

x  e6  403.428 66. ln x  3 Solution ln x  3

x  e3  20.0855 67. 6 + ln x  10 Solution 6  ln x  10

ln x  4 x  e4  54.5782 68. 3 + 4 ln x  9 Solution 3  4 ln x  9

4 ln x  6 3 ln x  2 3

x  e 2  4.4817 69. ln 2 x  5 Solution ln 2 x  5

2 x  e5 1 x  e5  74.2066 2 70. 2  ln 3 x  1 Solution 2  ln 3 x  1

ln 3 x  1 3x  e x

1 e  74.2066 3

71. ln 2 x  1  4

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1337


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

ln  2 x  7   4 2 x  7  e4 2 x  e4  7 x

e4  7  30.7991 2

72. ln 3 x  5  7 Solution

ln  3 x  5   7 3 x  5  e7 3 x  e7  5 x

x7  5  367.2111 3

Solve each logarithmic equation.

73. log 2 2 x  3  log 2 x  4

Solution

log 2  2 x  3   log 2  x  4  2x  3  x  4 x7

74. log 3 3 x  5  log 3 2 x  6  0 Solution

log 3  3 x  5   log 3  2 x  6   0

log 3  3 x  5   log 3  2 x  6  3x  5  2x  6 x1

75. log x  log x  48  2 Solution

log x  log  x  48   2 log x  x  48   2

x  x  48   102

x 2  48 x  100  0

 x  50 x  2  0 x  50  0 x  50

or

x20 x  2: extraneous

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1338


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

76. log x  log x  9  1 Solution

log x  log  x  9  1 log x  x  9  1

x  x  9  101

x 2  9 x  10  0

 x  1 x  10  0 x10

or

x  10  0

x1

x  10: extraneous

77. log x  log x  15  2 Solution

log x  log  x  15  2 log x  x  15   2

x  x  15  102

x 2  15 x  100  0

 x  20 x  5  0 x  20  0

or

x 5 0

x  20

x  5: extraneous

78. log x  log x  21  2 Solution

log x  log  x  21  2 log x  x  21  2

x  x  21  102

x 2  21x  100  0

 x  4  x  25  0 x 4  0

or

x4

x  25  0 x  25: extraneous

79. log x  90  3  log x Solution

log  x  90   3  log x

log x  log  x  90   3

log x  x  90   3

x  x  90   103

x 2  90 x  1000  0

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1339


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 x  10 x  100  0 x  10  0

x  100  0

or

x  10

x  100 extraneous

80. log x  3  log 6  2 Solution

log  x  3   log 6  2 log

x 3 2 6 x 3  102 6 x  3  600 x  603

81. log 5000  log x  2  3 Solution

log  5000   log  x  2   3 log

5000 3 x 2 5000  103 x 2 5000  1000  x  2  5000  1000 x  2000 7000  1000 x 7x

82. log 4 2x  3  log 4 x  1  0 Solution

log 4  2 x  3   log 4  x  1  0

log 4  2 x  3  log 4  x  1 2x  3  x  1 x2

83. log 7 x  log 7 x  5  log 7 6 Solution

log 7 x  log 7  x  5   log 7 6 log 7 x  x  5   log 7 6 x  x  5  6

x  5x  6  0 2

 x  6 x  1  0

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1340


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

x 6  0 x 6

or

x10

x  1 extraneous

84. ln x  ln x  2  ln 120 Solution

ln x  ln  x  2   ln 120 ln x  x  2   ln 120 x  x  2   120

x  2 x  120  0 2

 x  12 x  10   0 x  12  0 or x  12

x  10  0 x  10

extraneous

85. ln 15  ln x  2  ln x Solution

ln 15  ln  x  2   ln x ln

15  ln x x 2 15 x x 2 15  x  x  2  0  x 2  2 x  15

0   x  5  x  3  x 5  0 x 5

or

x30 x  3 extraneous

86. ln 10  ln x  3  ln x Solution

ln 10  ln  x  3   ln x ln

10  ln x x 3 10 x x 3 10  x  x  3  0  x 2  3 x  10

0   x  5  x  2 

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1341


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

x 5  0

x20

or

x 5

x  2 extraneous

87. log 6 8  log 6 x  log 6 x  2

Solution

log 6 8  log 6 x  log 6  x  2  8  log 6  x  2  x 8  x 2 x 8  x  x  2

log 6

0  x 2  2x  8

0   x  4  x  2  x  4  0 or

x20

x4

x  2 extraneous

88. log x  6  log x  2  log

5 x

Solution

5 x 5 x 6  log log x 2 x

log  x  6   log  x  2   log

x 6 5  x 2 x x  x  6  5  x  2 x 2  6 x  5 x  10 x 2  11x  10  0

 x  1 x  10  0 x10

or

x1

x  10  0 x  10

extraneous

89. log 8 x  1  log 8 6  log 8 x  2  log 8 x Solution

log 8  x  1  log 8 6  log 8  x  2  log 8 x log 8 x6 1  log 8 x x 2 x 1 6

 x x 2

x  x  1  6  x  2 

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1342


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

x 2  x  6 x  12 x 2  7 x  12  0

 x  3 x  4   0 x 3 0 x3 90. log x 2  log x 

or

x 4 0 x4

2

Solution

log x 2  log x 

2

2log x  log x 

2

0  log x   2log x 2

0  log x log x  2 

log x  0 or log x  2  0 x1

log x  2

x1

x  100

91. log log x  1 Solution

log log x   1 log x  101 log x  10 x  1010

92. log 3 log 3 x  1 Solution

log 3 log 3 x   1 log 3 x  31 log 3 x  3 x  33  27

93.

log  3 x  4  log x

2

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1343


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

log  3 x  4  log x

2

log  3 x  4   2log x log  3 x  4   log x 2 3x  4  x 2 0  x 2  3x  4 No real solutions

94.

ln  8x  7  ln x

2

Solution

ln  8 x  7  ln x

2

ln  8 x  7   2ln x ln  8 x  7   ln x 2 8x  7  x 2 0  x 2  8x  7 x  7  0 or x  1  0 x7 x1 extraneous 95.

ln  5x  6 2

 ln x

Solution

ln  5 x  6 

 ln x 2 ln  5 x  6   2ln x ln  5 x  6   ln x 2 5x  6  x 2 0  x 2  5x  6

0   x  6  x  1 x  6  0 or x 6

x10 x  1

extraneous 96.

1 log  4x  5  log x 2

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1344


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution 1 log  4 x  5  log x 2 log  4 x  5  2log x

log  4 x  5  log x 2 4x  5  x2

0  x2  4x  5

0   x  5 x  1 x  5  0 or x 5

x10 x  1

extraneous  1 97. log 3 x  log 3    4 x

Solution  1 log 3 x  log 3    4 x log 3 x  log 3 x 1  4 log 3 x   log 3 x  4 2log 3 x  4 log 3 x  2 x 9

98. log 5 7  x  log 5 8  x  log 5 2  2 Solution

log 5  7  x   log 5  8  x   log 5 2  2 log 5

 7  x 8  x   2 2  x 7  8  x  2

 52

 7  x 8  x   50  x 2  x  56  50 x2  x  6  0

 x  3 x  2  0 x  3  0 or x3

x20 x  2

99. 2log 2 x  3  log 2 x  2

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1345


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

2log 2 x  3  log 2  x  2

log 2 x  log 2  x  2  3 2

log 2

x2 3 x  x2 x2  23 x 2 x 2  8  x  2

x 2  8 x  16  0

 x  4 x  4  0 x  4  0 or

x 4 0

x4

x4

100. 2log 3 x  log 3 x  4  2  log 3 2 Solution

2log 3 x  log 3  x  4   2  log 3 2

log 3 x  log 3  x  4   log 3 2  2 2

log 3

x2

2  x  4

2

x2  32 2x  8 x 2  9  2 x  8 x 2  18 x  72  0

 x  12 x  6  0 x  12  0 or x  12

x 6  0 x 6

101. ln 7 y  1  2 ln y  3  ln 2 Solution

ln  7 y  1  2 ln  y  3  ln 2

 y  3 ln  7 y  1  ln 2

2

y2  6y  9 2 14 y  2  y 2  6 y  9 7y  1 

0  y2  8y  7

0   y  7  y  1 y  7  0 or y 7

y 10 y 1

102. 2 log y  2  log y  2  log 12

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1346


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

2 log  y  2  log  y  2  log 12 log  y  2  log 2

 y  2

y 2 y2  4y  4  12 12 y 2  48 y  48  y  2

12

12 y 2  47 y  46  0

 12 y  23 y  2  0 12 y  23  0 12 y  23 y 

y 20 y  2

or

23 12

extraneous

Use a graphing calculator to solve each equation. If an answer is not exact, give the result to the nearest hundredth.

103. log x  log x  15  2 Solution

Graph y  log x  log x  15 and y = 2 and find the x-coordinate of the point(s) of intersection: x = 20

104. log x  log x  3  1 Solution

Graph y  log x  log x  3 and y = 1 and find the x-coordinate of the point(s) of intersection: x = 2

105. 2x  1  7

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1347


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution x 1

Graph y  2 x

and y = 7 and find the x-coordinate of the point(s) of intersection:

1.81

106. ln 2 x  5  ln 3  ln x  1 Solution

Graph y  ln 2x  5  ln3 and y  ln x  1 and find the x-coordinate of the point(s) of intersection: x

8

Fix It In exercises 107 and 108, identify the step the first error is made and fix it. 107. Solve the equation: 5e7 x  1  11 Solution Step 5 was incorrect. Step 1: 5e4 x  10 Step 2: e 7 x  2

 

Step 3: ln e7 x  ln 2 Step 4: 7 x  ln 2 Step 5: x 

ln2 7

108. Solve the equation: log6 2x  log6 x  log6 10

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1348


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution Step 4 was incorrect.

 

Step 1: log 6 2 x 2  log 6 10 Step 2: 2 x 2  10 Step 3: x 2  5 Step 4: x  5 Applications

Use a calculator to help solve each problem. 109. Tritium decay The half-life of tritium is 12.4 years. How long will it take for 25% of a sample of tritium to decompose? Round to one decimal place. Solution A  A0 2 t / h 0.75 A0  A0  2

 t /  12.4 

0.75  2t /12.4

log  0.75   log 2t /12.4 log  0.75    

12.4log  0.75 

t log 2 12.4

t log 2 5.1 years  t

110. Radioactive decay In 2 years, 20% of a radioactive element decays. Find its half-life. Round to one decimal place. Solution A  A0 2t / h 0.80 A0  A0  2t / h 0.80  22/ h

log  0.80   log 22/ h

2 log  0.80    log 2 h h log  0.80   2log 2 h

2log 2

log  0.80 

h  6.2 years

111. Thorium decay An isotope of thorium, 227Th, has a half-life of 18.4 days. How long will it take 80% of the sample to decompose? Round to one decimal place.

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1349


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

A  A0 2t / h 0.20 A0  A0  2

 t /  18.4 

0.20  2t /18.4

log  0.20   log 2t /18.4 log  0.20    

18.4log  0.20 

t log 2 18.4

t log 2 42.7 days  t

112. Lead decay An isotope of lead, 201Pb, has a half-life of 8.4 hours. How many hours ago was there 30% more of the substance? Round to one decimal place. Solution

A  A0 2t / h 1.3 A0  A0  2

 t /  8.4 

1.3  2t /8.4

log  1.3  log 2t /8.4 log  1.3   

8.4log  1.3

t log 2 8.4

t log 2 3.2hours  t  About 3.2hours ago

113. Carbon-14 dating A cloth fragment is found in an ancient tomb. It contains 70% of the carbon-14 that it is assumed to have had initially. How old is the cloth? Round to the nearest hundred. Solution A  A0 2t / h 0.70 A0  A0  2t /5700 0.70  2t /5700

log  0.70   log 2t /5700 log  0.70    

5700log  0.70 

t log 2 5700

t log 2 2900 years  t

114. Carbon-14 dating Only 25% of the carbon-14 in a wooden bowl remains. How old is the bowl? Round to the nearest thousand.

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1350


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution A  A0 2t / h 0.25 A0  A0  2t /5700 0.25  2t /5700

log  0.25   log 2t /5700 log  0.25    

5700log  0.25 

t log 2 5700

t log 2 11, 000 years  t

115. Compound interest If $500 is deposited in an account paying 8.5% annual interest, compounded semiannually, how long will it take for the account to increase to $800? Round to the nearest tenth.

Solution  r A  P 1   k 

kt

 0.085  800  500  1   2   2t 8   1.0425  5 2t 8 log    log  1.0425  5

2t

log 8  log 5  2t log  1.0425  log 8  log 5

2log  1.0425 

t

5.6 years  t 116. Continuous compound interest In Exercise 115, how long will it take if the interest is compounded continuously? Round to the nearest tenth.

Solution

A  Pert 800  500e0.085t 8  e0.085t 5 8 ln    ln e0.085t 5 ln8  ln5  0.085t ln8  ln5

t 0.085 5.5 years  t

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1351


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

117. Compound interest If $1300 is deposited in a savings account paying 9% interest, compounded quarterly, how long will it take the account to increase to $2100? Round to the nearest tenth.

Solution  r A  P 1   k 

kt

 0.09  2100  1300  1   4   4t 21   1.0225  13 4t  21  log    log  1.0225   13 

4t

log 21  log 13  4t log  1.0225  log 21  log 13

4log  1.0225 

t

5.4 years  t 118. Compound interest A sum of $5000 deposited in an account grows to $7000 in 5 years. Assuming annual compounding, what interest rate is being paid? Round to two decimal places.

Solution  r A  P1  k 

kt

 r 7000  5000  1   1  5 7  1  r  5 5 7 5 5  1  r  5 5

5

1 5

7  1 r 5

7  1  r  r  0.696  6.96% 5

119. Rule of Seventy A common rule for finding how long it takes an investment to double is called the Rule of Seventy. To apply the rule, divide 70 by the interest rate

 14 years to double the investment. At 7%, (expressed as a percent). At 5%, it takes 70 5  10 years. Explain why this formula works. it takes 70 7

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1352


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution A  Pert 2P  Pert 2  ert ln2  ln ert ln2  rt ln2

t r 0.70 t r 70 t  100  r  %

120. Oceanography The intensity I of a light a distance x meters beneath the surface of a lake decreases exponentially. If the light intensity at 6 meters is 70% of the intensity at the surface, at what depth will the intensity be 20%? Round to the nearest meter.

Solution I  I0ekx 0.70I0  I0e

I  I0ekx

k 6

0.20I0  I0e0.05944 x

0.7  e6k

0.2  e0.05944 x

ln0.7  ln e6k

ln0.2  ln e0.05944 x

ln0.7  6k

ln0.2  0.05944 x

ln0.7

k

6 0.05944  k

ln0.2

x 0.05944 27 meters  x

121. Bacterial growth A staphylococcus bacterial culture grows according to the formula P = P0at. If it takes 5 days for the culture to triple in size, how long will it take to double in size? Round to one decimal place.

Solution P  P0at

P  P0at

3P0  P0a5

2P0  P0 31/5

 

3  a5

 

31/5  a5 31/5  a

1/5

t

2  3t /5 log 2  log 3t /5 log 2  5log 2 log 3

t log 3 5

t

3.2days  t

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1353


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

122. Rodent control The rodent population in a city is currently estimated at 30,000. If it is expected to double every 5 years, when will the population reach 1 million? Round to one decimal place.

Solution P  P0ekt 60, 000  30, 000e

k 5

2e

5k

ln2  ln e

5k

ln2  5k ln2 k 5

P  P0ekt In 2

1, 000, 000  30, 000e 5

t

In 2 100 t e5 3 In 2 100 t ln  ln e 5 3 ln2 t ln 100  ln3  5 5 ln 100  ln3   t  t  25.3 years ln 2

123. Temperature of coffee Refer to the section opener and find the time it takes for the white chocolate mocha to reach a temperature of 80°F. Round to the nearest minute.

Solution T  70  110e0.2t 80  70  110e0.2t 10  110e0.2t 1  e0.2t 11  1 ln    ln e0.2t  11   1 ln    0.2t  11  ln  111 

t 0.2 12  t  t  12 minutes

124. Time of death The exponential function T(t) = 17e–0.0626t + 20 models the temperature T in °C of a person’s body t hours after death. If a dead body is discovered at 8:30 a.m. and the body’s temperature is 30°C, what was the person’s approximate time of death?

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1354


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

T  t   17e0.0626t  20 30  17e0.0626t  20 10  17e0.0626t

10  e0.0626t 17  10  ln    ln e0.0626t  17   10  ln    0.0626t  17  ln  10 17  t 0.0626 8.5  t  timeof death: midnight 125. Newton’s Law of Cooling Water whose temperature is at 100°C is left to cool in a room where the temperature is 60°C. After 3 minutes, the water temperature is 90°. If the water temperature T is a function of time t given by T = 60 + 40ekt, find k.

Solution T  60  40ekt 90  60  40e

k  3

30  40e3k 0.75  e3k

ln  0.75   ln e3k ln  0.75   3k

ln  0.75  3

k

126. Newton’s Law of Cooling Refer to Exercise 125 and find the time for the water temperature to reach 70°C. Round to one decimal place.

Solution

From # 125, k 

ln  0.75 3

. ln 0.75 

T  60  40e 3

ln 0.75 

70  60  40e 3 ln  0.75

10  40e 3 ln  0.75

0.25  e 3

t t

t

t

ln  0.75

t

ln0.25  ln e 3 ln0.75 ln0.25  t 3 3 ln0.25  t  t  14.5 minutes ln0.75

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1355


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

127. Newton’s Law of Cooling A block of steel, initially at 0°C, is placed in an oven heated to 300°C. After 5 minutes, the temperature of the steel is 100°C. If the steel temperature T is a function of time t given by T = 300 – 300ekt, find the value of k.

Solution T  300  300ekt 100  300  300e

k  5

200  300e5k 2  e5k 3 2 ln    ln e5k 3 2 ln    5k 3 ln  23  k 5 128. Newton’s Law of Cooling Refer to Exercise 127 and find the time for the steel temperature to reach 200°C. Round to one decimal place.

Solution From # 127, k 

ln  23  4

. ln  2/3

T  300  300e 5

ln  2/3

200  300  300e 5 ln  2/3 

100  300e 5 ln  2/3  1 t e 5 3 ln  2/3 

ln  1/3   ln e 5 ln  1/3   5 ln  1/3  ln  2/3 

t

t

t

ln  2/3  5

t

t

 t  t  13.5minutes

Discovery and Writing 129. Explain how to solve the exponential equation 5x + 1 = 125. Solution Answers may vary. 130. Explain how to solve the exponential equation 5x + 1 = 126. Solution Answers may vary.

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1356


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

131. Explain why it is necessary to check the solutions of a logarithmic equation. Solution Answers may vary. 132. What is meant by the term “half-life”? Solution Answers may vary. 133. Use the population growth formula to show that the doubling time for population growth is given by

t

ln 2 r

.

Solution P  P0ert 2P0  P0ert 2  ert ln2  ln ert ln2  rt ln2 r

t

134. Use the population growth formula to show that the tripling time for population growth is given by

t

ln 3 r

.

Solution P  P0ert 3P0  P0ert 3  ert ln3  ln ert ln3  rt ln3 r

t

135. Can you solve x = log x algebraically? Can you find an approximate solution? Solution Answers may vary. 136. Can you solve x = ln x algebraically? Can you find an approximate solution? Solution Answers may vary.

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1357


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Find x.

137. log 2 log 5 log 7 x   2 Solution

log 2 log 5 log 7 x   2

log 5 log 7 x   22  4 log 7 x  54  625 x  7625 1 6

138. log 8  16 3 4096   x   Solution

log 8 16 3 4096

 x 8   16 4096  1/6

3

x

1/6

8x  24  212/3   

1/6

2   2 3

x

4/3

3x 

4 4 x 3 9

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 139. The exponential equation 8x = 0 has no solution. Solution True. 140. The exponential equation 8x = 1 has no solution. Solution False. The solution is x = 0. 141. The equations  32   27 and x 2  278 are exponential equations. 8 x

3

Solution False. x 3/2  278 is not an exponential equation. 142. The exponential equation 7 x 5   491 

3 x 8

can be solved by writing each side of the

equation as a power of the same base. Solution True.

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1358


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

143. The exponential equation e x 5  e3 x1 8 can be solved by writing each side of the equation as a power of the same base. Solution True. 144. The exponential equation 6 x  5  123 x  8 can be solved by writing each side of the equation as a power of the same base. Solution False. You will need to use logarithms. 145. To solve the exponential equation, ex = 15, take the common logarithm of both sides. Solution False. Use the natural logarithm.

146. The logarithmic equations log 3 2 x  7  4 and log 3 2x  7  log 3 4 can be solved using the same method. Solution

  To solve log  2 x  7   log 4, you should use the fact that the logarithm function is

False. To solve log 3 2 x  7  4, you should rewrite the equation in exponential form. 3

3

one-to-one. 147. To solve the logarithmic equation log2 8  log2 x  5, we combine the two logarithms on the left side of the equation and then write the equation in the exponential form 25 = 8 – x. Solution False. After combining the logarithms, you would end up with 25  8x . 148. Proposed solutions of logarithmic equations that are negative must be discarded. Solution False. Solutions are only discarded if they produce an undefined logarithm.

CHAPTER REVIEW SOLUTIONS Exercises Use properties of exponents to simplify. 1.

5 2 5 2

Solution 5 2 5 2  5 2  2  52 2

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1359


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

2.

2  5

2

Solution

2   2 5

2

5 2

 2 10

Evaluate each exponential function at the given values and simplify. 3.

 1 f x    4 a.

f 3

b.

f 0

c.

f  3

x

Solution 3

a.

 1 1 f  3     4 64  

b.

 1 f 0     1 4

c.

 1 f  3     4

0

4.

3

 43  64

f  x   3ex  5 a.

f 2

b.

f 0

c.

f  2

Solution a.

f  2  3e2  5

b.

f  0   3e0  5  3  5  2

c.

f  2   3e2  5 

3 5 e2

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1360


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

The graph of an exponential function is shown. Identify its domain, range, equation of its horizontal asymptote, and whether it is increasing or decreasing on its domain. 5.

Solution

domain: ,  ; range: 5,  ; horizontal asymptote: y  5; increasing on its domain 6.

Solution

domain: ,  ; range: 2,  ; horizontal asymptote: y  2; decreasing on its domain Graph the function defined by each equation. 7.

f  x   3x Solution

f  x   3x :  0, 1 ,  1, 3

8.

 1 f x    3

x

Solution x

 1  1 f  x     :  0, 1 ,  1,  3  3

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1361


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

9. The graph of f(x) = 7x will pass through the points (0, p) and (1, q). Find p and q. Solution

f  x   7 x : goes through (0, 1) and (1, 7)

p = 1, q = 7

 

10. Give the domain and range of the function f x  b x , with b  0 and b  1. Solution

y  b x : domain   ,   ; range   0,  

Use translations to help graph each function. x

 1 11. g  x      2 2 Solution x

 1 gx     2 2 x

 1 Shift y    down 2: 2

 1 12. g  x     2

x 2

Solution  1 gx    2

x 2

x

 1 Shift y    left 2: 2

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1362


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Graph each function.

 

13. f x  5x Solution

f  x   5x

Reflect y  5x about x-axis.

 

14. f x  5x  4 Solution

f  x   5 x  4; Reflect y  5 x about x . Shift U4.

 

15. f x  e x  1 Solution

f  x   ex  1 Shift y  e x up 1:

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1363


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

16. f x  e x 3 Solution

f  x   e x 3 Shift y  e x right 3:

17. Compound interest How much will $10,500 become if it earns 9% per year for 60 years, compounded quarterly? Solution

 r A  P1  k 

kt

 0.09   10,500  1   4    $2, 189, 703.45

4  60 

18. Continuous compound interest If $10,500 accumulates interest at an annual rate of 9%, compounded continuously, how much will be in the account in 60 years? Solution A  Pert 0.09 60 

 10,500e  2, 324, 767.37

19. The half-life of a radioactive material is about 34.2 years. How much of the material is left after 20 years?

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1364


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution A  A0 2t / h  A0  220/34.2  0.6667 A0

about 23 of the original 20. Find the intensity of light at a depth of 12 meters if I0 = 14 and k = 0.7. Round to two decimals. Solution I  I0 k x

 14  0.7 

12

 0.19 lumen

21. The population of the United States is approximately 330,000,000 people. Find the population in 50 years if k = 0.015. Round to the nearest million. Solution

P  P0ekt

0.015 50 

P  300,000,000e P  635, 000,000 people 22. Spread of hepatitis In a city with a population of 450,000, there are currently 1000 cases of hepatitis. If the spread of the disease is projected by the following logistic function, how many people will contract the hepatitis virus after 5 years? Round to the nearest whole number.

P t  

450,000

1   450  1 e0.2t

Solution

P t   

450, 000

1   450  1 e0.2t 450,000

1   450  1 e

0.2 5

 2708 people

 

23. Give the domain and range of the logarithmic function f x  log 3 x. Solution

domin   0,   ; range   ,  

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1365


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

24. Give the domain and range of the natural logarithm function, f(x) = ln x. Solution

domin   0,   ; range   ,  

Find each value. 25. log3 9 Solution log 3 9  ? 3?  9 log 3 9  2

26. log 9

1 3

Solution

1 ? 3 1 9?  3 1 1 log 9   3 2

log 9

27. log x 1 Solution log x 1  ? x?  1 log x 1  0

28. log5 0.04 Solution log 5 0.04  ? 5?  0.04  log 5 0.04  2

1 25

29. log a a Solution log a a  ? a?  a log a a 

1 2

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1366


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

30. log a 3 a Solution log a 3 a  ? a?  3 a log a 3 a 

1 3

Find x. 31. log2 x  5 Solution log 2 x  5

25  x 32  x 32. log 3 x  4 Solution log 3 x  4

 3  x 4

9x

33. log 2 x  6 Solution log 2 x  6

 2  x 6

8x

34. log0.1 10  x Solution log 0.1 10  x

0.1  10 x x

 1     10  10  x  1 35. log x 2  

1 3

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1367


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

1 3

log x 2   x 1/3  2

x   2 1/3

3

x

3

1 8

36. log x 32  5 Solution log x 32  5

x 5  32 x2 37. log0.25 x  1 Solution log 0.25 x  1

0.25  x 1 1

 1   x 4 4x 38. log0.125 x  

1 3

Solution log 0.125 x  

0.125 

1/3

 1   8

1/3

1 3

x x

2x

39. log 2 32  x Solution log 2 32  x

 2   32 x

2   2 1/2

x

5

1 x 5 2 x  10 © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1368


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

40. log 5 x  4 Solution log 5 x  4

 5  x 4

1 x 25

41. log 3 9 3  x Solution log 3 9 3  x

 3  9 3 x

3   3 1/2

x

5/2

1 5 x 2 2 x 5

42. log 5 5 5  x Solution log 5 5 5  x

 5  5 5 x

5   5 1/2

x

3/2

1 3 x 2 2 x3

Find the domain of each logarithmic function. Write the answer in interval notation.

 

43. f x  2 log 3 x  5

Solution x 5  0

x 5

5,  

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1369


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

44. f x  5  ln 2  x

Solution 2 x  0  x  2 x2

 , 2

Evaluate the logarithmic functions at each of the given x-values.

 

45. f x  2  log 4 x a.

f  64 

b.

f  1

c.

 1  f   16 

Solution a.

f  64   2  log 4  64   2  3  1

b.

f  1  2  log 4  1  2

c.

 1   1  f    2  log 4    2  2  4 16    16 

 

46. f x  log 5 3  x a.

f  22

b.

f 2

c.

 14  f   5 

Solution

a.

f  22   log 5 3   22   log 5  25   2

b.

f  2  log 5  3  2  log 5  1  0

c.

 14    1 14  f    log 5  3    log 5    1 5   5   5

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1370


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

For each logarithmic function graph shown, state the domain, range, and equation of the vertical asymptote, 47.

Solution

domain: 2,  ; range: ,  ; x  2 48.

Solution

domain: 8,  ; range: ,  ; x  8 Graph each function.

 

49. f x  log x  2

Solution

f  x   log  x  2 ; Shift y  log x right 2:

 

50. f x  3  log x Solution

f  x   3  log x; Shift y  log x up 3:

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1371


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Graph each pair of equations on one set of coordinate axes. 51. y  4 x and y  log 4 x Solution

y  4 x ; y  log 4 x

x

 1 52. y    and y  log 1/3 x 3 Solution y   31  ; y  log 1/3 x x

Use a calculator to find each value to four decimal places. 53. ln 452 Solution ln 452  6.1137

54. ln log 7.85

Solution

ln log 7.85   0.1111

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1372


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Use a calculator to solve each equation. Round each answer to four decimal places. 55. ln x  2.336 Solution ln x  2.336 x  10.3398

56. ln x  log 8.8 Solution ln x  log 8.8 x  2.5715

Graph each function.

 

57. f x  1  ln x Solution

y  f  x   1  ln x

Shift y  ln x up 1:

 

58. f x  ln x  1 Solution

y  f  x   ln  x  1

Shift y  ln x left 1:

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1373


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Simplify each expression.

 

59. ln e 12

Solution

 

ln e 12  12 ln e  12

60. eln 14 x Solution ln 14 x

e

 14 x

61. Decibel gain An amplifier has an output of 18 volts when the input is 0.04 volt. Find the dB gain. Round to the nearest decibel. Solution

dB gain  20 log

E0

E1 18  20 log 0.04  53 dB gain

62. Intensity of an earthquake An earthquake had a period of 0.3 second and an amplitude of 7500 micrometers. Find its measure on the Richter scale. Round to the nearest tenth. Solution A P 7500  log 0.3  4.4

R  log

63. Charging batteries How long will it take a dead battery to reach an 80% charge? (Assume k = 0.17.) Round to one decimal place. Solution

1  C t   ln  1   k  M  1 0.8M  ln  1  t  0.17  M  1 ln  1  0.8  9.5 minutes t 0.17 64. Doubling time How long will it take the population of the United States to double if the growth rate is 3% per year? Round to the nearest year. Solution ln 2 ln 2 t   23 years r 0.03

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1374


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

65. Isothermal energy Find the amount of energy that must be supplied to double the volume of 1 mole of gas at a constant temperature of 350K. (Hint: R = 8.314.) Round to the nearest joule. Solution

 

E  RT ln Vf   8.314  350  ln V

i

   8.314350 ln 2  2017 joules 2Vf

Vi

Simplify each expression. 66. log 7 1 Solution

log7 1  0 67. log7 7 Solution

log7 7  1 68. log 7 73 Solution

log 7 73  3 69. 7log7 4 Solution 7

log 7 4

4

70. ln e4 Solution

ln e4  4ln e  4 71. ln 1 Solution ln 1  0 log 10 7

72. 10

Solution log 10 7

10

7

73. eln 3 Solution ln 3

e

loge 3

e

3

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1375


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

74. log b b4 Solution

log b b4  4log b b  4 75. ln e9 Solution

ln e9  9ln e  9 Assume x, y, and z are positive numbers. Write each expression in terms of the logarithms of x, y, and z. 76. log 5

x2 y 3 z4

Solution

log 5

x2 y 3  log 5 x 2 y 3  log 5 z 4 z4  log 5 x 2  log 5 y 3  log 5 z 4

 

 2log 5 x  3log 5 y  4log 5 z 77. log 8

x yz 2

Solution 1/2

log 8

 x  x  log 8  2  2 yz  yz   x  1  log 8  2  2  yz  1  log 8 x  log 8 yz 2 2 1  log 8 x  log 8 y  2log 8 z  2

78. ln

x4 y 5 z6

Solution ln

x4 y 5 z6

 ln x 4  ln y 5 z 6

 ln x 4  ln y 5  ln z 6

 4ln x   5ln y  6ln z   4ln x  5ln y  6ln z

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1376


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

79. ln 3 xyz Solution

ln3 xyz  ln  xyz 

1/3

1 ln xyz 3 1  ln x  ln y  ln z  3 

Assume x, y, and z positive numbers. Write each expression as single logarithm. 80. 3 log4 x  5log4 y  7 log4 z Solution

3 log 4 x  5log 4 y  7 log 4 z  log 4 x 3  log 4 y 5  log 4 z 7  log 4

81.

x3 z7 y5

1 log7 x  3log7 y   7log7 z 2 Solution 1 log 7 x  3log 7 y   7log 7 z  21 log 7 x  log 7 y 3  log 7 z 7  21 log 7 xy 3  log 7 z 7 2  log 7 xy 3  log 7 z 7

 log 7

xy 3 z7

82. 4 ln x  5 ln y  6 ln z Solution

4ln x  5ln y  6ln z  ln x 4  ln y 5  ln z 6  ln

83.

x4 y 5 z6

1 1 ln x  3ln y  ln z 2 3 Solution 1 1 y3 x ln x  3ln y  ln z  ln x 1/2  ln y 3  ln z 1/3  ln 3 2 3 z

Given that log a

0.6, log b

0.36, and log c

2.4, approximate the value of each expression.

84. log abc Solution log abc  log a  log b  log c  0.6  0.36  2.4  3.36

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1377


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

85. log a2 b Solution

log a2 b  log a2  log b  2log a  log b  2  0.6  0.36  1.56

86. log

ac b

Solution

log 87. log

ac  log a  log c  log b  0.6  2.4  0.36  2.64 b a2 c3 b2

Solution

log

a2

 log a2  log c3 b2  log a2  log c3  log b2  2log a  3log c  2log b

c3 b2

 2  0.6  3  2.4   2  0.36   6.72

88. To four decimal places, find log5 17. Solution log 5 17 

log 17 log 5

 1.7604

89. pH of grapefruit The pH of grapefruit juice is about 3.1. Find its hydrogen ion concentration. Write the answer using scientific notation and round to two decimal places. Solution pH   log H  3.1   log H  3.1  log H  7.94  104  H  90. Loudness of sound Find the decrease in loudness if the intensity is cut in half. Solution L  k ln I I k ln  k ln I  ln 2   k ln I  k ln 2  The loudness decreases by k ln 2. 2 Solve each equation for x. 91. 81x  2  27 Solution

81x  2  27

3  4

x 2

 33

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1378


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

34 x 8  33 4x  8  3 4 x  5 x 2

92. 2x  4 x 

5 4

1 8

Solution

1 8 2 2x  4 x  23 2

2x  4 x 

x 2  4 x  3 x  4x  3  0 2

 x  1 x  3  0 x10

or

x  1

x30 x  3

93. e x  e 6 x  14 Solution e x  e6 x  14 x  6 x  14 7 x  14 x2

94. e 2 x 2  e 18 Solution

e2 x 2  e18 2 x 2  18 x2  9 x  3 95. 3 x  7 Solution

3x  7 log 3x  log 7 x log 3  log 7 x

log 7 log 3

x  1.7712

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1379


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

96. 2 x  3 x  1 Solution 2x  3x  1 log 2 x  log 3 x  1

x log 2   x  1 log 3 x log 2  x log 3  log 3 log 3  x log 3  x log 2 log 3  x log 3  log 2 

log 3 log 3  log 2

x

2.7095  x

97. 2e x  16 Solution

2e x  16 ex  8 ln e x  ln8 x ln e  ln8 x  ln8  2.0794 98.  5e x  35 Solution

5e x  35 ex  7 ln e x  ln7 x ln e  ln7 x  ln7  1.9459 Solve each equation for x.

99. log 7 7 x  2  log 7 3 x  32

Solution

log 7  7 x  2   log 7  3 x  32  7 x  2  3 x  32 30  10 x 3  x

100. ln x  3  ln 5 x  51

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1380


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

ln  x  3   ln  5 x  51 x  3  5 x  51 6 x  48 x 8

101. log x  log 29  x  2 Solution

log x  log  29  x   2 log x  29  x   2

x  29  x   102

 x 2  29 x  100  0 x 2  29 x  100  0

 x  25 x  4   0 x  25  0 or x  25

x 4 0 x4

102. log 2 x  log 2 x  2  3 Solution

log 2 x  log 2  x  2   3

log 2 x  x  2   3

x  x  2   23

x 2  2x  8  0

 x  4  x  2  0 x  4  0 or

x20

x4

x  2 extraneous

103. log 2 x  2  log 2 x  1  2 Solution

log 2  x  2  log 2  x  1  2

log 2  x  2 x  1  2

 x  2 x  1  2

2

x2  x  2  4 x2  x  6  0

 x  2 x  3  0 x  2  0 or x2

x30

x  3 extraneous

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1381


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

104.

log  7 x  12 log x

2

Solution

log  7 x  12 log x

2

log  7 x  12  2log x log  7 x  12  log x 2 7 x  12  x 2 0  x 2  7 x  12

0   x  3 x  4  x  3, x  4

105. ln x  ln x  5  ln6 Solution

ln x  ln  x  5  ln6

ln x  x  5  ln6 x  x  5  6 x 2  5x  6

x 2  5x  6  0

 x  6 x  1  0 x  6  0 or x 6

x10

x  1 extraneous

106. log 3  log x  1  1 Solution

log 3  log  x  1  1 log

3  1 x1 3  101 x1 3 1  x  1 10 30  x  1 31  x

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1382


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

107. ex In 2  9 Solution

e

x In 2

9

ln e

x In 2

 ln9

x ln 2 ln e  ln9 x ln2  ln9 ln9 x  3.1699 ln2

108. ln x  ln x  1 Solution

ln x  ln  x  1

x  x1 no solution 109. ln x  3  4 Solution ln x  3  4

ln x  7 x  e7  1096.6332

110. ln x  ln x  1  1 Solution

ln x  ln  x  1  1

ln x  ln  x  1  1 ln

x 1 x1 x  e1 x1 x e  x1 1 x  e  x  1 x  ex  e e  ex  x

e  x  e  1

e  x, or x  1.5820 e1

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1383


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

111. ln x  log10 x (Hint: Use the Change-of-Base Formula.) Solution ln x

Note: log 10 x 

ln 10

ln x  log 10 x ln x ln x  ln 10 ln x ln 10  ln x ln x ln 10  ln x  0 ln x ln 10  1  0

ln x  0  x  1 112. Carbon-14 dating A wooden statue found in Egypt has a carbon-14 content that is two-thirds of that found in living wood. If the half-life of carbon-14 is 5700 years, how old is the statue? Round to the nearest hundred year. Solution

A  A0 2t / h 2 A  A0  2t /5700 3 0 2  x t /5700 3 log  2/3   log 2t /5700

log  2/3    

5700log  2/3 log 2

t log 2 5700

 t  t  3300 years

CHAPTER TEST SOLUTIONS Graph each function.

1.

f  x   2x  1 Solution

f  x   2x  1  Shift y  2x up 1.

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1384


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

2.

f  x   ex 2 Solution

f  x   ex 2  Shift y  e x right 2.

Solve each problem.

3. Radioactive decay A radioactive material decays according to the formula A  A0  2 

t

How much of a 3-gram sample will be left in 6 years? Solution

A  3  2

6

 3

1 3 gram  64 64

4. Compound interest An initial deposit of $1000 earns 6% interest, compounded twice a year. How much will be in the account in one year? Solution

 0.06  A  1000  1   2  

2 1

 $1060.90

5. Continuous compound interest An account contains $2000 and has been earning 8% interest, compounded continuously. How much will be in the account in 10 years? Round to two decimal places. Solution 0.08 10

A  2000e

 $4451.08

Find each value.

6.

log7 343 Solution

log 7 343  log 7 73  3

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1385


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

7.

log3

1 27

Solution

log3 8.

1  log3 33  3 27 log 10 5

log 10 1012  10 Solution

log 10 5

log 10 1012  10 9.

log3/2

 12  5  17

9 4

Solution 2

log 3/2

10. log 2/3

3 9  log 3/2    2 4 2

27 8

Solution

log 2/3

2 27  log 2/3   8 3

3

 3

Graph each function.

 

11. f x  log x  1 Solution

f  x   log  x  1 ; Shift y  log x right 1.

 

12. f x  2  ln x Solution

f  x   2  ln x; Shift y  ln x up 2.

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1386


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Write each expression in terms of the logarithms of a, b, and c. Assume a, b, and c are positive numbers. 2

3

13. log a bc

Solution

log a2bc3  log a2  log b  log c3  2log a  log b  3log c 14. ln

a b2c

Solution

 a  a ln 2  ln  2  bc b c

1/2

1 a 1 1 ln 2  ln a  ln b2  ln c  ln a  2 ln b  ln c  2 bc 2 2

Write each expression as a logarithm of a single quantity. Assume a, b, and c are positive numbers.

15.

1 log  a  2  log b  2log c 2 Solution

1 b a2 log  a  2  log b  2log c  log a  2  log b  log c2  log 2 c2 16.

1 ln a  2 ln b  ln c 3 Solution 3 a

2 1 1 a a ln a  2 ln b   ln c  ln 2  ln c  ln 3 2  ln c  ln b  3 3 b c b

Given that log 2 ≈ 0.3010 and log 3 ≈ 0.4771, approximate each value. Do not use a calculator.

17. log 24 Solution

log 24  log 8  3  log 23  3  3log 2  log 3  3  0.3010   0.4771  1.3801

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1387


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

18. log

8 3

Solution

log

8 23  log  3log 2  log 3  3  0.3010   0.4771  0.4259 3 3

Use the Change-of-Base Formula to find each logarithm. Do not attempt to simplify the answer.

19. log7 3 Solution log 7 3 

ln 3 log 3 or log 7 ln 7

20. log e Solution log  e 

ln e log e 1 or  log  ln  ln 

Determine whether each statement is true or false.

21. loga ab  1  loga b Solution log a ab  1  log a a  log a b  1  log a b TRUE

22.

log a  log a  log b log b Solution a log  log a  log b b FALSE

Find the solution.

23. pH of a solution Find the pH of a solution with a hydrogen ion concentration of 3.7 10–7. (Hint: pH = –log(H+).) Round to one decimal place. Solution

pH   log H 

  log 3.7  107

 6.4

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1388


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

24. Decibel gain Find the dB gain of an amplifier when EO = 60 volts and EI = 0.3 volt. (Hint: dB gain = 20 log(EO/EI).) Round to the nearest decibel. Solution EO EI 60  20 log 0.3  46 dB gain

dB gain  20log

Solve each equation. 2

25. 3x  2 x  27 Solution 2

3x 2 x  27 2

3x 2 x  33 x 2  2x  3 x 2  2x  3  0

 x  3 x  1  0 x  3  0 or

x10

x3

x  1

26. 3 x  1  100 x Solution

3x  1  100 x log 3x  1  log 100x

 x  1 log 3  x log 100 x log 3  log 3  2 x x log 3  2 x  log 3 x log 3  2   log 3

log 3 log 3  2 x  0.3133 x

27. 5e x  45 Solution

5e x  45 ex  9 ln e x  ln9 x ln e  ln9 x  ln9  2.1972

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1389


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

28. ln 5 x  2  ln 2 x  5

Solution

ln  5 x  2   ln  2 x  5  5x  2  2x  5 3x  3 x1

29. log x  log x  9  1 Solution

log x  log  x  9   1 log x  x  9   1

x  x  9   101

x 2  9 x  10  0

 x  10 x  1  0 x  10  0 or x  10

x10 x  1

extraneous

30. log 6 18  log 6 x  3  log 6 3 Solution

log 6 18  log 6  x  3   log 6 3 log 6

18  log 6 3 x 3 18 3 x 3 18  3 x  9 27  3 x 9x

CUMULATIVE REVIEW SOLUTIONS Graph each function.

1.

f  x   2  x  5  8 2

Solution f  x   2  x  5  8 2

a  2  up, vertex:  5,  8 

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1390


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

0  2  x  5  8 2

8  2  x  5 4   x  5

2

2

2  x  5 5  2  2

x  3, x  7   3, 0  ,  7, 0 

f  0   42   0, 42 

axis of symmetry: x  5

2.

f  x    x 2  6x  5 Solution

f  x    x 2  6x  5 a  1, b  6, c  5 6 b   3 x 2a 2  1 y   x 2  6x  5    3  6  3  5  4 2

vertex:  3, 4  , a  1  down 0   x 2  6x  5 0  x 2  6x  5

0   x  1 x  5  x  1 or x  5   1, 0  ,  5, 0 

f  0   5   0,  5 

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1391


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

3.

f  x   x3  x Solution

y  f  x   x3  x

f x   x   x  3

 x3  x

 f  x   odd x-int .

x3  x  0

x x2  1  0

y -int . y  03  0 y 0

0, 0

x 0

0, 0

4.

f  x    x 4  2x 2  1 Solution

f  x    x 4  2x 2  1

f x    x   2 x   1 4

2

  x 4  2 x 2  1  even x-int .  x 4  2x 2  1  0 x 4  2x 2  1  0 not rational numbers

y -int . f 0  1 y 1

0, 1

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1392


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Let P(x) = 4x3 + 3x + 2. Use synthetic division to find each value.

5.

P  1 Solution

P  1  9

1 4 0 3 2 4 4 7 4 4 7 9 6.

P  2 Solution

2 4

7.

0

3

2

8

16  38

4 8

19  36

P  2   36

 1 P  2

Solution 4 0 3 2

1 2

P  21   4

2 1 2 4 2 4 4

8.

P i  Solution

i

4

0

3

2

4i

4

i

4 4i

P i   2  i

1 2i

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1393


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Determine whether each binomial is a factor of P(x) = x3 + 2x2 – x – 2. Use synthetic division.

9.

x1

Solution

1

1 1

2

1 2

1

3

2

3

2

0

factor

10. x  2 Solution

2

1

2

1 2

2

0

2

0

1

0

2

1

2

1

1

2

1

2

0

1

factor

11. x  1 Solution

1

1 1

factor

12. x  2 Solution

2

1 1

2

1

2

2

8

14

4

7

12

not a factor

Determine how many zeros each function has.

 

13. P x  x 12  4 x 8  2 x 4  12 Solution

P  x   x 12  4 x 8  2 x 4  12  0 12 zeros

 

14. P x  x 2000  1 Solution

P  x   x 2000  1  0  2000 zeros

Determine the number of possible positive, negative, and nonreal zeros of each function.

 

15. P x  x 4  2x 3  3x 2  x  2

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1394


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solution

P  x   x 4  2 x 3  3 x 2  x  2: 2 sign variations  2 or 0 position roots P x   x   2 x   3 x   x   2 4

3

2

 x 4  2 x 3  3 x 2 : 2 sign variations  2 or 0 negative roots

# pos

# neg

# nonreal

2

2

0

2

0

2

0

2

2

0

0

4

 

16. P x  x 4  3 x 3  2 x 2  3 x  5 Solution

P  x   x 4  3x 3  2x 2  3x  5 1 sign variations  1 position roots P x   x   3 x   2 x   3 x   5 4

3

2

 x 4  3x 3  2x 2  3x  5 3 sign variations  3 or 1 negative roots # pos

# neg

# nonreal

1

3

0

1

1

2

Find the zeros of each polynomial function.

 

17. P x  x 3  x 2  9 x  9 Solution Possible rational roots

 1,  3,  9 Descastes’ Rule of Signs

# pos

# neg

1

2

0

1

0

2

Test x  1 :  1

# nonreal

1

9

9

1

0

9

0

9

0

1 1

x 3  x 2  9x  9  0

 x  1  x  9  0  x  1 x  3 x  3  0 Solution set: 1,  3, 3 2

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1395


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

18. P x  x 3  2 x 2  x  2 Solution Possible rational roots

 1,  2 Descastes' Rule of Signs

# pos

# neg

2

1

0

0

1

2

Test x  1 :  1

# nonreal

1 1

2

1

1

1

3

9

3

2

0

x 3  2x 2  x  2  0

 x  1  x  3x  2  0  x  1 x  1 x  2  0 Solution set: 1, 1, 2 2

Graph each function. Show all asymptotes.

19. f  x  

x x 3

Solution

x x 3 Vertical: x = 3; Horizontal: y = 1 Slant: none; x-intercepts: (0, 0) y-intercepts: (0, 0); Symmetry: none y 

20. f  x  

x2  1 x2  9

Solution y 

x2  1 x 9 2

 x  1 x  1  x  3 x  3

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1396


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Vertical: x = –3, x = 3; Horizontal: y = 1 Slant: none; x-intercepts: (–1, 0), (1, 0) y-intercepts: (0, 91 ); Symmetry: y-axis

Graph the function defined by each equation.

 

21. f x  3x  2 Solution

f  x   3x  2; Shift y  3x D2.

 

22. f x  2e x Solution

f  x   2ex ; Stretch y  ex vertically by a factor of 2.

 

23. f x  log 3 x Solution

f  x   log 3 x

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1397


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

 

24. f x  ln x  2

Solution

f  x   ln  x  2 ; Shift y  ln x R2.

Find each value.

25. log2 64 Solution log 2 64  6

because 2  64  6

26. log 1/2 8 Solution log 1/2 8  3

because    8 1 2

3

27. ln e3 Solution

ln e3  3 ln e  3 28. 2log2 2 Solution log 2 2

2

2

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1398


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

Solve for x.

29. log5 x  3 Solution log 5 x  3 53  x 1 x 125

30. log x 72  2 Solution log x 72  2

x 2  72 x  72  6 2 Write each expression in terms of the logarithms of a, b, and c. Assume a, b, and c are positive numbers.

31. log abc Solution log abc  log a  log b  log c

32. log

a2 b c

Solution log

a2 b  log a2 b  log c c  log a2  log b  log c  2log a  log b  log c

33. log

ab c3

Solution 1/2

log

 ab  ab  log  3  3 c c   ab  1  log  3  2 c  1  log ab  log c 3 2 1  log a  log b  3log c  2

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1399


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

34. ln

ab2 c

Solution

2

 ln ab

ab ab2 ln  ln c c

1/2

1/2

 ln c

1 ln ab2  ln c 2 1  ln a  ln b2  ln c 2 1  ln a  2ln b   ln c 2 1  ln a  ln b  ln c 2 

Write each expression as the logarithm of a single quantity. Assume a, b, and c are positive numbers. 35. 3 ln a  3 ln b

Solution

3 ln a  3 ln b  ln a3  ln b3  ln

36.

a3 b3

1 2 log a  3 log b  log c 2 3 Solution

a b3 1 2 a 1/2 b3 log a  3 log b  log c  log a 1/2  log b3  log c 2/3  log 2/3  log 3 2 2 3 c c Solve each equation. 37. 3 x  1  8

Solution 3x  1  8 log 3 x  1  log 8

 x  1 log 3  log 8 log 8 log 3 log 8 x log 3

x1

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1400


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

38. 3 x  1  32 x

Solution

3x  1  32 x x  1  2x 1  x 39. log x  log 2  3

Solution log x  log 2  3

log 2 x  3 103  2 x 1000  2 x 500  x

40. log x  1  log x  1  1

Solution

log  x  1  log  x  1  1 log  x  1 x  1  1

log x 2  1  1 101  x 2  1 10  x 2  1 11  x 2 

11  x

Only the positive answer, x 

11, checks.

GROUP ACTIVITY SOLUTIONS Installment Loans What Are Fixed Installment Loans? A fixed installment loan is a loan that is repaid in equal payments. Sometimes part of the cost is paid at the time of purchase. This amount is the down payment. Real-World Example of Installment Loans Installment loans allow you the ability to purchase an item and use it now. This is called installment purchasing and being able to use the item is an advantage. The disadvantage is that interest is paid on the amount borrowed. A common example is purchasing an automobile.

Group Activity 1.

Suppose you graduate from college, obtain an amazing job, get married, and then purchase a new vehicle. Use the given information below to answer four questions.

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1401


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 5: Exponential and Logarithmic Functions

   

Cost of automobile: $26,500 Down payment: $2,000 Monthly payment: $550.25 Loan term: 48 months

a. What is the amount financed? Note that the amount financed is the price of the car minus the down payment. b. What is the total amount of all monthly payments? c. What is the total installment price? Note that the total installment price is the sum of all the monthly payments plus the down payment. d. What was the financial charge and explain what it represents? Note that the financial charge is the total installment price minus the purchase price of the automobile.

Solution a. $26,500  $2000  $24,500 b.

$550.25  48  $26, 412

c.

$26,500  $2000  $28,500

d.

$28, 412  $26,500  $1912 ; interest paid over 48 months

2. Prior to making an automobile purchase, it is often helpful to know in advance what the monthly payment will be. The monthly payment can be determined using the formula shown below, where M is the monthly payment, P is the principal value of the loan, r is the APR (annual percentage rate) in decimal form, and n is the total number of payments.

M

r  P  t   r  1 1  12  

n

a. Using the internet, identify the MSRP (Manufacturer’s Suggested Retail Price) of an automobile you would be interested in purchasing. A suggested website would be www.edmunds.com. b. Use the MSRP from part a and the given information below to determined the monthly payment for the automobile. Compare your calculated monthly payment with an online loan calculator. A suggested loan calculator is found at www.bankrate.com.  APR is 3.89%  60 months  Trade-in value of your current automobile is $4,250 c. What was the total installment price? d. What was the financial charge?

Solution (a–d) answers will vary.

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1402


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER 6: SYSTEMS OF LINEAR EQUATIONS, MATRICES, AND INEQUALITIES

TABLE OF CONTENTS End of Section Exercise Solutions ................................................................................ 1403 Exercises 6.1 ........................................................................................................................... 1403 Exercises 6.2 .......................................................................................................................... 1443 Exercises 6.3 .......................................................................................................................... 1485 Exercises 6.4 .......................................................................................................................... 1506 Exercises 6.5 .......................................................................................................................... 1530 Exercises 6.6 .......................................................................................................................... 1560 Exercises 6.7 .......................................................................................................................... 1585 Exercises 6.8 .......................................................................................................................... 1606 Chapter Review Solutions.............................................................................................. 1627 Chapter Test Solutions ................................................................................................... 1651 Group Activity Solutions ................................................................................................ 1660

END OF SECTION EXERCISE SOLUTIONS EXERCISES 6.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Graph the linear equation y  21 x  3. Solution

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1403


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2. Graph the linear equation 5 x  3 y  15. Solution

3. Given 3x  2 y  11. a. Does the ordered pair (3, 1) satisfy the equation? b. Does the ordered pair

 , 5 satisfy the equation? 1 3

Solution a. no b. yes 4. Given 2x  3 y  4z  1. a. Does x = 1, y = –1, and z = –1, satisfy the equation? b. Does x = –1, y = 1, and z = 1, satisfy the equation? Solution a. yes b. no 5. Given x – 2y = 2. Substitute 5x + 8 in for y and solve for x. Solution

x  2 5 x  8  2 x  10 x  16  2 9 x  18 x  2

6. Given 4x – 3y + z = 20. Substitute –4 in for y and 6 in for z and solve for x. Solution

4 x  3  4   6  20 4 x  12  6  20 4 x  18  20 4x  2 1 x 2

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1404


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A set of several equations with several variables is called a __________ of equations. Solution system 8. Any set of numbers that satisfies each equation of a system is called a __________ of the system. Solution solution 9. If a system of equations has a solution, the system is __________. Solution consistent 10. If a system of equations has no solution, the system is __________. Solution inconsistent 11. If a system of equations has only one solution, the equations of the system are __________. Solution independent 12. If a system of equations has infinitely many solutions, the equations of the system are __________. Solution dependent

 x  y  5 is __________ (consistent, inconsistent). 13. The linear system   x  y  1 Solution consistent

 x  y  5 is __________ (consistent, inconsistent). 14. The linear system   x  y  1 Solution inconsistent

 x  y  5 are __________ (dependent, independent). 15. The equations of the linear system  2 x  2 y  10

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1405


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution dependent

 x  y  5 are __________ (dependent, independent). 16. The equations of the linear system   x  y  1 Solution independent

 x  2 y  7 . 17. The ordered pair (1, 3) __________ (is, is not) a solution of the linear system  2x  y  1 Solution is

 3x  y  6 . 18. The ordered pair (1, 3) __________ (is, is not) a solution of the linear system   x  3 y  8 Solution is Practice Solve each system of linear equations by graphing. If the system has no solution, write no solution; inconsistent system. If the system has an infinite number of solutions, write dependent equations; infinite number of solutions.

 y  3 x  5 19.   x  2 y  3 Solution ìïï y = - 3 x + 5 í ïïî x - 2 y = - 3

solution: (1, 2)

 x  2 y  3 20.  3 x  y  9 Solution ìïï x - 2 y = -3 í ïïî3 x + y = -9

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1406


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

solution: (–3, 0)

3 x  2 y  2 21.  2 x  3 y  16 Solution ìïï 3 x + 2 y = 2 í ïïî 2 x + 3 y = 16

solution: (–2, 4)

 x  y  0 22.  5 x  2 y  14 Solution ìïï x + y = 0 í ïïî5 x - 2 y = 14

solution: (2, –2)

 y   x  5 23.   x  y  10 Solution ìïï y = - x + 5 í ïïî3 x + 3 y = 30

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1407


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

no solution inconsistent system

 x  3 y  3 24.  2 x  6 y  12 Solution ìïï x - 3 y = -3 í ïïî2 x - 6 y = 12

no solution inconsistent system

 y  x  6  25.  1 1  x y 3 2 2 Solution  y   x  6  5 x  5 y  30

dependent equations infinitely many solutions

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1408


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 2 x  y  3  26.  1 3 x  y    2 2 Solution 2 x  y  3  8 x  4 y  12

dependent equations infinitely many solutions Use a graphing calculator to approximate the solutions of each system of linear equations. Give answers to the nearest tenth.

 y  5.7 x  7.8 27.   y  37.2  19.1x Solution ìïï y = - 5.7 x + 7.8 í ïïî y = 37.2 - 19.1x

solution: (2.2, –4.7)

 y  3.4 x  1 28.   y  7.1x  3.1 Solution ìïï y = 3.4 x - 1 í ïïî y = -7.1x + 3.1

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1409


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

solution: (0.4, 0.3)

 5.5  2.7 x y  29.  3.5 5.3 x  9.2 y  6.0  Solution ìï ïï y = 5.5 - 2.7 x í 3.5 ïï ïî5.3 x - 9.2 y = 6.0

solution: (1.7, 0.3)

29x  17 y  7 30.  17 x  23 y  19 Solution ìïï29 x + 17 y = 7 í ïïî 17 x + 23 y = 19

solution: (–0.2, 0.7)

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1410


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solve each system of linear equations by substitution, if possible. If the system has no solution, write no solution; inconsistent system. If the system has an infinite number of solutions, write dependent equations and provide a general solution.

 y  x  1 31.   y  2 x Solution ìïï (1) y = x - 1 í ïïî (2) y = 2 x Substitute y = x - 1 from (1) into (2):

y = 2x x - 1 = 2x -1 = x Substitute and solve for y : y = 2x y = 2(-1) = -2 (-1, - 2)

 y  2 x  1 32.   x  y  5 Solution ìïï(1) y = 2 x - 1 í ïïî(2) x + y = 5 Substitute y = 2 x - 1 from (1) into (2):

x+ y =5 x + 2x - 1 = 5 3x = 6 x=2 Substitute and solve for y : y = 2x - 1 y = 2(2) - 1 = 3 (2, 3)

2 x  3 y  0 33.   y  3 x  11 Solution ìïï(1) 2 x + 3 y = 0 í ïïî(2) y = 3 x - 11 Substitute y = 3 x - 11 from (2) into (1):

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1411


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2x + 3 y = 0 2 x + 3(3 x - 11) = 0 2 x + 9 x - 33 = 0 11x = 33 x=3 Substitute and solve for y : y = 3 x - 11 y = 3(3) - 11 = -2 (3, - 2)

2x  y  3 34.   y  5x  11 Solution ìïï (1) 2 x + y = 3 í ïïî (2) y = 5 x - 11 Substitute y = 5 x - 11 from (2) into (1): 2x + y = 3 2 x + 5 x - 11 = 3 7 x = 14 x=2 Substitute and solve for y : y = 5 x - 11 y = 5(2) - 11 = -1 (2, - 1)

4 x  3 y  3 35.  2x  6 y  1 Solution ìïï (1) 4 x + 3 y = 3 í ïïî (2) 2 x - 6 y = -1 3 - 4x from (1) into (2): 3 2 x - 6 y = -1 3 - 4x = -1 2 x - 6. 3 2 x - 2(3 - 4 x ) = -1 2 x - 6 + 8 x = -1 10 x = 5 x = 21 Substitute y =

Substitute and solve for y :

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1412


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

4x + 3 y = 3 4 ( 21 ) + 3 y = 3 2 + 3y = 3 3y = 1 y = 31

Solution: ( 21 , 31 )

4 x  5 y  4 36.  8x  15 y  3 Solution ìïï(1) 4 x + 5 y = 4 í ïïî(2) 8 x - 15 y = 3 4 - 4x Substitute y = from (1) into (2): 5 8 x - 15 y = 3 4 - 4x 8 x - 15. =3 5 8 x - 3(4 - 4 x ) = 3 8 x - 12 + 12 x = 3 20 x = 15 x = 43 Substitute and solve for y : 4x + 5 y = 4 4 ( 43 ) + 5 y = 4 3 + 5y = 4 5y = 1 y = 51 Solution: ( 43 , 51 )

2x  3 y  6 37.  5x  4 y  8 Solution  (1) 2x  3 y  6  (2) 5 x  4 y  8 Substitute x  

3 y  3 from (1) into (2): 2

 3  5   y  3   4 y  8  2  15 y  15  4 y  8 2 7 y  7 2 y  2

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1413


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Substitute and solve for x: 2 x  3  2   6 2 x  6  6 2x  0 x 0

Solution: 0,  2

3x  2 y  3 38.  2x  3 y  2 Solution  (1) 3 x  2 y  3  (2) 2 x  3 y  2 Substitute x 

2 y  1 from (1) into (2): 3

2  2  y  1  3 y  2 3  4 y  2  3y  2 3 5 y 0 3 y 0

Substitute and solve for x: 3 x  2 0   3 x1

 

Solution: 1, 0

 x  3 y  1 39.  2 x  6 y  3 Solution ìïï (1) x + 3 y = 1 í ïïî (2) 2 x + 6 y = 3 Substitute x = 1 - 3 y from (1) into (2):

2x + 6 y = 3 2(1 - 3 y ) + 6 y = 3 2 - 6y + 6y = 3 2¹3 Inconsistent system  No solution

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1414


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 x  3 y  14 40.  3( x  12)  9 y Solution ìïï (1) x - 3 y = 14 í ïïî (2) 3( x - 12) = 9 y

Substitute x = 3 y + 14 from (1) into (2): 3( x - 12) = 9 y 3(3 y + 14 - 12) = 9 y 3(3 y + 2) = 9 y 9y + 6 = 9y 6¹0 Inconsistent system  No solution  y  3x  6  41.  1 x  y  2 3  Solution ìïï (1) y = 3 x - 6 í ïïî (2) x = 31 y + 2 Substitute x = 31 y + 2 from (2) into (1): y = 3x - 6 y = 3( 31 y + 2 ) - 6 y = y +6-6 0=0 Dependent equations General solution: (x , 3x - 6)

3 x  y  12 42.   y  3 x  12 Solution ìïï (1) 3 x - y = 12 í ïïî (2) y = 3 x - 12 Substitute y = 3 x - 12 from (2) into (1): 3 x - (3 x - 12) = 12 3 x - 3 x + 12 = 12 0=0 Dependent equations General solution: (x , 3x - 12)

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1415


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solve each system of linear equations by the elimination method, if possible. If the system has no solution, write no solution; inconsistent system. If the system has an infinite number of solutions, write dependent equations and provide a general solution.

5x  3 y  12 43.  2x  3 y  3 Solution 5 x - 3 y = 12  ´ (-1) 2x - 3 y = 3

-5 x + 3 y = -12 2x - 3 y = 3 -3 x x

= -9 = 3

2x - 3 y = 3 2(3) - 3 y = 3 6 - 3y = 3 - 3 y = -3 y=1

Solution: (3, 1)

2x + 3 y = 8 2(1) + 3 y = 8 2 + 3y = 8 3y = 6 y =2

Solution: (1, 2)

 2x  3 y  8 44.  5 x  y  3 Solution

2x + 3 y = 8 -5 x + y = -3  ´ (-3)

2x + 3 y = 8 15 x - 3 y = 9 17 x x

= 17 = 1

 x  7 y  11 45.  8x  2 y  28 Solution

x - 7 y = -11  ´ (-8) 8 x + 2 y = 28

-8 x + 56 y = 88 8 x + 2 y = 28 58 y = 116 y = 2

x - 7 y = -11 x - 7(2) = -11 x - 14 = -11 x=3

Solution: (3, 2)

3 x  9 y  9 46.   x  5 y  3 Solution

3x + 9 y = 9 -x + 5 y = -3  ´ (3)

3x + 9 y = 9 -3 x + 15 y = - 9 24 y = y =

0 0

3x + 9 y = 9 3 x + 9(0) = 9 3x + 0 = 9 3x = 9 x=3

Solution: (3, 0)

4 x  5 y  6 47.  5x  7 y  9

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1416


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

4 x  5 y  6   (5) 5 x  7 y  9   ( 4)

20 x - 25 y = 30 -20 x + 28 y = - 36 3y =- 6 y =-2

4 x - 5 (-2) = 6

Solution:

4 x + 10 = 6 4 x = -4 x = -1

(-1, - 2)

2 x  4 y  16 48.  7 x  3 y  15 Solution

2 x - 4 y =- 16  ´ (7) 7 x + 3 y =- 15  ´ (-2) 14 x - 28 y =- 112 -14 x - 6 y = 10

2 x - 4 (3) = -16 2 x - 12 = -16 2 x = -4 x = -2

Solution:

(-2, 3)

- 34 y =- 102 y =3

3( x  y )  y  9 49.  5( x  y )  15 Solution

3( x - y ) = y - 9  3 x - 3 y = y - 9  3 x - 4 y = - 9  ´ (5) 15 x - 20 y = -45 5( x + y ) = - 15  5 x + 5 y = - 15  5 x + 5 y = -15  ´ (4) 20 x + 20 y = -60 35 x x

5 x + 5 y = -15 5(-3) + 5 y = -15 -15 + 5 y = -15 5y = 0 y =0

= -105 = -3

Solution: (-3,0)

2( x  y )  y  1 50.   3( x  1)  y  3 Solution

2( x + y ) = y + 1  2 x + 2 y = y + 1  2 x + y = 1 3( x + 1) = y - 3  3 x + 3 = y - 3  3 x - y = -6 5x x

= -5 = -1

2x + y = 1 2(-1) + y = 1 -2 + y = 1 y =3

Solution: (-1, 3)

 1 2  xy  51.  2  3 xy 

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1417


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 1 x+y 3 2= x-y

2=

 2( x + y ) = 1  2 x + 2 y = 1  2( x - y ) = 3  2 x - 2 y = 3 4x x

=4 =1

2x + 2 y = 1 Solution: 2 (1) + 2 y = 1 (1, - 21 ) 2 + 2y = 1 2 y = -1 y = - 21

 1  12  x  y 52.   3 x  4  y Solution 1 -12 x - 12 y = -1 = 12  1 = 12 ( x + y ) -12 x - 12 y = -1 x+y 3x - 4 y  3 x + 4 y = 0  ´ (4) 12 x + 16 y = 0 = -4  3 x = y 4 y = -1 y = - 41

3 x = -4 y 3 x = -4 (- 41 ) 3x = 1 x = 31

Solution: ( 31 , - 41 )

 y  2x  5 53.  0.5 y  2.5  x Solution y + 2x = 5  2x + y= 5 2x + y = 5 0.5 y = 2.5 - x  x + 0.5 y = 2.5  ´ (-2) -2 x - y = -5 0 = 0

Dependent Equations 2x + y = 5 y = -2 x + 5 General solution: ( x, - 2 x + 5)

0.3x  0.1 y  0.1 54.  6 x  2 y  2 Solution -0.3 x + 0.1 y = -0.1  ´ (20) 6x - 2 y = 2

-6 x + 2 y = - 2 6x - 2 y = 2 0= 0

Dependent Equations 6x - 2 y = 2 -2 y = 2 - 6 x y = -1 + 3 x = 3 x - 1 General solution: ( x, 3 x - 1)

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1418


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 x  2( x  y )  2 55.  3( y  x )  y  5 Solution x + 2 ( x - y ) = 2  x + 2x - 2 y = 2  3x - 2 y = 2 3 ( y - x ) - y = 5  3 y - 3 x - y = 5  -3 x - 2 y = 5

No Solution Inconsistent system

0¹7

3 x  4(2  y ) 56.  3( x  2)  4 y  0 Solution 3 x = 4 (2 - y )  3x = 8 - 4 y  3x + 4 y = 8 3x + 4 y = 8 3 ( x - 2) + 4 y = 0  3 x + 4 y = 6  ´ (-1) -3 x - 4 y = - 6  3x - 6 + 4 y = 0 0 ¹

2

No Solution  Inconsistent System

 y 5  x   3 3 57.  x y    3 x  3 Solution

5 y  3x + y = 5  ´ (3) 3 x + y = 5 3x + y = 5 = 3 3 x+y = 3 - x  ´ (3) x + y = 9 - 3 x  4 x + y = 9  ´ (-1) -4 x - y = -9 3 -x =- 4 x = 4

x+

3 x + y = 5 Solution: 3 (4) + y = 5 (4, - 7) 12 + y = 5 y = -7 3 x  y  0.25  58.  3  x  y  2.375  2 Solution 3 x - y = 0.25  3 x - y = 0.25  ´(3) 9 x - 3 y = 0.75 x + 32 y = 2.375  ´(2) 2 x + 3 y = 4.75  2 x + 3 y = 4.75 11x x

= =

5.5 0.5

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1419


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3 x - y = 0.25 3 (0.5) - y = 0.25 1.5 - y = 0.25 1.25 = y

Solution: (0.5, 1.25)

3 1  x  y  2 3 59.  2 2 x  1 y  1  3 9 Solution 3 1 x + y = 2  ´ (6) 9 x + 2 y = 12  9 x + 2 y = 12 2 3 2 1 x + y = 1  ´ (9) 6 x + y = 9  ´ (-2) -12 x - 2 y = -18 3 9 -3 x = -6 x = 2

6x + y = 9 Solution: 6 (2) + y = 9 (2, - 3) 12 + y = 9 y = -3

x  y x  y  2  5 60.  2 x  y  1  2 Solution

x+y x-y + = 2 5 x =

2  ´ (10) 5 ( x + y ) + 2 ( x - y ) = y + 1  ´ (2) 2

7 x + 3 y = 20 2 x - y = 2  ´ (3)

5 x + 5 y + 2 x - 2 y = 20

2x = y + 2 

7 x + 3 y = 20 6x - 3 y = 6 13 x x

20 

= 26 = 2

7 x + 3 y = 20 7 (2) + 3 y = 20 14 + 3 y = 20 3y = 6 y =2

2x - y =

2

Solution: (2, 2)

x  y x  y  6  2 61.  5 x  y  x  y  3  2 4 Solution x- y x+y + = 6  ´ (10) 2 ( x - y ) + 5 ( x + y ) = 60  2 x - 2 y + 5 x + 5 y = 60 5 2 x-y x+ y = 3  ´ (8) 4 ( x - y ) - 2 ( x + y ) = 24  4 x - 4 y - 2 x - 2 y = 24 2 3

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1420


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

7 x + 3 y = 60  ´ (2) 2 x - 6 y = 24

7 x + 3 y = 60 Solution: 14 x + 6 y = 120 7 2 x - 6 y = 24 (9) + 3 y = 60 (9, - 1) 63 + 3 y = 60 16 x = 144 3 y = -3 x = 9 y = -1

x  2 y  3  5  2 62.  5 x  3  y  2  6  2 3 Solution x -2 y +3 + = 5  ´ (10) 2 ( x - 2) + 5 ( y + 3) = 50  2 x - 4 + 5 y + 15 = 50 5 2 x +3 y -2 + = 6  ´ (6) 3 ( x + 3) + 2 ( y - 2) = 36  3 x + 9 - 2 y - 4 = 36 2 3 2 x + 5 y = 39  ´ (-2) 3 x + 2 y = 31  ´ (5)

-4 x - 10 y = -78 15 x + 10 y = 155 = =

11x x

77 7

3 x + 2 y = 31 3 (7) + 2 y = 31 2 y = 10 y =5

Solution:

(7, 5)

Solve each system of linear equations by any method, if possible. If the system has no solution, write no solution; inconsistent system. If the system has an infinite number of solutions, write dependent equations and provide a general solution. x  y  z  3  63. 2 x  y  z  4 3 x  y  z  5 

Solution

Add equations (2) and (3) : (1) x + y + z = 3 Add (1) and (3) : (2) 2x + y + z = 4 (1) x + y + z = 3 (2) 2x + y + z = 4 (3) 3x + y - z = 5 (3) 3x + y - z = 5 (3) 3x + y - z = 5 =9 (4) 4 x + 2 y = 8 (5) 5x + 2 y Solve the system of two equations and two unknowns formed by equations (4) and (5) : 4 x + 2 y = 8  ´ (-1) -4 x - 2 y = -8 5x + 2 y = 9  5x + 2 y = 9 x

=

1

4x + 2 y = 8 4 (1) + 2 y = 8 2y = 4 y =2

x + y + z = 3 Solution: 1 + 2 + z = 3 (1, 2, 0) 3+ z = 3 z=0

x  y  z  0  64.  x  y  z  0 x  y  z  2 

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1421


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

Add equations (2) and (3) : (1) x - y - z = 0 Add (1) and (3): (2) x + y - z = 0 (1) x - y - z = 0 (2) x + y - z = 0 (3) x - y + z = 2 (3) x - y + z = 2 (3) x - y + z = 2 = 2 (5) 2 x =2 (4) 2x - 2 y =1

x

Solve the system of two equations and two unknowns formed by equations (4) and (5) : 2x - 2 y = 2 2 (1) - 2 y = 2 -2 y = 0 y =0

x-y +z =2 1-0 + z = 2 1+z=2 z=1

Solution: (1, 0, 1)

x  y  z  0  65.  x  y  2z  1  x  y  z  0 

Solution

(1) x - y + z = 0 Add (1) and (2): (2) x + y + 2z = -1 (1) x - y + z = 0 (3) -x - y + z = 0 (2) x + y + 2z = -1 (4) 2 x + 3z = -1

Add equations (2) and (3) : (2) x + y + 2z = -1 (3) -x - y + z = 0

(5)

3 z = -1 z = - 31

Solve the system of two equations and two unknowns formed by equations (4) and (5) : 2 x + 3 z = -1 2 x + 3 (- 31 ) = -1 2x = 0 x=0

x- y +z =0 0 - y + (- 31 ) = 0

Solution: (0, - 31 , - 31 )

- y = 31 y = - 31

2 x  y  z  7  66.  x  y  z  2  x  y  3z  2 

Solution

(1) 2 x + y - z = 7 Add (1) and (2): (2) x - y + z = 2 (1) 2 x + y - z = 7 (3) x + y - 3z = 2 (2) x - y + z = 2 =9 (4) 3 x x

=3

Add equations (2) and (3) :

(2) x - y + z = 2 (3) x + y - 3z = 2 - 2z = 4 (5) 2x

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1422


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solve the system of two equations and two unknowns formed by equations (4) and (5) 2 x - 2z = 4 2 (3) - 2z = 4 - 2z = -2 z= 1

x + y - 3z = 2 3 + y - 3 (1) = 2 y =2

Solution: (3, 2, 1)

x  y  z  6  67. 2 x  y  3z  17  x  y  2z  11 

Solution (1) x + y + z = 6

(2) 2 x + y + 3z = 17 (3) x + y + 2z = 11

Add (1) and (2) :

Add equations (1) and (3) :

(1) x + y + z = 6 (2) -2x - y - 3z = -17 - 2z = -11 (4) - x

(2) (3) (5)

x+ y+z = 6 - x - y - 2z = -11 - z= -5 z= 5

Solve the system of two equations and two unknowns formed by equations (4) and (5) : - x - 2z = -11 - x - 2 (5) = -11 - x = -1 x=1

x+ y+ z=6 1+ y + 5 = 6 y =0

Solution: (1, 0, 5)

x  y  z  3  68. 2 x  y  z  6  x  2 y  3z  2 

Solution (1) x + y + z = 3

(2) 2x + y + z = 6 (3) x + 2 y + 3z = 2

Add - (1) and (2) :

Add equations - 2 ⋅ (1) and (3) :

- (1) - x - y - z = -3 (2) 2x + y + z = 6

-2 ⋅ (1)

(4)

x

=

3

(3)

-2 x - 2 y - 2z =- 6 x + 2 y + 3z = 2

(5)

-x

+ z = -4

Solve the system of two equations and two unknowns formed by equations (4) and (5) : -x + z = -4 - 3 + z = -4 z =- 1

x+y+ z=3 3 + y + (-1) = 3 y=1

Solution: (3, 1, - 1)

3 x  4 y  2 z  4  69. 6 x  2 y  z  4 3 x  8 y  6 z   3 

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1423


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

(1) 3 x + 4 y + 2z = 4 (2) 6 x - 2 y + z = 4 (3) 3 x - 8 y - 6z = -3

Add (1) and 2 ⋅ (2) :

Add equations 2 ⋅ (1) and (3) :

(1) 3x + 4 y + 2z = 4 2 ⋅ (2) 12 x - 4 y + 2z = 8 + 4z = 12 (4) 15x

2 ⋅ (1) 6 x + 8 y + 4z = 8 (3) 3x - 8 y - 6z = -3

(5) 9x

- 2z = 5

Solve the system of two equations and two unknowns formed by equations (4) and (5) : 15 x + 4 z = 12  9 x - 2z = 5  ´(2)

15 x + 4 z = 12 18 x - 4 z = 10 = 22 33 x = 23 x

Solution: ( 23 , 41 , 21 )

15 x + 4 z = 12 15 ( 23 ) + 4 z = 12 4z = 2 z = 21

3 x + 4 y + 2z = 4 3 ( ) + 4 y + 2 ( 21 ) = 4 4y = 1 y = 41 2 3

2 x  y  z  0  70.  x  2 y  z  1  x  y  2 z  1 

Solution

(1) 2 x - y - z = 0 (2) x - 2 y - z = -1 (3) x - y - 2z = -1

Add (1) and - 2 ⋅ (2) :

Add equations (1) and - 2 ⋅ (3) :

(1) 2x - y - z = 0 -2 ⋅ (2) -2 x + 4 y + 2z = 2 3y + z =2 (4)

(1) 2x - y - z = 0 (3) -2x + 2 y + 4z = 2 y + 3z = 2 (5)

Solve the system of two equations and two unknowns formed by equations (4) and (5) : 3 y + z = 12 3y + z = 2  y + 3z = 2  ´(-3) -3 y - 9z =- 6 -8 z = -4 z = 21

y + 3z = 2 y + 3 ( 21 ) = 2 y = 21

x - y - 2 z = -1 x - 21 - 2 ( 21 ) = -1 x = 21

Solution: ( 21 , 21 , 21 )

3 x  y  z  0  71. 2 x  y  z  0 2 x  y  z  0 

Solution

(1) 3 x + y + z = 0 Add (1) and (2): (2) 2 x - y + z = 0 (1) 3 x + y + z = 0 (3) 2 x + y + z = 0 (2) 2 x - y + z = 0 (4) 5 x + 2z = 0

Add equations (2) and (3) :

(2) 2x - y + z = 0 (3) 2x + y + z = 0 (5) 4 x + 2z = 0

Solve the system of two equations and two unknowns formed by equations (4) and (5):

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1424


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

5x + 2z = 0  ´ (-1) 4 x + 2z = 0 

-5 x - 2 z = 0 4 x + 2z = 0 -x x

=0 =0

4 x + 2z = 0 4 (0) + 2z = 0 2z = 0 z=0

3x + y + z = 0 3 (0) + y + 0 = 0 y =0

Solution: (0, 0, 0)

2 x  y  4  72.  x  z  2 y  z  1 

Solution

(1) 2 x + y = 4 (2) x - z = 2 (3) y + z = 1

Add (2) and (3) :

-z = 2 (2) x y + z=1 3 () =3 (4) x + y

Solve the system of two equations and two unknowns formed by equations (1) and (4):

2 x + y = 4  ´ (-1) x+y =3 

-2 x - y = -4 x+y = 3 -x x

= -1 = 1

x+y =3 1+ y = 3 y =2

y+z =1 2+ z = 1 z = -1

Solution:

(1, 2, - 1)

x  2 y  z  2  73. 2 x  y  1 3 x  y  z  1 

Solution

(1) x + 2 y - z = 2 Add (1) and (3): 2 x - y = -1 (1) x + 2 y - z = 2 (2) (3) 3 x + y + z = 1 (3) 3x + y + z = 1 (4) 4 x + 3 y = 3 Solve the system of two equations and two unknowns formed by equations (2) and (4):

2 x - y = -1  ´(3) 4x + 3 y = 3 

6 x - 3 y =- 3 4x + 3 y = 3 10 x = 0 x = 0

2 x - y =- 1 3 x + y + z = 1 Solution: 2 (0) - y = -1 3 (0) + 1 + z = 1 (0, 1, 0) - y = -1 z=0 y=1

( x  y )  ( y  z )  1  74 ( x  z )  ( x  z )  3 ( x  y )  ( x  z )  1 

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1425


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

(1) ( x + y ) + ( y + z ) = 1  (2) ( x + z ) + ( x + z ) = 3  (3) ( x - y ) - ( x - z ) = -1 

x + 2y + z = 1 2 x + 2z = 3 - y + z = -1

Add - 2 ⋅ (1) and (2) : -2 ⋅ (1) - 2 x - 4 y - 2z = -2 + 2z = 3 (2) 2 x

(4) - y + z = -1 - (- 41 ) + z = -1 z = - 45

2 x + 2z = 3 2 x + 2 (- 45 ) = 3 2 x = 22 4 x = 114

-4 y y

= 1 = - 41

Solution: ( 114 , - 41 , - 45 )

x  y  z  3  75.  x  z  2 2 x  2 y  2z  3 

Solution

x+ y +z =3 Add - 2 ⋅ (1) and (3) : (1) 2 ⋅ (1) - 2 x - 2 y - 2z = -6 2 x+z =2 () (3) 2 x + 2 y + 2z = 3 (3) 2 x + 2 y + 2z = 3 0 ¹- 3 (4)

No solution; inconsistent system

x  y  2  76.  y  z  2 3 x  3 y  2 

Solution

(1) x + y = 2 (2) y + z = 2 (3) 3x + 3 y = 2

Add - 3 ⋅ (1) and (3) :

Inconsistent system  No solution

-3 ⋅ (1) -3 x - 3 y = -6 (3) 3x + 3 y = 2

(4)

0 ¹ -4

x  3 y  z  5  77. 3 x  y  z  2 2 x  y  1 

Solution

(1) x + 3 y - z = 5 (2) 3 x - y + z = 2 2x + y = 1 (3)

Add (1) and (2) :

(1) x + 3 y - z = 5 (2) 3x - y + z = 2 =7 (4) 4 x + 2 y

Add equations - 2 ⋅ (3) and (4) : -2 ⋅ (3) -4 x - 2 y =- 2 (4) 4 x + 2 y = 7

(5)

5

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No solution Inconsistent System

1426


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

x  y  z  3  78.  x  z  2 2 x  y  2 z  5 

Solution

(1) x + y + z = 3 (2) x+ z=2 (3) 2 x + y + 2z = 5

Add (1) and - (2) :

(1) x + y + z = 3 - (2) -x - z = -2 (4) y = 1

Add equations - 2 ⋅ (1) and (3) : -2 (1) -2 x - 2 y - 2z =- 6 (3) 2x + y + 2z = 5

(5)

-y

= -1

Since both additions resulted in the same equation, the equations are dependent. Thus, y must equal 1, but x and z can be any real numbers that satisfy the equations. Notice that if the value y = 1 is substituted into any of the equations, x + z = 2 is the result. So y = 1, and x and z must satisfy x + z = 2. Solution: x = any real number, y = 1, z = 2 - x  ( x, 1, 2 - x ) x  y  2  79.  y  z  2 x  z  0 

Solution (1) x + y = 2

(2) y + z = 2 (3) x - z = 0

Add (1) and - (3) :

(1) x + y = 2 +z =0 - (3) -x (4)

y+z =2

Since (4) is the same as (2), the equations are dependent. Let x = any real number. Then, from (1), y = 2 – x. Finally, substituting for y in (2) yields z = 2 – y = (2 – x) = x. Solution: ( x, 2 - x, x )

( x  y )  ( y  z )  ( z  x )  6  80. ( x  y )  ( y  z )  ( z  x )  0  x  y  2z  4 

Solution

(1) ( x + y ) + ( y + z ) + (z + x ) = 6  (2) ( x - y ) + ( y - z ) + ( z - x ) = 0  x + y + 2z = 4 (3)

2 x + 2 y + 2z = 6 0=0 x + y + 2z = 4

Add (1) and - 2 ⋅ (3) :

(1) 2 x + 2 y + 2z = 6 -2 ⋅ (3) -2 x - 2 y - 4 z = -8 (4)

- 2 z = -2 z= 1

Since (2) is always true, the equations are dependent. z must equal 1. Then, from (3), x + y = 2, Let x = any real #. Then y = 2 – x. Solution: ( x, 2 - x, 1)

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1427


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Fix It In exercises 81 and 82, identify the step the first error is made and fix it.

3 x  y  1 . 81. Use the substitution method to solve the linear system   x  2 y  8 Solution Step 3 was incorrect. Step 1: y  3 x  1

Step 2:  x  2 3 x  1  8 Step 3: x  2 Step 4: Use back substitution and solve for y: y  5

3 x  4 y  11 . 82. Use the elimination method to solve the linear system  5 x  6 y  12 Solution Step 4 was incorrect.

 15 x  20 y  55 Step 1:  15 x  18 y  36 Step 2: 38 y  19 Step 3: y  

1 2

Step 4: Use back substitution and solve for x: x  3

Let x represent the first number and y represent the second number. Use a system of linear equations to solve and find the numbers. 83. The sum or two numbers is 92 and their difference is 14. Find the numbers.

Solution

 (1) x  y  92 Let x be one number and y be another number. Then  (2) x  y  14 x  y  92 x  y  14 2x x

 106  53

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1428


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

53  y  14  y  39 y  39

The two numbers are 53 and 39.

84. The sum of a first number and eight times a second number is 15. Three times the first number decreased by the second number is –5. Determine the numbers.

Solution

 (1) x  8 y  15 . Let x be one number and y be another number. Then  (2) 3 x  y  5 x  8 y  15    3 3 x  y  5

3 x  24 y  45 3 x  y  5 25 y  50 y 2 x  8  2   15 x  16  15

The two numbers are  1 and 2.

x  1

85. Four times the first number added to three times the second number is 14. Seven times the first number decreased by two times the second number is 39. Determine the two numbers.

Solution

 (1) 4 x  3 y  14 Let x be one number and y be another number. Then  (2) 7 x  2 y  39 4 x  3 y  14   2 7 x  2 y  39   3 8 x  6 y  28 21x  6 y  117 29 x x

 145 5

4  5   3 y  14 20  3 y  14

The two numbers are 5 and  2.

3 y  6 y  2 86. Half of the first number added to four times the second number is –16. Three times the first number decreased by one-fifth of the second number is 25. Identify the numbers.

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1429


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

 1 x  4 y  16 (1) 2 Let x be one number and y be another number. Then  (2) 3 x  1 y  25  5 1 x  4 y  16   2 2 1 5 3 x  y  25 5 x  8 y  32    15 15 x  y  125 15 x  120 y  480 15 x  y  125 121 y  605 y  5

3x 

1  5  25 5 3 x  1  25

The two numbers are  5 and 8.

3 x  24 x 8 Applications Use systems of linear equations to solve each problem. 87. Price of food items If Ivan purchases two hamburgers and four orders of french fries for $8 and Hannah purchases three hamburgers and two orders of fries for $8, what is the price of each item?

Solution

ìï (1) 2 x + 4 y = 8 Let x = cost of a hamburger and let y = cost of the fries. Then ï í ïï(2) 3 x + 2 y = 8 î 2x + 4 y = 8 2x + 4 y = 8 3 x + 2 y = 8  ´ (-2) -6 x - 4 y = -16 -4 x = -8 x = 2

2x + 4 y = 8 2 (2) + 4 y = 8 4y = 4 y=1

A hamburger costs $2, while an order of fries costs $1.

88. Price of tennis equipment Rafael purchases two tennis rackets and four cans of tennis balls for $102. Jana purchases three tennis rackets and two cans of tennis balls for $141. What is the price of each item?

Solution

ìï(1) 2 x + 4 y = 102 Let x = cost of a racket and let y = cost of a can of balls. Then ï í ïï(2) 3 x + 2 y = 141 î

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1430


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2 x + 4 y = 102 2 x + 4 y = 102 3 x + 2 y = 141  ´(-2) 6 x - 4 y = -282 -4 x = - 180 x = 45

2 x + 4 y = 102 2 (45) + 4 y = 102 4 y = 12 y =3

A racket costs $45, while a can of balls costs $3.

89. Gourmet popcorn sales Alejandro sells two flavors or gourmet popcorn, caramel corn and white cheddar, online. If 440 bags of popcorn were sold in one week, and the number of white cheddar bags sold tripled the number of caramel corn bags sold, how many of each favor popcorn did he sell during that week?

Solution Let x = cost of caramel corn popcorn bag and let y = cost of cost of white cheddar popcorn ìï (1) x + y = 440 bag. Then ï í ïï(2) 3x = y î Substitute y = 3 x from (2) into (1):

x + 3x = 440 4 x = 440 x = 110 Substitute and solve for y:

110 + y = 440 y = 330

Alejandro sells 110 caramel corn and 330 white cheddar

90. Trail mix sales Ingrid sells two sizes of her gourmet trail mix, regular and large, online. During a two-day sale, she sold 150 bags of trail mix. If the regular size was priced at $8 per bag and the large size at $11 per bag, how many of each size did she sell if her revenue for the two days totaled $1350?

Solution

ìï (1) r + l = 150 Let r = regular trail mix and let y = large trail mix. Then ï í ïï(2) 8r + 11l = 1350 î r + l = 150  ´- 8 8r + 11l = 1350

-8r + -8l = -1200 8r + 11l = 1350 3l = 150 l = 50 Substitute and solve for r: r + 50 = 150 r = 100

Ingrid sold 100 regular trail mixes and 50 large trail mixes.

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1431


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

91. Planning for harvest A farmer raises corn and soybeans on 350 acres of land. Because of expected prices at harvest time, he thinks it would be wise to plant 100 more acres of corn than of soybeans. How many acres of each does he plant?

Solution

ìï (1) x + y = 350 Let x = acres of corn and let y = acres of soybeans. Then ï í ïï(2) x = y + 100 î Substitute x = y + 100 from (2) into (1): x + y = 350 y + 100 + y = 350 2 y = 250 y = 125

Substitute and solve for x: x = y + 100

= 125 + 100 = 225 The farmer should plant 225 acres of corn and 125 acres of soybeans.

92. Club memberships There is an initiation fee to join the recreation club, as well as monthly dues. The total cost after 7 months’ membership will be $3025, and after 1 21 years, $3850. Find both the initiation fee and the monthly dues.

Solution Let x = the initiation fee and let y = the monthly dues. Then the following system applies: x + 7 y = 3025  ´(-1) -x - 7 y = -3025 x + 18 y = 3850 x + 18 y = 3850 11 y = y=

825 75

x + 7 y = 3025 x + 7 (75) = 3025 x = 2500

The initiation fee is $2500 and dues are $75 per month.

93. Boating A riverboat can travel 30 kilometers downstream in 3 hours and can make the return trip in 5 hours. Find the speed of the boat in still water.

Solution Let b = speed in still water. Let c = the speed of the current. Then the following system applies: 3 (b + c ) = 30 3b + 3c = 30  ´(5) 15b + 15c = 150 5 (b - c ) = 30 5b - 5c = 30  ´(3) 15b - 15c = 90 30b b

The boat has a speed of 8 kilometers per hour in still water.

= 240 = 8

94. Framing pictures A rectangular picture frame has a perimeter of 1900 centimeters and a width that is 250 centimeters less than its length. Find the area of the picture.

Solution

ìï( 1) 2w + 2l = 1900 Let w = width and let l = length. Then ï í ïï(2) w = l - 250 î

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1432


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Substitute w = l – 250 from (2) into (1): 2w + 2l = 1900 2 (l - 250) + 2l = 1900 2l - 500 + 2l = 1900 4l = 2400 l = 600

Substitute and solve for w: w = l - 250 = 600 - 250 = 350 Area = lw = (600 cm)(350 cm) The area is 210,000 cm2 .

95. Making an alloy A metallurgist wants to make 60 grams of an alloy that is to be 34% copper. She has samples that are 9% copper and 84% copper. How many grams of each must she use?

Solution Let x = grams of 9% alloy. Let y = grams of 84% alloy. Then the following system applies: (note: 34% of 60 is 0.34 × 60 = 20.4) x+ y = 60  ´(-9) -9 x - 9 y = -540 9 x + 84 y = 2040 0.09 x + 0.84 y = 20.4  ´(100) 75 y = 1500 y = 20

She must use 40 grams of the 9% and 20 grams of the 84% alloy.

96. Archimedes’ law of the lever The two weights shown will be in balance if the product of one weight and its distance from the fulcrum is equal to the product of the other weight and its distance from the fulcrum. Two weights are in balance when one is two meters and the other three meters from the fulcrum. If the fulcrum remained in the same spot and the weights were interchanged, the closer weight would need to be increased by six pounds to maintain balance. Find the weights.

Solution 2w 1 = 3w 2  2w 1 - 3w 2 = 0  ´(-2) -4w 1 + 6w 2 = 0 9w 1 - 6w 2 = 30 3w 1 = 2 (w 2 + 5)  3w 1 - 2w 2 = 10  ´(3) 5w 1 = 30 = 6 w1

The weights are 6 and 4 pounds.

97. Lifting weights A 112-pound force can lift the 448-pound load shown. If the fulcrum is moved 1 additional foot away from the load, a 192-pound force is required. Find the length of the lever.

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1433


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

448 x = 112 y 0  ´(-1) -448 x + 112 y = 0  448 x - 112 y = 448 x - 192 y = -640 448 ( x + 1) = 192 ( y - 1)  448 x - 192 y = -640  -80 y = -640 y= 8 448 x = 112 y The lever has a length of 10 feet. 448 x = 112 (8) x=2

98. Writing test questions For a test question, a mathematics teacher wants to find two constants a and b such that the test item “Simplify a(x + 2y) – b(2x – y)” will have an answer of –3x + 9y. What constants a and b should the teacher use?

Solution a ( x + 2 y ) - b (2 x - y ) = ax + 2ay - 2bx + by

= (a - 2b) x + (2a + b) y

a - 2b = -3  ´(-2) -2a + 4b = 6 a - 2b = -3 2a + b = 9 2a + b = 9 a - 2 (3) = -3 5b = 15 a=3 b= 3

Solution: a = 3, b = 3

99. Break-even point A company can manufacture a pair of rollerblades for $43.53. Daily fixed costs of manufacturing rollerblades amount to $742.72. A pair of rollerblades can be sold for $89.95. Find equations expressing the Costs C and the revenue R as functions of x, the number of pairs manufactured and sold. At what production level will costs equal revenues?

Solution E ( x ) = 43.53 x + 742.72 R ( x ) = 89.95 x

E (x) = R (x) 43.53 x + 742.72 = 89.95 x 742.72 = 46.42 x 16 = x  Daily production should be 16 pairs.

100. Choosing salary options For its sales staff, a company offers two salary options. One is $326 per week plus a commission of 3 21 % of sales. The other is $200 per week plus 4 41 %

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1434


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

of sales. Find equations that express incomes I1 and I2 as functions of sales x, and find the weekly sales level that produces equal salaries.

Solution S1 ( x ) = 326 + 0.035 x

S1 ( x ) = S2 ( x ) S2 ( x ) = 200 + 0.0425 x 326 + 0.035 x = 200 + 0.0425 x 126 = 0.0075 x 16, 800 = x

The salary would be the same at a sales level of $16,800.

Let x represent the first number, y represent the second number, and z represent the third number. Use system of linear equations to solve and find the three numbers. 101. The sum or three numbers is 6. Twice the first number decreased by three times the second number and decreased by the third number is –7. Three times the first number decreased by the second number and added to the third number is 4. Determine the three numbers.

Solution Let x represent the first number, y represent the second number and z represent the third ìï(1) x + y + z = 6 ïï number. Then ï í(2) 2x - 3 y - z = -7 ïï ïï(3) 3x - y + z = 4 î Add (1) and (2). (1) x + y + z = 6

(2) 2x - 3 y - z = -7 3 x - 2 y = -1 (4)

Add (1) and –(3)

x+ y +z =6 (1) -(3) -3x + y - z = -4 (5) - 2x + 2 y = 2 Add (4) and (5) and solve for x. 3x - 2 y = -1 -2x + 2 y = 2

x =1 Substitute x = 1 in (4) and solve for y:

3 (1) - 2 y = -1 3 - 2 y = -1 -2 y = -4 y =2 Substitute x = 1 and y = 2 into (1) and solve for z:

1+ 2+ z = 6 z=3 Solution: (1, 2, 3)

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1435


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

102. Two times the first number combined with three times the second number then decreased by the third number is –14. Negative two times the first number combined with the second and third numbers is –2. The first number decreased by the second number and decreased by the third number is 0. Determine the three numbers.

Solution Let x represent the first number, y represent the second number and z represent the third ìï(1) 2 x + 3 y - z = -14 ïï number. Then ï í(2) -2 x + y + z = -2 ïï ïï(3) x - y - z = 0 î Add (2) and (3): -2 x + y + z = - 2 x- y -z =0 -x

= -2 x =2

Add (1) and (2)

2 x + 3 y - z = -14 -2 x + y + z = -2 4y

= -16 y = -4

Substitute x = 2 and y = −4 into (3) and solve for z.

2 - (-4) - z = 0 6-z = 0 -z = -6 z=6 Solution: (2, - 4, 6)

Use systems of three equations in three variables to solve each problem. 103. Work schedules A college student earns $196 per week working three part-time jobs. Half of his 20-hour work week is spent cooking hamburgers at a fast-food chain, earning $10 per hour. In addition, the student earns $8 per hour working at a convenience store and $12 per hour doing handyman work. How many hours per week does the student work at each job?

Solution Let x = hours at fast food restaurant, y = hours at gas station and z = janitorial hours. Substitute x = 15 into (2) and (3): x = 15 (1) x + y + z = 30 y + z = 30 15 + (2) (2) + y + 10z = 198.5 5.7 15 6.3 3 ( ) (3) 5.7 x + 6.3 y + 10z = 198.50 ( ) Solve the system of two equations and two unknowns formed by equations (2) and (3):

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1436


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

y + z = 15  ´(-10) 6.3 y + 10z = 113 

-10 y - 10 z = -150 6.3 y + 10z = 113 - 3.7 y = - 37 y = 10

y + z = 15 He spends 15 hours 10 + z = 15 cooking, 10 hours at the z = 5 gas station and 5 hours doing janitorial work.

104. Investment income A woman invested a $22,000 rollover IRA account in three banks paying 5%, 6%, and 7% annual interest. She invested $2000 more at 6% than at 5%. The total annual interest she earned was $1370. How much did she invest at each rate?

Solution Let x = amount at 5%, y = amount at 6% and z = amount at 7%.

Add ( 1) and (2) : (1) x + y + z = 22000 = + y x 2 2000 () (1) x + y + z = 22000 (3) 0.05 x + 0.06 y + 0.07 z = 1370 (2) -x + y = 2000 2 y + z = 24000 (4) Add - 0.05 (1) and (3) : -0.05 (1) -0.05 x - 0.05 y - 0.05z = -1100 (3) 0.05x + 0.06 y + 0.07 z = 1370 (5) 0.01 y + 0.02z = 270 Solve the system of two equations and two unknowns formed by equations (4) and (5) :

z = 24000  2y + 2 y + z = 24000  ´ (-2) -4 y - 2z = -48000 y + 2z = 27000 0.01 y + 0.02z = 270  ´( 100) y + 2z = 27000  -3 y = - 21000 = y 7000 She invested $5000 at 5%, $7000 at 6% and $10,000 at 7%.

105. Hiking socks Stefan makes and sells small, medium, and large hiking socks for charity. At a one-day charity event at his church, he sold 90 pairs of socks. The number of large pairs sold were twice the combined number of small and medium pairs sold. The number of pairs of small socks sold were half the number of medium pairs sold. How many pairs of each size did he sell?

Solution Let S = small hiking socks, M = medium hiking socks, and L = large hiking socks. Then ìï (1) S + M + L = 90 ïï ïï (2) L = 2S + 2M í ïï 1 ïï (3) M = S 2 îï From (3), S = 2M Substitute (3) and (2) into (1).

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1437


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2M + M + 2 (2M ) + 2M = 90 9M = 90 M = 10 Also, if S = 2M , then S = 2 (10) = 20 Substitute S and M into (1) and solve for L.

20 + 10 + L = 90 L = 60 Solution: 10 Small, 20 Medium, 60 Large 106. Hot chocolate bombs Angelina makes and sells three flavors of hot chocolate bombs: dark chocolate, white chocolate, and milk chocolate. At a recent weekend craft fair she sold 440 bombs. The number of white chocolate bombs sold were twice the number of dark chocolate bombs sold. The number of milk chocolate bombs sold were four times the number of white chocolate bombs sold. How many of each flavor did she sell?

Solution Let d = dark chocolate, w = white chocolate, and m = milk chocolate Then, ìï (1) d + w + m = 440 ïï ïï 1 í (2) w = 2d  d = w ïï 2 ïï 3 m = 4w ( ) ïî Substitute (2) and (3) into (1). 1 w + w + 4w = 440 2 11 w = 440 2 11w = 880 w = 80 Substitute w = 80 into (2) and (3). 1 (2) 2 (80) = d (40) = d And, (3) m = 4 (80) = 320

Solution: 40 dark chocolate, 80 white chocolate, 320 milk chocolate 107. Age distribution Approximately 3 million people live in a country. 2.61 million are younger than 50 years, and 1.95 million are older than 14 years. How many people are in each of the categories 0–14 years, 15–49 years, and 50 years and older?

Solution Let x = # between 0-14, y = # between 15-49 and z = # 50 or over.

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1438


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

(1) x + y + z = 3 x + y = 2.61 (2) y + z = 1.95 (3)

Add (1) and - (2) :

(1) x + y + z = 3 =- 2.61 - (2) - x - y z=

Substitute z = 0.39 into (3) : y + z = 1.95 y + 0.39 = 1.95 y = 1.56

0.39

Substitute y = 1.56 into (2) : x + y = 2.61 x + 1.56 = 2.61 x = 1.05

There are 1.05 million between 0-14, 1.56 million between 15-49 and 0.39 million over 50.

108. Designing arches The engineer designing a parabolic arch knows that its equation has the form y = ax2 + bx + c. Use the information in the illustration to find a, b, and c. Assume that the distances are given in feet. (Hint: The coordinates of points on the parabola satisfy its equation.)

Solution Points on the parabola: (0, 0), (10, 22.5) and (40, 0). Substitute each into the equation: y = ax 2 + bx + c

y = ax 2 + bx + c

2

22.5 = a ( 10) + b ( 10) + c 2 ( ) 22.5 = 100a + 10b + c

0 = a (0) + b (0) + c 1 0 ( ) =c

2

y = ax 2 + bx + c 2

0 = a (40) + b (40) + c 3 ( ) 0 = 1600a + 40b + c

Substitute c = 0 into (2) and (3) and solve the resulting system of equations:

100a + 10b = 22.5  ´(-4) 1600a + 40b = 0

Solution:

-400a - 40b = -90 1600a + 40b = 0 = -90 1200a 3 =- 40 a

1600a + 40b = 0 3 1600 (- 40 ) + 40b = 0 - 120 + 40b = 0 b=3

3 a = - 40 , b = 3, c = 0

109. Geometry The sum of the angles of a triangle is 180°. In a certain triangle, the largest angle is 20° greater than the sum of the other two and is 10° greater than 3 times the smallest. How large is each angle?

Solution Let x = the smallest angle, y = the middle angle and z = the largest angle.

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1439


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

(1) x + y + z = 180 z = x + y + 20 (2) z = 3 x + 10 (3)

Substitute (3) into (1) : x + y + 3 x + 10 = 180 4 x + y = 170 (4)

Substitute (3) into (2) : 3 x + 10 = x + y + 20 2 x - y = 10 (5)

Solve the system of two equations and two unknowns formed by equations (4) and (5) : 4 x + y = 170 2 x - y = 10 6 x = 180 x = 30

4 x + y = 170 4 (30) + y = 170 120 + y = 170 y = 50

z = 3 x + 10 z = 3 (30) + 10 z = 90 + 10 z = 100

Solution: The angles have measures of 30°, 50°, and 100°.

110. Ballistics The path of a thrown object is a parabola with the equation f(x) = ax2 + bx + c. Use the information in the illustration to find a, b, and c. (Distances are in feet.)

Solution Points on the parabola: (0, 0), (8, 12) and (12, 15). Substitute each into the equation: y = ax 2 + bx + c

y = ax 2 + bx + c

2

(1)

0 = a (0) + b (0) + c 0=c

y = ax 2 + bx + c

2

(2)

12 = a (8) + b (8) + c 12 = 64a + 8b + c

2

(3)

15 = a (12) + b (12) + c 15 = 144a + 12b + c

Substitute c = 0 into (2) and (3) and solve the resulting system of equations: 64a + 8b = 12  ´(-3) -192a - 24b = -36 64a + 8b = 12 288a + 24b = 30 64 (- 161 ) + 8b = 12 144a + 12b = 15  ´(2) = -6 96a -4 + 8b = 12 = - 161 a 8b = 16 b= 2 Solution: a = - 161 , b = 2, c = 0

Discovery and Writing 111. If no method is stated, describe how you would determine the most efficient method to use to solve a linear system.

Solution Answers may vary. 112. Describe how a system of three equations in three variables can be reduced to a system of two equations and two variables.

Solution Answers may vary.

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1440


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

113. When using the elimination method, how can you tell whether the system has no solution?

Solution Answers may vary. 114. When using the elimination method, how can you tell whether the system has infinitely many solutions?

Solution Answers may vary.

 x  8 y  51 . 115. Use a graphing calculator to attempt to find the solution of the system  3 x  25 y  160 Solution Answers may vary. 116. Solve the system of Exercise 115 algebraically. Which method is easier, and why?

Solution Answers may vary.

17 x  23 y  76 . 117. Use a graphing calculator to attempt to find the solution of the system  29 x  19 y  278 Solution Answers may vary. 118. Solve the system of Exercise 117 algebraically. Which method is easier, and why?

Solution Answers may vary. 119. Write a system of two equations in two variables with the solution (–2, 5).

Solution One example: ìï (1) x + y = 3 ï í ïï(2) x - y = -7 î 120. Write a system of three equations in three variables with the solution (–4, 5, 1).

Solution One example: ìï ( 1) x + y = 1 ïï ïí (2) y + z = 6 ïï ïï (3) x + z = -3 î 121. Write a system of two equations in two variables with no solution.

Solution One example: © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1441


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

ìï(1) x + y = 3 ï í ïï(2) x + y = 7 î 122. Write a system of three equations in three variables with an infinite number of solutions.

Solution One example: ìï (1) x+ y+z =1 ïï ïí(2) 2 x + 2 y + 2z = 2 ïï ïï(3) 3 x + 3 y + 3z = 3 î Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 123. If a system of two equations in two variables is represented by two lines with the same slope and different y-intercepts, then the system has an infinite number of solutions.

Solution False. The lines are parallel and the system has no solutions. 124. If a system of two equations in two variables is represented by two lines with negative reciprocal slopes, then the system has an infinite number of solutions.

Solution False. The lines are perpendicular and intersect in a single point. There is one solution. 125. If a linear system of three equations in three variables has infinitely many solutions, then any ordered triple is a solution of the system.

Solution False. The solution would consist only of ordered triples that satisfy the system. 126. When using the graphing method, a system of two equations in two variables can appear to have no solution and yet have a unique one.

Solution True. 127. A linear system of two equations in three variables cannot have a unique solution.

Solution True. 128. If a linear system of two equations in two variables has a solution set involving fractions, then use the graphing method to ensure accuracy.

Solution False. Use the substitution or elimination method to ensure accuracy. 129. To solve a linear system of three equations in three variables, we use the graphing method.

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1442


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution False. Use the substitution or elimination method. 130. The system of equations 999x – 999y = 999 and –999x + 999y = –999 has an infinite number of solutions.

Solution True. 131. The system of equations –777x + 777y = –777 and 777x – 777y = –777 has no solution.

Solution True. 132. The system of equations 555x + 555y = 555 and 555x – 555y = –555 has no solution.

Solution False. The system has one solution, (1, 0).

EXERCISES 6.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Solve the system. Write the solution in the form (x, y).  x  5 y  19   y  4

Solution  x  5 y  19   y  4 Substitute y  4 into equation (1), x  5  4   19 x  20  19 x  1

Solution: 1, 4

2. Solve the system. Write the solution in the form (x, y, z).  x  3 y  2z  4   y  4 y  11  z  2 

Solution  x  3 y  2z  4   y  4 z  11  z  2 

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1443


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Substitute z = –2 into equation (2). y  4  2   11 y  8  11 y  3

Substitute z  2 and y  3 into equation (1).

x  3  3   2  2  4 x 94  4 x 5  4 z  1

Solution: 1,  3,  2

3. Solve the system. Write the solution in the form (w, x, y, z). w  5 x  y  z  5   x  3 y  4 z  1   y  2z  12 z  5 

Solution w  5 x  y  z  5   x  3 y  4 z  13   y  2z  12 z  5  Substitute z  5 into equation (3). y  2  5   12 y  10  12 y 2

Substitute y  2 and z  5 into equation (2).

x  3  2   4  5   13 x  6  20  13 x  14  13 x1 Substitute x  1, y  2, and z  5 into equation (1). w  5  1  2  5  5 w 525  5 w 3

Solution: 3, 1, 2, 5

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1444


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3 6 9 12   4. Given the array of numbers  1 3 1 0 . 2 2 1 2 

Multiply the first row of numbers by 31 and rewrite the array of numbers.

Solution  1 2 3 4     1 3 1 0   2 2 1 2    3 6 9 12   5. Given the array of numbers  1 3 1 0 .  2 2 1 2 

Interchange the first and second rows of numbers and rewrite the array of numbers.

Solution  1 3 1 0    3 6 9 12  2 2 1 2 

1 1 2 6. Given the array of numbers  . 4 2 3 a. Multiply the first row of numbers by –4 and add each product to the corresponding number below in the second row. What three numbers do you obtain? 0, –6, and –5 b. Rewrite the array as follows. Keep the first row of numbers the same and replace the second row of number with the numbers obtained in part a.

Solution a. 0, 6,  5 b.

1 1 2   0 6 5

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A rectangular array of numbers is called a __________.

Solution matrix

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1445


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

8. A 3 × 5 matrix has __________ rows and __________ columns.

Solution 3, 5 9. The matrix containing the coefficients of the variables is called the __________ matrix.

Solution coefficient 10. The coefficient matrix joined to the column of constants is called the __________ matrix or the __________ matrix.

Solution system, augmented 11. Each row of a system matrix represents one __________.

Solution equation 12. The rows of the system matrix are changed using elementary __________.

Solution row operations 13. If one augmented matrix is changed to another using row operations, the matrices are __________.

Solution row equivalent 14. If two augmented matrices are row equivalent, then the systems have the __________.

Solution same solutions 15. In a type 1 row operation, two rows of a matrix can be __________.

Solution interchanged 16. In a type 2 row operation, one entire row can be __________ by a nonzero constant.

Solution multiplied 17. In a type 3 row operation, any row can be changed by __________ to it any __________ of another row.

Solution adding, multiple

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1446


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

18. The first nonzero entry in a row is called that row’s __________.

Solution lead entry Write the augmented matrix for each system of linear equations.

 3 x  5 y  7 19.  4 x  y  3 Solution  3 5 7     4 1 3 

  5 x  6 y  2 20.   2 x  11 y  8 Solution  5 6  2    2 11 8  2 x  y  5 z  6  21.  4 x  y  3 3 x  2 y  9 z  1 

Solution  2 1 5 6    1 0 3  4  3 2 9 1     x  3 y  z  0  22. 4 x  y  4 3 x  14 z  1 

Solution  1 3 1 0     4 1 0 4   3 0 14 1    3w  2 x  4 y  z  5  5w  y  7 z  10 23.  4 x  y  z  1 w  x  y  4 z  6 

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1447


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

 3 2 4 1  5    5 0 1 7 10  0 4 1 1 1    1 1 4 6   1 w  5 x  4 y  z  5  2w  3 y  7 z  8 24.  3w  10 x  y  z  1 2w  z  16 

Solution  1 5 4 1  5    2 0 3 7 8   3 10 1 1 1     2 0 0 1 16  Write the system of linear equations represented by the augmented matrix. For 25 and 26, use x and y. For 27 and 28, use x, y, and z. For 29 and 30, use w, x, y and z.

5 1  2 25.   1 3 4 Solution 5 x  y  2   3 x  4 y  1

 2 5 7 26.    3 0  6 Solution 2x  5 y  7  3x  6 2 1 5  27.  1 2 0 0  1 2 

0  3 7 

Solution 2 x  y  5 z  0  x  2 y  3   y  2 z  7 

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1448


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

0 3 2  28.  1 4 0 0 3 4 

1  2 11

Solution 3 y  2z  1    x  4 y  2 3 y  4 z  11 

 1 2 3 4  0 6 7 8 29.  4 3 0 1  5  8 0 6

1  2 3  4 

Solution w  2 x  3 y  4 z  1  6 x  7 y  8z  2   4w  3 x  z  3   8w  6 y  5z  4 

 1 1  2 0 30.  0 3  1  2

5 4 3 2 0 1 0 0

1  7 1  5

Solution w  x  5 y  4z  1   2w  3 y  2z  7  3 x  z  1   2w  x  5  Use the given matrix and perform the indicated row operation.

 1 2 31.  3 6

4  15

a. R1  R2 b.

 1   R2 3

c. (–3)R1 + R2

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1449


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 3 6 a.   1 2

15  4

b.

 1 2   1 2

4  5

c.

1 3  0 0

4  3

 1 1 32.  4 8

2  24 

a. R1  R2 b.

 1   R2 4

c. (–4)R1 + R2

Solution 4 8 a.   1 1

24  2

b.

 1 1   1 2

2  6

c.

 1 1  0 4

2  16

 1 2 4  1 1 33. 4  2 8 10 

1  2  4 

a. R2  R3 b.

 1   R3 2

c. (–2)R1 + R3

Solution a.

 1 2 4   2 8 10  4 1 1 

1  4 2

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1450


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

b.

 1 2 4  1 1 4  1 4 5 

1  2  2 

c.

 1 2 4  1 1 4 0 12 18 

1  2 2 

 5 2 4 1  0 1 2 3 34.  0 6 12 18  1 0  1 2

1  5 24   0

a. R1  R4 b.

 1    R3  6

c. (6)R2 + R3

Solution 1 2 1 0  0 1 2 3 a.  0 6 12 18  4 1  5 2

b.

 5 2 4 1  1 2 3 0 0 1 2 3   1 2 1 0

1  5 4  0

c.

 5 2 4 1  1 2 3 0 0 0 0 0   1 2 1 0

1  5 6  0

0  5 24   1

Determine whether each matrix is in row-echelon form, reduced row-echelon form, or neither.  1 3 0 5   35. 0 1 2 7  0 0 1 0   

Solution row echelon form

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1451


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1 3 0 5   36. 0 1 2 7  0 0 0 0   

Solution row echelon form  1 0 1   0 1 5  37. 0 0 0   0 0 0

Solution reduced row echelon form  1 0 1   0 1 5 38.  0 0 1    0 0 0 Solution reduced row echelon form Practice Use Gaussian elimination to solve each system of linear equations. If the system has no solution, write no solution; inconsistent system. If the system has infinitely many solutions, write dependent equations and show a general solution.

 x  y  5 39.   x  2 y  4 Solution

  1 x  y  5  1 x  y  5   1 x  y  5      y 3  2  x  2 y  4   2   3 y  9  2  1   1   2    2   3  2   2 From  2  : y  3

From  1 : x  y  5 x35

Solution:

 2, 3

x2

 x  3 y  8 40.  2x  5 y  5

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1452


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

  1 x  3 y  8  1 x  3 y  8  1 x  3 y  8      2x  5 y  5 y  1   2   11 y  11  2   2  2  1   2    2   111  2    2  From  2  : y  1

From  1 : x  3 y  8 x  3  1  8

Solution:

5, 1

x5

 x  y  1 41.  2 x  y  8 Solution

 1 x  y  1   1 x  y  1     2   y  6  2  2 x  y  8 2  1   2    2  From  2  : y  6

From  1 : x  y  1 x 6  1

Solution:

 7, 6

x7

 x  5 y  4 42.  2 x  3 y  21 Solution

  1 x  5 y  4   1 x  5 y  4   1 x  5 y  4      13 y  13 y  1  2   2  2 x  3 y  21   2  1  2  1   2    2  2  2 13   From  2  : y  1

From  1 : x  5 y  4

x  5  1  4

Solution:

9, 1

x 9

2 x  y  3 43.   x  3 y  5 Solution 2 1 3   1 3 5   1 3  1 3 5  5             1 3 5  2 1 3 0 7  7  0 1 1 1 R1  R2 R  R2  2R1  R2  R2 7 2 From R2 : y  1 From R1 :

x  3y  5

x  3  1  5

Solution:  2,  1

x35 x2

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1453


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 x  2 y  1 44.  3 x  5 y  19 Solution  1 2  1  1 2  1 1 2 1        3 5 19  0 11 22 0 1  2  3R1  R2  R2  111 R2  R2 From R2 : y  2 From R1 :

x  2 y  1

x  2  2   1

Solution:  3,  2 

x3

 x  7 y  2 45.  5 x  2 y  10 Solution  1 7  2   1 7  2  1 7 2         5 2  10 0 33 0  0 1 0 1  5R1  R2  R2 R  R2 33 2 From R2 : y  0 From R1 : x  7 y  2 x  7  0   2

Solution:  2, 0 

x  2

3x  y  3 46.  2x  y  3 Solution

3 1 3    2 1 3

3 1 3      5 15 0  2 2   32 R2  R1  R2

 1  1 1 3   0 1 3 1 R  R1 3 1  52 R2  R2

From R2 : y  3 From R1 : x  31 y  1 x

1 3

 3  1

Solution:  0,  3 

x1 1 x 0

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1454


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2 x  y  5 47.   x  3 y  6 Solution 2 1 5   1 3 6  1 3 6  1 3 6             1 3 6 0 1 1  2 1 5  0 7 7   R1  R2  R2  71 R2  R2 R1  R2 Solution:  3, 1

From R2 : y  1 From R1 : x  3 y  6 x  3  1  6 x3

3 x  5 y  25 48.   2 x  y  5 Solution

3 5 25    2 1 5 

 1  5  25  3 5  25 3 3      13 65 5  0  2  2  0 1 1 R  R1  23 R2  R1  R2 3 1  132 R2  R2

From R2 : y  5 From R1 : x  53 y   25 3 x

5 3

 5  

Solution:  0, 5

25 3

x  25   25 3 3 x 0  2x  y  7  49.  1 7  x  y   3 3  Solution 1  1 7  2 1 7   1  1 73  3 3 3          1 7 1 7 1 0    2  1 7    3 3 3 3    R1  R2  2R1  R2  R2 From R2 : y  7 From R1 :

x  31 y  73

x  31  7   73

 1  31 73    0 1  7   3R2  R2

Solution:  0, 7 

x  73  73 x 0

 45 x  6 y  60 50.  30 x  15 y  63.75

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1455


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 45 6 60   1 0.5 2.125 1  1 0.5 2.125 0.5 2.125             60  1.25  30 15 63.75 45 6 0 28.5 35.625 0 1 1  45R1  R2  R2  28.5 R1  301 R2 R2  R2 From R2 : y  1.25

From R1 :

x  0.5 y  2.125

x  0.5  1.25   2.125

Solution:  1.5, 1.25

x  0.625  2.125 x  1.5

 x  2 y  3 51.  2 x  4 y  6 Solution  1 2 3   1 2 3        2 4 6 0 0 12 2R1  R2  R2

 From R2 , 0 x  0 y  12. This is impossible. No solution  inconsistent system

 x  2 y  5 52.  3 x  6 y  18 Solution

 1 1 2 5  2 5        3 6 18 0 0 33 3R1  R2  R2 From R2 , 0 x  0 y  33. This is impossible. No solution  inconsistent system

 x  3 y  3 53.  3  x  3 y   9 Solution  1 3 3   1 3 3        0 0 0  3 9 9 3R1  R2  R2 Dependent equations From R1 , x  3 y  3 3 y  3  x 1 y  1  x 3

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1456


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1  Solution:  x,  1  x  3   2(2 y  x )  6 54.  4 y  2( x  3) Solution

2 2 y  x   6

 4 y  2x  6

4 y  2  x  3 

 2 x  4 y  6

4 y  2 x  6  2 x  4 y  6

 2 4 6  2 4 6      Dependent equations  2 4 6  0 0 0   R1  R2  R2 From R1 :  2 x  4 y  6

Solution:  x , 21 x  23 

4 y  2x  6 y  21 x  32  x  2y  z  2  55.  x  3 y  2z  1  x  y  3 z  6 

Solution

 1 x  2 y  z  2  1 x  2 y  z  2  1 x  2 y  z  2     5 y  3z  1   2  5 y  3z   1   2 x  3 y  2z  1   2     y  2z  8 13z  39  3  3  3 x  y  3z  6   1   2   2  5  3   2   3   1   3   3

 1 x  2 y  z  2 y z  2 z3  3   2   2  3   3 3 5

1 5

1 13

1 5

From  3 : z  3 3 1 z 5 5 3 1 y   3  5 5 y 2

From  2 : y 

From  1 :

x  2y  z  2

x  2  2   3  2

Solution:  1, 2, 3 

x1

x  5y  z  2  56.  x  2 y  z  3  x yz2 

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1457


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution   1 x  5 y  z  2  1 x  5 y  z  2   1 x  5 y  z  2    y  23 z   31  3 y  2z   1    2   2 x  2 y  z  3   2    4 y  2z  0 4 y  2z  0   3    3  x  y  z  2   3  1   2    1   2  2  2 3    3    1   3 

  1 x  5 y  z  2  y  23 z   31   2  2 z  43   3  3  4  2   3   3

  1 x  5 y  z  2     2 y  23 z   31  z 2   3  3 3   3 2 

From  3  : z  2 From  2  :

y  23 z   31

y  23  2    31

From  1 :

x  5y  z  2

x  5  1  2  2

y 1

x  1

Solution:  1, 1, 2  x  y  z  3  57. 5 x  y  6 y  z  4 

Solution   1   1 x  y  z  3   1 x  y  z  3 x  y  z  3     6    2 y  65 z  72  6 y  5z  21    2    2 5 x  y     z  3 yz  4 y z 4   3    3    3  1  5  1   2    2  2  2 6   6  3   2   3

  1 x  y  z  3  From  3  : z  3 y  65 z  72    2 From  2  : y  65 z  72  3 3 z     y  65  3   72   3   3 y 1

From  1 : x  y  z  3 x   1   3   3 x1 Solution:  1, 1, 3 

x  y  1  58.  x  z  3 y  z  2 

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1458


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution   1 x  y  1 x  y  1 x  y 1 1  1     z  3   2   y  z  2   2  y  z  2    2    yz2 yz2 2z  4  3   3   3    1   2    2    2  2

 2   3   3

 1 x  y 1  y  z  2   2   z2  3  1 3   3 2 

From  3  : z  2

From  1 :

From  2  : y  z  2

y   2   2

xy 1

x  0  1 x1

Solution:  1, 0, 2

y 0

 x yz 3  59. 2 x  y  z  4  x  2 y  z  1 

Solution  1 1 1 3   1 1 1 3   1 1 1 3        2 1 1 4   0 1 1 2   0 1 1 2   1 2 1 1 0 3 2 4  0 0 1 2         2R1  R2  R2  3R2  R3  R3  R1  R3  R3 From R1 :

From R3 : z  2 From R2 :

y  z  2

y   2   2 y 0

xyz3

x  0    2  3 x1

Solution:  1, 0, 2 

2 x  y  z  1  60.  x  y  z  0 3 x  y  2 z  2 

Solution  2 1 1 1   2 1 1 1   2 1 1 1   1 21  21 21           1 1 1 0   0 1 1 1   0 1 1 1  0 1 1 1 3 1 2 2  0  2 5 2  0 0 3 0  0 0 1 0          1 R R  2R2  R1  R2  R2  R2  3 3 3  3R2  R3  R3

 2R2  R3  R3

1 2

R1  R1

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1459


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

From  3  : z  0 From  2  :

From  1 :

y  z  1

y   0   1

x  21 y  21 z  21

x  21  1  21  0   21

Solution:

 1,  1, 0 

x  21  21

y  1

x1

 x  y  z  1  61. 3 x  y  4  y  2 z  4 

Solution  1 1 1 1   1 1 1 1   1 1 1 1   1 1 1 1          3 3 7 7 3 1 0 4   0 2 3 7   0 1  2  2   0 1  2  2  0 1 2 4  0 1 2 4  0 0 1 1  0 0 1 1           3R1  R2  R2  21 R2  R2  R3  R3 From  3 : z  1

From  1 :

3 7 z  2 2 3 7 y   1   2 2 y  2

From  2  : y 

R2  2R3  R3 x  y  z  1

x   2    1  1

Solution:

 2,  2, 1

x2

3 x  y  7  62.  x  z  0  y  2z  8 

Solution 7  3 1 0 7  1 1 0 1 1 0 7  1 1 0 7 3 3 3 3 3 3         7   0 1 3 7   0 1 3 7   1 0 1 0   0 1 3 0 1 2 8  0 1 2 8  0 0 5 15 0 0 1 3          1  3R2  R1  R2  R3  R2  R3  R R 3 5 3 1 3

From  3 : z  3

From  2 : y  3z  7 y  3  3  7

R1  R1

From  1 :

x  31 y  73

x  31  2   73

Solution:

 3,  2, 3

x3

y  2 x  y  z  2  63. 2 x  y  z  5 3 x  4 z   5 

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1460


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

 1 1 1 2   1 1 1 2   1 1 1 2        1 5   0 3 1 1   0 1  31 31   2 1 3 0 4 5 0 3 7 11 0 0 6 12       1 R R  2R1  R2  R2  2 3 2  3R1  R3  R3  1 1 1 2   1 1 0 1  3 3  0 0 1 2  1 R  R3 6 3

 R3  R2  R3

From  3  : z  2 From  2  :

From  1 :

y  31 z  31 y

1 3

 2 

1 3

y 1

xyz2

x   1   2  2 x1

Solution:

 1, 1, 2

 x  z  1  64. 3 x  y  2 2 x  y  5 z  3 

Solution 1 0 1  3 1 0 2 1 5 

 1 0 1 1  1 0 1 1  1 0 1 1 1        2   0 1 3 5   0 1 3 5   0 1 3 5  0 1 3 0 0 6 0  0 0 1 0  3  5       1  3R1  R2  R2  R2  R3  R3 R  R 3 3 6  2R1  R3  R3

From  3  : z  0 From  2  :

From  1 :

y  3z  5

y  3 0  5

x  z  1

x   0   1

Solution:

 1, 5, 0 

x  1

y 5  x  y  2z  4  65.   x  y  3z  5 2 x  y  z  2 

Solution 1 1 2 4 1 1 2 4  1 1 2 4   1 1 2 4          1 1 3 5  0 0 1 1   0 1 3 6  0 1 3 6  2 1 0 1 3 6 0 0 1 1  0 0 1 1  1 2         R1  R2  R2 R2  R3  R2  R2  2R1  R3  R3

 R3  R3

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1461


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

From  3  : z  1 From  2  :

From  1 :

y  3z  6

y  3  1  6

x  y  2z  4

x   3   2  1  4

Solution:

 1, 3, 1

x  1

y 3  x  3 y  2z  10  66. 3 x  2 y  2z  7  2 x  y  z  10 

Solution  1 3 2 10  1 3 2 10   1 3 2 10        1 2    3 2 2 7    0 7 4 23  0 1  2 1 1 10  0 5 5 10  0 7 4 23       3R1  R2  R2  R1  R1  2R1  R3  R3

R2  R3  R3  R2 1 5

 1 3 2 10   1 3 2 10      1  2   0 1 1 2  0 1 0 0 3 9 0 0 1 3      7R2  R3  R3  31 R3  R3

From  3  : z  3

From  2  : y  z  2 y  3  2 y  5

From  1 :

x  3 y  2z  10

x  3  5   2  3   10

Solution:

 1,  5, 3

x  15  6  10 x1

2 x  y  z  6  67. 3 x  y  z  2   x  3 y  3z  8 

Solution

 2 1 1 6  1 3 3 8  1 3 3 8         3 1 1 2   3 1 1 2   0 10 10 26   1 3 3 8  2 1 1 6  0 5 5 22       R1  R3 3R1  R2  R2 2R1  R3  R3

 1 3 3 8    R3 indicates 0x  0 y  0z  18. This is impossible.  0 10 10 26  No solution  inconsistent system 0 0 0 18   2R3  R2  R3

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1462


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

x  y  z  5  68.  x  y  2z  5 2 x  2 y  4 z  6 

Solution  1 1 1 5 1 1 1 5       1 1 2 5    1 1 2 5   2 2 4 6 0 0 0 4       2R2  R3  R3 R3 indicates 0x  0 y  0z  4. This is impossible. No solution  inconsistent system  x  y  2z  4  69.  x  y  z  5 2 x  2 y  4 z  8 

Solution

 1 1 2 4    1 5  1 1  2 2 4 8   

 1 1 2 4    0 2 1 1   0 0 0 0    1R1  R2  R2 2R1  R3  R3

From (2):

 1 1 2 4   0 1  1 1    2 2 0 0 0 0    1 R  R2 2 2

 3 9 1 0  2 2  0 1  1 1   2 2   0 0 0 0   R1  R2  R1

From (1):

1 1 y z 2 2 1 1 y  z 2 2

x

3 9 z 2 2 3 9 x z 2 2

 3 9 1 1  Solution:   z  , z  , z  2 2 2   2 x  y  z  5  70.  x  y  z  2 3 x  3 y  z  12 

Solution  1 1 1 5   1 1 1 5   1 1 1 5        3  1 1 1 2   0 0 2 3  0 1 1  2  3 3 1 12 0 0 2 3  0 0 1  3  2      1  R1  R2  R2  R R 2 2 2  3R1  R3  R3

1 2

R3  R3

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1463


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

From R2 : z   23 From R1 :

x yz 5

x  y    32   5

Solution:  x,  x  72 ,  23 

y   x  72 Use Gauss–Jordan elimination to solve each system of linear equations. If the system has no solution, write no solution; inconsistent system. If the system has infinitely many solutions, write dependent equations and show a general solution. w  x  y  2z  8  2w  x  2 y  z  2 71.  w  2 x  y  3z  10  w  x  y  2z  4 

Solution  1 1 1 2 8     2 1 2 1 2    1 2 1 3 10    1 1 1 2 4 

 1 1 1 2 8    0 3 4 5 14   0 1 0 5 18    0 2 0 0 4   2R1  R2  R2  1R1  R3  R3

 1 1 1 2 8    0 1 0 5 18  0 3 4 5 14    0 2 0 0 4  R2  R3

R1  R4  R4  1 0 1 7 26    0 1 0 5 18   0 0 4 20 68   0 0 0 10 40   1R2  R1  R1 3R2  R3  R3

 1 0 1 7 26    0 1 0 5 18  0 0 1 5 17    0 0 0 10 40  1 R  R3 4 3

 2R2  R4  R4

1 0 0 2 9    0 1 0 5 18  0 0 1 5 17    0 0 0 1 4  R3  R1  R1 1 R  R4 10 4

 1 0 0 0 1   0 1 0 0 2  0 0 1 0 3    0 0 0 1 4  5R4  R3  R3 5R4  R2  R2 2R4  R1  R1

Solution:  1, 2, 3, 4  w  x  2 y  z  4  w  2 x  3 y  z  15 72.   w  x  y  2z  11 2w  x  2 y  3z  10 

Solution  1 1 2 1 4    1 15  1 2 3   1 1 1 2 11    2 1 2 3 10 

1 2 3 1 15     1 1 2 1 4    1 1 1 2 11    2 1 2 3 10  R1  R2

1 2 3 1 15   5 0 19 0 3  0 1 2 3 4   0 3 8 5 20 R1  R2  R2

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1464


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

R1  R3  R3  2R1  R4  R4

1 2 3 1 15    2 3 4  0 1  0 3 5 0 19    0 3 8 5 20 R2  R3

 1 0  1 5 7    0 1 2 3 4   0 0 1 9 7    0 0 2 4 8  2R2  R1  R1  3R2  R3  R3

 1 0 1 5 7    0 1 2 3 4   0 0 1 9 7    0 0 2 4 8  1R2  R3

3R2  R4  R4

1 0 0 4 0    0 1 0 15 18   0 0 1 9 7    0 0 0 22 22 R3  R1  R1

1 0 0 4 0    0 1 0 15 18   0 0 1 9 7    1 1  0 0 0 1 R  R4 22 4

2R3  R2  R2

15R4  R2  R2 9R4  R3  R3

2R3  R4  R4

1 0 0 0 4   0 1 0 0 3  0 0 1 0 2    0 0 0 1 1 4R4  R1  R1

Solution: 4, 3, 2,  1

 x  2 y  7 73.   y  3 Solution  1 2 7   1 0 13       Solution:  13, 3 0 1 3  0 1 3  2R2  R1  R1

 x  2 y  7 74.   y  8 Solution  1 2 7   1 0 23       Solution:  23, 8 0 1 8 0 1 8  2R2  R1  R1

 x  y  7 75.   x  y  13 Solution  1 1 7   1 1 7         1 1 13 0 2 6  R1  R2  R2

 1 1 7   1 0 10      0 1 3 0 1 3  1 R  R2 R2  R1  R1 2 2

Solution:

 10, 3

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1465


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 x  2 y  7 76.  2 x  y  1 Solution 1 2 7  1 2  1 2 7  1 0 1  Solution 7             0 5 15 0 1 3  1, 3  2 1 1 0 1 3   2R1  R2  R2  51 R2  R2  2R2  R1  R1

 1 x  y  0 77.  2  x  2y  0  Solution  1  1 0  1  21 0 2       5  1 2 0 0 2 0  R1  R2  R2

 1  1 0  1 0 0 Solution: 2      0 1 0 0 1 0  0, 0  2 1 R  R2 R  R1  R1 5 2 2 2

x  y  5  78.  1   x  y  9 5  Solution  1 1 5   1 1 5   1 1 5   1 0 10            1 4  1 5 9 0  5 4  0 1 5 0 1 5  R1  R2  R2  45 R2  R2 R2  R1  R1

Solution:

 10, 5

 x  2 y  8 79.  5 x  10 y  8 Solution  1 2 8   1 2 8        No solution  inconsistent system 0 0 32 5 10 8  5R1  R2  R2

2 x  4 y  8 80.   x  2 y  4 Solution  2 4 8  1 2 4  1 2 4           No solution  inconsistent system 2 4 8 0 0 16  1 2 4  R1  R2  2R1  R2  R2

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1466


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 x  2 y  8 81.  5 x  10 y  40 Solution  1 2 8   1 2 8       5 10 40 0 0 0  5R1  R2  R2 From (1):

  1 Solution:  x, x  4  2  

x  2y  8 2 y   x  8 1 y  x 4 2

3 x  y  6 82.  6 x  2 y  12 Solution     1 1 3 1 6  2  2  1  1     3 3       6 2 12 12   6 2   0 0 0  1 R  R1 6R1  R2  R2 3 1 From (1): x

1 y 2 3 1 y  x  2 3 y  3 x  6

Solution:  x ,  3 x  6 

x  2 y  z  3  83.  y  3z  1  z  2 

Solution

 1 2 1 3   1 1 7 1   1 0 0 13       Solution: 1   0 1 0 7  0 1 3 1   0 1 3 0 0 1 2 0 0 1 2 0 0 1 2   13, 7,  2         2R2  R1  R1 7R3  R1  R1  3R3  R2  R2

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1467


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 x  3 y  2 z  1  84.  y  2z  3 z  5 

Solution

 1 3 2 1  1 0 4 8  1 0 0 28       0 1 2 3   0 1 2 3  0 1 0 13  0 0 0 0 1 5  0 0 1 5  1 5       3R2  R1  R1 4R3  R1  R1

Solution:

 28, 13, 5

2R3  R2  R2  x  y  2z  0  85.  x  y  z  2 x  z  1 

Solution 1 1 2 0   1 1 2 0  1 1 2 0 1 1 2 0           1 1 1 2  0 0 1 2   0 1 1 1   0 1 1 1  1 0 1 1  0  1  1 1  0 0 1 2  0 0 1 2         R2  R3  R1  R2  R2  R2  R2  R1  R3  R3  R3  R3 1 0 1 1  1 0 0 3      0 1 1  1  0 1 0 1  Solution:  3, 1, 2  0 0 1 2  0 0 1 2      R3  R1  R1 R2  R1  R1  R3  R2  R2  x  2 y  3  86.  x  4 y  2 2 x  z   8 

Solution  1 2 0 3   1 2 0 3  1 0 0 4   1 0 0 4          1  1 4 0 2   0 2 0 1   0 2 0 1   0 1 0 2  2 0 1 8 0 4 1 2 0 0 1 0  0 0 1 0         

Solution:  4, , 0 

 R1  R2  R2

 R2  R1  R1

 2R1  R3  R3

2R2  R3  R3

1 2

R2  R2

1 2

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1468


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2 x  y  2z  1  87.   x  y  3z  0 4 x  3 y  4 

Solution  2 1 2 1   2 1 2 1   6 0 2 2  6 0 2 2          1    0 3 8 1   1 1 3 0   0 3 8 1    0 3 8  4 3 0 4 0 1 4 2  0 0 20 5  0 0 1  1 4        1  3R1  R2  R1  20 R3  R3 R1  2R2  R2  2R1  R3  R3

 3R3  R2  R3

 6 0 0  32   1 0 0 41      1 1  0 3 0 3   0 1 0 1  Solution:  4 , 1, 4  0 0 1 1  0 0 1 1  4  4    61 R1  R1 2R3  R1  R1

8R3  R2  R2

1 3

R2  R2

3 x  y  3  88. 3 x  y  z  2 6 x  z  5 

Solution 3 1 0 3  3 1 0 3  3 1 0 3  6 0 1 5          3 1 1 2  0 0 1 1  0 2 1 1  0 2 1 1 6 0 1 5  0 2 1 1 0 0 1 1 0 0 1 1           R1  R2  R2 2R1  R2  R1 R2  R3  2R1  R3  R3  R3  R3 6 0 0 4   1 0 0 23      2 0 2 0  2  0 1 0 1  Solution:  3 , 1, 1 0 0 1 1  0 0 1 1      1  R3  R1  R1  R R 1 6 1  R3  R2  R2

 21 R2  R2

2 x  y  z  10  89.  x  2 y  2z  3 4 x  2 y  2 z  5 

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1469


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution  1 2 2 3   2 1 1 10   4 2 2 5    R1  R2

 1 2 2 3   0 5 5 16   0 10 10 17     2R1  R2  R2  4R1  R3  R3

 1 2 2 3     16  1 1 0  5     0 10 10 17  1 R  R2 5 2

 17  1 0 0  5     0 1 1 16   5      0 0 0 15     2R2  R1  R1 10R2  R3  R3

From (3): no solution  inconsistent system 2 x  y  z  3  90.  2 x  4 y  2z  4  x  2 y  z  2 

Solution  1 2 1 2   1  2 1 2        2 4  2 4    0 0 0  8   2 1 1 3    2 1 1 3      R1  R3 2R1  R2  R2

From (2): no solution  inconsistent system

3 x  6 y  9z  18  91. 2 x  4 y  3z  12  x  2 y  3z  6 

Solution 3 6 9 18  1 2 3 6   1 2 3 6   1 2 3 6          2 4 3 12   2 4 3 12   0 0 3 0   0 0 1 0   1 2 3 6  3 6 9 18 0 0 0 0  0 0 0 0          R1  R3  2R1  R2  R2  31 R2  R2 From  2  :

z 0

From  1 :

 3R1  R3  R3 x  2 y  3z  6

x  2 y  3 0  6

Solution:

 x,  x  3, 0 1 2

2 y  x  6 y   21 x  3 x  2 y  z  7  92. 2 x  y  z  2 3 x  4 y  3z  3 

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1470


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution  1 2 1 7   1 2 1 7   1 2 1 7        2 1 1 2   0 5 3 12   0 5 3 12  3 4 3 3 0 10 6 24  0 0 0 0         2R1  R2  R2  2R2  R3  R3  3R1  R3  R3

 1 2 1 7   1 2 1 7      3 12 0 5 3 12  0 1  5 5   0 0 0 0  0 0 0 0       51 R2  R2  2R2  R1  R1

From  2  : y  53 z  125

From  1 : x  51 z  115

y  z 3 5

 1 0 1 11  5 5   3 12 0 1   5 5  0 0 0 0   

12 5

Solution:

x z 1 5

11 5

 z  , z  1 5

11 5

3 5

12 5

, z

1 3 2  x  y  z  2 3 4 3  1 1  93.  x  y  z  1 2 3  1 1 6 x  8 y  z  0 

Solution 9  1 3  2 2  1 9 1 2 6 2 6 3 4 4  3 41      1 21 1   6 3 2 6   0  2 14 42  1 2 3  1  1 1 0  4 3 24 0  0 12 16 24  8 6       6R1  R2  R2 3R1  R1

6R2  R2 24R3  R3

 4R1  R3  R3

 1 9 2 6 1 0 1 1 0 0 9  3 4 4       4 4 0 1   4  0 1   4  0 1 0  3      3 3 0 3 4 6 0 0 8 6  0 0 1 3  4        212 R2  R2  94 R2  R1  R1  81 R3  R1  R1  41 R3  R3

 3R2  R3  R3

1 6

Solution:

 ,  3,  9 4

3 4

R3  R2  R2 1 8

R3  R3

1  x  y  3z  1 4 1 94.  x  4 y  6z  1 2 1  3 x  2 y  2z  1 

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1471


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

1  41 2 1 3

 1 4 12 4  1 3 1    4 6  1   1 8 12  2   1 6 6 3  2 2 1   4R1  R1 , 2R2  R2 3R3  R3

 1 4 12 4   1 0 12 2      1 1 0 1 0 2   0 1 0 2   0 10 18 7  0 0 18 2      1 R R 4 R R R1     2 2 1 12 2  10R2  R3  R3

 1 4 12 4    0 12 0 6   0 10 18 7     R2  R1  R2  R3  R1  R3

1 0 0  0 1 0 0 0 1 

  Solution:  2, 1, 1  3 2 9   23 R3  R1  R1 1 18

2 3 1 2 1 9

R3  R3

1 1  x y z 2 2 4  1 1 3 2 95.  x  y  z  4 2 2 3 1 2 3 x  z   3  Solution

 1 1 1 2   1 1 2 4  2  22 41 1    3 8 3 6 18   3 4 2  2  2 0 1  1  2 0 3 1  3 3   2R1  R1 , 12R2  R2 3R3  R3

1 1 4  2 2   0 1 22 14   0 1 7 9    8R1  R2  R2  2R1  R3  R3

 1 21  1 0 9 3  1 0 9 3  2 4        0 1 22 14   0 1 22 14   0 1 22 14   0 1 7 9 0 0 15 5  0 0 1  31        R2  R2  21 R2  R1  R1  151 R3  R3 R2  R3  R3 1 0 0 0    20 20 1 0 1 0 3  Solution:  0, 3 ,  3  0 0 1  1  3  9R3  R1  R1 22R3  R2  R2

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1472


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

5 1  x yz 0 3 7 1 2 96.  x  y  z  9 7 8  27  6 x  4 y  4 z  20 

Solution

5  1 3  72 1 7 6 4 

1 

1 8 27 4

7 7  1 7 1  7 0 0  0  15 5 15 5      952 77 9    16 56 7 504   0 15  5 504   24 16 27 80  0 408  303 80  20 15 5     7 R R , 56 R R 16 R R R      1 2 2 1 2 2 5 1

4R3  R3  1  157  0 1 0 1 

 24R1  R3  R3

175 63  175 63   1 0 136  1 0 136 0 17 17      33 135 33 135   0 1   0 1      136 17  136 17  135 68       5 1 27   0 0 68 0 0 15 68 7 R  R R  R  R R  R 2 1 1 3 952 2 15 2 135 3 15 408

7 5 33 136 303 136

135 17 150 51

R3  R3

1 0 0  0 1 0 0 0 1 

 R3  R2  R3

  427 2617 68  Solution:  918 , 306 , 27    175  136 R3  R1  R1 33 136

427 918 2617 306 68 27

R3  R2  R2

2w  2 x  3 y  z  2  w  x  y  z  5 97.  w  2 x  3 y  2z  2 w  x  2 y  z  4 

Solution  2 2 3 1  1 1 1 1  1 2 3 2  1 2 1  1

 2 2 3  4 0 5 3 12 2 1 2       8  5 0 4 1 1 8  0 4 1 1     0 2 3 5  0 0 5 9 2 6  4       6  2  4  0 2 4 0 4 1 3  0  2R2  R1  R2  2R1  R2  R1 2R3  R1  R3 2R3  R2  R3  2R4  R1  R4  R4  R2  R4

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1473


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 4 0 5 3   0 4 1 1  0 0 5 9  1 2  0 0 1 2

R4  R4

1 0 0 0  0 5 0 0  0 0 5 0  0 0 0 1

4 0 12 0 12   8   0 20 0 4     5 9 4 0 2   1  0 1 0 0  R1  R3  R1

1 0 0 3 16    36  0 5 0 1     4 0 0 5 9   1  0 0 0 1 1 R  R1 4 1

5R2  R3  R2

1 4

5R4  R3  R4

 R4  R4

1 0 0 0 1    10  0 1 0 0     5 0 0 1 0   1  0 0 0 1

 3R4  R1  R1

 51 R2  R2

 R4  R2  R2

 51 R3  R3

1  2 1  1 

4   9  4   1 

R2  R2

Solution:

 1, 2, 1, 1

 9R4  R3  R3

w  x  2 y  z  1  w  2 x  y  z  2 98.  2w  x  y  z  4 w  x  y  2z  3 

Solution 1 1 2 1  1 2 1 1 2 1 1 1   1 1 1 2

1 1 2  1 0 1 0 1 1 1 3      2 0 1 1 0 1 0 1 1 0 1      0 1  3 1 2  0 0  4 1 3  4      3  2 2 0 0 1 1 0 0 1 1  R1  R2  R2 R3  R1  R1  2R1  R3  R3 R3  R2  R3  R1  R4  R4

 1 0 1 0  0 1 1 0 0 0 1 1 4  0 0 1 1  41 R3  R3

9 1 0 0 1   5 0 0 0 10  3  4 4      1 1 0  1  0 1 0 4    0 5 0 0 4   0 0 1 1  0 0 5 0  43  5   43  4      5 5 2  0 0 0 45   0 0 0 45 4 4    5R1  R4  R1 R1  R3  R1

R2  R3  R2 R4  R3  R4

 51 R1  R1  R2  R2 1 5 1 5 4 5

 R3  R3 R4  R4

1 0 0 0  0 1 0 0 0 0 1 0  0 0 0 1

2  0 1  1 

 5R2  R4  R2  5R3  R4  R3

Solution:

 2, 0,  1, 1

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1474


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

w  x  z  4  w  y  z  2 99.  2w  2 x  y  2z  8 w  x  y  z  2 

Solution 1 1 0 1  1 0 1 1 2 2 1 2   1 1 1 1

1 1 0 1 1 0 1 1 4  4 2      2  0 1 1 0 2  0 1 1 0 2       0 0 1 0 0  0 0 1 0 0  8       2 0 2 1 2 6  0 0 1 2 2   R2  R1  R2  R2  R1  R1  2R1  R3  R3  2R2  R4  R4  R4  R1  R4

1 0 0 1  0 1 0 0 0 0 1 0  0 0 0 2

1 0 0 0 2 1    2 0 1 0 0 2   Solution:  1, 2, 0, 1 0 0 1 0 0  0    2  1  0 0 0 1 1  2 R4  R1  R1

R3  R1  R1 R3  R2  R2

1 2

R4  R4

R3  R4  R4

w  x  2 y  z  3  3w  2 x  y  z  4 100.  2w  x  2 y  z  10 w  2 x  y  3z  8 

Solution  1 1 2 1   3 2 1 1  2 1 2 1  1 3  1 2

 1 1 2  1 0 5 3 2  3  1 3       5  5  4  0 1 7 4 0 1 7 4      0 3 2 3 4  0 0 19 9 10  19       8  20  0 3 1 4 5  0 0 20 8  3R1  R2  R2 R2  R1  R1  2R1  R3  R3  3R2  R3  R3  R1  R4  R4  3R2  R4  R4

 1 0  5 3   0 1  7 4 9 0 0 1 19  2 0 0 1 5

12  1 0 0  19 2    13 0 1 0  19 5    9 0 0 1 1  19   7 1  0 0 0 95 5R3  R1  R1

1 19 1 20

R3  R3 R4  R4

12  1 0 0  19 3   13 2 0 1 0  19   9 0 0 1 1 19   0 1 0 0 0 95 R4  R4 7

3  2  1  0

7R3  R2  R2  R4  R3  R4

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1475


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

1 0 0 0  0 1 0 0   0 0 1 0  0 0 0 1

12 19 13 19 9 19

3  2 1  0 

Solution:

 3, 2, 1, 0 

R4  R1  R1 R4  R2  R2 R4  R3  R3

Each system contains a different number of equations than variables. Solve each system using Gauss–Jordan elimination. If a system has no solution, write no solution; inconsistent system. If a system has infinitely many solutions, write dependent equations and show a general solution.  x  y  2  3 x  y  6 101.  2 x  2 y  4 x  y  4 

Solution  1 1 2   1 1 2   1 1 2  1 0 1          3 1 6   0 4 12   0 1 3  0 1 3  2 2 4  0 0 0  0 2 6  0 0 0           1 1 4  0 2 6  0 0 0  0 0 0   3R1  R2  R2  41 R2  R2  R2  R1  R1  2R1  R3  R3  R1  R4  R4

R3  R4

Solution:

 1,  3

2R2  R3  R3

 x  y  3   2 x  y  3 102.  3 x  y   7  4 x  y  7 

Solution  1 1 3   1 1  3       2 1 3   0 3 3    3 1  7  0 2 2      4 1 7  0 5 5   2R1  R2  R2  3R1  R3  R3  4R1  R4  R4

 1 1 3   1 0 2      Solution: 0 1 1   0 1 1  0 2 2  0 0 0   2, 1     0 5 5  0 0 0  1 R  R2 R2  R1  R1 3 2 R3  R4

 2R2  R3  R3  5R2  R4  R4

 x  2 y  z  4 103.  3 x  y  z  2

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1476


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 1 2 1 4  1 2 1 4  1 2 1 4           10 4 3 1 1 2 0 7 4  10 0 1 7 7   3R1  R2  R2  71 R2  R2

From R1 : x  71 z  87 x

8 7

From R2 : y  47 z  107  z

Solution:

y  107  47 z

1 7

1 2  1 8  7 7   10 4 0 1 7 7    2R2  R1  R1

x  87  71 z, y  107  47 z z  any real number

 x  2 y  3z  5 104.  5 x  y  z  11 Solution 1  1 2 3 5  1 0  1 2 3 5   1 2 3 5  179  9            14 14 14 14 5 1 1 11 0 9 14 14  0 1  9  9  0 1  9  9   5R1  R2  R2  91 R2  R2  2R2  R1  R1

From R1 : x  91 z   179

From R2 : y  149 z   149

x   179  91 z

y   149  149 z

Solution: x   179  91 z y   149  149 z z  any real number

w  x  1  105. w  y  0 x  z  0 

Solution

 1 1 0 0 1  1 1 0 0 1 1 0 1 0 0        1 0 1 0 0  0 1 1 0 1   0 1 1 0 1   0 1 0 1 0 0 1 0 1 0 0 0 1 1 1        R2  R1  R2  R2  R1  R1  R2  R3  R3  1 0 0 1 1    0 1 0 1 0   0 0 1 1 1  

From R1 :

From R2 :

From R3 :

wz 1

xz0

y  z  1

w  1 z

x  z

y  1  z

Solution: w  1  z, x   z, y  1  z , z  any real #

w  x  y  z  2  106. 2w  x  2 y  z  0 w  2 x  y  z  1 

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1477


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution  1 1 1 1 2   1 1 1 1 2   3 0 3 2 2        2 1 2 1 0   0 3 0 1 4   0 0 0 1 1   1 2 1 1 1 0 3 0 0 3  0 1 0 0 1         2R1  R2  R2 3R1  R2  R1  R3  R1  R3

R2  R3  R2 1 R  R3 3 3

 3 0 3 0 0   1 0 1 0 0      0 0 0 1 1   0 0 0 1 1  From R1 : 0 1 0 0 1  0 1 0 0 1  w  y  0     1 w y 2R2  R1  R1 R  R 1 3 1

Solution: w  y , x  1, y  any real #, z  1

 R2  R2 x  y  3  107. 2 x  y  1 3 x  2 y  2 

Solution  1 1 3 1 1 3  1 1 3        2 1 1   0 1 5  0 1 5 3 2 2  0 1 7  0 0 2        2R1  R2  R2  R2  R3  R3

R3 indicates that 0 x  0 y  2. This is impossible. The system is inconsistent.  no solution

 3R1  R3  R3 x  2 y  z  4  x  y  z  1 108.  2 x  y  2 z  2 3 x  3 z  6 

Solution 1 2 1 4  1 2 1 4  1 2 1 4         1 1 1 1   0 3 0 3   0  3 0 3  2 1 2 2  0 3 0 6  0 0 0 3        3 0 3 6  0 6 0 6  0 6 0 6   R1  R2  R2  R2  R3  R3  2R1  R3  R3

R3 indicates that 0x  0 y  0z  3. This is impossible. The system is inconsistent.  no solution

 3R1  R4  R4

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1478


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Fix It In exercises 109 and 110, identify the step the first error is made and fix it.  x  2 y  3z  4  109. Use the Gaussian elimination to solve the linear system 3 y  6z  3 . 2 y  5 z   3 

Solution Step 4 was incorrect.

Step 4: x  1, y  1, z  1; 1, 1,  1

 y  2 z  3  110. Use Gauss–Jordan elimination method to solve the linear system  x  3 y  6z  11.  z  2 

Solution Step 4 was incorrect.

1 0 0  Step 4:  2  R3  R2  0 1 0 0 0 1 

2  1 2

Step 5: x  2, y  1, z  2; 2, 1,  2

Applications Use matrices to solve each problem. 111. Flight range The speed of an airplane with a tailwind is 300 miles per hour and with a headwind is 220 miles per hour. On a day with no wind, how far could the plane travel on a 5-hour fuel supply?

Solution Let p  speed with no wind. Let w = the speed of the wind. Then the following system applies:

 1 1 300  1 1 300  1 0 260  p  w  300  1 1 300              p  w  220  1 1 220  0 2 80  0 1 40  0 1 40  1  R2  R1  R2 R  R2  R2  R1  R1 2 2 The plane has a speed of 260 miles per hour with no wind, so it could travel 1300 miles in 5 hours. 112. Resource allocation 120,000 gallons of fuel are to be divided between two airlines. Triple A Airways requires twice as much as UnityAir. How much fuel should be allocated to Triple A?

Solution Let T  gallons for Triple A. Let U  gallons for UnityAir. Then the following system applies:

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1479


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

T  U  120000  T  2U

 1 1 120000 T  U  120000  1 1 120000        0  T  2U  0  1 2 0 3 120000  R2  R1  R2

 1 1 120000  1 0 80000       Triple A should be allocated 80,000 gallons. 0 1 40000 0 1 4000  1 R  R2  R2  R1  R1 3 2 113. Library shelving To use space effectively, librarians like to fill shelves completely. One 35inch shelf can hold 3 dictionaries, 5 atlases, and 1 thesaurus; or 6 dictionaries and 2 thesauruses; or 2 dictionaries, 4 atlases, and 3 thesauruses. How wide is one copy of each book?

Solution Let d = width of a dictionary, a = width of an atlas and t = width of a thesaurus. 3d  5a  t  35  6d  2t  35 2d  4a  3t  35 

3 5 1 35 3 5 3 5 1 35  1 35        6 0 2 35  0 10 0 35  0 1 0 3.5 35  2 7 2 4 3 35 0 0 2 7 35  3 3 3        2R1  R2  R2  101 R2  R2  23 R1  R3  R3

 3 0 1 17.5  3 0 0 13.5  1 0 0 4.5        0 1 0 3.5   0 1 0 3.5   0 1 0 3.5  0 0 7 28  0 0 1 0 0 1 4  4       1 R R  5R2  R1  R1  71 R3  R1  R1  1 3 1  2R2  R3  R3

1 7

R3  R3

3R3  R3

Dictonaries are 4.5 in. wide. Atlases are 3.5 in. wide. Thesauruses are 4 in. wide

114. Copying machine productivity When both copying machines A and B are working, an office assistant can make 100 copies in one minute. In one minute’s time, copiers A and C together produce 140 copies, and all three working together produce 180 copies. How many copies per minute can each machine produce separately?

Solution Let A, B and C represent the number of copies per minute each copier can make.

 A  B  100   A  C  140  A  B  C  180 

 1 1 0 100  1 1 0 100      1 0 1 140  0 1 1 40    1 1 1 180  0 0 1 80       R1  R2  R2  R1  R3  R3

 1 0 1 140   1 0 0 60     0 1 1  40  0 1 0 40 0 0 1 80  0 0 1 80      R3  R1  R1 R2  R1  R1  R2  R2

R3  R2  R2

Copier A can make 60 copies per minute. Copier B can make 40 copies per minute. Copier C can make 80 copies per minute.

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1480


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

115. Nutritional planning One ounce of each of three foods has the vitamin and mineral content shown in the table. How many ounces of each must be used to provide exactly 22 milligrams (mg) of niacin, 12 mg of zinc, and 20 mg of vitamin C?

Food

Niacin

Zinc

Vitamin C

A

1 mg

1 mg

2 mg

B

2 mg

1 mg

1 mg

C

2 mg

1 mg

2 mg

Solution Let A, B and C represent the number of ounces of each food.

 A  2B  2C  22   A  B  C  12  2 A  B  2C  20

 1 2 2 22   1 2 2 22       1 1 1 12   0 1 1 10       2 1 2 20 0 3 2 24   R1  R2  R2  2R1  R3  R3

1 0 0 2    0 1 1 10  0 0 1 6    2R2  R1  R1  3R2  R3  R3  R2  R2

 1 0 0 2   0 1 0 4  0 0 1 6     R3  R2  R2

2 ounces of Food A, 4 ounces of Food B, and 6 ounces of Food C should be used.

116. Chainsaw sculpting A wood sculptor carves three types of statues with a chainsaw. The number of hours required for carving, sanding, and painting a totem pole, a bear, and a deer are shown in the table. How many of each should be produced to use all available labor hours?

Totem Pole

Bear

Deer

Time Available

Carving

2 hr

2 hr

1 hr

14 hr

Sanding

1 hr

2 hr

2 hr

15 hr

Painting

3 hr

2 hr

2 hr

21 hr

Solution Let A, B and C represent the numbers of poles, bears, and deer made. 2 A  2B  C  14   A  2B  2C  15 3 A  2B  2C  21 

2 2 1 14  1 2 2 15       1 2 2 15   0 2 3 16   3 2 2 21  0 4 4 24       2R2  R1  R2  3R2  R3  R3 R2  R1

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1481


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1 0 1 1   1 0 0 3  1 0 0 3       0 2 3 16  0 2 0 4   0 1 0 2  0 0 2 8  0 0 1 4  0 0 1 4        1 1    R2  R1  R1 R R R R R 1 1 2 2 3 2 2 2R2  R3  R3

 32 R3  R2  R2

 R2  R2

1 2

R3  R3

3 poles, 2 bears, and 4 deer should be made.

Discovery and Writing 117. What is a matrix?

Solution Answers may vary. 118. Describe the three elementary row operations that can be used to produce an equivalent system.

Solution Answers may vary. 119. Explain the difference between the row-echelon form and the reduced row-echelon form of a matrix.

Solution Answers may vary. 120. Explain the differences between Gaussian elimination and Gauss–Jordan elimination.

Solution Answers may vary. 121. Describe the steps you would use to solve a system of linear equations using Gaussian elimination.

Solution Answers may vary. 122. Describe the steps you would use to solve a system of linear equations using Gauss– Jordan elimination.

Solution Answers may vary. 123. If the upper-left corner entry of a matrix is zero, what row operation might you do first?

Solution Answers may vary. 124. What characteristics of a row-reduced matrix would let you conclude that the system is inconsistent?

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1482


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution Answers may vary. Use matrices to solve each system.  x 2  y 2  z 2  14  125. 2 x 2  3 y 2  2z 2  7  x2  5 y 2  z2  8  (Hint: Solve first as a system in x2, y2, and z2.)

Solution

1 1 1 1 1 1 1 14  1 14  1 14        2 3 2 7   0 1 4 35  0 6 0 6    1 5 1 8  0 6 0 6  0 1 4 35       R2  R3  2R1  R2  R2  R1  R3  R3 1 0  1 0 0 4 1 13      0 6 0 6   0 1 0 1  0 0 4 36 0 0 1 9      1 1 R R R R R R     1 1 1 1 6 2 4 3 1 6

R2  R3  R3

 61 R2  R2

x 2  4  x  2 y 2  1  y  1 z 2  9  z  3

 41 R3  R3

5 x  2 x  z  22  126.  x  y  z  5  3 x  2 y  3 z  10 Solution 5 2  1 1 1 5   1 1 1 5  1 22       1 22  0 3 6 3   1 1 1 5    5 2 3 2 3 10  3 2 3 10  0 5 0 5        5R1  R2  R2 R1  R2  3R1  R3  R3  1 1 1 5   1 1 1 5   1 0 1 4   1 0 0 4         0 5 0  5  0 1 0 1   0 1 0 1   0 1 0 1  0 3 6 3  0 1 2 1 0 0 2 0  0 0 1 0          1  51 R2  R2  R2  R1  R1   R2  R3 R R R 1 1 2 3 1 5

x  4  x  16 ,

R3  R3

y  1  y  1,

R2  R3  R3

1 2

R3  R3

z 0 z 0

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1483


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 127. A 7 × 5 matrix has 5 rows.

Solution False. It has 7 rows. 128. 2 × 9 matrix has 2 columns.

Solution False. It has 9 columns. 129. Adding a multiple of a row to another row produces a new augmented matrix corresponding to an equivalent system of linear equations.

Solution True. 130. Multiplying a row by any constant produces a new augmented matrix corresponding to an equivalent system of linear equations.

Solution False. The constant must be nonzero. 131. Every matrix has a unique reduced row-echelon form.

Solution True.  1 0 6 10   132. If the augmented matrix is 0 1 8 2  , then the system has a unique solution. 0 0 4 0   

Solution True.  1 0 6 10   133. If the augmented matrix is 0 1 8 2  , then the system has an infinite number of 0 0 0 4    solutions.

Solution False. It has no solution.  1 0 6 10   134. If the augmented matrix is 0 1 8 2  , then the system has an infinite number of 0 0 0 0    solutions.

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1484


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution True. 135. If the row-echelon form of the augumented matrix contains the row [1 0 0 | 0], then the original system is consistent.

Solution True. 136. If an augmented matrix is square and contains one row of zeros, then the linear system has infinitely many solutions.

Solution False. The system may have a unique solution or no solution.

EXERCISES 6.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

 3 4 5 Let A    . How many rows and columns does matrix A have?  2 7 11 Solution 2 rows and 3 columns

 4 6   2. Let B   2 9 . What number occurs in the second row and second column of matrix B?  3 5  

Solution −9

 4 7   . Write the additive inverse of each entry of the matrix. 3. Let A    11 1  3   Solution The additive inverse of 4 is −4. The additive inverse of −7 is 7. The additive inverse of −11 is 11. The additive inverse of

1 1 is  . 3 3

4. Perform the indicated operation: –7 – (–5)

Solution 7  5  2 5. Perform the indicated operation: 3(–2) + 4(–2) + 4(5).

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1485


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 6  8  20  6 6. Solve 2x – a = b for x.

Solution 2x – a  b 2x  a  b 1 x  a  b 2 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. In a matrix A, the symbol aij is the entry in row __________ and column __________.

Solution i, j 8. For matrices A and B to be equal, they must be the same __________, and corresponding entries must be __________.

Solution size, equal 9. To find the sum of matrices A and B, we add the __________ entries.

Solution corresponding 10. To multiply a matrix by a scalar, we multiply __________ by that scalar. Solution every element 11. The product of a 3 × 2 matrix A and a 2 × 4 matrix B will exist because the number of __________ of A is equal to the number of __________ of B.

Solution columns, rows 12. The product of the matrices A and B in Exercise 5 will be a __________ matrix.

Solution 3×4

0 0 13. Among 2 × 2 matrices,   is the __________ or zero matrix. 0 0 Solution additive identity

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1486


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1 0 14. Among 2 × 2 matrices,   is the __________ matrix. 0 1  Solution multiplicative identity Practice Identify the size (or order) of each matrix.

 4  2 15.  .  1 0 7  Solution 2×3

 2 3e 3    1   7 4   16.  5 12 9  .   1  11 2 4.8    6 5.3 2  Solution 5×3 17. [1 3 5 –7 –9 –11]

Solution 1×6  3   6 18.   9    12

Solution 4×1  2  3 2    Let A   5 4  , B   4 3  11 6     4

7 1.5   5 10    . Identify the entry of the matrix that is 6 8  , and C   2    4    3 0 5e  represented by the given notation.

19. a21

Solution 5

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1487


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

20. a32

Solution –6 21. b31

Solution 3

4

22. b33

Solution 5e 23. c12

Solution 10 24. c21

Solution

2 3 Find values of x and y, if any, that will make the matrices equal.

x 25.  1

y  2 5    3   1 3

Solution x = 2, y = 5

x 26.  3

5  0 5    y   3 2

Solution x = 0, y = 2

x  y 27.   2

3  x   3 4   5 y   2 10

Solution xy 3

3 x  4 x  1 5 y  10  y  2

x  y 28.   2x

x  y   x x  2   3 y   y 8  y 

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1488


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution x  y  x

x  1

x  y  x 2 2x   y 3y  8  y

y 2

Find A + B.

 2 29. A    3

1 2

 3 1 2 1 , B    5  3 2 5

Solution é 2 1 -1ù é-3 1 2ùú éê-1 2 1 ùú ú+ê A+B = ê = ê-3 2 5ú ê-3 -2 -5ú ê-6 0 0ú ë û ë û ë û  3  2 2 1 6 2      30. A   2 3 3  , B   5 7 1  4 2 1  4  6 7    

Solution é 3 2 1ùú éê -2 6 -2ùú éê 1 8 -1ùú ê A + B = ê -2 3 -3ú + ê 5 7 -1ú = ê 3 10 -4ú ê ú ê ú ê ú êë-4 -2 -1úû êë-4 -6 7úû êë-8 -8 6úû Find the additive inverse of each matrix.  5 2 7    31. A   5 0 3  2 3 5   

Solution

é-5 2 -7ùú ê 0 -3ú additive inverse of A = ê 5 ê ú êë 2 -3 5úû

 2 1 5 32. A  3   3 2   Solution additive inverse of A = éêë-3

2 3

5 - 21 ùúû

Find A – B.

 3 33. A    1

 3 3 2 2 2 , B    4 5  2 5 5

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1489


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é-3 2 -2ù é 3 -3 -2ù é-6 5 0ù ú-ê ú=ê ú A-B = ê ê -1 4 -5ú ê-2 ú ê 1 -1 0ú 5 5 ë û ë û ë û  2  4 2 0    34. A   2 8 1 , B   1  3 3  8   1   

3 2 4

7  0 1

Solution é 2 2 0ùú éê-4 3 7ùú éê 6 -1 -7ùú ê A - B = ê-2 8 1ú - ê -1 2 0ú = ê-1 6 1ú ê ú ê ú ê ú êë 3 -3 -8úû êë 1 4 -1úû êë 2 -7 -7úû Find 5A.

3 3 35. A    0 2 Solution é 3 -3ù é 15 -15ù ú=ê ú 5A = 5 ê ê0 -2ú ê 0 -10ú ë û ë û

 3 3   36. A  5  0 1 Solution 3ù é3 é 15 3ùú 5ú = ê 5A = 5 ê ê0 -1ú ê 0 -5ú û ë û ë

 5 15 2 37. A    1  2 5 Solution é 5 15 -2ù ú 5A = 5 ê ê-2 -5 1úû ë é 25 75 -10ùú =ê ê-10 -25 5úû ë

 3 1 2 38. A     8 2 5

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1490


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é -3 1 2ùú 5A = 5 ê ê-8 -2 -5ú ë û é -15 ù 5 10 ú =ê ê-40 -10 -25ú ë û Find 5A + 3B.

 3 39. A    4

 1 2 1 2 , B   3 2  5 5

2  3

Solution é 3 1 -2ù é ù é ù é ù ú + 3 ê 1 -2 2ú = ê 15 5 -10ú + ê 3 -6 6ú 5 A + 3B = 5 ê ê-4 3 -2ú ê-5 -5 3ú ê-20 15 -10ú ê-15 -15 9ú ë û ë û ë û ë û é 18 -1 -4ù ú =ê ê-35 0 -1ú ë û

 2 5 5 2 40. A   , B     5 2 2 5 Solution é 2 -5ù é ù é ù é ù é ù ú + 3 ê5 -2ú = ê 10 -25ú + ê 15 -6 ú = ê 25 -31ú 5 A + 3B = 5 ê ê-5 ê2 -5ú ê-25 2úû 10úû êë 6 -15úû êë-19 -5úû ë ë û ë

 1 3 4   5 2 1  Let A    and B    . Solve each matrix equation for X. 2 1 2     4 1 2  41. X + A = B

Solution X+A=B

é 5 2 -1ù é 1 -3 4ù é4 5 -5ù ú-ê ú=ê ú X = B-A= ê ê4 1 -2ú ê2 -1 2ú ê 2 2 -4ú ë û ë û ë û 42. X + B = A

Solution X +B = A

é 1 -3 4ù é 5 2 -1ù é-4 -5 5ù ú-ê ú=ê ú X = A-B = ê ê2 -1 2ú ê4 1 -2ú ê -2 -2 4ú ë û ë û ë û 43. 2B – X = A

Solution 2B - X = A -X = -2B + A

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1491


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é 5 2 -1ù é 1 -3 4ù é9 7 -6ù ú-ê ú=ê ú X = 2B - A = 2 ê ê4 1 -2ú ê2 -1 2ú ê6 3 -6ú ë û ë û ë û 44. 2A – X = B

Solution 2A - X = B

-X = -2 A + B é 1 -3 4ù é 5 2 -1ù é-3 -8 9ù ú-ê ú=ê ú X = 2A - B = 2 ê ê ú ê ú ê ú ë2 -1 2û ë4 1 -2û ë 0 -3 6û 45. X + 2A = 3B

Solution X + 2 A = 3B

é 5 2 -1ù é ù é ù ú - 2 ê 1 -3 4ú = ê 13 12 -11ú X = 3B - 2 A = 3 ê ê4 1 -2ú ê2 -1 2ú ê 8 5 -10ú ë û ë û ë û 46. X + 3B = 2A

Solution X + 3B = 2 A

é 1 -3 4ù é ù é ù ú - 3 ê 5 2 -1ú = ê-13 -12 11ú X = 2 A - 3B = 2 ê ê2 -1 2ú ê4 1 -2ú ê -8 -5 10ú ë û ë û ë û 47. 2X – 3A = B

Solution 2X - 3A = B 2X = 3A + B X=

3 1 3 é 1 -3 4ùú 1 éê 5 2 -1ùú éê4 - 72 112 ùú A+ B = ê + = 2 2 2 ëê2 -1 2úû 2 êë4 1 -2úû êë 5 -1 2úû

48. 2X – 3B = A

Solution 2 X - 3B = A 2 X = 3B + A X=

3 1 3 é 5 2 -1ùú 1 éê 1 -3 4ùú éê8 B+ A= ê + = 2 2 2 ëê4 1 -2úû 2 êë2 -1 2úû êë7

ù ú 1 -2úû

3 2

1 2

49. 3A + 5B = –3X

Solution 3 A + 5B = -3 X

é 1 -3 4ù 5 é 5 2 -1ù é- 28 - 1 - 7 ù 5 3 3ú ú- ê ú=ê 3 X = -A - B = - ê 2 4ú ê2 -1 2ú 3 ê4 1 -2ú ê- 26 3 3 3 3 ë û ë û ë û

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1492


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

50. 5A + 3B = –3X

Solution 5 A + 3B = -3 X

X =-

3 - 173 ùú 5 5 é 1 -3 4ùú éê 5 2 -1ùú éê- 20 3 = 22 A-B = - ê 3 3 êë2 -1 2úû êë4 1 -2úû êë- 3 23 - 43 úû

Find each product, if possible.

 2 3  1 2 51.    3 2 0 2 Solution é (2)(1) + (3)(0) é 3úù êé 1 2úù ê2 = êê ê ú ê ú ë3 -2û 2´2 ë0 -2û 2´2 êë(3)(1) + (-2)(0)

(2)(2) + (3)(-2)úù = êé2 -2ùú (3)(2) + (-2)(-2)ûúú 2´2 ëê3 10ûú

 2 3  2 4 52.     3 2  5 7  Solution é(-2)(2) + (3)(-5) é-2 3ùú éê 2 4ùú ê = êê ê 3 -2ú ê-5 7ú ë û 2´2 ë û 2´2 ëê(3)(2) + (-2)(-5)

(-2)(4) + (3)(7)ùú = éê-19 13ùú (3)(4) + (-2)(7)úûú 2´2 êë 16 -2úû

 4 2  5 6 53.     21 0  21 1 Solution é -4 -5 + -2 21 é-4 -2ù é-5 6ù ê ú ê ú = êê( )( ) ( )( ) ê 21 0úû 2´2 êë 21 -1úû 2´2 ê (21)(-5) + (0)(21) ë ë é -22 -22ù ú =ê ê-105 126ú ë û

(-4)(6) + (-2)(-1)ùú (21)(6) + (0)(-1)úûú 2´2

 5 4 6 2 54.     4 5  1 3 Solution é(-5)(6) + (4)(1) é 4úù êé6 -2úù ê-5 = êê ê 4 -5ú ê 1 3úû 2´2 ê(4)(6) + (-5)(1) ë û 2´2 ë ë

(-5)(-2) + (4)(3)úù = êé-26 22úù (4)(-2) + (-5)(3)úûú 2´2 êë 19 -23úû

 2 1 3  1 2 3    55.  1 2 1  2 2 1 0 1 0  0 0 1   

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1493


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 2 1 3ù ê ú ê 1 2 -1ú ê ú êë0 1 0úû

é1 2 3ùú ê ê 2 -2 1ú ê ú ê0 0 1úû 3´3 ë 3´3 é (2)(1) + (1)(2) + (3)(0) (2)(2) + (1)(-2) + (3)(0) ê = êê(1)(1) + (2)(2) + (-1)(0) (1)(2) + (2)(-2) + (-1)(0) ê 0 1 + 1 2 + 0 0 êë ( )( ) ( )( ) ( )( ) (0)(2) + (1)(-2) + (0)(0) é4 2 10ùú ê ê 5 = - 2 4ú ê ú êë 2 -2 1úû

(2)(3) + (1)(1) + (3)(1)ùú (1)(3) + (2)(1) + (-1)(1)úú (0)(3) + (1)(1) + (0)(1)úúû 3´3

2 1 1  1 2 3    56.  1 1 2  1 2 3  1 2 1  1 1 3   

Solution é2 1 1ùú ê ê1 1 2ú ê ú êë 1 -2 -1úû

é 1 2 3ùú ê ê 1 2 -3ú ê ú ê-1 -1 3úû 3´3 ë 3´3 é (2)(1) + (1)(1) + (1)(-1) (2)(2) + (1)(2) + (1)(-1) (2)(3) + (1)(-3) + (1)(3)ùú ê = êê (1)(1) + (1)(1) + (2)(-1) (1)(2) + (1)(2) + (2)(-1) (1)(3) + (1)(-3) + (2)(3)úú ê 1 1 + - 2 1 + - 1 -1 1 2 + - 2 2 + -1 - 1 1 3 + - 2 - 3 + -1 3 ú ëê ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )úû é 2 5 6ù ê ú = ê 0 2 6ú ê ú êë0 -1 6úû

 1   57.  2 4 5 6  3  

Solution é é ù ê (1)(4) ê 1ú ê ê-2ú é4 -5 -6úù = ê(-2)(4) û 1´3 ê ú êë ê êë(-3)(4) ëê-3ûú 3´1

é 4 -5 -6ù (1)(-5) (1)(-6)ùú ú = êê -8 10 12úú 2 5 2 6 ( )( ) ( )( )ú ê ú (-3)(-5) (-3)(-6)úûú 3´3 ëê-12 15 18ûú

 4  1 2 3    58.   5 1  2 0  6   

Solution é ù é 1 4 + - 2 -5 + -3 -6 ù é 1 -2 -3ù ê 4ú é ù ê ú ê-5ú = ê( )( ) ( )( ) ( )( )ú = ê32ú ê ú ê2 ú ê2ú 0 1û 2´3 ê ú 2 4 + 0 - 5 + 1 -6 ë êë-6úû ëê ( )( ) ( )( ) ( )( )ûú 2´1 ë û 3´1

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1494


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

4 5 6  59.  1 2 3    7 8 9 Solution

é ù é 1 2 3ù ê4 5 6ú êë úû 1´3 ê 7 8 9ú ë û 2´3 Not possible  3 5 8   2 5   60.  5   2 7  1 7   3 6 2 

Solution

é 5 -8ùú é 2 5ù ê 3 ê ú ê-2 7 5ú ê-1 7ú ê ú ë û 2´2 ê 3 -6 2úû ë 3´3 Not possible  2 3 4   1    61.  1 2 3  2  2 2 2   3    

Solution é (2)(-1) + (3)(2) + (4)(3)ù é 2 3 4ù é-1ù é 16ù ê ú ê ú ê ú ê ú ê ú ê 1 2 3ú ê 2ú = ê (1)(-1) + (2)(2) + (3)(3)ú = ê 12ú ê ú ê ú ê ú ê -2 -1 + 2 2 + 2 3 ú êë-2 2 2úû êë 3úû ê 12ú 3´3 3´1 ëê( )( ) ( )( ) ( )( )ûú 3´1 ë û  2 5   1  3 2 4  3 62.   0 2  2 3 1     1 5

Solution é 2 5ùú ê ê-3 1ú éê 3 -2 4ùú ê ú ê 0 -2ú êë-2 -3 1úû 2´3 ê ú êë 1 -5úû 4´2 é (2)(3) + (5)(-2) (2)(-2) + (5)(-3) ê ê -3 3 + 1 -2 ( )( ) ( )( ) (-3)(-2) + (1)(-3) = êê ê(0)(3) + (-2)(-2) (0)(-2) + (-2)(-3) ê êë (1)(3) + (-5)(-2) (1)(-2) + (-5)(-3)

(2)(4) + (5)(1)ùú é -4 -19 13ù ê ú ú (-3)(4) + (1)(1)ú = êê-11 3 -11úú ê 4 6 -2ú (0)(4) + (-2)(1)úú ê ú ú (1)(4) + (-5)(1)úû 4´3 êë 13 13 -1úû

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1495


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1 2 3    1 2 63. 4 5 6    3 4  7 8 9   

Solution é 1 2 3ù ê ú ê4 5 6ú ê ú êë7 8 9úû

3´3

é 1 2ù ê ú ê 3 4ú ë û 2´2

Not possible  1 4 0 0  1    4 1 0 2  2 64.   0 0 1 0  2    1  1  0 2 0

Solution é ù é 1 4 0 é 9ù 0ùú éê 1ùú ê (1)(1) + (4)(2) + (0)(-2) + (0)(-1)ú ê ê ú ê ú -4)(1) + (1)(2) + (0)(-2) + (-2)(-1)ú ê-4 1 0 -2ú ê 2ú ê 0ú ( ê ê ú ê ú =ê ú = êê-2úú ê 0 0 1 0ú ê-2ú ê (0)(1) + (0)(2) + (1)(-2) + (0)(-1)ú ê ú ê ú ê ú 1ûú 4´4 ëê -1ûú 4´1 êê (0)(1) + (2)(2) + (0)(-2) + (1)(-1)úú ê 3ú ëê 0 2 0 ë û 4´1 ë û  2.3 1.7 3.1  2.5   5.8        Let A   2 3.5 1 , B   5.2  , and C   2.9  . Use a graphing calculator to find each result.  8 4.7 9.1  7   4.1      

65. AB

Solution é2.3 -1.7 3.1ù é-2.5ù ê úê ú AB = ê -2 3.5 1ú ê 5.2ú ê úê ú êë -8 4.7 9.1úû êë -7 úû é-36.29ù ê ú 16.2ú =ê ê ú êë -19.26úû 66. B + C

Solution é-2.5ù é-5.8ù ê ú ê ú B + C = ê 5.2ú + ê 2.9ú ê ú ê ú êë -7úû êë 4.1úû é-8.3ù ê ú = ê 8.1ú ê ú êë-2.9úû

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1496


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

67. A2

Solution é2.3 -1.7 3.1ù é2.3 -1.7 3.1ù é -16.11 4.71 33.64ùú ê úê ú ê 2 ê ú ê ú ê A = -2 3.5 1 -2 3.5 1 = -19.6 20.35 6.4ú ê úê ú ê ú êë -8 4.7 9.1úû êë -8 4.7 9.1úû êë-100.6 72.82 62.71úû 68. AB + C

Solution

é2.3 -1.7 3.1ù é-2.5ù é-5.8ù é-42.09ù ê úê ú ê ú ê ú AB + C = ê -2 3.5 1ú ê 5.2ú + ê 2.9ú = ê 19.1ú ê úê ú ê ú ê ú êë -8 4.7 9.1úû êë -7úû êë 4.1úû êë -15.16úû

2 3  2 1 5   2  1 6  1 2  1 2  Let A   , B   , C   , D    , and E    . Verify each property 3  1 3  1 1 2  0 1 1  1 3 2 by doing the operations on each side of the equation and comparing the results. 69. Distributive Property: A(B + C) = AB + AC

Solution

é2 3ù æç é2 1 -5ù é-2 -1 6ù ö÷ é2 3ù é0 0 1ù é3 0 5ù ú çê ú+ê ú÷ = ê úê ú=ê ú A (B + C ) = ê ê 1 3ú ççè ê 1 1 2úû êë 0 -1 -1úû ÷÷ø êë 1 3úû êë 1 0 1úû êë3 0 4úû ë û ë é2 AB + AC = ê ê1 ë é7 =ê ê5 ë

3ùú éê2 1 -5ùú éê2 3ùú éê-2 -1 6ùú + 3úû êë 1 1 2úû êë 1 3úû êë 0 -1 -1úû 5 -4ùú éê-4 -5 9ùú éê3 0 5ùú + = 4 1úû êë -2 -4 3úû êë3 0 4úû

70. Associative Property of Scalar Multiplication: 5(6A) = (5 · 6)A

Solution æ é2 3ù ö÷ é ù é ù ú ÷÷ = 5 ê 12 18ú = ê60 90ú 5 (6 A) = 5 ççç6 ê ê 6 18ú ê30 90ú çè êë 1 3úû ÷ø ë û ë û é2 3ù é60 90ù (5 ⋅ 6) A = 30 A = 30 êê 1 3úú = êê30 90úú ë û ë û

71. Associative Property of Scalar Multiplication: 3(AB) = (3A)B

Solution æ é2 3 ( AB) = 3 ççç ê çè êë 1 æ é2 (3 A) B = çççç3 êê 1 è ë

é7 5 -4ù é 21 15 -12ù 3ùú éê2 1 -5ùú ö÷ ú=ê ú ÷ = 3ê + ê5 4 3úû êë 1 1 2úû ÷÷ø 1úû êë 15 12 3úû ë 3úù ÷ö êé2 1 -5úù êé6 9úù êé2 1 -5úù êé 21 15 -12úù ÷ + = 3úû ÷÷ø êë 1 1 2úû êë3 9úû êë 1 1 2úû êë 15 12 3úû

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1497


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

72. Associative Property of Multiplication: A(DE) = (AD)E

Solution é2 3ù æç é 1 ú çê A (DE ) = ê ê 1 3ú çèç ê 1 ë û ë æ é2 3ù é 1 ( AD) E = çççç êê 1 3úú êê 1 èë ûë

2ùú éê 1 -2ùú ö÷ éê2 3ùú éê5 4ùú éê 31 29ùú ÷= = 3úû êë2 3úû ÷÷ø êë 1 3úû êë7 7 úû êë26 25úû 2ùú ÷ö éê 1 -2ùú éê 5 13ùú éê 1 -2ùú éê 31 29ùú ÷ = = 3úû ÷÷ø êë2 3úû êë4 11úû êë2 3úû êë26 25úû

 1 3  1  Let A    , B    , and C   3 2  . Perform the operations, if possible. 2 5  3 73. A – BC

Solution é 1 3ù é-1ù é ù é ù é ù ú - ê ú éê3 2ùú = ê 1 3ú - ê-3 -2ú = ê 4 5ú A - BC = ê û ê2 5ú ê 9 ê2 5ú ê 3ú ë 6úû êë-7 -1úû ë û ë û ë û ë

74. AB + B

Solution

é 1 3ù é-1ù é-1ù é 8ù é-1ù é 7ù úê ú+ê ú = ê ú+ê ú = ê ú AB + B = ê ê2 5ú ê 3ú ê 3ú ê 13ú ê 3ú ê 16ú ë ûë û ë û ë û ë û ë û 75. CB – AB

Solution é-1ù é 1 3ù é-1ù é ù ú ê ú = éê3ùú - ê 8 ú  not possible CB - AB = éëê3 2ùûú ê ú - ê ë û ê 3ú ê2 5ú ê 3ú ê 13ú ë û ë ûë û ë û

76. CAB

Solution

é 1 3ù é-1ù é ù ú ê ú = éê7 19ùú ê-1ú = éê50ùú CAB = éëê3 2ùûú ê ë û ê2 5ú ê 3ú ê 3ú ë û ë ûë û ë û 77. ABC

Solution é 1 3ù é-1ù é ù é ù ú ê ú éê3 2ùú = ê 8 ú éê3 2ùú = ê24 16 ú ABC = ê û ê 13ú ë û ê39 26ú ê2 5ú ê 3ú ë ë ûë û ë û ë û 78. CA + C

Solution é 1 3ù ú + é3 2ùú = éê7 19ùú + éê3 2ùú = éê 10 21ùú CA + C = éëê3 2ùûú ê û ë û ë û ë û ê2 5ú êë ë û

79. A2B

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1498


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 1 3ù é 1 3ù é-1ù é 7 18ù é-1ù é47ù úê úê ú = ê úê ú = ê ú A2B = ê ê2 5ú ê2 5ú ê 3ú ê 12 31ú ê 3ú ê 81 ú ë ûë ûë û ë ûë û ë û 80. (BC)2

Solution 2

æ é-1ù

ö

2

èë

ø

û

2

æ é-3 -2ù ö÷ é ùé ù é ù ú ÷÷ = ê-3 -2ú ê-3 -2ú = ê-9 -6ú ê 9 6úû ø÷ 6úû êë 9 6úû êë 27 18 úû èë ë

(BC ) = çççç êê 3úú éëê3 2ùûú ÷÷÷÷ = çççç êê 9 Fix It

In exercises 81 and 82, identify the step the first error is made and fix it.

 1 2 3  2 4 6 81. Let A    and B    . Solve the matrix equation 3X + B = A for x.  1 2 3  2 4 6 Solution Step 4 was incorrect. Step 4: X 

1  3 6 9   3  3 6 9

 1 2 3 Step 5: X     1 2 3 2 4   1 2  1 2 82. Let A   , B    , and C    . Determine AB + C. 1 3 1  1  1 2 Solution Step 2 was incorrect.  2  1  4  1 2  2   4  1   1 2  Step 2: AB  C    3  1  1  1 3  2    1 1   1 2

Step 3:

 6 0  1 2 AB  C      2 7   1 2

 5 2 Step 4: AB  C     3 5 Applications 83. Sporting goods Two suppliers manufactured footballs, baseballs, and basketballs in the quantities and costs given in the tables. Find matrices Q and C that represent the quantities and costs, find the product QC, and interpret the result.

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1499


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Quantities Footballs

Baseballs

Basketballs

Supplier 1

200

300

100

Supplier 2

100

200

200

Unit Costs

(in $)

Footballs

8

Baseballs

4

Basketballs

7

Solution é 5ù ê ú é200 300 100ù ú , C = ê 2ú Q = êê ê ú ú êë 100 200 200úû ê 4ú êë úû é 5ù é200 300 100ù ê ú é2000ù Cost of balls from Supplier 1 ú ê 2ú = ê ú QC = êê úê ú ê ú ëê 100 200 200ûú ê4ú ëê 1700 ûú Cost of balls from Supplier 2 ëê ûú 84. Retailing Three ice cream stores sold cones, sundaes, and milkshakes in the quantities and prices given in the tables. Find matrices Q and P that represent the quantities and prices, find the product QP, and interpret the results.

Quantities Cones

Sundaes

Shakes

Store 1

75

75

32

Store 2

80

69

27

Store 3

62

40

30

Unit Price Cones

$3.00

Sundaes

$4.00

Shakes

$5.00

Solution é 75 75 32ù é 1.50 ù ê ú ê ú ê ú Q = ê80 69 27 ú , P = êê 1.75 úú ê62 40 30ú ê3.00ú úû ëê ëê ûú

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1500


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é 75 75 32ù é 1.50 ù é 339.75ù $ made by Store 1 ê úê ú ê ú QC = êê80 69 27 úú êê 1.75 úú = êê 321.75 úú $ made by Store 2 ê62 40 30ú ê3.00ú ê253.00ú $ made by Store 3 êë úû êë úû êë úû 85. Beverage sales Beverages were sold to senior adults, adults, and students at a school football game in the quantities and prices given in the tables. Find matrices Q and P that represent the quantities and prices, find the product QP, and interpret the result.

Quantities Coffee

Bottled Water

Cola

Senior Adults

217

23

319

Adults

347

24

340

Students

3

97

750

Price Coffee

$1.00

Bottled Water

$1.50

Cola

$2.00

Solution é 217 23 319 ù é0.75ù ê ú ê ú ê ú Q = ê347 24 340ú , P = êê 1.00 úú ê 3 97 750ú ê 1.25 ú ëê ûú ëê ûú é 217 23 319 ù é0.75ù é 584.50 ù $ spent by adult males ê úê ú ê ú QC = êê347 24 340úú êê 1.00 úú = êê 709.25 úú $ spent by adult females ê 3 97 750ú ê 1.25 ú ê 1036.75ú $ spent by children êë úû êë úû êë úû 86. Production costs Each of four factories manufactures three products in the daily quantities and unit costs given in the tables. Find a suitable matrix product to represent production costs.

Production Quantities Factory

Product A

Product B

Product C

Phoenix

19

23

27

Boston

17

21

22

Chicago

21

18

20

Denver

27

25

22

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Unit Production Costs Day Shift

Night Shift

Product A

$1.20

$1.35

Product B

$0.75

$0.85

Product C

$3.50

$3.70

Solution é 19 23 27 ù é ù ê ú é 1.20 1.35 ù ê 134.55 145.10 ú Day/Night costs in Ashtabula ê 17 21 22 ú ê ú ê 113.15 122.20ú Day/Night costs in Boston ê ú ê ú ú ê ê21 18 20ú ê0.75 0.85ú = ê 108.70 117.65 ú Day/Night costs in Chicago ê ú ê3.50 3.70ú ê ú ê ú ê ûú êê 128.15 139.10 úú Day/Night costs in Denver êë27 25 22 úû ë ë û 87. Connectivity matrix An entry of 1 in the following connectivity matrix A indicates that the person associated with that row knows the address of the person associated with that column. For example, the 1 in Victor’s row and Alan’s column indicates that Victor can write to Alan. The 0 in Victor’s row and Carlotta’s column indicates that Bill cannot write to Carlotta. However, Victor could ask Alan to forward his letter to Carlotta. The matrix A2 indicates the number of ways that one person can write to another with a letter that is forwarded exactly once. Find A2.

Alan Victor Carlotta Alan 0  Victor  1 Carlotta 0

1 0 1

1  0  A 0

Solution 2

é0 1 1 ù é0 1 1 ù é0 1 1 ù é 1 1 0ù ê ú ê úê ú ê ú ê ú A = 1 0 0 = ê 1 0 0ú ê 1 0 0ú = ê0 1 1 ú ê ú ê úê ú ê ú êë0 1 0úû êë0 1 0úû êë0 1 0úû êë 1 0 0úû 2

88. Communication routing Refer to Exercise 87. Find the matrix A + A2. Can everyone receive a letter from everyone else with at most one forwarding?

Solution 2

é0 1 1 ù é0 1 1 ù é0 1 1 ù é 1 1 0ù é 1 2 1 ù ê ú ê ú ê ú ê ú ê ú 2 ê ú ê ú A + A = 1 0 0 + 1 0 0 = ê 1 0 0ú + ê0 1 1 ú = ê 1 1 1 ú ê ú ê ú ê ú ê ú ê ú êë0 1 0úû êë0 1 0úû êë0 1 0úû êë 1 0 0úû êë 1 1 0úû The matrix A + A2 contains the number of ways for a message to get from one person to another either directly or through one other person. The only 0 in the matrix is in the location which represents a message from Carl to himself. Thus, everybody is able to receive a letter from everyone else with at most one forwarding. 89. Routing telephone calls A long-distance telephone carrier has established several direct links among four cities. In the following connectivity matrix, entries aij and aji indicate the

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1502


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

number of direct links between cities i and j. For example, cities 2 and 4 are not linked directly but could be connected through city 3. Find and interpret matrix A2. 0 2 1 0   2 0 1 0 A  1 1 0 2   0 0 2 0

Solution 2

é0 2 1 0ù é0 2 1 0ù é0 2 1 0ù é5 1 2 2ù ê ú ê úê ú ê ú ê ú ê 2 0 1 0ú ê 2 0 1 0ú ê 1 5 2 2ú 2 0 1 0 2 ú =ê úê ú=ê ú A =ê ê 1 1 0 2ú ê 1 1 0 2ú ê 1 1 0 2ú ê 2 2 6 0 ú ê ú ê úê ú ê ú êë0 0 2 0úû êë0 0 2 0úû êë0 0 2 0úû êë2 2 0 4úû

A2 represents the number of ways two cities can be linked through one intermediary. 90. Communication on one-way channels Three communication centers are linked as indicated in the illustration, with communication only in the direction of the arrows. Thus, location 1 can send a message directly to location 2 along two paths, but location 2 can return a message directly on only one path. Entry cij of matrix C indicates the number of channels from i to j. Find and interpret C2. 0 2 2    C   1 0 1  1 0 0  

Solution 2

é0 2 2ù é0 2 2ù é0 2 2ù é4 0 2ù ê ú ê úê ú ê ú C 2 = ê 1 0 1 ú = ê 1 0 1 ú ê 1 0 1 ú = ê 1 2 2ú ê ú ê úê ú ê ú êë 1 0 0úû êë 1 0 0úû êë 1 0 0úû êë0 2 2úû C 2 represents the number of ways two cities can be linked through one intermediary.

Discovery and Writing 91. Explain how to add two matrices. Give an example.

Solution Answers may vary. 92. Explain how to subtract two matrices. Give an example.

Solution Answers may vary. 93. What is scalar multiplication and how is it performed? Give an example.

Solution Answers may vary.

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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

94. Explain the steps you would use to multiply two matrices. Give an example.

Solution Answers may vary. 95. If A and B are 2 × 2 matrices, is (AB)2 equal to A2B2? Support your answer.

Solution é 1 1ù é ù ú and B = ê 1 0ú . Let A = ê ê 1 1ú ê0 0ú ë û ë û 2 2 æ ö é 1 1ù é 1 0ù ÷ é 1 0ù é 1 0ù é 1 2 ç ê ú ê ú ê ú ÷ ( AB) = çççê 1 1ú ê0 0ú ÷÷ = ê 1 0ú = êê 1 0úú êê 1 èë ûë ûø ë û ë ûë 2 2 é 1 1ù é 1 0ù é ùé ù é ú ê ú = ê2 2ú ê 1 0ú = ê2 A2B2 = ê ê 1 1ú ê0 0ú ê2 2ú ê0 0ú ê2 ë û ë û ë ûë û ë

0úù êé 1 0úù = 0ûú ëê 1 0ûú 2 0ùú ⋅ ( AB) ¹ A2B2 0úû

96. Let a, b, and c be real numbers. If ab = ac and a ≠ 0, then b = c. Find 2 × 2 matrices A, B, and C, where A ≠ 0 to show that such a law does not hold for all matrices.

Solution

é1 Let A = ê ê0 ë é1 AB = ê ê0 ë

é0 1ù é 0ùú ú and C = ê0 ,B=ê ê 1 1ú ê1 0úû ë û ë ù é ù é ù é1 0ú ê0 1ú ê0 1 ú = ; AC = ê ú ê ú ê ú ê0 0û ë 1 1û ë0 0û ë

1 ùú . 2úû 0ùú éê0 1 ùú éê0 1 ùú = 0úû êë 1 2úû êë0 0úû

So AB = AC, but B ¹ C. 97. Another property of the real numbers is that if ab = 0, then either a = 0 or b = 0. To show that this property is not true for matrices, find two nonzero 2 × 2 matrices A and B, such that AB = 0.

Solution é 1 2ù é ù é ùé ù é ù ú and B = ê 2 2ú . AB = ê 1 2ú ê 2 2ú = ê0 0ú Let A = ê ê 1 2ú ê-1 -1ú ê 1 2ú ê-1 -1ú ê0 0ú ë û ë û ë ûë û ë û 98. Find 2 × 2 matrices to show that (A + B)(A – B) ≠ A2 – B2.

Solution

é 1 0ù é ù ú and B = ê0 1 ú . Let A = ê ê0 0ú ê0 0ú ë û ë û æ é 1 0ù é0 1 ù ÷ö æ é 1 0ù é0 1 ù ÷ö ( A + B)( A - B) = ççççêê0 0úú + êê0 0úú ÷÷÷ ççççêê0 0úú - êê0 0úú ÷÷÷ èë û ë ûø èë û ë ûø é 1 1 ù é 1 -1ù é 1 -1ù úê ú=ê ú =ê ê0 0ú ê0 0ú ê0 0ú ë ûë û ë û 2 2 é ù é ù é 1 0ú 0 1ú 1 0ùú éê 1 0ùú éê0 1 ùú éê0 1 ùú -ê =ê A2 - B2 = ê ê0 0ú ê0 0ú ê0 0ú ê0 0ú ê0 0ú ê0 0ú ë û ë û ë ûë û ë ûë û é 1 0ù é0 0ù é 1 0ù ú-ê ú=ê ú =ê ê0 0ú ê0 0ú ê0 0ú ë û ë û ë û

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1504


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true.

2 2  5 5  10 10 99. If A    and B    , then AB   . 2 2 5 5      10 10 Solution False. é2 2ù é5 5ù é20 20ù úê ú=ê ú. AB = êê úê ú ê ú êë2 2úû êë5 5úû êë20 20úû

2 4  4 16 2 100. If A    , then A   . 6 8 36 64 Solution False. é2 4ù é2 4ù é28 40ù úê ú=ê ú. A2 = êê úê ú ê ú ëê6 8ûú ëê6 8úû êë60 88ûú 101. Matrix addition is commutative.

Solution True. 102. Matrix multiplication is commutative.

Solution False. Matrix multiplication is not commutative. 103. For the product of two matrices to be defined, the number of rows of the first matrix must equal the number of columns of the second matrix.

Solution False. The number of columns of the 1st matrix must equal the number of rows of the 2nd. 104. If A is a 20 × 30 matrix and B is a 30 × 20 matrix, then AB is a 600 × 600 matrix.

Solution False. AB is a 20 × 20 matrix.

 1 2 3 4 5  1 2 3 4 5 105. The additive inverse of   is  .  6 7 8 9 10 6 7 8 8 10 Solution True.

 1 0 0 106.   is an example of an identity matrix. 0 1 0

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1505


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution False. An identity matrix must be square. 107. The associative property of multiplication holds for matrices.

Solution True. 108. If AC = BC, then A = B.

Solution False. See #78 for a similar result.

EXERCISES 6.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

2 1 Identify the multiplicative inverse of each the given real numbers. –4,  ,  , 1, 7, and 3 2e 2π Solution 1 3 1 1 - , - , - 2e, 1, , 4 2 7 2p

2. When you multiply a real number by its multiplicative inverse what real number do you obtain?

Solution 1

 3 5  12 5 3. Determine the product:   .  7 12  7 3 Solution é 36 - 35 15 - 15ùú éê 1 0ùú ê ê84 - 84 -35 + 36ú = ê0 1ú úû ëê ûú ëê 4. Determine the product:  3   1 2 1  8    1  1 2 1   4 0 1 2     1  8

5 8 1 4 1 8

1  2 0 .  1  2 

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1506


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é3 2 1 ê + + ê8 4 8 ê ê3 2 1 ê - + ê8 4 8 ê ê ê0+ 1 - 1 ê 4 4 ëê

5 2 1 - 8 4 8 5 4 1 + 8 8 8 1 1 0- + 4 4

1 1ù +0- ú 2 2 úú é 1 ê 1 1ú ê + 0 - úú = ê0 2 2ú ê ú êëê0 0 + 0 + 1úú ûú

0 1 0

0ùú ú 0ú ú 1úûú

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 5. Matrices A and B are multiplicative inverses if __________.

Solution AB = BA = I 6. A nonsingular matrix __________ (is, is not) invertible.

Solution is 7. If A is invertible, elementary row operations can change [A/I] into __________.

Solution I A1   

8. If A is invertible, the solution of AX = B is __________.

Solution

X  A1B Verify that A and B are multiplicative inverses. 9.

 4 3  7 3 A B   9 7   9 4 Solution é 4 3ùú éê 7 3ùú éê 28 - 27 12 - 12ùú éê 1 0ùú = = AB = ê ê-9 -7ú ê-9 -4ú ê-63 + 63 -27 + 28ú ê0 1ú ë ûë û ë û ë û é 7 ùé 4 ù é 28 - 27 ù é 1 0ù 3 3 21 21 úê ú=ê ú=ê ú BA = ê ê-9 -4ú ê-9 -7 ú ê-36 + 36 -27 + 28ú ê0 1ú ë ûë û ë û ë û

 7   2 4  2   10. A   B 2  4 7   2 1

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1507


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é ù é 2 -4ù ê- 7 2ú é -7 + 8 4 - 4ù é 1 0ù ê ú ú=ê ú ê ú=ê AB = ê úê 2 ú ê ú ú ê ëê4 -7ûú ê -2 1ú ëê-14 + 14 8 - 7ûú ëê0 1ûú ë û é 7 ù ê2ú éê 2 -4ùú éê -7 + 8 14 - 14ùú éê 1 0ùú ú = = BA = ê 2 ê ú ê4 -7úú êê-4 + 4 8 - 7úûú êëê0 1úûú û ë êë -2 1úû ëê  1 1 1  1 1 0     11. A  2 1 1 B   1 1 1  1 1 0  3 2 1    

Solution é 1 -1 1ù é-1 ê úê AB = ê2 -1 1ú ê 1 ê úê êë 1 1 0úû êë 3 é-1 1 0ùú éê 1 ê ê BA = 1 -1 1ú ê2 ê úê êë 3 -2 1úû êë 1

1 0ùú éê -1 - 1 + 3 1 + 1 - 2 0 - 1 + 1ùú éê 1 0 0ùú -1 1ú = ê-2 - 1 + 3 2 + 1 - 2 0 - 1 + 1ú = ê0 1 0ú ú ê ú ê ú -2 1úû êë -1 + 1 + 0 1 - 1 + 0 0 + 1 + 0úû êë0 0 1úû -1 1ùú éê-1 + 2 + 0 1 - 1 + 0 -1 + 1 + 0ùú éê 1 0 0ùú ú ê -1 1 = 1 - 2 + 1 -1 + 1 + 1 1 - 1 + 0ú = ê0 1 0ú ú ê ú ê ú 1 0úû êë 3 - 4 + 1 -3 + 2 + 1 3 - 2 + 0úû êë0 0 1úû

 1 0  1 2 1  3 4   12. A   1 0 0 B    5 5  1 1 3 1   3   5 5

 0  1 5 2  5

Solution é ù é 6 1 ê0 ú ê 1 0 ê ú ê0 + é 1 2 -1ù 5 5 ê ú ê ú 3 4 1ú ê AB = ê 1 0 0ú êê 0 0+0 = + ê ê ú 5 5 úú ê êë 1 -1 3úû ê 51 3 3 2 ú ê0 - + 3 ê ê ú ê 5 5 êë 5 5 5 úû ë é é ù ê ê0 ú 1 0 ê ú é 1 2 -1ù ê 0 + 1 + 0 ê3 ú ê3 4 1 4 1 úú ê ê 1 0 0ú = ê - + BA = êê ú ê ú ê5 5 5 5 5ú ê ê5 1 -1 3úû ê 1 3 2 1 3 2 ë ú ê ê ú ê - + ê êë 5 5 5 ûú ëê 5 5 5

8 3 2 2ù 0 + - ú é 1 0 0ù + 5 5 5 5 úú ê ú 1 + 0 + 0 0 + 0 + 0ú = ê0 1 0ú ê ú 4 9 1 6 ú ê0 0 1úû 1+ 0 - + úú ë 5 5 5 5û ù 0+0+0 0 + 0 + 0úú é 1 0 0ùú 6 1 3 3 úú ê +0- + 0 + ú = ê0 1 0ú ú 5 5 5 5ú ê 2 2 1 6 ú êë0 0 1úû +0- +0+ ú 5 5 5 5 ûú 1-

Practice Find the multiplicative inverse of each matrix, if possible.

 3 4  13.    2 3

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1508


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é ù ê 3 -4 1 0ú  ê-2 ú 3 0 1û ë é 1 0 3 4ù ê ú ê0 1 2 1 ú  êë 3 3 ûú 4R2 + R1  R1

4 1 4 1 é ù é ù ê 1 - 3 3 0ú  ê 1 - 3 3 0ú  1 2 ê ú ê-2 ú 1ú 3 0 1 úû êë êë0 3 3 û 1  +  R R R R R 2 1 1 2 2 3 1 é 1 0 3 4ù é 3 4ù ê ú ê ú ê0 1 2 3ú  Inverse: ê2 3ú ë û ë û 3R2  R2

2 3 14.   3 5  Solution é2 3 1 0ù ê ú ê3 5 0 1 ú  ë û

3 1 é é 1 3 1 0ù 0ùú 2 2 2 ê ú ê1 2 ê0 1 - 3 1ú  ê3 5 0 1 ú 2 2 ëê ûú ëê ûú 1  +  R R R R R 3 1 1 2 2 2 1 é1 0 ù é1 0 ù é ù 5 3 5 3 ê ú ê ú  Inverse: ê 5 -3ú ê0 1 - 3 ú ê0 1 -3 ú ê ú 1 2 3 2 êë úû 2 2 ë û ë û -3R2 + R1  R1 2R2  R2

3 7  15.   2 5  Solution é3 7 1 0ù ê ú ê2 5 0 1 ú  ë û

7 1 é é 1 7 1 0ù 0ùú 3 3 3 ê ú ê1 3 ê0 1 - 2 1ú  ê2 5 0 1 ú êë úû êë úû 3 3 1 - 2R1 + R2  R2 R  R1 3 1 é 1 0 5 -7 ù é1 0 é ù 5 -7 úù ê ú ê ê 5 -7 ú  Inverse: 1 2 ê0 ê0 1 - 2 ê 1úú -3 3ú 3ûú 3 ë-2 ë û ëê û 3R2  R2 -7R2 + R1  R1

 1 2 16.   2 5 Solution é ù ê 1 -2 1 0ú  ê ú ë2 -5 0 1 û

é é ù é ù 1 0úù ê 1 -2 ê 1 0 5 -2ú  Inverse: ê5 -2ú  ê0 -1 -2 1ú ê0 1 2 -1ú ê2 -1ú ë û ë û ë û - 2R1 + R2  R2 - 2R2 + R1  R1 - R2  R2

 1 2 17.    3 4 

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1509


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

é ù é ù é-2 -1ù é 1 é 1 2 1 0ù ê 1 2 1 0ú ê 1 0 -2 -1ú ê ú 2 1 0ùú ê ê ú ê ú ê ú 1  Inverse: ê 3 1ú ê-3 -4 0 1ú  ê0 2 3 1ú  ê0 1 3 1 ú  ê0 1 3 ú ê ú ë û ë û êë êë 2 2 2 ûú 2 2 úû 2 úû ëê 1 3R1 + R2  R2 R  R2 - 2R2 + R1  R1 2 2

5 10 18.    1 2 Solution é 1 -2 é5 10 1 0ù é ù é ê 1ùú ê ú  ê 1 -2 0 1ú  ê 1 -2 0 ê ê 1 -2 0 1ú ê5 10 1 0ú ê0 20 1 -5ú  ê0 1 ë û ë û ë û êë R1  R2

- 5R1 + R2  R2

é 1 1 ùú ê ê 10 2 úú Inverse: ê 1 1 ê - úú ê 4û ë 20

é 1 1 ùú ê 0 1ùú ê 1 0 10 2 úú . 1 1ú  ê ú 1 1ú ê - ú ê0 1 20 4 úû 20 4 úû êë

1 R  R2 20 2

2R2 + R1

 4 8  19.    1 2 Solution é-4 é 1 -2 0 1ù é ù 8 1 0ùú ê ê ú  ê 1 -2 0 1ú ê 1 -2 0 1ú  ê-4 ú ê 8 1 0 0 0 1 4ú ë û ë û ë û R1  R2 4R1 + R2  R2 Since the original matrix cannot be changed into an identity, there is no inverse matrix.

 3 6 20.    1 2 Solution é 3 - 6 1 0ù é ù é ù ê ú  ê 1 -2 0 -1ú  ê 1 -2 0 1ú ê-1 ê3 -6 1 0ú ê0 2 0 1ú 0 1 3ú ë û ë û ë û R1  -1R2 - 3R1 + R2  R2 Since the original matrix cannot be changed into an identity, there is no inverse matrix.  1 0 3   21.  1 1 3  2 1 1  

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1510


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 1 0 3 1 0 0ù ê ú ê - 1 1 3 0 1 0ú  ê ú ê-2 1 1 0 0 1 ú ë û

é 1 0 3 1 0 0ù ê ú ê0 1 6 1 1 0ú  ê ú ê0 1 7 2 0 1 ú ë û R1 + R2  R2 2R1 + R3  R3

é 1 0 3 1 0 0ù ê ú ê0 1 6 1 1 0ú  ê ú ê0 0 1 1 -1 1ú ë û - R2 + R3  R3

é 1 0 0 -2 3 -3ù é-2 3 -3ù ê ú ê ú ê0 1 0 -5 7 -6ú  Inverse: ê-5 7 -6ú ê ú ê ú ê0 0 1 êë 1 -1 1 -1 1ú 1úû ë û -3R3 + R1  R1 -6R3 + R2  R2  2 1 1   22.  2 2 1  1 1 1  

Solution é 2 é 1 1 -1 0 0 -1ù é1 1 -1 1 0 0ùú 1 -1 0 0 -1ùú ê ê ú ê ê 2 2 -1 0 1 0ú  ê2 2 -1 0 1 0ú  ê0 0 1 0 1 2ú  ê ú ê ú ê ú ê-1 -1 ê2 1 -1 1 0 0ú ê0 -1 1 0 0 1ú 1 1 0 2ú ë û ë û ë û - R3  R1 - 2R1 + R2  R2 - 2R1 + R3  R3 é 1 1 -1 0 0 -1ù é1 0 0 ù 1 0 1ú ê ú ê ê0 1 -1 -1 0 -2ú  ê0 1 -1 -1 0 -2ú  ê ú ê ú ê0 0 ê0 0 1 0 1 1 0 1 2ú 2ú ë û ë û R2  -R3 - R2 + R1  R1 é1 0 0 é 1 0 1ù 1 0 1ùú ê ê ú ê0 1 0 -1 1 0ú  Inverse = ê-1 1 0ú ê ú ê ú ê0 0 1 0 1 2ú êë 0 1 2úû ë û R2 + R3  R2  3 2 1   23.  1 1 1 4 3 1  

Solution é3 2 é 1 1 -1 0 1 0ù é1 1 1 0 0ùú 1 -1 0 1 0ùú ê ê ú ê ê 1 1 -1 0 1 0ú  ê 3 2 1 1 0 0ú  ê0 -1 4 1 -3 0ú  ê ú ê ú ê ú ê4 3 ú ê ê0 -1 5 0 -4 1ú 1 0 0 1 4 3 1 0 0 1ú ë û ë û ë û R1  R2 -3R1 + R2  R2 -4R1 + R3  R3

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1511


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é1 é1 0 1 -1 0 1 0ùú 3 1 -2 0ùú ê ê ê0 ú ê 1 -4 -1 3 0  0 1 -4 -1 3 0ú  ê ú ê ú ê0 -1 ê0 0 5 0 -4 1ú 1 -1 -1 1ú ë û ë û - R2 + R1  R1 -R2  R2 R2 + R3  R3 é1 0 0 4 é 4 1 -3ùú 1 -3ùú ê ê ê0 1 0 -5 -1 ú ê 4  Inverse = -5 -1 4ú ê ú ê ú ê0 0 1 -1 -1 êë -1 -1 1ú 1úû ë û -3R3 + R1  R1 4R3 + R2  R2  2 1 3    24.  2 3 0   1 0 1 

Solution é-2 1 -3 1 0 0ù é 1 0 é1 0 1 0 0 1 0 0 1 ùú 1ùú ê ú ê ê ê 2 3 0 0 1 0ú  ê 2 3 0 0 1 0ú  ê0 3 -2 0 1 -2ú  ê ú ê ú ê ú ê 1 0 ê-2 1 -3 1 0 0ú ê0 1 - 1 1 0 1 0 0 1ú 2ú ë û ë û ë û R3  R1 -2R1 + R2  R2 2R1 + R3  R3 é1 0 ù é ù 1 0 0 1ú 1 0 0 1ú ê ê1 0 ê0 1 -1 1 0 2ú  ê0 1 -1 1 0 2ú  ê ú ê ú ê0 3 -2 0 1 -2ú ê0 0 1 -3 1 -8ú ë û ë û R2  R3 - 3R2 + R3  R3 é é 3 -1 9úù 9úù ê1 0 0 ê 3 -1 ê0 1 0 -2 1 -6ú  Inverse = ê-2 1 -6ú ê ú ê ú ê0 0 1 -3 êë-3 1 -8ú 1 -8úû ë û - R3 + R1  R1 R3 + R2  R2  1 2 3   25. 0 1 2  0 0 1  

Solution é 1 2 3 1 0 0ù é 1 0 -1 1 -2 0ù é 1 0 0 1 -2 1ùú ê ú ê ú ê ê0 1 2 0 1 0ú  ê0 1 2 0 1 0ú  ê0 1 0 0 1 -2ú  ê ú ê ú ê ú ê0 0 1 0 0 1 ú ê0 0 ê0 0 1 0 1 0 0 1ú 0 2ú ë û ë û ë û - 2R2 + R1  R1 R3 + R1  R1 -2R3 + R2  R2

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1512


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é 1 -2 1ùú ê Inverse = ê0 1 -2ú ê ú êë0 0 1úû  1 2 3   26. 0 1 1  0 1 0   

Solution é 1 2 3 1 0 0ù é 1 2 3 1 0 0ù é 1 0 3 1 0 2ù ê ú ê ú ê ú ê0 ú ê ú 1 1 0 1 0  0 -1 0 0 0 1  ê0 1 0 0 0 -1ú  ê ú ê ú ê ú ê0 -1 0 0 0 1 ú ê0 ê0 0 1 0 1 1 1 0 1 0ú 1ú ë û ë û ë û R2  R3 2R2 + R1  R1 R2 + R3  R3 -R2  R2 é 1 0 0 1 -3 -1ù é 1 -3 -1ù ê ú ê ú ê0 1 0 0 0 -1ú  Inverse = ê0 0 -1ú ê ú ê ú ê0 0 1 0 êë0 1 1ú 1 1úû ë û - 3R3 + R1  R1    1 1 1 1 1  27.  1 2 2  3  1 1    2

Solution

é 8 -2 -6ù ê ú Inverse = ê-5 2 4ú ê ú êë 2 0 -2úû   1  2 1  1 3 1 28.      2 2 2  1 1  0   2

Solution

é-0.2 1.2 1.6ùú ê Inverse = ê-0.2 -0.8 -0.4ú ê ú êë 0.4 1.6 2.8úû

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1513


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1 3 5   29. 0 1 6   1 4 11  

Solution é 1 3 5 1 0 0ù é1 3 5 é1 3 5 1 0 0ùú 1 0 0ùú ê ú ê ê ê0 1 6 0 1 0ú  ê0 1 6 0 1 0ú  ê0 1 6 0 1 0ú ê ú ê ú ê ú ê 1 4 11 0 0 1 ú ê0 1 6 -1 0 1ú ê0 0 0 -1 -1 1ú ë û ë û ë û - R1 + R3  R3 - R2 + R3  R3

Since the original matrix cannot be changed into the identity, there is no inverse matrix  1 2 3   30. 4 5 6   7 8 9  

Solution é 1 2 3 1 0 0ù é1 é1 2 3 1 0 0ùú 2 3 1 0 0ùú ê ú ê ê ê4 5 6 0 1 0ú  ê0 -3 -6 -4 1 0ú  ê0 -3 -6 -4 1 0ú ê ú ê ú ê ú ê7 8 9 0 0 1 ú ê0 -6 -12 -7 0 1ú ê0 0 0 1 -2 1ú ë û ë û ë û - 4R1 + R2  R2 -2R2 + R3  R3 - 7R1 + R3  R3 Since the original matrix cannot be changed into the identity, there is no inverse matrix  1 6 4   31.  1 2 5  2 4 1  

Solution é1 é1 é1 6 4 1 0 0ùú 6 4 1 0 0ùú 6 4 1 0 0ùú ê ê ê ê 1 -2 -5 0 1 0ú  ê0 -8 -9 -1 1 0ú  ê0 -8 -9 -1 1 0ú ê ú ê ú ê ú ê2 ê0 -8 -9 -2 0 1ú ê0 4 -1 0 0 1 ú 0 0 -1 -1 1ú ë û ë û ë û - R1 + R2  R2 -R2 + R3  R3 - 2R1 + R3  R3

Since the original matrix cannot be changed into the identity, there is no inverse matrix  1 1 1   32.  1 0 1  1 2 3  

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1514


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é1 1 é1 é1 1 1 0 0ùú 1 1 1 0 0ùú 1 1 1 0 0ùú ê ê ê ê 1 0 -1 0 1 0ú  ê0 -1 -2 -1 1 0ú  ê0 -1 -2 -1 1 0ú ê ú ê ú ê ú ê1 2 3 0 0 1ú ê0 ê0 0 1 2 -1 0 1ú 0 -2 1 1ú ë û ë û ë û - R1 + R2  R2 R2 + R3  R3 - R1 + R3  R3

Since the original matrix cannot be changed into the identity, there is no inverse matrix  1 2 3 4   0 1 2 3  33. 0 0 1 2    0 0 0 1

Solution é 1 -2 1 0ùú ê ê0 1 -2 1ú ú Inverse = ê ê0 0 1 -2ú ê ú 0 0 1úû êë0

 1 0 0 0   1 1 0 0 34.   1 1 1 0    1 2 2 1

Solution é 1 0 0 0ùú ê ê-1 1 0 0ú ú Inverse = ê ê 0 -1 1 0ú ê ú êë 1 0 -2 1úû

 1 0 0 0   2 1 0 0  35.  3 2 1 0   4 3 2 1

Solution  1 0 0 0   2 1 0 0 Inverse    1 2 1 0   1 2 1  0

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1515


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3 3 3 2   1 4 3 5  36.  3 0 2 1    1 5 3 6

Solution é-2.5 5 3 5.5ùú ê ê 5.5 -8 -6 -9.5ú ú Inverse = ê ê -1 3 1 3ú ê ú 9 6 10.5úû êë-5.5

 d b a b  1 1 If A =    . Use this alternate method of finding the multiplicative  , then A  ad  bc   c a  c d  inverse of a 2 × 2 matrix to determine A–1.

 2 7  37. A     2 6 Solution

é 7ù é6 7 ù 1 éê6 7ùú ê-3 - ú ê ú A = =- ê =ê 2 úú 2 êë2 2úûú ê 2 (6) - (-7)(-2) êëê2 2úûú êë -1 -1úû 1

-1

3 2 38. A    9 5 Solution -1

A

é 5 2ù é-5 2ù ú 1 éê-5 2ùú êê ú = = = ê 3 3ú ê ú ê ú 9 3 9 3 ê ú 3 (-5) - (-2)(9) ë û 3ë û ê -3 1ú ë û 1

1  3 39. A   2    4 6 Solution é é 6 3ù é 6 3ù ê 2 ê ú ú ê 5 1 ê A = ê ê 1ú = 1ú = ê æ 1 ö÷ -4 çç ÷ 6 - (-3)(4) êêë-4 2 úúû 15 êêë-4 2 úúû êê ÷ ë 15 èç 2 ø÷ -1

1

1 ùú 5 úú 1 ú ú 30 û

 1 0   2 40. A    1 1  5 5 

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1516


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 1 1ù é 1 1ù ê ú ê ú 1 ê 5 2ú ê 5 2 ú éê 2 5ùú A = ê ú = 10 ê ú= ê -1 0ú êë-2 0úû 1 æç 1 öæ 1 ö÷ ê -1 0ú ÷ ç ÷ ÷ 0 ê ú ê ú ç ç ( ) 5 ç 2 ÷÷ç 5 ÷÷ ë 5 û ë5 û è øè ø -1

Write each system of linear equations as a matrix equation of the form AX = B, where A is the coefficient matrix of the system, X is the column matrix of variables, and B is the column matrix of constants.

2x  3 y  2 41.  5x  2 y  14 Solution é2 -3ù é x ù é-2ù ê úê ú = ê ú ê5 2úû êë y úû êë 14úû ë

5x  4 y  1 42.  6 x  y  5 Solution é 5 4ù é x ù é 1ù ê úê ú = ê ú ê-6 -1ú ê y ú ê5ú ë ûë û ë û  x  3 y  2z  0  43.  2 x  y  3z  4  4 x  5 y  z  3 

Solution é 1 -3 2ùú éê x ùú éê 0ùú ê ê -2 1 -3ú ê y ú = ê 4ú ê úê ú ê ú êë-4 5 1úû êë z úû êë-3úû 2 x  3 y  2z  2  44.  5 x  2 y  3z  5 7 x  2 y  8z  11 

Solution é 2 -3 2ùú éê x ùú éê-2ùú ê ê-5 2 -3ú ê y ú = ê-5ú ê úê ú ê ú êë 7 -2 8úû êë z úû êë 11úû Write each matrix equation as a system of linear equations.

 3 2  x   6 45.       2 5  y   1

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1517


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution ìïï 3 x - 2 y = 6 í ïïî 2 x + 5 y = -1

 5 4  x   9 46.      0 7   y   3 Solution ìïï5 x + 4 y = -9 í ïïî -7 y = 3 4 3 1  x   1       47. A   2 5 2  y    4   1 3 6   z   10       

Solution ìï 4 x + 3 y + z = - 1 ïï í-2 x - 5 y + 2z = 4 ïï ïïî -x - 3 y - 6z = -10  1 4 2   x   2        48. A   9 5 3  y    12  2 1 0   z   3       

Solution ìï -x + 4 y + 2z = 2 ïï í-9 x + 5 y - 3z = -12 ïï = 3 ïïî 2 x - y Solve each system of linear equations given the multiplicative inverse of the coefficient matrix.

 x  2 y  7 49.  3 x  4 y  17 Multiplicative inverse of the coefficient matrix:

 2 1    3  1   2 2  Solution é 1 -2ù é x ù é 7ù ê úê ú = ê ú ê-3 4úû êë y úû êë-17úû ë -1 é x ù é 1 -2ù é 7ù ê ú=ê ú ê ú ê y ú ê-3 4úû êë-17úû ë û ë

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1518


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é ù é x ù ê -2 -1ú é 7ù é 3ù ê ú=ê 3 ú=ê ú úê 1 ê y ú êê ú ê ú ë û ê 2 - 2 úú ë-17û ë-2û ë û

 x  4 y  21 50.  2 x  5 y  23 Multiplicative inverse of the coefficient matrix:

5 4   13 13   2  1  13 13  Solution é1 4ùú éê x ùú éê 21ùú ê = ê2 -5ú ê y ú ê-23ú ë ûë û ë û -1 é xù é 1 4ùú éê 21ùú ê ú=ê ê y ú ê2 -5ú ê-23ú ë û ë û ë û é5 4 ùú é x ù êê é ù é ù 13 úú ê 21ú = ê 1ú ê ú = ê 13 ê yú ê 2 ê 1 -23ûú ëê5ûú ë û ê - úú ë 13 û ë 13  x  2 z  3  51.  x  2 y  3z  5  2 x  5 y  2 

Multiplicative inverse of the coefficient matrix:  10  5 3  4  2  3   3 5  3

4   3 1   3 2  3 

Solution é 1 0 2ù é x ù é-3ù ê úê ú ê ú ê 1 2 3ú ê y ú = ê-5ú ê úê ú ê ú êë-2 5 0úû êë z úû êë-2úû

é x ù é 1 0 2ù ê ú ê ú ê y ú = ê 1 2 3ú ê ú ê ú êë z úû êë-2 5 0úû

-1

é-3ù ê ú ê-5ú ê ú êë-2úû

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1519


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é ù ê-5 10 - 4 ú ê 3 3 úú é-3ù é ù ê x ú êê 4 1 ê ú ê yú = 2 - úú ê-5ú ê ú êê 3 3 ú êê úú 5 2 ú ë-2û ëê z ûú ê ê 3 ú êë 3 3 úû é x ù é 1ù ê ú ê ú ê y ú = ê 0ú ê ú ê ú êë z úû êë-2úû  x  2 y  z  3  52.   y  z  1   x  2z  1 

Multiplicative inverse of the coefficient matrix:  2 4 3     5 5 5    1  3  1   5 5 5  1 2 1     5 5 5 

Solution é 1 -2 1ùú éê x ùú éê-3ùú ê ê 0 -1 -1ú ê y ú = ê -1ú ê úê ú ê ú êë-1 0 2úû êë z úû êë 1úû -1

é x ù é 1 -2 1ùú éê-3ùú ê ú ê ê y ú = ê 0 -1 -1ú ê -1ú ê ú ê ú ê ú êë z úû êë-1 0 2úû êë 1úû é 2 4 3ù ê - ú ê 5 5 úú é-3ù éxù ê 5 ê ú ê 1 3 1 ê ú ê yú = - úú ê -1ú ê ú êê 5 5 5 ú êê úú êë z úû ê 1 2 1 ú ë 1û ê ú êë 5 5 5 úû é x ù é-1ù ê ú ê ú ê y ú = ê 1ú ê ú ê ú êë z úû êë 0úû

Solve each system of linear equations using the multiplicative inverse of the coefficient matrix. Write the solution as an ordered pair or ordered triple.

3x  4 y  1 53.  2x  3 y  5

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1520


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 3 -4 ù é x ù é 1 ù ê úê ú = ê ú ê-2 3úû êë y úû êë5úû ë -1 é x ù é 3 -4 ù é 1 ù ê ú=ê ú ê ú ê ú ê 3ûú ëê5ûú ë y û ë-2 é x ù é 3 4ù é 1 ù ê ú=ê úê ú ê y ú ê2 3ú ê5ú ë û ë ûë û é x ù é23ù ê ú=ê ú ê y ú ê 17 ú ë û ë û

3x  4 y  1 54.  2x  3 y  3 Solution é 3 -4ù é x ù é-1ù ê úê ú = ê ú ê-2 3úû êë y úû êë 3úû ë -1 é x ù é 3 -4ù é-1ù ê ú=ê ú ê ú ê y ú ê-2 3úû êë 3úû ë û ë é x ù é3 4ù é-1ù ê ú=ê úê ú ê y ú ê2 3ú ê 3ú ë û ë ûë û é x ù é9ù ê ú=ê ú ê y ú ê7ú ë û ë û

3x  4 y  0 55.  2x  3 y  0 Solution é 3 -4ù é x ù é0ù ê úê ú = ê ú ê-2 3úû êë y úû êë0úû ë -1 é x ù é 3 -4ù é0ù ê ú=ê ú ê ú ê ú ê 3ûú ëê0ûú ë y û ë-2 é x ù é3 4ù é0ù ê ú=ê úê ú ê y ú ê2 3ú ê0ú ë û ë ûë û é x ù é0ù ê ú=ê ú ê y ú ê0ú ë û ë û

3 x  4 y  3 56.  2 x  3 y  2 Solution é 3 -4ù é x ù é-3ù ê úê ú = ê ú ê-2 3úû êë y úû êë-2úû ë

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1521


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

-1

é x ù é 3 -4ù é-3ù ê ú=ê ú ê ú ê y ú ê-2 3úû êë-2úû ë û ë é x ù é3 4ù é-3ù ê ú=ê úê ú ê y ú ê2 3ú ê-2ú ë û ë ûë û é x ù é-17ù ê ú=ê ú ê y ú ê-12ú ë û ë û

5 x  3 y  13 57.  7 x  5 y  9 Solution é 5 3ù é x ù é 13ù ê úê ú = ê ú ê-7 5ú ê y ú ê-9ú ë ûë û ë û

é x ù é 5 3ù ê ú=ê ú ê y ú ê-7 5ú ë û ë û é x ù é2ù ê ú=ê ú ê y ú ê 1ú ë û ë û

-1

é 13ù ê ú ê-9ú ë û

8x  3 y  7 58.  3x  2 y  0 Solution é 8 - 3ù é x ù é 7 ù ê úê ú = ê ú ê-3 2úû êë y úû êë0úû ë -1

é x ù é 8 -3ù é7ù ê ú=ê ú ê ú ê y ú ê-3 2úû êë0úû ë û ë é x ù é2ù ê ú=ê ú ê y ú ê3ú ë û ë û 2 x  y  z  2  59. 2 x  2 y  z  4   x  y  z  1 

Solution é 2 1 -1ùú éê x ùú éê 2ùú ê ê 2 2 -1ú ê y ú = ê 4ú ê úê ú ê ú êë-1 -1 1úû êë z úû êë-1úû -1

éxù é 2 1 -1ùú éê 2ùú ê ú ê ê y ú = ê 2 2 -1ú ê 4ú ê ú ê ú ê ú êë z úû êë-1 -1 1úû êë-1úû é x ù é 1 0 1ù é 2ù é 1 ù ê ú ê úê ú ê ú ê y ú = ê-1 1 0ú ê 4ú = ê2ú ê ú ê úê ú ê ú êë z úû êë 0 1 2úû êë-1úû êë2úû

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1522


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2 x  y  z  3  60. 2 x  2 y  z  1  x  y  z  4 

Solution é 2 1 -1ùú éê x ùú éê 3ùú ê ê 2 2 -1ú ê y ú = ê-1ú ê úê ú ê ú êë-1 -1 1úû êë z úû êë 4úû éxù é 2 1 -1ùú ê ú ê ê y ú = ê 2 2 -1ú ê ú ê ú êë z úû êë-1 -1 1úû

-1

é 3ù ê ú ê-1ú ê ú êë 4úû

é x ù é 1 0 1ù é 3ù é 7ù ê ú ê úê ú ê ú ê y ú = ê-1 1 0ú ê-1ú = ê-4ú ê ú ê úê ú ê ú êë z úû êë 0 1 2úû êë 4úû êë 7úû 2 x  y  3z  5  61. 2 x  3 y  1  x  z  2 

Solution é-2 1 -3ù é x ù é 5ù ê úê ú ê ú ê 2 3 0ú ê y ú = ê 1ú ê úê ú ê ú êë 1 0 1úû êë z úû êë-2úû -1

é x ù é-2 1 -3ù é 5ù ê ú ê ú ê ú ê yú = ê 2 3 0ú ê 1ú ê ú ê ú ê ú êë z úû êë 1 0 1úû êë-2úû é x ù é 3 -1 9ùú éê 5ùú éê-4ùú ê ú ê ê y ú = ê-2 1 -6ú ê 1ú = ê 3ú ê ú ê úê ú ê ú êë z úû êë-3 1 -8úû êë-2úû êë 2úû 5 x  2 y  3z  12  62. 2 x  5z  7 3 x  z  4 

Solution é5 2 3ù é x ù é 12ù ê úê ú ê ú ê2 0 5ú ê y ú = ê 7 ú ê úê ú ê ú êë3 0 1 úû êë z úû êë 4 úû

é x ù é5 2 3ù ê ú ê ú ê y ú = ê2 0 5ú ê ú ê ú êë z úû êë3 0 1 úû

-1

é 12ù é 1 ù ê ú ê ú ê 7 ú = ê2ú ê ú ê ú êë 4 úû êë 1 úû

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1523


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3 x  2 y  z  0  63. 5 x  2 y  5 3 x  y  z  6 

Solution é3 2 -1ùú éê x ùú éê0ùú ê ê5 -2 0ú ê y ú = ê5ú ê úê ú ê ú êë3 1 1úû êë z úû êë6úû -1

é x ù é3 2 -1ùú éê0ùú éê 1 ùú ê ú ê ê y ú = ê5 -2 0ú ê5ú = ê0ú ê ú ê ú ê ú ê ú êë z úû êë3 1 1úû êë6úû êë3úû 2 x  y  3z  2  64. 2 x  3 y  3 x  z  5 

Solution é-2 1 -3ù é x ù é 2ù ê úê ú ê ú ê 2 3 0ú ê y ú = ê-3ú ê úê ú ê ú êë 1 0 1úû êë z úû êë 5úû -1

é x ù é-2 1 -3ù é 2ù ê ú ê ú ê ú ê yú = ê 2 3 0ú ê-3ú ê ú ê ú ê ú êë z úû êë 1 0 1úû êë 5úû é x ù é 3 -1 9ùú éê 2ùú éê 54ùú ê ú ê ê y ú = ê-2 1 -6ú ê-3ú = ê -37 ú ê ú ê úê ú ê ú êë z úû êë-3 1 -8úû êë 5úû êë-49úû

Fix It In exercises 65 and 66, identify the step where the first error is made and fix it.

2 5  65. Find the multiplicative inverse of A    using row operations.  1 2 Solution Step 4 was incorrect.

é 1 0 -2 5ùú Step 4: (-2) R2 + R1  êê 0 1 1 -2ú ë û é-2 5ùú Step 5: A-1 = ê ê 1 -2ú ë û

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1524


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3x  2 y  3 66. Write the system of linear equations  in the form AX = B and solve the system  x  3 y  8 3  of linear equations given A   7 1  7

2  7  . Write the solution as an ordered pair. 3 7 

1

Solution Step 3 was incorrect. é 1ù Step 3: X = ê ú ê3ú ë û Step 4: x = 1 and y = 3 Step 5: (1, 3)

Applications 67. Manufacturing and testing The numbers of hours required to manufacture and test each of two models of heart monitor are given in the first table, and the numbers of hours available each week for manufacturing and testing are given in the second table.

Hours Required per Unit Model A

Model B

Manufacturing

23

27

Testing

21

22

Hours Available Manufacturing

127

Testing

108

How many of each model can be manufactured each week?

Solution é23 27ù é x ù é 127ù ê úê ú ê ú ê 21 22ú ê y ú = ê 108ú êë úû êë úû êë úû -1 é x ù é23 27ù é 127ù ê ú=ê ú ê ú ê y ú ê 21 22ú ê 108ú ëê ûú ëê ûú ëê ûú é x ù é 2ù ê ú = ê ú  2 of model A and 3 of model B can be made ê y ú ê 3ú ë û ë û

68. Making clothes A clothing manufacturer makes coats, shirts, and slacks. The times required for cutting, sewing, and packaging each item are shown in the table. How many of each should be made to use all available labor hours? © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1525


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Coats

Shirts

Slacks

Cutting

20 min

15 min

10 min

Sewing

60 min

30 min

24 min

Packaging

5 min

12 min

6 min

Time Available

Solution é 1 1 1 ù é x ù é 115ù ê 3 41 62 ú ê ú ê ú ê 1 ú ê y ú = ê280ú 2 5 ê 1 1 1úê ú ê ú ê 12 5 10 ú ê z ú ê 65ú û ë ûë û ë éxù é 1 1 ê ú ê 3 41 ê yú = ê 1 2 ê ú ê1 1 êë z úû ê 12 5 ë

1 6 2 5 1 10

ù ú ú ú ú û

-1

Cutting

115 hr

Sewing

280 hr

Packaging

65 hr

é 115ù é 120 ù ê ú ê ú ê280ú = ê200ú  120 coats, 200 shrits, and ê ú ê ú êë 65úû êë 150 úû 150 slacks should be made.

69. Cryptography The letters of a message, called plain text, are assigned values 1–26 (for a-z) and are written in groups of 2 as 2 × 1 matrices. To write the message in cipher text, each 2 × 1 matrix B is multiplied by a matrix A, where

 1 1 A  2 3  Find the plain text if the cipher text of one message is

 17  AB    43 Solution é 17 ù AB = ê ú ê43ú ë û A-1 AB = IB = B -1

é 1 1 ù é 17 ù ú ê ú B = A AB = ê ê2 3ú ê43ú ë û ë û é8ù = ê ú  ''HI'' ê9ú ë û -1

70. Cryptography The letters of a message, called plain text, are assigned values 1–26 (for a–z) and are written in groups of 3 as 3 × 1 matrices. To write the message in cipher text, each 3 × 1 matrix Y is multiplied by matrix A, where

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1526


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1 1 0   A  2 3 3   1 1 1  

Find the plain text if the cipher text of one message is  30    AY   122  49  

Solution é 30ù ê ú AY = ê 122ú ê ú êë 49úû A-1 AY = IY = Y -1

é 1 1 0ù é 30ù ê ú ê ú -1 Y = A AY = ê2 3 3ú ê 122ú ê ú ê ú êë 1 1 1 úû êë 49úû é25ù ê ú = ê 5 ú  ''Yes'' ê ú êë 19 úû

Discovery and Writing 71. Explain what is meant by “multiplicative inverse of a matrix.”

Solution Answers may vary. 72. Describe a strategy to use to determine the multiplicative inverse of an invertible matrix.

Solution Answers may vary. 73. Once you have applied the steps to find the multiplicative inverse of a matrix, how can you check your answer?

Solution Answers may vary. 74. Describe a strategy you would use to solve the matrix equation AX = B.

Solution Answers may vary.

 1  1 Let A =  .  1 1 75. Show that A2 = 0.

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1527


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é-1 -1ù é-1 -1ù é0 0ù úê ú=ê ú A2 = ê ê 1 úê 1 ú ê0 0ú 1 1 ë ûë û ë û 76. Show that the multiplicative inverse of I – A is I + A.

Solution é 2 1ù é ù ú , I + A = ê0 -1ú ; I-A= ê ê-1 0ú ê 1 2ú ë û ë û é 2 1ù é0 -1ù é 1 0ù (I - A)(I + A) = êê-1 0úú êê 1 2úú = êê0 1úú ë ûë û ë û Since the product is the identity, they are inverse matrices.  3 0 0 x     Let A   2 1 2  and X   y  . Solve each equation. Each solution is called an eigenvector of  3 6  z 3     the matrix A.

77. (A – 2I)X = 0

Solution é 1 0 0ùú éê x ùú éê0ùú ê ê-2 -3 -2ú ê y ú = ê0ú ê úê ú ê ú êë 3 6 1úû êë z úû êë0úû -1

é xù é 1 0 0ùú éê0ùú éê0ùú ê ú ê ê y ú = ê-2 -3 -2ú ê0ú = ê0ú ê ú ê ú ê ú ê ú êë z úû êë 3 6 1úû êë0úû êë0úû 78. (A – 3I)X = 0

Solution Cannot be solved using inverse matrix. é0ù ê ú Solution is ê0ú . ê ú êë0úû 79. Suppose that A, B, and C are n × n matrices and A is invertible. If AB = AC, prove that B = C.

Solution AB = AC A-1 AB = A-1 AC IB = IC B=C

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1528


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

a b  80. Prove that   has an multiplicative inverse if and only if ad – bc ≠ 0. (Hint: Try to find c d  the multiplicative inverse and see what happens.) Solution b 1 é1 é 1 b 1 0ù éa b 1 0ù 0ùú a a a a ê ê ú ê ú êc d 0 1 ú  êc d 0 1 ú  ê0 ad -bc - c 1ú  êë a a ë û ûú ëê ûú 1  +  R R cR R R 1 1 2 2 a 1

é ê1 ê0 ëê

d d é é - ad -b bc ùú - ad -b bc úù 0 úù ad -bc ad -bc ê1 0 ê   Inverse: a ú c a ú ê0 1 - c ê 1 - ad -c bc ad -a bc úú ad -bc ad -bc û ad -bc û ë- ad -bc ë û a - ab R2 + R1  R1 R  R2 ad -bc 2 b a

1 a

The inverse will be defined if and only if the denominator, ad - bc, is not equal to 0. 81. Suppose that B is any matrix for which B2 = 0. Show that I – B is invertible by showing that the multiplicative inverse of I – B is I + B.

Solution

(I - B)(I + B) = I 2 + IB - BI - B2 = I + B - B - B2 = I - B2 = I - 0 = I  Thus, I - B and I + B are inverses.

82. Suppose that C is any matrix for which C3 = 0. Show that I – C is invertible by showing that the inverse of I – C is I + C + C2.

Solution

(I - C )(I + C + C 2 ) = I 2 + IC + IC 2 - CI - C 2 - C 3 = I2 + C + C2 - C - C2 - C3 = I - C3 = I - 0 = I  Thus, I - C and I + C + C 2 are inverses.

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 83. All matrices have multiplicative inverses.

Solution False. Some square matrices have inverses.

 1 2  1 2 1 84. If   , then A   . 3 4   3 4

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1529


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

 2 1  False. A1   3 1  2  2  1  2 4  1 2 85. If A    , then A   6 8 1    6

1  4. 1 8 

Solution

 1 21  False. A1   3 1  4  4  86. (A–1)–1 = A

Solution True. 87. The multiplicative inverse of a matrix is unique.

Solution True. 88. (5A)–1 = 5A–1

Solution False.

5A  A 1

1 5

1

89. (AB)–1 = B–1A–1

Solution True. 90. If C is invertible, then CA = CB implies that A = B.

Solution True.

EXERCISES 6.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Simplify: 5(–6) – 3(–7)

Solution –30 + 21 = –9

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1530


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2. Simplify:

4  3  2  5

4  2   1 3 

Solution -12 - 10 -22 = = -2 8+3 11 3. If i = 3 and j = 2, is i + j even or odd?

Solution 3+2=5

odd

4. If i = 1 and j = 3, is i + j even or odd?

Solution 1+3=4

even

5. Simplify: –[3(2) – 4(–1)] + 0[2(–1) – 2(2)] + 3[–2(2) – 3(–1)]

Solution –10 + 0 + – 3 = –13 6. Find an equation of the line in standard form that passes through (1, –1) and (2, –3).

Solution m = -2 y + 1 = -2 ( x - 1) y + 1 = -2 x + 2 2x + y = 1

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The determinant of a square matrix A is written as __________ or __________.

Solution

A , det A 8.

a b     __________ c d  Solution ad  bc

9. If every entry in one row or one column of square matrix A is zero, then A  __________.

Solution 0

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1531


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

10. If A is square matrix and matrix B is obtained from matrix A by adding one row to another, then B  __________.

Solution

A 11. If two columns of the square matrix A are identical, then A  __________.

Solution 0 12. In Cramer’s Rule, the denominator is the determinant of the __________.

Solution coefficient matrix Practice Evaluate each determinant. 13.

2 1 2 3 Solution é 2 1ù ê ú = 2 3 - 1 -2 ê-2 3ú ( )( ) ( )( ) ë û = 6 - (-2) = 8

14.

3 6 2 5 Solution é-3 -6ù ê ú = -3 -5 - -6 2 ê 2 -5ú ( )( ) ( )( ) ë û

= 15 - (-12) = 27 15.

2 3 3 5 Solution é 2 -3ù ê ú = (2)(5) - (-3)(-3) ê-3 5úû ë = 10 - 9 = 1

16.

5 8 6 2

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1532


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 5 8ùú ê = 5 -2 - 8 - 6 ê-6 -2ú ( )( ) ( )( ) ë û

= -10 - (-48) = 38

17.

4 2 5 3 Solution é4 2ùú ê = 4 (-3) - 2 (5) ê 5 -3ú ë û = -12 - 10 = -22

18.

6 1 7 2 Solution é-6 -1ù ê ú = -6 (2) - (-1)(7) ê 7 2ú ë û = -12 + 7 = -5

19.

9 6 3 2 Solution é-9 -6ù ê ú = -9 (2) - (-6)(3) ê 3 2úû ë = -18 + 18 = 0

20.

2 8 5 20 Solution é2p 8p ù ê ú = 2p (20) - 8p (5) ê 5 20ú ë û = 40p - 40p = 0

2 3 5 21. 5 1 1  2 2 Solution é2 3 ùú ê æ öæ ö æ öæ ö ê5 5 úú = çç 2 ÷÷ çç- 1 ÷÷ - çç 3 ÷÷ çç 1 ÷÷ ê ÷ ÷÷ ç 2 ÷÷ ç ç ÷ 2 ÷ø÷ èç 5 øè ø ê 1 - 1 ú è 5 øè ê ú 2û ë2

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1533


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

1 3 1 =- =5 10 2  22.

1 2 3 3 1 1  3 9

Solution é 1 2 ùú êæ ö æ ö ê 3 3 úú = - 1 çç- 1 ÷÷ - 2 çç 1 ÷÷ ê ÷ 3 çè 9 ÷ø 3 çè 3 ÷÷ø ê 1 - 1ú ê ú 9û ë 3 1 2 5 = - =27 9 27  1 2 3   In Exercises 23–30, A   4 5 6  . Find each minor or cofactor.  7 8 9  

23. M21

Solution é-2 3ù ú = -2 9 - 3 8 M21 = ê ê 8 9ú ( )( ) ( )( ) ë û = -18 - 24 = -42 24. M13

Solution é 4 5ù ú = 4 8 - 5 -7 M13 = ê ê-7 8ú ( )( ) ( )( ) ë û = 32 - (-35) = 67 25. M33

Solution é 1 -2ù ú = (1)(5) - (-2)(4) M33 = ê ê4 5úû ë = 5 + 8 = 13 26. M32

Solution é1 3ùú M32 = ê = 1 -6 - 3 4 ê4 -6ú ( )( ) ( )( ) ë û = -6 - 12 = -18

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1534


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

27. C21

Solution é-2 3ù ú = - é(-2)(9) - (3)(8)ù C21 = - ê ëê ûú ê 8 9ú ë û = - éëê-18 - 24ùûú = 42 28. C13

Solution é 4 5ù ú = 4 8 - 5 -7 C13 = ê ê-7 8ú ( )( ) ( )( ) ë û = 32 - (-35) = 67 29. C33

Solution é 1 -2ù ú = (1)(5) - (-2)(4) C33 = ê ê4 5úû ë = 5 + 8 = 13 30. C32

Solution é1 3ùú C32 = - ê = - éê(1)(-6) - (3)(4)ùú ë û ê4 -6ú ë û = - éêë-6 - 12ùúû = 18 1 0 1 2 3 1 1 0 In Exercises 31–38, A  . Find each minor or cofactor. 2 1 0 3 1 2 1 0

31. M31

Solution é 0 -1 2ù ê ú M31 = ê 1 -1 0ú = 0 (0) - (-1)(0) + 2 (-3) = -6 ê ú êë-2 -1 0úû 32. M24

Solution é1 0 -1ùú ê M24 = ê2 -1 0ú = 1(1) - 0 (-2) + (-1)(-3) = 4 ê ú êë 1 -2 -1úû

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1535


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

33. M13

Solution é3 1 0ùú ê ê M13 = 2 -1 3ú = 3 (6) - (1)(-3) + (0)(-3) = 21 ê ú êë 1 -2 0úû 34. M41

Solution é 0 -1 2ù ê ú M41 = ê 1 -1 0ú = 0 (-3) - (-1)(3) + 2 (-1) = 1 ê ú êë-1 0 3úû 35. C32

Solution é 1 -1 2ù ê ú C32 = - ê3 -1 0ú = - éê 1(0) - (-1)(0) + 2 (-2)ùú = 4 ë û ê ú êë 1 -1 0úû 36. C43

Solution é 1 0 2ù ê ú C43 = - ê3 1 0ú = - éê 1(3) - 0 (9) + 2 (-5)ùú = 7 ë û ê ú êë2 -1 3úû 37. C22

Solution é 1 -1 2ù ê ú C22 = ê2 0 3ú = 1(3) - (-1)(-3) + 2 (-2) = -4 ê ú êë 1 -1 0úû 38. C11

Solution é 1 -1 0ù ê ú C11 = ê -1 0 3ú = 1(3) - (-1)(6) + 0 (1) = 9 ê ú êë-2 -1 0úû Evaluate each determinant by expanding by cofactors. 2 3 5 1 3 39. 2 1 3 2

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1536


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 2 -3 5 1 3 -2 3 -2 1 -2 1 3 =2 - (-3) +5 3 -2 1 -2 1 3 1 3 -2 = 2 (-11) + 3 (1) + 5 (-7) = -22 + 3 - 35 = -54

1

3

1

40. 2 5 3 3 2 2

Solution 1 3 1 -2 -2 5 3 3 5 -2 -3 +1 5 3 =1 -2 - 2 3 -2 3 -2 3 - 2 -2

= 1(-4) - 3 (-5) + 1(-11) = -4 + 15 - 11 = 0 1 1 2 41. 2 1 3 1 1 1

Solution 1 -1 2 1 3 2 3 2 1 2 1 3 =1 - (-1) +2 1 -1 1 -1 1 1 1 1 -1 = 1(-4) + 1(-5) + 2 (1) = -4 - 5 + 2 = -7

1 2 42.

3 1 1 1

2 1

1

Solution 1 3 1 1 -1 2 -1 2 1 -3 +1 2 1 -1 = 1 -1 1 2 1 2 -1 2 -1 1

= 1(0) - 3 (4) + 1(-4) = 0 - 12 - 4 = -16 3 1 2 1 43. 3 2 1 3 0

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1537


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 3 1 -2 2 1 -3 1 -3 2 -3 2 1 =3 -1 + (-2) 3 0 1 0 1 3 1 3 0 = 3 (-3) - 1(-1) - 2 (-11) = -9 + 1 + 22 = 14

2 44. 1

1 1 3 5

2 5

3

Solution 2 1 -1 3 5 1 5 1 3 -1 + (-1) 1 3 5 =2 -5 3 2 3 2 -5 2 -5 3

= 2 (34) - 1(-7) - 1(-11) = 68 + 7 + 11 = 86 0 1 3 45. 3 5 2 2 5 3

Solution 0 1 -3 5 2 -3 2 -3 5 5 2 =0 -3 -1 + (-3) -5 3 2 3 2 -5 2 -5 3 = 0 - 1(-13) - 3 (5) = 0 + 13 - 15 = -2

1 7 2 46. 2 0 3 1

7

1

Solution 1 - 7 -2 -2 3 -2 0 0 3 -2 - (-7) + (-2) 0 3 =1 -1 1 -1 7 7 1 -1 7 1

= 1(-21) + 7 (1) - 2 (-14) = -21 + 7 + 28 = 14

0 0 1 0 2 1 0 1 47. 1 0 1 2 2 0 1 2

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1538


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 0 0 1 0 -2 1 1 -2 1 0 1 = 0 (* * *) - 0 (* * *) + 1 1 0 2 - 0 (* * *) 1 0 1 2 2 0 2 2 0 1 2 æ 0 2 1 2 1 0 ÷ö ÷ = -2 (0) - 1(-2) + 1(0) = 2 = 1ççç-2 -1 +1 0 2 2 2 2 0 ÷÷ø çè

1 0 2 1 0 1 0 1 48. 0 3 1 2 0 1 0 1 Solution Expand along 1st column... 1 0 2 1 1 0 1 0 1 0 1  1 3 1 2  0       0       0      0 3 1 2 1 0 1 0 1 0 1  1 2 3 2 3 1   1 1 0 1   1  1  0  1  1  2  0 1 1 1 1 0  

1 1 2 3 0 1 2 0 49. 1 3 0 1 2 2 1 1

Solution 1 -1 2 3 1 -1 2 3 0 1 -2 0 0 1 -2 0 = 1 -3 0 1 0 - 2 -2 -2 2 -2 -1 1 0 0 -5 -5

- 1R1 + R3 -2R1 + R4

1 -2 0 = 1 -2 -2 -2 (expanded along first column of previous matrix) 0 -5 -5 æ -2 - 2 - 2 -2 -2 -2 ö÷ ÷ = 1ççç1 - (-2) +0 0 -5 0 -5 ø÷÷ èç -5 -5 = 1(0) + 2 (10) + 0 (10) = 20

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1539


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

1 3 2 5 2 1 0 1 50. 1 3 2 5 2 1 0 1 Solution 1 3 2 5 1 2 1 0 1 0  1 3 2 5 0 2 1 0 1 0

3 7 6 5

2 5 4 11 2R1  R2 4 10 R1  R3 4 9 2R1  R4

7 4 11  1 6 4 10 5 4 9

expanded along first column of previous matrix 

 4 11 6 10 6 4   1 7  4  11   4 10   5 9 5 4    7  4   4  4   11  4   0

Determine whether each statement is true. Do not evaluate the determinants. 1 3 4 2 1 3 51. 2 1 3   1 3 4 1 3 2 1 3 2

Solution R1 and R2 have been switched. This multiplies the determinant by –1. TRUE 4 6 8 2 3 4 52. 10 5 15  10 5 15 20 5 10 20 5 10

Solution R1 has been multiplied by 21 . This multiplies the determinant by 21 . FALSE 2 3  4 2 3 4 53. 5 1 2   5 1 2

1

2

3

1 2

3

Solution R1 and R2 have both been multiplied by –1. This multiplies the determinant by –1 twice. FALSE 1 2 3 5 7 9 54. 4 5 6  4 5 6 7 8 9 7 8 9

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1540


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution R1 has been added to R2. This leaves the determinant unchanged. TRUE a If d g

b c e f  3, find the value of each determinant. h i

55.

d e f a b c  g h i

Solution R1 and R2 have been switched. This multiplies the determinant by –1. However R3 has been multiplied by –1, which also multiplies the determinant by –1. Thus, the determinant remains equal to 3. 5a 5b 5c 56. d e f 3g 3h 3i

Solution R1 has been multiplied by 5, which multiplies the determinant by 5. R3 has been multiplied by 3, which multiplies the determinant by 3. R2 has been multiplied by –1, which multiplies

  

the determinant by –1. Thus, the determinant  5 3 1  3  45.

57.

ag bh ci d e f g h i

Solution R1 has been added to R3. This leaves the determinant equal to 3. g h i 58. a b c d e f

Solution R1 and R3 were switched, and then R2 and the new R3 were switched. Both switches multiply the determinant by –1. Thus, the determinant remains equal to 3. Evaluate each determinant. Use row and/or column operations to help and save steps. 5 10 5 59. 1 2 1 2 1 0

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1541


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

1 2 -1 1 2 -1 - 5 10 5 = -1 0 0 10 -5R1 + R2 2 -1 0 0 -5 2 -2R1 + R3 0 10 (expanded along first column of previous matrix) -5 2 = -1 éê0 (2) - 10 (-5)ùú = -1(50) = -50 ë û = -1

2 1 8 12 1  1 1

1 60. 4

Solution 1 2 -1 1 2 -1 4 8 12 = 0 0 16 -4R1 + R2 1 - 1 -1 0 -3 0 -1R1 + R2 0 16 (expanded along first column of previous matrix) -3 0 = 1 éê0 (0) - 16 (-3)ùú = 1(48) = 48 ë û =1

2 2 2 61. 0 1 1 6 12 6

Solution 1

2 2 -2 1 1 12 0 1 -1 = 2 0 1 -1 6 -12 6 6 -12 6

R1

1 1 -1 1 -1 =20 0 -18 12 -6R1 + R3 é 1 -1 ùú = 2 êê 1 (expanded along first column of previous matrix) 18 12 úû ë = 2 éê 1(12) - (-1)(-18)ùú = 2 (-6) = -12 ë û

62.

10 10 20 10 20 40 0 10 10

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1542


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution -10 10 20 -1 1 2 10 20 40 = 10 ⋅ 10 1 2 4 0 10 -10 0 10 -10

1 1

10

R1

10

R2

1 2 4 R1  R2 = -100 0 3 6 R1 + R2 0 10 -10

é 3 6 ùú = -100 êê 1 (expanded on first column of previous matrix) 10 -10 ú ë û = -100 éê3 (-10) - 6 (10)ùú = -100 (-90) = 9000 ë û 3 3 3 3 1 0 1 0 63. 2 0 2 1 2 1 1 1

Solution 3 -3 3 3 1 0 1 0 R1  R2 1 0 1 0 3 -3 3 3 = -1 2 0 2 1 2 0 2 1 2 -1 1 2 2 -1 1 1 1 0 1 0 1 0 1 0 1 -1 1 1 0 -1 0 1 -1R1 + R2 = -3 = -3 2 0 2 1 0 0 0 1 -2R1 + R3 2 -1 1 1 0 -1 -1 1 é -1 0 1 ù ê ú = -3 ê 1 0 0 1 ú (expanded along first column of previous matrix) ê ú ê -1 -1 1 ú ë û é 0 1 0 1 0 0 ùú = -3 êê-1 -0 +1 -1 1 -1 -1 úû ë -1 1 = -3 éê-1(1) - 0 (1) + 1(0)ùú = -3 (-1) = 3 ë û 10 20 10 30 2 1 3 1 64. 1 0 1 2 2 1 1 3

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1543


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 10 20 10 30 10 20 10 30 1 3 1 0 0 4 4 2  1 2 0 2 2 1 1 0 2 1 1 3 0 5 3 3

4 4 1  10 2 2 5 3 3 0

R2  R4 1 10 1 5

R1  R3

 R1  R4

expanded along first column of previous matrix 

 2 1 2 1 2 2   10  0  4 4   3 3   5 3 5 3    10 0  4  1  4  4    10  12   120

Use Cramer’s Rule to find the solution of each system, if possible. Write the solution as an ordered pair, ordered triple or ordered quadruple of real numbers.

3x  2 y  7 65.  2x  3 y  4 Solution 7 2 -4 -3

-13 = =1 x= -13 3 2 2 -3

3 7 2 -4

y=

3 2 2 -3

=

-26 =2 -13

 x  5 y  6 66.  3 x  2 y  1 Solution x=

-6 - 5 -1 2

1 -5 3 2

=

-17 = -1 17

y=

1 -6 3 -1 1 -5 3 2

=

17 =1 17

 x  y  3 67.  3x  7 y  9 Solution x=

3 -1 9 -7 1 -1 3 -7

=

-12 =3 -4

y=

1 3 3 9 1 -1 3 -7

=

0 =0 -4

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1544


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2 x  y  6 68.   x  y  0 Solution x=

-6 -1 0 1

2 -1 1 1

-6 = -2 3

=

y=

2 -6 1 0 2 -1 1 1

=

6 =2 3

4 x  5 y  13 69.  3 x  2 y  4 Solution x=

13 -5 4 2

=

4 -5 3 2

46 =2 23

y=

4 13 3 4 4 -5 3 2

=

-23 = -1 23

2x  4 y  3 70.   x  y  0 Solution x=

3 -4 0 1 2 -4 1 1

=

3 1 = 6 2

y=

2 3 1 0 2 -4 1 1

=

-3 -1 = 6 2

5x  2 y  3 71.  10 x  3 y  13 Solution 3 2 13 -3

-35 = = -1 x= 35 -5 2 -10 -3

y=

-5 3 -10 13 -5 2 -10 -3

=

-35 = -1 35

3 x  4 y  9 72.  2 x  5 y  13 Solution - 9 -4 13 5

7 = = -1 x= -7 -3 -4 2 5

y=

-3 -9 2 13 -3 -4 2 5

=

-21 =3 -7

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1545


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

x  2 y  z  2  73.  x  y  z  2  x  y  3z  4  Solution

x=

2 2 1 2 -1 1 4 1 3 1 2 1 1 -1 1 1 1 3

=

-6 =1 -6

y=

1 2 1 1 2 1 1 4 3 1 2 1 1 -1 1 1 1 3

=

0 =0 -6

z=

1 2 2 1 -1 2 1 1 4 1 2 1 1 -1 1 1 1 3

=

-6 =1 -6

2 x  y  z  5  74. 3 x  3 y  2z  10 x  3 y  z  0 

Solution

x=

5 -1 1 10 -3 2 0 3 1 2 -1 1 3 -3 2 1 3 1

=

-5 =1 -5

y=

2 5 1 3 10 2 1 0 1 2 -1 1 3 -3 2 1 3 1

=

5 = -1 -5

z=

2 -1 5 3 -3 10 1 3 0 2 -1 1 3 -3 2 1 3 1

=

-10 =2 -5

=

-4 = -1 4

x  y  z  2  75.  x  y  z  2   x  y  z  4 

Solution

x=

2 -1 - 1 2 1 1 1 - 4 -1 1 - 1 -1 1 1 1 -1 - 1 1

=

8 =2 4

y=

1 2 -1 1 2 1 1 -1 - 4 1 -1 -1 1 1 1 -1 -1 1

=

4 =1 4

z=

1 -1 2 1 1 2 -1 -1 -4 1 -1 -1 1 1 1 -1 -1 1

x  2 y  z  6  76.  x  3 y  z  7 2 x  y  z  0 

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1546


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

x=

6 -2 1 -7 3 -1 0 1 -1 1 -2 1 1 3 -1 2 1 -1

=

-5 =1 -5

y=

1 6 1 1 -7 -1 2 0 1 1 -2 1 1 3 -1 2 1 -1

=

15 =3 -5

z=

1 -2 6 1 3 -7 2 1 0 1 -2 1 1 3 -1 2 1 -1

=

5 -5

= -1

2 x  y  z  9  77.  x  y  2z  4 x  3 y  z  9 

Solution

x=

- 9 -1 -1 4 1 2 9 3 -1 2 -1 -1 1 1 2 1 3 -1

=

38 = -2 -19

y=

2 - 9 -1 1 4 2 1 9 -1 2 -1 -1 1 1 2 1 3 -1

=

2 -1 -9 1 1 4 1 3 9

-76 -19 =4 z= = =1 -19 -19 2 -1 -1 1 1 2 1 3 -1

 x  2 y  z  1  78. 2 x  y  z  1  x  3 y  5z  17 

Solution

x=

-1 2 -1 1 1 -1 17 -3 -5 1 2 -1 2 1 -1 1 - 3 -5

=

4 17

y=

1 -1 -1 2 1 -1 1 17 -5 1 2 -1 2 1 -1 1 - 3 -5

=

-30 17

z=

1 2 -1 2 1 1 1 -3 17 1 2 -1 2 1 -1 1 -3 -5

=

-39 17

x y z     11 2 3 2 z x 79.   y   6 3 6  x y  2  6  z  16 

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1547


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution Rewrite system: 3 x + 2 y + 3z = 66 2 x + 6 y - z = 36 3 x + 1 y + 6z = 96

y=

3 66 3 2 36 -1 3 96 6 3 2 3 2 6 -1 3 1 6

=

x=

198 =6 33

z=

66 2 3 36 6 -1 96 1 6 3 2 3 2 6 -1 3 1 6 3 2 66 2 6 36 3 1 96 3 2 3 2 6 -1 3 1 6

=

=

198 =6 33

396 = 12 33

x y z     17 2 5 3 x y z 80.     32 5 2 5 y z   x  3  2  30 

Solution Rewrite system: 15 x + 6 y + 10z = 510 2 x + 5 y + 2z = 320 6 x + 2 y + 3z = 180

y=

15 510 10 2 320 2 6 180 3 15 6 10 2 5 2 6 2 3

=

-3540 = 60 -59

x=

z=

510 6 10 320 5 2 180 2 3 15 6 10 2 5 2 6 2 3 15 6 510 2 5 320 6 2 180 15 6 10 2 5 2 6 2 3

=

-590 = 10 -59

=

0 =0 -59

w  x  y  z  8  w  x  y  2z  7 81.  w  x  2 y  3z  3 w  2 x  3 y  4 z  4 

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1548


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

a=

c=

8 1 1 1 7 1 1 2 3 1 2 3 4 2 3 4 1 1 1 1 1 1 1 2 1 1 2 3 1 2 3 4 1 1 1 1 1 1 1 2

8 1 7 2 3 3 4 4

1 1 1 1 1 1 1 2 1 1 2 3 1 2 3 4

=

=

-7 =7 -1

3 = -3 -1

b=

d=

1 1 1 1

8 1 1 7 1 2 3 2 3 4 3 4

1 1 1 1 1 1 1 2 1 1 2 3 1 2 3 4 1 1 1 8 1 1 1 7 1 1 2 3 1 2 3 4 1 1 1 1 1 1 1 2 1 1 2 3 1 2 3 4

=

-5 =5 -1

=

1 = -1 -1

2w  x  3 y  z  0  w  x  z  1 82.  3w  y  2 w  2 x  3z  7 

Solution

p=

r=

0 -1 3 -1 -1 1 0 -1 2 0 -1 0 7 -2 0 3 2 -1 3 -1 1 1 0 -1 3 0 -1 0 1 -2 0 3 2 -1 0 -1 1 1 -1 -1 3 0 2 0 1 -2 7 3 2 -1 3 -1 1 1 0 -1 3 0 -1 0 1 -2 0 3

=

=

-15 5 = -18 6

-9 1 = -18 2

q=

s=

2 0 3 -1 1 -1 0 -1 3 2 -1 0 1 7 0 3 2 -1 3 -1 1 1 0 -1 3 0 -1 0 1 -2 0 3 2 -1 3 0 1 1 0 -1 3 0 -1 2 1 -2 0 7 2 -1 3 -1 1 1 0 -1 3 0 -1 0 1 -2 0 3

=

-12 2 = -18 3

=

-45 5 = 2 -18

Use the determinant to find an equation of the line in standard form that passes through the given points. 83. P(0, 0), Q(4, 6)

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1549


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution x 0 4 x

y 1 0 1 =0 6 1

0 1 0 1 0 0 -y +1 =0 6 1 4 1 4 6 x (-6) - y (-4) + 1(0) = 0 -6 x + 4 y = 0 3x - 2 y = 0

84. P(2, 3), Q(6, 8)

Solution x 2 6 x

y 1 3 1 =0 8 1

3 1 2 1 2 3 -y +1 =0 8 1 6 1 6 8

x (-5) - y (-4) + 1(-2) = 0 -5 x + 4 y - 2 = 0 5 x - 4 y = -2

85. P(–2, 3), Q(5, –3)

Solution x y 1 -2 3 1 =0 5 -3 1 x

-2 1 -2 3 1 3 -y +1 =0 -3 1 5 1 5 -3

x (6) - y (-7 ) + 1(-9) = 0 6x + 7 y - 9 = 0 6x + 7 y = 9

86. P(1, –2), Q(–4, 3)

Solution x y 1 1 -2 1 = 0 -4 3 1 x

-2 1 1 1 1 -2 -y +1 =0 3 1 -4 1 -4 3

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1550


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

x (-5) - y (5) + 1(-5) = 0 -5 x - 5 y - 5 = 0 x + y = -1 87. P(4, 1), Q(2, –3)

Solution x y 1 4 1 1 =0 2 -3 1 é 1 1ù é4 1ù é 1ùú ú- yê ú + 1 ê4 =0 xê ê-3 1ú ê 2 1ú ê 2 -3ú ë û ë û ë û x (4) - y (2) + 1(-14) = 0 4 x - 2 y - 14 = 0 2x - y = 7

88. P(3, –1), Q(2, 3)

Solution

x y 1 3 -1 1 = 0 2 3 1 é-1 1ù é ù é ù ú - y ê3 1ú + 1 ê3 -1ú = 0 xê ê 3 1ú ê2 1ú ê2 3ú ë û ë û ë û x (-4) - y (1) + 1(11) = 0 4 x + y = 11 Use the determinant to find the area of each triangle with vertices at the given points. 89. P(0, 0), Q(12, 0), R(12, 5)

Solution 

0 0 1 1 1 12 0 1 =  (60) 2 2 12 5 1 = 30 square units

90. P(0, 0), Q(0, 5), R(12, 5)

Solution 

0 0 1 1 1 0 5 1 =  (-60) 2 2 12 5 1 = 30 square units

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1551


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

91. P(2, 3), Q(10, 8), R(0, 20)

Solution 

2 3 1 1 1 10 8 1 =  (146) 2 2 0 20 1 = 73 square units

92. P(1, 1), Q(6, 6), R(2, 10)

Solution 

1 1 1 1 1 6 6 1 =  (40) 2 2 2 10 1 = 20 square units

93. P(2, –3), Q(3, 1), R(–1, 4)

Solution

A= =

2 -3 1 1 1 3 1 1 =  (19) 2 2 4 1 -1

19 square units 2

94. P(–2, –2), Q(0, 5), R(3, –1)

Solution

1 A= 2 =

-2 -2 1 1 0 5 1 =  (-33) 2 3 -1 1

33 square units 2

In Exercises 95–98, illustrate each column operation by showing that it is true for the determinant

a b . c d 95. Interchanging two columns

Solution a b = ad - bc c d b a = bc - ad = - (ad - bc ) d c

96. Multiplying each element in a column by k

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1552


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution a b = ad - bc c d ka b = kad - kbc = k (ad - bc ) kc d

97. Adding k times any column to another column

Solution a b = ad - bc c d a b + ka = a (d + kc ) - (b + ka) c = ad + akc - bc - akc = ad - bc c d + kc

af  ec ax  by  e . 98. Use the method of addition to solve  for y, and thereby show that y  ad  bc cx  dy  f Solution ax + by = e  ´(-c) cx + dy = f  ´ a

-acx acx +

bcy = ady =

- ec fa

(ad - bc) y = af - ec y=

af - ec ad - bc

Expand the determinants and solve for x. 99.

3 x 2 1  1 2 x 5 Solution 3 x 2 -1 = 1 2 x -5 6 - x = -10 + x -2 x = -16 x=8

100.

x 4 4 x2  1 1 2 3 Solution x 4 4 x2 = 2 3 1 -1 -4 - x 2 = 3 x - 8 0 = x2 + 3x - 4 0 = ( x - 1)( x + 4) x = 1 or x = -4

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1553


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3 x 1 2 x 101. x 0 2  x 4 4 0 1

Solution

3 x 1 2 x x 0 -2 = x 4 4 0 1

-x

x -2 = 8 - x2 4 1

-x ( x + 8) = 8 - x 2

- x 2 - 8 x = -x 2 + 8 x = -1 x 1 2 2 2 102. 2 x 3  5 x 4 3 1

Solution

x -1 2 2 2 -2 x 3 = 5 x 4 -3 -1 x

x 3 -2 3 -2 x - (-1) +2 = 2 x - 10 -3 - 1 4 -1 4 -3 x (-x + 9) + 1(2 - 12) + 2 (6 - 4 x ) = 2 x - 10 -x 2 + 9 x - 10 + 12 - 8 x = 2 x - 10 0 = x 2 + x - 12 0 = ( x + 4)( x - 3)  x = -4 or x = 3

Use a graphing calculator to evaluate each determinant. 2.3 5.7 6.1 103. 3.4 6.2 8.3 5.8 8.2 9.2

Solution

2.3 5.7 6.1 3.4 6.2 8.3 = 21.468 5.8 8.2 9.2

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1554


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

0.32 7.4 6.7 104. 3.3 5.5 0.27 8 0.13 5.47

Solution

0.32 -7.4 -6.7 3.3 5.5 -0.27 = -164.716332 -8 -0.13 5.47 Fix It In exercises 105 and 106, identify the step the first error is made and fix it. 105. Evaluate

1 2 1 3 5 0 . Use expansion along row 2. 2 1 1

Solution Step 1 was incorrect. Step 1: -3

-2 1 1 1 +5 -1 1 -2 1

Step 2: -3 éê-2 (1) - 1 (-1)ùú + 5 éê 1(1) - 1(-2)ùú ë û ë û Step 3: -3 (-1) + 5 (3) Step 4: 18

2 x  3 y  7 106. Use Cramer’s Rule to solve the linear system  for x and y. Write the solution 4 x  y  4 as an ordered pair. Solution Step 2 was incorrect.

Step 1: x =

7 -3 -4 1 2 -3 -4 1

;y=

2 7 -4 -4 2 -3 -4 1

Step 2: x =

7 - 12 -8 + 28 ;y= 2 - 12 2 - 12

Step 3: x =

-5 20 ; y= -10 -10

æ1 ö Step 4: çç , - 2÷÷÷ çè 2 ø÷

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1555


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Applications 107. Investing A student wants to average a 6.6% return by investing $20,000 in the three stocks listed in the table. Because HiTech is a high-risk investment, he wants to invest three times as much in SaveTel and OilCo combined as he invests in HiTech. How much should he invest in each stock?

Stock

Rate of Return

HiTech

10%

SaveTel

5%

OilCo

6%

Solution Let x = $ invested in HiTech, y = $ invested in SaveTel, and z = $ invested in OilCo.

ìï x + y + z = 20, 000 x + y + z = 20000 ïï ïí y + z = 3 x -3 x + y + z = 0 ïï 10 x + 5 y + 6z = 132000 ïïî0.10 x + 0.05 y + 0.06z = 0.066 (20, 000) 20000 1 1 1 20000 1 -3 0 1 1 0 1 132000 5 6 10 132000 6 20000 32000 = = 5000, y = = = 8000 x= 4 4 1 1 1 1 1 1 -3 1 1 -3 1 1 10 5 6 10 5 6

z=

1 1 20000 -3 1 0 10 5 132000 1 1 1 -3 1 1 10 5 6

=

28000 = 7000  He should invest $5000 in HiTech, $8000 in 4 SaveTel, and $7000 in OilCo.

108. Ice skating The illustration shows three circles traced out by a figure skater during her performance. If the centers of the circles are the given distances apart, find the radius of each circle.

Solution Let x, y, and z represent the radii of the circles.

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1556


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

ìï x + y = 10 ïï í x + z = 18 x = ïï ïïî y + z = 14

z=

1 1 10 1 0 18 0 1 14 1 1 0 1 0 1 0 1 1

=

10 1 0 18 0 1 14 1 1 1 1 0 1 0 1 0 1 1

=

-14 = 7, y = -2

1 10 0 1 18 1 0 14 1 1 1 0 1 0 1 0 1 1

=

-6 =3 -2

-22 = 11  The radii are 7 yd, 3 yd, and 11 yd. -2

Discovery and Writing 109. Explain how to find the determinant of a 2 × 2 matrix.

Solution Answers may vary. 110. Explain how to find the determinant of a 3 × 3 matrix using cofactor expansion.

Solution Answers may vary. 111. Explain why applying row or column operations can help evaluate the determinant.

Solution Answers may vary. 112. What is Cramer’s Rule? Describe how it can be used to solve systems of equations.

Solution Answers may vary. In Exercises 113–116, evaluate each determinant. What do you discover? 1 3 4 113. 0 5 2 0 0 2

Solution 1 3 4 0 5 2 = 10 0 0 2 1 ⋅ 5 ⋅ 2 = 10 2 1 2 114. 0 3 4 0 0 1

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1557


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 2 1 -2 0 3 4 = -6 0 0 -1 2 ⋅ 3 ⋅ (-1) = -6 1 0 115. 0 0

2 2 0 0

4 3 2 1 3 2 0 4

Solution

1 0 0 0

2 2 0 0

4 3 2 1 = 24 3 2 0 4

1 ⋅ 2 ⋅ 3 ⋅ 4 = 24 2 1 2 1 0 2 2 1 116. 0 0 3 1 0 0 0 2

Solution

2 1 -2 1 0 2 2 -1 = 24 0 0 3 1 0 0 0 2 2 ⋅ 2 ⋅ 3 ⋅ 2 = 24 117. Alternate determinant method Another way to evaluate a 3 × 3 determinant is to copy its first two columns to the right of the determinant as shown. Then find the product of the numbers on each red diagonal and find their sum. Then find the product of the numbers on each blue diagonal and find their sum. Then subtract the sum of the products on the blue diagonals from the sum of the products on the red diagonals. Find the value of the determinant.

Solution 3 (1)(1) + 2 (-2)(1) + (-1)(2)(3) = -7 1 (1)(-1) + 3 (-2)(3) + 1(2)(2) = -15 -7 - (-15) = 8

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1558


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

0 1 3 118. Use the method of Exercise 117 to evaluate the determinant 3 5 2 . 2 5 3

Solution 0 (5)(3) + 1(2)(2) + (-3)(-3)(-5) = -41 2 (5)(-3) + (-5)(2)(0) + 3 (-3)(1) = -39 -41 - (-39) = -2

119. A determinant is a function that associates a number with every square matrix. Give the domain and the range of that function.

Solution domain: n × n matrices range: all real numbers 120. Use an example chosen from 2 × 2 matrices to show that for n × n matrices A and B, AB ≠ BA but AB  BA .

Solution Answers may vary. 121. If A and B are matrices and AB  0, must A  0 or B  0 ? Explain.

Solution Yes. 122. If A and B are matrices and AB  0, must A = 0 or B = 0? Explain.

Solution No. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 123. If A and B are square matrices of the same order, then A  B  A  B .

Solution False. In general, A  B  A  B . 124. If A and B are square matrices of the same order, then AB  A B .

Solution True.

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1559


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

125.

999 888 777 666  777 666 999 888 Solution

999 888 777 666 .  777 666 999 888

False.

111 222 0 126. 333 444 0  0 555 666 0

Solution True. 111 555 127. 111 555

222 666 222 666

333 777 333 777

444 888 0 444 888

Solution True. 128. The transpose of a matrix is formed by writing its columns as rows. If A is a square matrix and AT denotes its transpose, then A  AT .

Solution True.

EXERCISES 6.6 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Factor completely: x4 – 4x3 + 3x2 – 12x

Solution x ( x 3 - 4 x 2 + 3 x - 12) = x ( x 2 ( x - 4) + 3( x - 4)) = x ( x - 4)( x 2 + 3)

2. Add:

3 5  2x  1 x  3

Solution 3 ( x - 3)

(2 x + 1)( x - 3)

+

5 (2 x + 1)

(2 x + 1)( x - 3)

=

3 x - 9 + 10 x + 5

(2 x + 1)( x - 3)

=

13 x - 4

(2 x + 1)( x - 3)

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1560


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3. Subtract:

2 3  x  1  x  12

Solution

2 ( x - 1) 2

( x - 1)

-

3 2

( x - 1)

=

2x - 2 - 3 2

( x - 1)

=

2x - 5 2

( x - 1)

4 A  B  7 4. Solve the system  for A and B. 2A  5B  9 Solution ìï 4A + B = 7 4A + B = 7 ïí  ïï-2 (2 A - 5B = 9) -4 A + 10B = -18 î 11B = -11 B = -1 Substitute B = -1 into (1) . 4A - 1 = 7 4A = 8 A=2  A  2B  C  1  5. Solve the system B  2C  3 for A, B, and C. A  B  5 

Solution A - 2B + C = 1 B - 2C = 3  3B - 6C = 9 A + B = 5  - A - B = -5 A - 2B + C = 1 -A - B = -5 - 3B + C = -4 -3B + C = -4 3B - 6C = 9 -5C = 5 C = -1 Substitute C = -1 into (2) . B - 2 (-1) = 3 B+2= 3 B=1

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1561


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Substitute B = 1 into (3) . A+1= 5 A=4 So, A = 4, B = 1, C = -1 6. Use long division and divide:

x3  3 x2  x

Solution x-1 x 2 + x x3 + 0x 2 + 0x + 3 - (x3 + x2 ) - x2 + 0x - (-x 2 - x ) x +3 Answer: x - 1 +

x +3 x2 + x

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A polynomial with real coefficients factors as the product of __________ and __________ factors or powers of those.

Solution first-degree, second-degree 8. The second-degree factors of a polynomial with real coefficients are __________, which means they don’t factor further over the real numbers.

Solution prime Practice Decompose each fraction into partial fractions. 9.

3x  1 x( x  1) Solution

3x - 1 x ( x - 1)

=

A B + x x-1

ïíìï A + B = 3  A = 1, B = 2 = -1 ïîï-A

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1562


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3x - 1 x ( x - 1) 3x - 1 x ( x - 1)

= =

A ( x - 1) x ( x - 1)

+

Bx

3x - 1

x ( x - 1)

x ( x - 1)

=

1 2 + x x-1

Ax - A + Bx x ( x - 1)

( A + B) x - A x ( x - 1) x ( x - 1) 3x - 1

10.

=

4x  6 x ( x  2) Solution

4x + 6 x ( x + 2) 4x + 6 x ( x + 2) 4x + 6 x ( x + 2)

= = =

ïíìï A + B = 4  A = 3, B = 1 ïïî2 A =6

A B + x x +2 A ( x + 2) x ( x + 2)

+

Bx

4x + 6

x ( x + 2)

x ( x + 2)

=

3 1 + x x +2

Ax + 2 A + Bx x ( x + 2)

( A + B) x + 2 A x ( x + 2) x ( x + 2) 4x + 6

11.

=

2 x  15 x ( x  3) Solution

2 x - 15 x ( x - 3) 2 x - 15 x ( x - 3) 2 x - 15 x ( x - 3)

= = =

A B + x x -3 A ( x - 3) x ( x - 3)

+

Bx x ( x - 3)

Ax - 3 A + Bx x ( x - 3)

( A + B) x - 3 A x ( x - 3) x ( x - 3) 2 x - 15

=

ìïï A + B = 2  A = 5, B = -3 í ï= -15 ïî 3 A 2 x - 15 x ( x - 3)

=

5 3 x x -3

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1563


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

12.

5 x  21 x ( x  7) Solution 5 x + 21 A B = + x x 7 + x ( x + 7) 5 x + 21 x ( x + 7) 5 x + 21 x ( x + 7)

= =

A ( x + 7) x ( x + 7)

+

Bx x ( x + 7)

Ax + 7 A + Bx x ( x + 7)

( A + B) x + 7 A x ( x + 7) x ( x + 7) 5 x + 21

=

ìïï A + B = 5  A = 3, B = 2 í ïïî7 A = 21 5 x + 21 3 2 = + x ( x + 7) x x + 7 13.

3x  1 ( x  1)( x  1) Solution

3x + 1

( x + 1)( x - 1) 3x + 1

( x + 1)( x - 1) 3x + 1

( x + 1)( x - 1)

= = =

ìïï A + B = 3  A = 1, B = 2 í ïïî A + B = 1

A B + x +1 x-1 A ( x - 1)

( x + 1)( x - 1)

+

B ( x + 1)

3x + 1

( x + 1)( x - 1)

( x + 1)( x - 1)

=

1 2 + x +1 x-1

Ax - A + Bx + B

( x + 1)( x - 1) ( A + B) x + (-A + B) 3x + 1 = ( x + 1)( x - 1) ( x + 1)( x - 1) 14.

9x  3 ( x  1)( x  2) Solution

9x - 3

( x + 1)( x - 2) 9x - 3

= =

ìïï A + B = 9  A = 4, B = 5 í ïïî 2 A + B = -3

A B + x + 1 x -2 A ( x - 2)

+

B ( x + 1)

( x + 1)( x - 2) ( x + 1)( x - 2) ( x + 1)( x - 2)

9x - 3

( x + 1)( x - 2)

=

4 5 + x + 1 x -2

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1564


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

9x - 3

( x + 1)( x - 2)

=

Ax - 2 A + Bx + B

( x + 1)( x - 2) ( A + B) x + (-2 A + B) 9x - 3 = ( x + 1)( x - 2) ( x + 1)( x - 2) 15.

4 x  2x 2

Solution -4 = x - 2x -4 A B = + x x -2 x ( x - 2) 2

-4 x ( x - 2) -4 x ( x - 2)

= =

A ( x - 2) x ( x - 2)

+

Bx x ( x - 2)

Ax - 2 A + Bx x ( x - 2)

( A + B) x - 2 A x ( x - 2) x ( x - 2) -4

=

ìïï A + B = 0  A = 2, B = -2 í ï= -4 ïî 2 A -4 2 2 = 2 x x -2 x - 2x 16.

1 P  300P 2

Solution 1

P 2 - 300P 1 P (P - 300) 1 P (P - 300) 1 P (P - 300)

= = = =

A B + P P - 300 A (P - 300) P (P - 300)

+

BP P (P - 300)

AP - 300 A + BP P (P - 300)

( A + B) P - 300 A P (P - 300) P (P - 300) 1

=

ìïï 1 1 A+B = 0  A=,B= í ï=1 300 300 ïî 300 A 1 1 1 300 = 300 + 2 P P 300 P - 300P

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1565


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

17.

2 x  11 x2  x  6 Solution -2 x + 11

( x + 2)( x - 3) -2 x + 11

= =

-2 x + 11 -2 x + 11 = x 2 - x - 6 ( x + 2)( x - 3)

A B + x +2 x -3 A ( x - 3)

B ( x + 2)

+

( x + 2)( x - 3) ( x + 2)( x - 3) ( x + 2)( x - 3) -2 x + 11

( x + 2)( x - 3)

=

Ax - 3 A + Bx + 2B

( x + 2)( x - 3) ( A + B) x + (-3 A + 2B) -2 x + 11 = ( x + 2)( x - 3) ( x + 2)( x - 3) 18.

7x  2 x  x 2 2

Solution 7x + 2

( x + 2)( x - 1) 7x + 2

= =

7x + 2

A B + x +2 x-1 A ( x - 1)

3

x + x -2 +

B ( x + 2)

( x + 2)( x - 1) ( x + 2)( x - 1) ( x + 2)( x - 1) 7x + 2

( x + 2)( x - 1)

=

Ax - A + Bx + 2B

( x + 2)( x - 1) ( A + B) x + (-A + 2B) 7x + 2 = ( x + 2)( x - 1) ( x + 2)( x - 1) 19.

ìïï A + B = - 2  A = -3, B = 1 í ïïî 3 A + 2B = 11 -2 x + 11 -3 1 = + ( x + 2)( x - 3) x + 2 x - 3

=

7x + 2

( x + 2)( x - 1)

ïíìï A + B = 7  A = 4, B = 3 ïïî A + 2B = 2 7x + 2 4 3 = + + -1 x x 2 + x x 2 1 ( )( )

3 x  23 x  2x  3 2

Solution 3 x - 23

( x + 3)( x - 1) 3 x - 23

= =

3 x - 23

A B + x +3 x-1 A ( x - 1)

x 2 + 2x - 3 +

B ( x + 3)

( x + 3)( x - 1) ( x + 3)( x - 1) ( x + 3)( x - 1) 3 x - 23

( x + 3)( x - 1)

=

Ax - A + Bx + 3B

( x + 3)( x - 1) ( A + B) x + (-A + 3B) 3 x - 23 = ( x + 3)( x - 1) ( x + 3)( x - 1)

=

3 x - 23

( x + 3)( x - 1)

3 ïíìï A + B =  A = 8, B = -5 ïïî A + 3B = - 23 3 x + 23 8 5 = + ( x + 3)( x - 1) x + 3 x - 1

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1566


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

20.

 x  17 x2  x  6 Solution -x - 17

( x + 2)( x - 3) -x - 17

= =

-x - 17

A B + x +2 x -3 A ( x - 3)

2

x - x -6 B ( x + 2)

+

-x - 17

=

Ax - 3 A + Bx + 2B

( x + 2)( x - 3) ( A + B) x + (-3A + 2B) -x - 17 = ( x + 2)( x - 3) ( x + 2)( x - 3) 21.

-x - 17

( x + 2)( x - 3)

ìïï A + B = - 1  A = 3, B = -4 í ïïî 3 A + 2B = - 17 -x - 17 3 4 = + x x 2 3 + x x 2 3 ( )( )

( x + 2)( x - 3) ( x + 2)( x - 3) ( x + 2)( x - 3) ( x + 2)( x - 3)

=

9 x  31 2 x  13 x  15 2

Solution 9 x - 31

2 x 2 - 13 x + 15

=

9 x - 31

(2x - 3)( x - 5) 9 x - 31

= =

A B + 2x - 3 x - 5 A ( x - 5)

B (2 x - 3)

+

(2x - 3)( x - 5) (2x - 3)( x - 5) (2x - 3)( x - 5) 9 x - 31

(2x - 3)( x - 5)

=

Ax - 5 A + 2Bx - 3B

(2x - 3)( x - 5) ( A + 2B) x + (-5A - 3B) 9 x - 31 = (2x - 3)( x - 5) (2x - 3)( x - 5) ìïï A + 2B = 9  A = 5, B = 2 í ïîï-5 A - 3B = -31 22.

9 x - 31

(2x - 3)( x - 5)

=

5 2 + 2x - 3 x - 5

2 x  6 3x 2  7 x  2 Solution -2 x - 6 2

3x - 7 x + 2

=

-2 x - 6

(3x - 1)( x - 2) -2 x - 6

= =

A B + 3x - 1 x - 2 A ( x - 2)

+

B (3 x - 1)

(3x - 1)( x - 2) (3x - 1)( x - 2) (3x - 1)( x - 2) -2 x - 6

(3x - 1)( x - 2)

=

Ax - 2 A + 3Bx - B

(3x - 1)( x - 2) ( A + 3B) x + (-2A - B) -2 x - 6 = (3x - 1)( x - 2) (3x - 1)( x - 2)

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1567


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

ìïï A + 3B = - 2  A = 4, B = -2 í ïîï-2 A - B = -6 23.

-2 x - 6

(3x - 1)( x - 2)

=

4 2 3x - 1 x - 2

4x2  4x  2 x ( x 2  1)

Solution 4x2 + 4x - 2 x ( x - 1) 2

=

4x2 + 4x - 2 x ( x + 1)( x - 1) 4x2 + 4x - 2 x ( x + 1)( x - 1)

= =

2

4x + 4x - 2 x ( x + 1)( x - 1)

A B C + + x x +1 x-1 A ( x + 1)( x - 1) x ( x + 1)( x - 1) 2

=

+

Bx ( x - 1) x ( x + 1)( x - 1)

2

+

Cx ( x + 1) x ( x + 1)( x - 1)

2

Ax - A + Bx - Bx + Cx + Cx x ( x + 1)( x - 1)

( A + B + C ) x 2 + (-B + C ) x + (-A) x ( x + 1)( x - 1) x ( x + 1)( x - 1) 4x2 + 4x - 2

=

ìï A + B + C = 4 A= 2 ïï B C B + =  = -1 4 í ïï C= 3 = -2 ïïî-A

24.

4x2 + 4x - 2 x ( x + 1)( x - 1)

=

2 1 3 + x x +1 x-1

x 2  6 x  13 ( x  2)( x 2  1)

Solution

x 2 - 6 x - 13

( x + 2)( x 2 - 1)

= = =

x 2 - 6 x - 13

( x + 2)( x + 1)( x - 1) A ( x + 1)( x - 1)

= +

A B C + + x +2 x +1 x-1 B ( x + 2)( x - 1)

+

C ( x + 2)( x + 1)

( x + 2)( x + 1)( x - 1) ( x + 2)( x + 1)( x - 1) ( x + 2)( x + 1)( x - 1) Ax 2 - A + Bx 2 + Bx - 2B + Cx 2 + 3Cx + 2C

( x + 2)( x + 1)( x - 1) ( A + B + C ) x 2 + (B + 3C ) x + (-A - 2B + 2C) = ( x + 2)( x + 1)( x - 1)

ìï A + B + C = 1 A= 1 ïï B + 3C = - 6  B = 3 í ïï ïîï-A - 2B + 2C = -13 C = -3 25.

x 2 - 6 x - 13

( x + 2)( x + 1)( x - 1)

=

1 3 3 + x +2 x +1 x-1

x2  x  3 x ( x 2  3)

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1568


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution x2 + x + 3

=

x ( x + 3) 2

= = =

A Bx + C + 2 x x +3 A ( x 2 + 3) x ( x 2 + 3)

+

(Bx + C ) x x ( x 2 + 3)

Ax 2 + 3 A + Bx 2 + Cx x ( x 2 + 3)

( A + B) x 2 + Cx + 3 A x ( x 2 + 3)

ìï A + B =1 A= 1 ïï C=1 B=0 í ïï =3 C=1 ïïî3 A 2 1 1 x + x +3 = + 2 2 x x +3 x ( x + 3)

26.

5x 2  2x  2 x3  x Solution 5x 2 + 2x + 2 x ( x + 1) 2

=

= = =

A Bx + C + 2 x x +1 A ( x 2 + 1) x ( x + 1) 2

+

(Bx + C ) x x ( x 2 + 1)

Ax 2 + A + Bx 2 + Cx x ( x 2 + 1)

( A + B) x 2 + Cx + A x ( x 2 + 1)

ìï A + B =5 A=2 ïï =  =3 C 2 B í ïï =2 C=2 ïïî A 2 5x + 2x + 2 2 3x + 2 = + 2 2 x x +1 x ( x + 1)

27.

3 x 2  8 x  11 ( x  1)( x 2  2 x  3)

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1569


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 3 x 2 + 8 x + 11

( x + 1)( x + 2x + 3) 2

=

A Bx + C + x + 1 x 2 + 2x + 3 A ( x 2 + 2 x + 3)

(Bx + C )( x + 1) ( x + 1)( x 2 + 2x + 3) ( x + 1)( x 2 + 2 x + 3) ( x + 1)( x 2 + 2 x + 3) 3 x 2 + 8 x + 11 3 x 2 + 8 x + 11

=

( x + 1)( x 2 + 2x + 3)

=

+

Ax 2 + 2 Ax + 3 A + Bx 2 + Bx + Cx + C

( x + 1)( x 2 + 2x + 3) ( A + B) x 2 + (2A + B + C ) x + (3A + C ) 3 x 2 + 8 x + 11 = ( x + 1)( x 2 + 2 x + 3) ( x + 1)( x 2 + 2x + 3) ìï A + B A=3 = 3 ïï í 2A + B + C = 8  B = 0 ïï C=2 + C = 11 ïïî3 A 28.

3 x 2 + 8 x + 11

( x + 1)( x + 2x + 3) 2

=

3 2 + 2 x + 1 x + 2x + 3

3 x 2  x  5 ( x  1)( x 2  2)

Solution

-3 x 2 + x - 5

( x + 1)( x + 2) 2

=

A Bx + C + 2 x +1 x +2 A ( x 2 + 2)

(Bx + C )( x + 1) ( x + 1)( x 2 + 2) ( x + 1)( x 2 + 2) ( x + 1)( x 2 + 2) -3 x 2 + x - 5 -3 x 2 + x - 5

( x + 1)( x 2 + 2)

= =

+

Ax 2 + 2 A + Bx 2 + Bx + Cx + C

( x + 1)( x 2 + 2) ( A + B) x 2 + (B + C ) x + (2 A + C ) -3 x 2 + x - 5 = ( x + 1)( x 2 + 2) ( x + 1)( x 2 + 2) ìï A + B = -3 A = -3 ïï B+C = 1  B = 0 í ïï + C = -5 C= 1 ïïî2 A 29.

-3 x 2 + x - 5

( x + 1)( x + 2) 2

=

-3 1 + 2 x +1 x +2

5x 2  9x  3 x ( x  1)2

Solution 5x 2 + 9x + 3 2

x ( x + 1)

=

A B C + + x x + 1 ( x + 1)2 2

=

A ( x + 1)

Bx ( x + 1)

x ( x + 1)

2

+ 2

x ( x + 1)

+

Cx 2

x ( x + 1)

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1570


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

=

=

Ax 2 + 2 Ax + A + Bx 2 + Bx + Cx 2

x ( x + 1)

( A + B) x 2 + (2A + B + C ) x + A 2 x ( x + 1)

ìï A + B A=3 =5 ïï í2 A + B + C = 9  B = 2 ï =3 C=1 ïïîï A 30.

5x 2 + 9x + 3 2

x ( x + 1)

=

3 2 1 + + x x + 1 ( x + 1)2

2x2  7 x  2 x ( x  1)2

Solution

2x 2 - 7 x + 2 2

x ( x - 1)

=

A B C + + x x - 1 ( x - 1)2 2

= =

A ( x - 1)

+

x ( x - 1)

2

Bx ( x - 1) x ( x - 1)

2

+

Cx x ( x - 1)

2

Ax 2 - 2 Ax + A + Bx 2 - Bx + Cx 2

x ( x - 1)

( A + B) x + (-2A - B + C ) x + A x ( x - 1) 2

=

2

ìï A + B = 2 A= 2 ïï í-2 A - B + C = - 7  B = 0 ïï = 2 C = -3 ïïî A 31.

2x 2 - 7 x + 2 2

x ( x - 1)

=

2 3 x ( x - 1)2

2 x 2  x  2 x 2 ( x  1)

Solution

-2 x 2 + x - 2 x ( x - 1) 2

= = =

A B C + 2 + x x x-1 Ax ( x - 1) x 2 ( x - 1)

+

B ( x - 1) x 2 ( x - 1)

+

Cx 2 x 2 ( x - 1)

Ax 2 - Ax + Bx - B + Cx 2 x 2 ( x - 1)

( A + C ) x + (-A + B) x + (-B) x ( x - 1) 2

=

2

ìï A + C = -2 A= 1 ïï = 1 B= 2 í-A + B ïï = -2 C = -3 ïïî - B

-2 x 2 + x - 2 x ( x - 1) 2

=

1 2 3 + 2x x x-1

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1571


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

32.

x2  x  1 x3 Solution

x2 + x + 1 A B C = + 2 + 3 3 x x x x Ax 2 Bx C = 3 + 3 + 3 x x x Ax 2 + Bx + C = x3 ìï A =1 ïï 1 1 1 x2 + x + 1 B =1 = + 2+ 3 í 3 ïï x x x x C=1 ïïî 33.

3 x 2  13 x  18 x 3  6x 2  9x Solution

3 x 2 - 13 x + 18 3

2

x - 6x + 9x

=

3 x 2 - 13 x + 18 2

x ( x - 3)

A B C + + x x - 3 ( x - 3)2

=

2

A ( x - 3)

=

x ( x - 3)

ìï A + B = 3 A=2 ïï + =  =1 A B C B 6 3 13 í ïï = 18 C = 2 ïïî 9 A 34.

Bx ( x - 3) 2

x ( x - 3)

+

Cx 2

x ( x - 3)

Ax 2 - 6 Ax + 9 A + Bx 2 - 3Bx + Cx

=

=

+

2

2

x ( x - 3)

( A + B) x 2 + (-6 A - 3B + C ) x + 9 A 2 x ( x - 3) 3 x 2 - 13 x + 18 2

x ( x - 3)

=

2 1 2 + + x x - 3 ( x - 3)2

3 x 2  13 x  20 x3  4x2  4x Solution

3 x 2 + 13 x + 20 x3 - 4x2 + 4x

=

3 x 2 + 13 x + 20 2

x ( x + 2)

=

A B C + + x x + 2 ( x + 2)2 2

= =

A ( x + 2)

2

x ( x + 2)

+

Bx ( x + 2) 2

x ( x + 2)

+

Cx 2

x ( x + 2)

Ax 2 + 4 Ax + 4 A + Bx 2 + 2Bx + Cx 2

x ( x + 2)

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1572


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

( A + B) x 2 + (4 A + 2B + C ) x + 4 A = 2 x ( x + 2) ìï A + B = 3 A= 5 ïï í4 A + 2B + C = 13  B = -2 ïï = 20 C = -3 ïïî 4 A 35.

3 x 2 + 13 x + 20 2

x ( x + 2)

=

5 2 3 x x + 2 ( x + 2)2

x 2  2x  3 ( x  1)3

Solution

x 2 - 2x - 3 3

( x - 1)

=

A B C + + 2 3 x - 1 ( x - 1) ( x - 1) 2

= =

A ( x - 1)

+

( x - 1)

3

B ( x - 1)

( x - 1)

3

+

C

( x - 1)

3

Ax 2 - 2 Ax + A + Bx - B + C 3

( x - 1) Ax + (-2 A + B) x + ( A - B + C ) = ( x - 1) 2

3

ìï A = 1 A= 1 ïï 2 2 A B B + =  = 0 í ïï ïïî A - B + C = -3 C = -4 36.

x 2 - 2x - 3

=

3

( x - 1)

1 4 x - 1 ( x - 1)3

x 2  8 x  18 ( x  3)3

Solution

x 2 + 8 x + 18 3

( x + 3)

=

A B C + + x + 3 ( x + 3)2 ( x + 3)3 2

= =

A ( x + 3) 3

( x + 3)

+

B ( x + 3) 3

( x + 3)

+

C 3

( x + 3)

Ax 2 + 6 Ax + 9 A + Bx + 3B + C 3

( x + 3) 2 Ax + (6 A + B) x + (9 A + 3B + C ) = 3 ( x + 3) ìï A = 1 A= 1 ïï = 8B=2 í6 A + B ïï ïïî9 A + 3B + C = 18 C = 3

x 2 + 8 x + 18 3

( x + 3)

=

1 2 3 + + 2 3 x + 3 ( x + 3) ( x + 3)

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1573


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

37.

x 3  4 x 2  2x  1 x4  x3  x2 Solution

x 3 + 4 x 2 + 2x + 1 = x4 + x3 + x2 x 3 + 4 x 2 + 2x + 1 A B Cx + D = + 2 + 2 2 2 x x x +x+1 x ( x + x + 1) = = =

Ax ( x 2 + x + 1) x 2 ( x 2 + x + 1)

B ( x 2 + x + 1) x 2 ( x 2 + x + 1)

+

(Cx + D) x 2 x 2 ( x 2 + x + 1)

Ax 3 + Ax 2 + Ax + Bx 2 + Bx + B + Cx 3 + Dx 2 x 2 ( x 2 + x + 1)

( A + C ) x 3 + ( A + B + D) x 2 + ( A + B) x + B x 2 ( x 2 + x + 1)

ìï A +C = 1 A=1 ïï ïï A + B +D = 4 B=1  í ïï A + B C=0 =2 ï =1 B D=2 îïï 38.

+

x 3 + 4 x 2 + 2x + 1 x ( x + x + 1) 2

2

=

1 1 2 + + 2 x x2 x +x+1

3x 3  5x 2  3x  1 x 2 ( x 2  x  1)

Solution

3x 3 + 5x 2 + 3x + 1 x ( x + x + 1) 2

2

=

= = =

A B Cx + D + 2+ 2 x x x +x+1 Ax ( x 2 + x + 1) x 2 ( x 2 + x + 1)

x 2 ( x 2 + x + 1)

+

(Cx + D) x 2 x 2 ( x 2 + x + 1)

Ax 3 + Ax 2 + Ax + Bx 2 + Bx + B + Cx 3 + Dx 2 x 2 ( x 2 + x + 1)

( A + C) x 3 + ( A + B + D) x 2 + ( A + B) x + B

ìï A +C =3 A=2 ïï ïï A + B +D =5 B=1  í ïï A + B =3 C=1 ïï = =2 B D 1 ïî 39.

+

B ( x 2 + x + 1)

x 2 ( x 2 + x + 1) 3x 3 + 5x 2 + 3x + 1 x ( x + x + 1) 2

2

=

x +2 2 1 + 2+ 2 x x x +x+1

4 x 3  5x 2  3x  4 x 2 ( x 2  1)

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1574


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

4 x 3 + 5x 2 + 3x + 4 x ( x + 1) 2

2

=

= = = ìï A ïï ïíï ïïï A ïïî 40.

+C B B

A B Cx + D + + 2 x x2 x +1 Ax ( x 2 + 1) x 2 ( x 2 + 1)

+

B ( x 2 + 1) x 2 ( x 2 + 1)

+

(Cx + D) x 2 x 2 ( x 2 + 1)

Ax 3 + Ax + Bx 2 + B + Cx 3 + Dx 2 x 2 ( x 2 + 1)

( A + C ) x 3 + (B + D) x 2 + Ax + B x 2 ( x 2 + 1)

A=3 =4 B=4 +D = 5  C=1 =3 D=1 =4

4 x 3 + 5x 2 + 3x + 4 x ( x + 1) 2

2

=

3 4 x+1 + 2+ 2 x x x +1

2x 2  1 x4  x2 Solution

2x 2 + 1 4

x +x

2

=

2x 2 + 1

x ( x + 1) 2

2

=

= = = ìï A ïï ïï í ïïï A ïïî 41.

+C B B

A B Cx + D + + 2 x x2 x +1 Ax ( x 2 + 1) x 2 ( x 2 + 1)

+

B ( x 2 + 1) x 2 ( x 2 + 1)

+

(Cx + D) x 2 x 2 ( x 2 + 1)

Ax 3 + Ax + Bx 2 + B + Cx 3 + Dx 2 x 2 ( x 2 + 1)

( A + C ) x 3 + (B + D) x 2 + Ax + B x 2 ( x 2 + 1)

A=0 =0 B=1 +D = 2  =0 C=0 =1 D=1

2x 2 + 1

x ( x + 1) 2

2

=

1 x

2

+

1 2

x +1

 x 2  3x  5 x 3  x 2  2x  2 Solution

-x 2 - 3 x - 5 x 3 + x 2 + 2x + 2 -x 2 - 3 x - 5

( x + 1)( x + 2) 2

= =

-x 2 - 3 x - 5 x 2 ( x + 1) + 2 ( x + 1)

=

-x 2 - 3 x - 5

( x + 1)( x 2 + 2)

A Bx + C + 2 x +1 x +2

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1575


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

= =

A ( x 2 + 2)

(Bx + C )( x + 1) ( x + 1)( x + 2) ( x + 1)( x 2 + 2) +

2

Ax 2 + 2 A + Bx 2 + Bx + Cx + C

( x + 1)( x 2 + 2) ( A + B) x 2 + (B + C ) x + (2A + C ) = ( x + 1)( x 2 + 2) ìï A + B A = -1 = -1 ïï B + C = -3  B = 0 í ïïï 2 A + C = -5 C = -3 ïî 42.

-x 2 - 3 x - 5

( x + 1)( x + 2) 2

=

3 -1 x + 1 x2 + 2

2 x 3  7 x 2  6 x 2 ( x 2  2)

Solution

-2 x 3 + 7 x 2 + 6 x 2 ( x 2 + 2)

A B Cx + D + + 2 x x2 x +2

=

Ax ( x 2 + 2)

=

x 2 ( x 2 + 2)

B ( x 2 + 2) x 2 ( x 2 + 2)

(Cx + D) x 2 x 2 ( x 2 + 2)

x 2 ( x 2 + 2)

( A + C ) x 3 + (B + D) x 2 + 2Ax + 2B x 2 ( x 2 + 2)

ìï A A= 0 +C = -2 ïï B B= 3 +D = 7 ïíï  ïï 2 A = 0 C = -2 ïï 2 6 B = D = 4 ïî 43.

+

Ax 3 + 2 Ax + Bx 2 + 2B + Cx 3 + Dx 2

= =

+

-2 x 3 + 7 x 2 + 6 x ( x + 2) 2

2

=

3 -2 x + 4 + 2 x2 x +2

x 3  4 x 2  3x  6 ( x 2  2)( x 2  x  2)

Solution

x 3 + 4 x 2 + 3x + 6

( x + 2)( x + x + 2) 2

2

=

= = =

Ax + B 2

x +2

+

Cx + D 2

x + x +2

( Ax + B)( x 2 + x + 2)

+

(Cx + D)( x 2 + 2)

( x + 2)( x + x + 2) ( x + 2)( x + x + 2) 2

2

2

2

Ax 3 + Ax 2 + 2 Ax + Bx 2 + Bx + 2B + Cx 3 + 2Cx + Dx 2 + 2D

( x + 2)( x + x + 2) 2

2

( A + C) x 3 + ( A + B + D) x 2 + (2A + B + 2C ) x + (2B + 2D)

( x + 2)( x + x + 2) 2

2

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1576


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

ìï A A=1 +C =1 ïï ïï A + B B=1 +D = 4  í ïï2 A + B + 2C =3 C=0 ïï 2 2 6 + = B D D=2 ïî 44.

x 3 + 4x 2 + 3x + 6

( x + 2)( x + x + 2) 2

2

=

x+1 2

x +2

+

2 2

x + x +2

x 3  3x 2  2x  4 ( x 2  1)( x 2  x  2)

Solution

x 3 + 3x 2 + 2x + 4

( x + 1)( x + x + 2) 2

2

=

= = =

Ax + B x2 + 1

+

Cx + D x2 + x + 2

( Ax + B)( x 2 + x + 2)

( x + 1)( x + x + 2) ( x + 1)( x + x + 2) 2

2

2

2

Ax 3 + Ax 2 + 2 Ax + Bx 2 + Bx + 2B + Cx 3 + Cx + Dx 2 + D

( x + 1)( x + x + 2) 2

2

( A + C) x 3 + ( A + B + D) x 2 + (2A + B + C ) x + (2B + D)

( x + 1)( x + x + 2) 2

ìï A A=0 +C =1 ïï ïï A + B B=1 +D = 3  í ïï2 A + B + C =2 C=1 ïï 2B + D= 4 D=2 ïî 45.

+

(Cx + D)( x 2 + 1)

2

x 3 + 3x 2 + 2x + 4

( x + 1)( x + x + 2) 2

2

=

1 2

x +1

+

x +2 2

x + x +2

2 x 4  6 x 3  20 x 2  22 x  25 x ( x 2  2 x  5)2

Solution 2 x 4 + 6 x 3 + 20 x 2 + 22 x + 25 x ( x 2 + 2 x + 5)

2

=

A Bx + C Dx + E + 2 + 2 2 x x + 2x + 5 x + 2x + 5

(

) A ( x + 2 x + 5) (Bx + C )( x )( x + 2 x + 5) (Dx + E ) x = + + x ( x + 2 x + 5) x ( x + 2 x + 5) x ( x + 2 x + 5) 2

2

=

2

2

2

2

2

2

2

( A + B) x 4 + (4 A + 2B + C ) x 3 + (14 A + 5B + 2C + D) x 2 + (20 A + 5C + E ) x + (25 A) x ( x 2 + 2 x + 5)

2

ìï A + B A=1 = 2 ïï ïï 4 A + 2B + C B=1 = 6 ï = 20  C = 0 í 14 A + 5B + 2C + D ïï D=1 + 5C + E = 22 ïï20 A ïï 25 A E = =2 25 ïî

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1577


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2 x 4 + 6 x 3 + 20 x 2 + 22 x + 25 x ( x 2 + 2 x + 5)

2

46.

=

1 x x +2 + 2 + 2 x x + 2x + 5 x 2 + 2x + 5

(

)

x 3  3x 2  6x  6 ( x 2  x  5)( x 2  1)

Solution x 3 + 3x 2 + 6x + 6

( x + 1)( x + x + 5) 2

2

=

= = =

Ax + B 2

x +1

+

Cx + D 2

x + x +5

( Ax + B)( x 2 + x + 5)

(Cx + D)( x 2 + 1)

( x + 1)( x + x + 5) ( x + 1)( x + x + 5) 2

2

2

2

Ax 3 + Ax 2 + 5 Ax + Bx 2 + Bx + 5B + Cx 3 + Cx + Dx 2 + D

( x + 1)( x + x + 5) 2

2

( A + C ) x 3 + ( A + B + D) x 2 + (5 A + B + C ) x + (5B + D)

( x + 1)( x + x + 5) 2

ìï A A=1 +C =1 ïï ïï A + B B=1 + D= 3  í =6 C=0 ïïï5 A + B + C ïïî 5B + D= 6 D=1 47.

+

2

x 3 + 3x 2 + 6x + 6

( x + 1)( x + x + 5) 2

2

=

x+1 2

x +1

+

1 2

x + x +5

x3 ( x 2  3 x  2)

Solution

Use long division first: 7x + 6

( x + 1)( x + 2)

= = =

x3 7x + 6 7x + 6 = x -3+ 2 = x -3+ x 2 + 3x + 2 x + 3x + 2 x + ( 1)( x + 2)

A B + x +1 x +2 A ( x + 2)

+

B ( x + 1)

( x + 1)( x + 2) ( x + 1)( x + 2) Ax + 2 A + Bx + B

( x + 1)( x + 2) ( A + B) x + (2A + B) = ( x + 1)( x + 2) ìïï A + B = 7  A = -1, B = 8 í ïïî2 A + B = 6

x - 3 + ( x +71x)(+x6+2) = x - 3 - x +1 1 + x +8 2

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1578


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

48.

2x 3  6x 2  3x  2 x3  x2 Solution

Use long division first: 4 x 2 + 3x + 2 x 2 ( x + 1)

= = =

2x 3 + 6x 2 + 3x + 2 x3 + x2

= 2+

4 x 2 + 3x + 2 x3 + x2

= 2+

4 x 2 + 3x + 2 x 2 ( x + 1)

A B C + + x x2 x+1 Ax ( x + 1) x 2 ( x + 1)

+

B ( x + 1) x 2 ( x + 1)

+

Cx 2 x 2 ( x + 1)

Ax 2 + Ax + Bx + B + Cx 2 x 2 ( x + 1)

( A + C ) x 2 + ( A + B) x + (B) = x 2 ( x + 1) ìï A A=1 +C = 4 ïï A + B = 3  B =2 í ïï =2 B C=3 ïïî 49.

2+

4 x 2 + 3x + 2 x ( x + 1) 2

= 2+

1 2 3 + + x x2 x+1

3x 3  3x 2  6x  4 3x 3  x 2  3x  1 Solution

Use long division first:

2x 2 + 3x + 3

(3x + 1)( x 2 + 1)

=

= =

3x 3 + 3x 2 + 6x + 4 2x 2 + 3x + 3 2x 2 + 3x + 3 = 1 + = 1 + 3x 3 + x 2 + 3x + 1 3x 3 + x 2 + 3x + 1 (3x + 1)( x 2 + 1)

A Bx + C + 2 3x + 1 x +1 A ( x 2 + 1)

(Bx + C )(3 x + 1) (3x + 1)( x 2 + 1) (3x + 1)( x 2 + 1) +

Ax 2 + A + 3Bx 2 + Bx + 3Cx + C

(3x + 1)( x 2 + 1) ( A + 3B) x 2 + (B + 3C ) x + ( A + C ) = (3x + 1)( x 2 + 1)

ìï A + 3B =2 A=2 ïï B + 3C = 3  B = 0 í ïï + C=3 C=1 ïïî A

1+

2x 2 + 3x + 3

(3 x + 1)( x + 1) 2

= 1+

2 1 + 2 3x + 1 x + 1

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1579


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

50.

x 4  x 3  3x 2  x  4 ( x 2  1)2

Solution

Use long division first: x3 + x2 + x + 3

( x + 1)

2

2

=

=

x 4 + x 3 + 3x 2 + x + 4

( x + 1)

2

2

+

( x + 1)

2

51.

2

2

( x + 1)

2

2

Ax 3 + Ax + Bx 2 + B + Cx + D

( x + 1)

2

Ax 3 + Bx 2 + ( A + C ) x + (B + D)

( x + 1)

2

2

ìï A ïï ïï í ïï A ïï ïî

( x + 1)

( Ax + B)( x 2 + 1) (Cx + D)

2

=

x3 + x2 + x + 3

Ax + B Cx + D + 2 x + 1 ( x 2 + 1)2

2

=

= 1+

A=1 =1 B=1 =1  C=0 +C =1 D=2 B +D = 3 B

1+

x3 + x2 + x + 3

( x + 1)

2

2

= 1+

2 x+1 + 2 x + 1 ( x 2 + 1)2

x 3  3x 2  2x  1 x3  x2  x Solution

Use long division first: 2x 2 + x + 1

x ( x + x + 1) 2

=

= = = 2x 2 + x + 1

x ( x 2 + x + 1)

=

x 3 + 3x 2 + 2x + 1 x3 + x2 + x

= 1+

2x 2 + x + 1 x3 + x2 + x

= 1+

2x 2 + x + 1

x ( x 2 + x + 1)

A Bx + C + x x2 + x + 1 A ( x 2 + x + 1) x ( x 2 + x + 1)

+

(Bx + C ) x x ( x 2 + x + 1)

Ax 2 + Ax + A + Bx 2 + Cx x ( x 2 + x + 1)

( A + B) x 2 + ( A + C ) x + ( A) x ( x 2 + x + 1)

( A + B) x 2 + ( A + C ) x + ( A) x ( x 2 + x + 1)

ìï A + B =2 A=1 ïï +C= 1  B = 1 íA ïï C=0 =1 ïïî A

1+

2x 2 + x + 1

x ( x + x + 1) 2

= 1+

1 x + 2 x x +x+1

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1580


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

52.

x 4  x 3  3x 2  x  1 ( x 2  1)2

Solution

Use long division first: -x 3 - x 2 - x

( x + 1)

2

2

=

= =

x4 - x3 + x2 - x + 1

( x + 1)

2

2

( x + 1)

2

2

Ax + B Cx + D + 2 x + 1 ( x 2 + 1)2

( Ax + B)( x 2 + 1) (Cx + D) +

( x 2 + 1)

2

( x + 1)

2

2

Ax 3 + Ax + Bx 2 + B + Cx + D

( x + 1)

2

2

=

-x 3 - x 2 - x

= 1+

Ax 3 + Bx 2 + ( A + C ) x + (B + D)

( x + 1)

2

2

ìï A A = -1 = -1 ïï ïï B B = -1 = -1  í ïï A +C = -1 C= 0 ïï + = B D D= 1 0 ïî

1+

-x 3 - x 2 - x

( x + 1) 2

2

= 1+

= 1-

53.

-x - 1 1 + 2 2 x + 1 ( x + 1)2 x+1 2

x +1

+

1

( x + 1) 2

2

2x 4  2x 3  3x 2  1 ( x 2  x )( x 2  1)

Solution

Use long division first: 4 x 3 + x 2 + 2x - 1 x ( x - 1)( x + 1) 2

=

= =

2x 4 + 2x 3 + 3x 2 - 1

( x - x)( x + 1) 2

2

= 2+

4 x 3 + x 2 + 2x - 1 x ( x - 1)( x 2 + 1)

A B Cx + D + + 2 x x-1 x +1 A ( x - 1)( x 2 + 1)

Bx ( x 2 + 1)

(Cx + D)( x )( x - 1) x ( x - 1)( x 2 + 1) x ( x - 1)( x 2 + 1) x ( x - 1)( x 2 + 1) +

+

Ax 3 - Ax 2 + Ax - A + Bx 3 + Bx + Cx 3 - Cx 2 + Dx 2 - Dx x ( x - 1)( x 2 + 1)

( A + B + C ) x 3 + (-A - C + D) x 2 + ( A + B - D) x + (-A) = x ( x - 1)( x 2 + 1)

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1581


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

ìï A + B + C = 4 A=1 ïï ïï-A -C + D= 1 B=3  í -D = 2 C=0 ïïï A + B ïïî-A = -1 D=2 54.

2+

4 x 3 + x 2 + 2x - 1 x ( x - 1)( x + 1) 2

= 2+

1 3 2 + + 2 x x-1 x +1

x 4  x3  5x2  x  6 ( x 2  3)( x 2  1)

Solution

Use long division first: -x 3 + x 2 + x + 3

( x + 3)( x + 1) 2

2

=

= = = ìï A ïï ïíï ïï A ïï ïî

x 4 - x 3 + 5x 2 + x + 6

( x + 3)( x + 1) 2

2

= 1+

-x 3 + x 2 + x + 3

( x + 3)( x + 1) 2

2

Ax + B Cx + D + 2 x2 + 3 x +1

( Ax + B)( x 2 + 1) (Cx + D)( x 2 + 3)

( x + 3)( x + 1) 2

2

+

( x + 3)( x + 1) 2

2

Ax 3 + Ax + Bx 2 + B + Cx 3 + 3Cx + Dx 2 + 3D

( x + 3)( x + 1) 2

2

( A + C ) x 3 + (B + D) x 2 + ( A + 3C ) x + (B + 3D)

( x + 3)( x + 1) 2

A = -2 = -1 B B= 0 + D= 1  + 3C = 1 C= 1 B D= 1 + 3D = 3

2

+ C

1+

-x 3 + x 2 + x + 3

( x 2 + 3)( x 2 + 1)

= 1-

2x x+1 + x2 + 3 x2 + 1

Fix It In exercises 55 and 56, identify the step where the first error is made and fix it. 55. Decompose

2 x  29

into partial fractions.

 x  2 x  3

Solution Step 4 was incorrect. Step 4: A = 5, B = -7 Step 5:

5 7 x -2 x +3

56. Decompose

7 x 2  18 x  17

 x  3 x  2

2

into partial fractions.

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1582


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution Step 5 was incorrect. Step 5:

4 3 5 + x + 3 x - 2 ( x - 2)2

Discovery and Writing 57. Describe what is meant by partial fraction decomposition.

Solution Answers may vary. 58. How can you check your result of partial fraction decomposition?

Solution Answers may vary. 59. Explain how to use partial fraction decomposition when the denominator of a rational expression has distinct linear factors.

Solution Answers may vary. 60. Explain how to use partial fraction decomposition when the denominator of a rational expression has repeated linear factors.

Solution Answers may vary. 61. Explain how to use partial fraction decomposition when the denominator of a rational expression has a prime quadratic factor.

Solution Answers may vary. 62. Explain how to use partial fraction decomposition when the denominator of a rational expression has a repeated prime quadratic factor.

Solution Answers may vary. 63. Is the polynomial x3 + 1 prime?

Solution x 3 + 1 = ( x + 1)( x 2 - x + 1)  not prime

64. Decompose

1 into partial fractions. x 1 3

Solution

1 1 A Bx + C = = + x 3 + 1 ( x + 1)( x 2 - x + 1) x + 1 x 2 - x + 1

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1583


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

= =

A ( x 2 - x + 1)

(Bx + C)( x + 1) ( x + 1)( x - x + 1) ( x + 1)( x 2 - x + 1) 2

+

Ax 2 - Ax + A + Bx 2 + Bx + Cx + C

( x + 1)( x 2 - x + 1) ( A + B) x 2 + (-A + B + C) x + ( A + C) = ( x + 1)( x 2 - x + 1)

ìï A + B A = 31 =0 ï ïí-A + B + C = 0  B = - 1 3 2 ïïï A + C = 1 C = 3 ïî

1 - 31 x + 23 1 1 3 = = + x + 1 x2 - x + 1 x 3 + 1 ( x + 1)( x 2 - x + 1)

Critical Thinking Match the rational expression on the left with the correct partial fraction decomposition form on the right. 65.

x 2  2x  3 x ( x  4)( x 2  5)

a.

A B C D E Fx  G      x x2 x3 x4 x  4 x2  5

66.

x 2  2x  3 x 2 ( x  4)( x 2  5)

b.

A B C D Fx  G Hx  I     2  2 2 3 x x  4 ( x  4) ( x  4) x  5 ( x  5)2

c.

A B C Dx 2  Ex  F  2   x x x 4 x3  5

67.

x 2  2x  3 x 3 ( x  4)2 ( x 2  5)

68.

x 2  2x  3 x ( x  4)3 ( x 2  5)2

d.

A B C D E Fx  G      2 x x 2 x 3 x  4 ( x  4)2 x 5

69.

x 2  2x  3 x ( x  4)( x 2  5)

e.

A B Cx  D   x x  4 x2  5

f.

A B C Dx  E    x x2 x  4 x2  5

70.

4

x 2  2x  3 x 2 ( x  4)( x 3  5)

Solution 65. e 66. f 67. d 68. b 69. a 70. c

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1584


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

EXERCISES 6.7 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Graph the line: x = –4

Solution

2. Graph the line: y = –3

Solution

3. Graph the line: 3x + 2y = 8

Solution

4. If you substitute the ordered pair (0, 0) into the inequality y ≤ – 3x – 1, would the inequality be true or false?

Solution False 5. Graph the parabola: y  

1 2 x 2 2

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1585


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

6. Graph the circle: (x + 2)2 + (y + 1)2 = 9

Solution

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. The graph of Ax + By = C is a line. The graph of Ax + By ≤ C is a __________. The line is its __________.

Solution half-plane, boundary 8. The boundary of the graph Ax + By < C is __________ (included, excluded) from the graph.

Solution excluded 9. The origin __________ (is, is not) included in the graph of 3x – 4y > 4.

Solution is not 10. The origin __________ (is, is not) included in the graph of 4x + 3y ≤ 5.

Solution is Practice Graph the solution set of each inequality. 11. x  3

Solution x3

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1586


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

12. x  

5 2

Solution 5 x 2

13. y 

7 2

Solution 7 y  2

14. y > –1

Solution y > –1

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1587


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

15. 2x + 3y < 12

Solution 2x + 3y < 12

16. 4x – 3y > 6

Solution 4x – 3y > 6

17. 4x – y > 4

Solution 4x – y > 4

18. x – 2y < 5

Solution x – 2y < 5

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1588


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

19. y > 2x

Solution y > 2x

20. y < 3x

Solution y < 3x

21. y 

1 x1 2

Solution 1 y  x1 2

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1589


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

22. y 

1 x1 3

Solution

y

1 x1 3

23. 2 y  3 x  2

Solution 2 y  3x  2

24. 3 y  2 x  3

Solution 3 y  2x  3

25. y < x2

Solution y < x2

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1590


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

26. y   x 2  4

Solution y  x2  4

27. y  x

Solution

y  x

28. y  x  1

Solution

y  x1

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1591


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

29. x 2  y 2  4

Solution x2  y 2  4

30. x 2  y 2  4

Solution x2  y 2  4

Graph the solution set of each system of inequalities. If there is no solution, indicate so.

 y  3 31.   x  2 Solution  y  3   x  2

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1592


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 y  2 32.   x  0 Solution  y  2   x  0

 y  1 33.   x  2 Solution  y  1   x  2

 y  1 34.   x  1 Solution  y  1   x  1

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1593


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 y  x  2 35.   y  2 x  1 Solution  y  x  2   y  2 x  1

 y  3x  2 36.   y  2x  3 Solution  y  3x  2   y  2x  3

 x  y  2 37.   x  y  1 Solution  x  y  2   x  y  1

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1594


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3x  2 y  6 38.   x  3 y  2 Solution 3x  2 y  6   x  3 y  2

 x  2 y  3 39.  2x  4 y  8 Solution  x  2 y  3  2x  4 y  8

3x  y  1 40.   x  2 y  9 Solution 3x  y  1   x  2 y  9

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1595


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2x  3 y  6 41.  3x  2 y  6 Solution 2x  3 y  6  3x  2 y  6

4 x  2 y  6 42.  2 x  4 y  10 Solution 4 x  2 y  6  2 x  4 y  10

3  y   2 x  3 43.  3 x  2 y  2

Solution no solution 1  y  4 x  3 44.   x  4 y  4

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1596


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution no solution

 y  x2  4  45.  1 y  x  2 Solution  y  x2  4   1 y  x 2 

 y   x 2  4 46.   y   x  1 Solution 2  y   x  4   y   x  1

2  y   x  3 47.   y  1

Solution no solution

 y  x 2  4 48.   y  3 Solution no solution

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1597


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 y  x 2 49.  2  y  4  x Solution 2  y  x  2  y  4  x

2  x  y  1 50.  2  y  x  1

Solution  x 2  y  1  2  y  x  1

2 x  y  0  51.  x  2 y  10 y  0 

Solution 2 x  y  0   x  2 y  10 y  0 

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1598


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

x  2 y  0  52.  x  y  2 x  0 

Solution x  2 y  0  x  y  2 x  0 

3 x  2 y  5  53. 2 x  y  8 x  5 

Solution 3 x  2 y  5  2 x  y  8 x  5 

2 x  3 y  6  54.  x  y  4  y  4 

Solution 2 x  3 y  6  x  y  4  y  4 

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1599


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

x  y  4  x  y  4 55.  x  0 y  0 

Solution x  y  4  x  y  4  x  0, y  0 

2 x  3 y  12  2 x  3 y  6 56.  x  0 y  4 

Solution 2 x  3 y  12  2 x  3 y  6  x  0, y  4 

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1600


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

3 x  2 y  6   x  2 y  10 57.  x  0 y  0 

Solution 3 x  2 y  6   x  2 y  10  x  0, y  0 

3 x  2 y  12  5 x  y  15 58.  x  0 y  4 

Solution 3 x  2 y  12  5 x  y  15  x  0, y  4 

Fix It In exercises 59 and 60, identify the step where the first error is made and fix it. 59. Solve the linear inequality x – 2y > –4 by graphing.

Solution Step 3 was incorrect. Step 3:

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1601


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 y  1 by graphing. 60. Solve the system of inequalities  2 2  x  y  9 Solution Step 2 was incorrect. Step 2:

Step 3:

Applications 61. Building furniture A furniture maker has 60 hours of labor to make sofas (s) and loveseats (l). It takes 6 hours to make a sofa and 4 hours to make a loveseat. Write a system of inequalities that provides the restrictions on the variables. (Hint: Remember that a negative number of pieces of furniture cannot be made.)

Solution 6s  4l  60  s  0, l  0

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1602


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

62. Installing video Each week, Prime Time Video and Audio has 90 hours of labor to install satellite dishes (d) and home theater systems (t). On average, it takes 5 hours to install a satellite dish and 6 hours to install a home theater system. Write a system of inequalities that provides the restrictions on the variables. (Hint: Remember that a negative number of units cannot be installed.)

Solution 5d  6t  90  d  0, t  0 63. Fundraising A college club is selling baskets of fruit and blocks of cheese to raise at least $600 for a local children’s hospital. a. If the profit for selling a basket of fruit is $5 and for selling a block of cheese is $6, write a system of inequalities that describes when x boxes of fruit and y blocks of cheese will cause the fundraising goal to be reached. (Hint: Remember that a negative number of baskets of fruit or blocks of cheese cannot be sold.) b. Graph the system of inequalities.

Solution 5x  6 y  600 a.   x  0, y  0 b.

64. Fundraising A group of international exchange students are selling cookie dough and pizza kits to raise at least $3600 for an upcoming trip. a. If the profit for selling a tub of cookie dough is $6 and for selling a pizza kit is $8, write a system of inequalities that describes when x tubs of cookie dough and y pizza kits will cause the fundraising goal to be reached. (Hint: Remember that a negative number of cookie dough or pizza kits cannot be sold.) b. Graph the system of inequalities.

Solution 6 x  8 y  3600 a.   x  0, y  0

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1603


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

b.

Discovery and Writing 65. Explain how to graph an inequality in two variables.

Solution Answers may vary. 66. When graphing an inequality in two variables, explain how you decide which side to shade.

Solution Answers may vary. 67. Explain how you determine whether the graph of a linear inequality in two variables is a dashed or solid line.

Solution Answers may vary. 68. What is a system of inequalities in two variables?

Solution Answers may vary. 69. Explain how to graph the solution set of a system of inequalities in two variables.

Solution Answers may vary. 70. Explain why it is possible for a system of inequalities in two variables to have no solution.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 71. The origin (0, 0) is always used as a test point when graphing a linear inequality in two variables.

Solution

False. If 0, 0 is on the boundary, it cannot be used as the test point. 72. The solution of a linear inequality in two variables is always a half-plane.

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1604


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution True. 73. A system of inequalities in two variables always has a solution.

Solution False. A system of inequalities can have no solution. 74. If the inequality in two variables contains a > or < symbol, a dashed curve is drawn.

Solution True. 75. If the inequality in two variables contains a ≤ or ≥ symbol, a solid curve is drawn.

Solution True. 76. An inequality representing the graph shown is 3x + 5y ≥ 15.

Solution False. The inequality is 3x + 5y ൑ 15.

 x  5 . 77. A system of inequalities that represents the graph shown is   y  2

Solution True.

 x 2  y 2  36 . 78. A system of inequalities that represents the graph shown is  2 2  x  y  25

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1605


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 2 2  x  y  36 . False. The system is  2  x  y  25

EXERCISES 6.8 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

4 x  5 y  20  Graph the system of inequalities  x  0 and identify the corner points of the shaded y  0  region.

Solution 4 x  5 y  20  x  0 y  0  4 x  5 y  20 5 y  4 x  20 4 x4 5 Corner points: 4 x  5 y  20   x  0 y 

4  0   5 y  20 5 y  20 y 4

0, 4 

4 x  5 y  20   y  0

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1606


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

4 x  5  0   20 4 x  20 x 5

5, 0 

 x  0   y  0 0, 0

4 x  5 y  20  and identify the corner points of the shaded 2. Graph the system of inequalities  x  1 y  1 

region.

Solution 4 x  5 y  20  x  1 y  1  4 x  5 y  20 5 y  4 x  20 4 y  x4 5 Corner points: 4 x  5 y  20   x  1

4  1  5 y  20 5 y  16

16 5  16   1,   5

y 

4 x  5 y  20   y  1

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1607


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

4 x  5  1  20 4 x  15 15 x 4  15   , 1 4   x  1   y  1  1, 1

3. Evaluate the expression 10x + 20y at the following points: (5, 3), (10, 6), and (15, 2).

Solution

10  5   20  3   50  60  110

10  10   20  6   100  120  220 10  15   20  2   150  40  190

4. Stefan has $40 to spend at an amusement park. If tickets for the amusement park rides cost $1.50 each and tickets to play games cost $2 each, write an inequality that models Stefan’s budget restriction. Let x represent the number of rides and y represents the number of games.

Solution 1.5 x  2 y  40 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 5. In a linear programming problem, the inequalities are called __________.

Solution constraints 6. Ordered pairs that satisfy the constraints of a linear programming problem are called __________ solutions.

Solution feasible

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1608


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

7. The function to be maximized (or minimized) in a linear programming problem is called the __________ function.

Solution objective 8. The objective function of a linear programming problem attains a maximum (or minimum), subject to the constraints, at a __________ or along an __________ of the feasibility region.

Solution corner, edge Practice Maximize P subject to the following constraints. 9.

P  2x  3 y x  0  y  0 x  y  4  Solution

Point

P  2x  3 y

0, 0 0, 4  4, 0

 2  0  3  0  0  2  0  3  4  12  2  4   3  0  8

Max: P  12 at  0,4 10. P  3 x  2 y x  0  y  0 x  y  4 

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1609


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

Point

P  3x  2 y

0, 0 0, 4  4, 0

 3  0  2  0  0  3  0  2  4   8  3  4  2  0  12

Max: P  12 at  4,0 11. P  y 

1 x 2

x  0  y  0  2 y  x  1  y  2 x  2  Solution

Point

0, 0 0,   ,   1, 0

P  y  21 x

1 2

5 3

4 3

Max: P  136 at  53 , 43 

 0  21  0  0  21  21  0  21

 43  21  53   136

 0  21  1  21

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1610


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

12. P  4 y  x x  2  y  0  x  y  1 2 y  x  1 

Solution

Point

P  4y  x

 1, 0  2, 0  2,   , 

 4  0  1  1  4  0  2  2  4  32   2  4

3 2

1 3

2 3

Max: P  4 at  2, 32 

 4  23   31  73

13. P  2x  y y  0  y  x  2  2x  3 y  6 3x  y  3 

Solution

Point

 2, 0  1, 0  ,  0, 2 3 7

12 7

P  2x  y

 2  2  0  4  2  1  0  2  2  73   127  187  2  0  2  2

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1611


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Max: P  187 at  73 , 127  14. P  x  2 y x  y  5  y  3  x  2 x  0   y  0

Solution

Point

P  x  2y

0, 0  2, 0  2, 3 0, 3

 0  2  0  0  2  2  0  2  2  2  3  4  0  2  3  6

Max: P  2 at  2, 0 15. P  3 x  2 y x  1   x  1  y  x  1 x  y  1 

Solution

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1612


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

P  3x  2 y

 1, 0  1, 2  1, 0

 3  1  2  0  3  3  1  2  0  3

 1,  2

 3  1  2  2  1

 3  1  2  2  1

Max: P  3 at  1, 0 16. P  x  y 5 x  4 y  20  y  5  x  0 y  0 

Solution

Point

0, 0  4, 0 0, 5

Pxy

 00  0  40  4  0  5  5

Min: P  4 at  4, 0 Minimize P subject to the following constraints. 17. P  5x  12 y x  0  y  0 x  y  4 

Solution

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1613


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

0, 0 0, 4  4, 0

P  5x  12 y

 5  0  12  0  0  5  0  12  4  48  5  4  12  0  20

Min: P  0 at  0, 0 18. P  3 x  6 y x  0  y  0 x  y  4 

Solution

Point

0, 0 0, 4  4, 0

P  3x  6 y

 3  0   6  0  0  3  0  6  4  24  3  4  6  0  12

Min: P  0 at  0, 0 19. P  3 y  x x  0  y  0  2 y  x  1  y  2x  2 

Solution

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1614


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

P  3y  x

Point

0, 0 0,   ,   1, 0

 3  0  0  0  3  21   0  32

1 2

5 3

 3  43   53  173

4 3

 3  0  1  1

Min: P  0 at  0, 0 20. P  5 y  x x  0  y  0  x  y  1 2 y  x  1 

Solution

Point

 1, 0  2, 0  2,   ,  3 2

1 3

2 3

Min: P  1 at  1, 0

P  5y  x

 5  0  1  1  5  0  2  2  5  32   2  192

 5  23   31  113

21. P  6 x  2 y y  0  y  x  2  2x  3 y  6 3x  y  3 

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1615


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

Point

P  6x  2 y

 2, 0  1, 0  ,  0, 2 3 7

 6  2  2  0  12  6  1  2  0  6  6  73   2  127   6

12 7

 6  0  2  2  4

Min: P  12 at  2, 0 22. P  2 y  x x  0  y  0  x  y  5 x  2 y  2 

Solution

Point

P  2y  x

0, 1  2, 0 5, 0 0, 5

 2  1  0  2  2  0  2  2  2  0  5  5  2  5  0  10

Min: P  5 at  5,0

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1616


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

23. P  2 x  2 y x  1   x  1  y  x  1 x  y  1 

Solution

Point

P  2x  2 y

 1, 0  1, 2  1, 0

 2  1  2  0  2  2  1  2  0  2

 1,  2

 2  1  2  2  2

 2  1  2  2  2

Min: P  2 on the edge joining

 1, 2 and  1, 0

24. P  y  2 x x  2 y  4  2 x  y  4  x  2 y  2 2 x  y  2 

Solution

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1617


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

P  y  2x

2 3

2 3

 ,   2, 0  ,  0, 2

 23  2  23    23

4 3

4 3

 43  2  43    43

 0  2  2  4  2  2  0  2

Min: P  4 at  2, 0 Fix It In exercises 25 and 26, identify the step where the first error is made and fix it. x  0  25. Maximize P = 3x + 6y subject to the following constraints  y  0 . x  y  5 

Solution Step 4 was incorrect.

Step 4: Maximum is 30 at 0, 5

x  1  26. Minimize P = 5x – 6y subject to the following constraints  y  1 . x  y  4 

Solution Step 2 was incorrect.

     

Step 2: 1, 1 , 1, 3 , 3, 1 are the corner points. Step 3: Evaluate P at the corner points.

P at  1, 1 is  1; P at  1, 3 is  13; P at  3, 1 is 9

 

Step 4: Minimum is 13 at 1, 3

Applications Write the objective function and the inequalities that describe the constraints in each problem. Graph the feasibility region, showing the corner points. Then find the maximum or minimum value of the objective function. 27. Making furniture Two woodworkers, Chase and Devin, get $100 for making a table and $80 for making a chair. On average, Chase must work 3 hours and Devin 2 hours to make a chair. Chase must work 2 hours and Devin 6 hours to make a table. If neither wishes to work more than 42 hours per week, how many tables and how many chairs should they make each week to maximize their income? Find the maximum income.

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1618


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Table

Chair

Time Available

Devin’s Time (hr)

6

2

42

Chase’s Time (hr)

2

3

42

Income ($)

100

80

Solution Let x = # tables and y = # chairs. Maximize P  100 x  80 y

2 x  3 y  42  subject to 6 x  2 y  42  x  0, y  0 

Point

 7, 0  3, 12 0, 14

P  100 x  80 y

 100  7   80  0  700  100  3  80  12  1260  100  0  80  14  1120

They should make 3 tables and 12 chairs, for a maximum profit of $1260. 28. Making crafts Two artists, Nina and Rob, make yard ornaments. They get $80 for each wooden snowman they make and $64 for each wooden Santa Claus. On average, Nina must work 4 hours and Rob 2 hours to make a snowman. Nina must work 3 hours and Rob 4 hours to make a Santa Claus. If neither wishes to work more than 20 hours per week, how many of each ornament should they make each week to maximize their income? Find the maximum income.

Snowman

Santa Claus

Time Available

Rob’s Time (hr)

2

4

20

Nina’s Time (hr)

4

3

20

Income ($)

80

64

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1619


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution Let x = # snowman and y = # Santa Claus ornaments. Maximize P  80 x  64 y

4 x  3 y  20  subject to 2 x  4 y  20  x  0, y  0 

Point

5, 0  2, 4 0, 5

P  80 x  64 y

 80  5  64  0  400  80  2  64  4  416  80  0  64  5  320

They should make 2 snowman and 4 Santa Claus ornaments, for a maximum profit of $416. 29. Inventories An electronics store manager stocks from 20 to 30 IBM-compatible computers and from 30 to 50 Apple computers. There is room in the store to stock up to 60 computers. The manager receives a commission of $50 on the sale of each IBM-compatible computer and $40 on the sale of each Apple computer. If the manager can sell all of the computers, how many should she stock to maximize her commissions? Find the maximum commission.

Inventory

IBM

Apple

Minimum

20

30

Maximum

30

50

Commission

$50

$40

Solution Let x = # IBM and y = # Apple. Maximize P  50 x  40 y

 x  y  60  subject to 20  x  30 30  y  50 

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1620


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

 20, 30  30, 30  20, 40

P  50 x  40 y

 50  20  40  30  2200  50  30  40  30  2700  50  20  40  40  2600

She should stock 30 IBM and 30 Apple computers, for a maximum commission of $2700. 30. Diet problems A diet requires at least 16 units of vitamin C and at least 34 units of vitamin B complex. Two food supplements are available that provide these nutrients in the amounts and costs shown in the table. How much of each should be used to minimize the cost?

Supplement

Vitamin C

Vitamin B

Cost

A

3 units/g

2 units/g

3¢/g

B

2 units/g

6 units/g

4¢/g

Solution Let x = grams of A and y = grams of B. Maximize P  3 x  4 y

3 x  2 y  16  subject to 2 x  6 y  34  x  0, y  0 

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1621


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

0, 8  2, 5  17, 0

P  3x  4 y

 3  0  4  8  32  3  2  4  5  26  3  17  4  0  51

2 grams of A and 5 grams of B should be used, for a minimum cost of 26¢. 31. Production A company manufactures two types of digital tablets for children. These are designated as Tablet A and Tablet B. Each requires the use of the electronics, assembly, and finishing departments of a factory, according to the following schedule:

Hours for Digital Tablet A

Hours for Digital Tablet B

Hours Available per Week

Electronics

3

4

180

Assembly

2

3

120

Finishing

2

1

60

Each Tablet A has a profit of $40, and each Tablet B has a profit of $32. How many of each should be manufactured weekly to maximize profit? Find the maximum profit.

Solution Let x = # DVRs and y = # TVs. Maximize P  40 x  32 y 3 x  4 y  180  2 x  3 y  120 subject to  2 x  y  60  x  0, y  0 

Point

0, 0 0, 40  15, 30  30, 0

P  40 x  32 y

 40  0  32  0  0  40  0  32  40  1280  40  15  32  30  1560  40  30  32  0  1200

15 DVRs and 30 TVs should be made, for a maximum profit of $1560. © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1622


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

32. Production problems A company manufactures one type of computer chip that runs at 3.5 GHz and another that runs at 4.2 GHz. The company can make a maximum of 50 fast chips per day and a maximum of 100 slow chips per day. It takes 6 hours to make a fast chip and 3 hours to make a slow chip, and the company’s employees can provide up to 360 hours of labor per day. If the company makes a profit of $20 on each 4.2-GHz chip and $27 on each 3.5-GHz chip, how many of each type should be manufactured to earn the maximum profit?

Solution Let x = # slow chips (2.0) and y = # fast chips (2.8). Maximize P  27 x  20 y

 y  50, x  100  subject to 3 x  6 y  360  x  0, y  0 

Point

0, 0  100, 0  100, 10  20, 50 0, 50

P  27 x  20 y

 27  0  20  0  0  27  100  20  0  2700  27  100  20  10  2900  27  20  20  50  1540  27  0  20  50  1000

100 slow chips and 10 fast chips should be made, for a maximum profit of $2900. 33. Financial planning A stockbroker has $200,000 to invest in stocks and bonds. She wants to invest at least $100,000 in stocks and at least $50,000 in bonds. If stocks have an annual yield of 9% and bonds have an annual yield of 7%, how much should she invest in each to maximize her income? Find the maximum return.

Solution Let x = $ in stocks and y = $ in bonds. Maximize P = 0.09x + 0.07y

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1623


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 x  y  200000  subject to  x  100000  y  50000 

Point

 100000, 50000  150000, 50000  100000, 100000

P  0.09x  0.07 y

 12500  17000  16000

She should invest $150,000 in stocks and $50,000 in bonds, for a maximum return of $17,000. 34. Production A small country exports soybeans and flowers. Soybeans require 8 workers per acre, flowers require 12 workers per acre, and 100,000 workers are available. Government contracts require that there be at least 3 times as many acres of soybeans as flowers planted. It costs $250 per acre to plant soybeans and $300 per acre to plant flowers, and there is a budget of $3 million. If the profit from soybeans is $1600 per acre and the profit from flowers is $2000 per acre, how many acres of each crop should be planted to maximize profit? Find the maximum profit.

Solution Let x = acres of beans and y = acres of flowers. Maximize P = 1600x + 2000y 8 x  12 y  100, 000  x  3 y  0 subject to  250 x  300 y  3, 000, 000  x  0, y  0 

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1624


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

P  1600 x  2000 y

Point

0, 0

0

 12000, 0

 19, 200,000

 10000,  ,

5000 3

 19, 333, 333

25000 9

 18,888,889

25000 3

 

The country should plant 10,000 acres of beans and 1667 acres of flowers, for a maximum profit of $19,333,333. 35. Band trips A college band trip will require renting buses and trucks to transport no fewer than 100 students and 18 or more large instruments. Each bus can accommodate 40 students plus three large instruments; it costs $350 to rent. Each truck can accommodate 10 students plus 6 large instruments and costs $200 to rent. How many of each type of vehicle should be rented for the cost to be minimum? Find the minimum cost.

Solution Let x = # buses and y = # trucks. Minimize P  350 x  200 y

40 x  10 y  100  subject to 3 x  6 y  18  x  0, y  0 

Point

0, 10  2, 2 6, 0

P  350 x  200 y

 350  0  200  10  2000  350  2  200  2  1100  350  6  200  0  2100

2 buses and 2 trucks should be rented, for a minimum cost of $1100. 36. Making ice cream An ice cream store sells two new flavors: Fantasy and Excess. Each barrel of Fantasy requires 4 pounds of nuts and 3 pounds of chocolate and has a profit of $500. Each barrel of Excess requires 4 pounds of nuts and 2 pounds of chocolate and has a profit of $400. There are 16 pounds of nuts and 18 pounds of chocolate in stock, and the owner does not want to buy more for this batch. How many barrels of each should be made for a maximum profit? Find the maximum profit.

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1625


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution Let x = barrels of Fantasy and y = barrels of Excess. Maximize P  500 x  400 y

4 x  4 y  16  subject to 3 x  2 y  18  x  0, y  0  Point

0, 0 0, 4  4, 0

P  500 x  400 y

 500  0  400  0  0  500  0  400  4  1600  500  4  400  0  2000

4 barrels of Fantasy and 0 barrels of Excess should be made, for a maximum profit of $2000.

Discovery and Writing 37. Describe what linear programming is and some of the types of problems it can be used to solve.

Solution Answers may vary. 38. Explain what an objective function is in a linear programming problem.

Solution Answers may vary. 39. In a linear programming problem, describe what constraints are and how they are represented.

Solution Answers may vary. 40. Describe a strategy that can be used to solve a linear programming problem.

Solution Answers may vary. 41. Does the objective function attain a maximum at the corners of a region defined by following nonlinear inequalities? Attempt to maximize P(x) = x + y on the region and write a paragraph on your findings. x  0  y  0  y  4  x2 

Solution Answers may vary.

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1626


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

42. Attempt to minimize the objective function of Exercise 35.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 43. An objective function always has a maximum or minimum.

Solution False. There must be constraints that describe a bounded region. 44. A system of linear equations is used to write constraints.

Solution False. A system of linear inequalities is used to write the constraints. 45. The minimum value of objective function occurs at exactly one point.

Solution False. The minimum value can occur at more than one point. 46. If the feasibility region is unbounded, then it is possible that no maximum value of the objective function exists.

Solution True.

CHAPTER REVIEW SOLUTIONS Exercises Solve each system of linear equations in two variables by graphing. 1.

2 x  y  1   x  y  7 Solution 2 x  y  1   x  y  7

solution: (2, 5)

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1627


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2.

5 x  2 y  1  2 x  y  5 Solution 5 x  2 y  1  2 x  y  5

solution: (–1, 3) 3.

 y  5 x  7   x  y  7 Solution  y  5 x  7   x  y  7

solution: (0, 7) 4.

3 x  2 y  6   3 y   x  3 2  Solution

3 x  2 y  6   3 y   x  3 2 

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1628


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

infinitely many solutions dependent equations 5.

4 x  y  4   y  4  x  2  Solution 4 x  y  4   y  4  x  2 

no solutions inconsistent system

Solve each system of linear equations in two variables by substitution. If the system has no solution, write no solution; inconsistent system. If the system has an infinite number of solutions, write dependent equations and provide a general solution. 6.

2 y  x  0   x  y  3 Solution ìï (1) 2 y + x = 0 ï í ïï (2) x = y + 3 ïî Substitute x = y + 3 into (1):

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1629


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2y + x = 0 2 y + ( y + 3) = 0 3y + 3 = 0 3 y = -3 y =- 1 Substitute and solve for x: x = y +3 x = -1 + 3 = 2 x = 2, y = -1

7.

2 x  y  3   x  y  3 Solution ìï(1) 2 x + y = -3 ï í ïï(2) x - y = 3 ïî Substitute x = y + 3 into (1):

2 x + y = -3 2 ( y + 3) + y = -3 2 y + 6 + y = -3 3 y = -9 y = -3 Substitute and solve for x: x = y +3 x = -3 + 3 = 0 x = 0, y = -3

8.

x  y x  y  1  3  2  y  3x  2  Solution ìï ïï (1) x + y + x - y = 1 ïí 2 3 ïï ïïî (2) y = 3 x - 2 Substitute y = 3x – 2 into (1): x + 3 x - 2 x - (3 x - 2) + =1 2 3

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1630


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

4 x - 2 -2 x + 2 + =1 2 3 3 (4 x - 2) + 2 (-2 x + 2) = 6 12 x - 6 - 4 x + 4 = 6 8x = 8 x=1

Substitute and solve for y: y = 3x - 2 y = 3 (1) - 2 = 1 x = 1, y = 1 9.

 y  3 x  4  9 x  3 y  12 Solution ìï(1) y = 3 x - 4 ï í ïï(2) 9 x - 3 y = 12 ïî Substitute y = 3x – 4 into (2):

9 x - 3 y = 12 9 x - 3 (3 x - 4) = 12 9 x - 9 x + 12 = 12 0=0 Dependent equations General solution: (x, 3x – 4)

 3 x   y  3 10.  2 2 x  3 y  4  Solution ìï(1) x = - 3 y + 3 ï 2 í ïï(2) 2 x + 3 y = 4 ïî Substitute x = - 32 y + 3 into (2):

2x + 3 y = 4 2 (- y + 3) + 3 y = 4 3 2

-3 y + 6 + 3 y = 4 6¹4 Inconsistent system  No solution

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1631


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solve each system of linear equations in three variables by elimination, if possible. If the system has no solution, write no solution; inconsistent system. If the system has an infinite number of solutions, write dependent equations and provide a general solution by elimination.

 x  5 y  7 11.  3 x  y  7 Solution x + 5y = 7 3 x + y = -7  ´ (-5)

x + 5y = 7 -15 x - 5 y = 35 -14 x x

= 42 = -3

x + 5y = 7 -3 + 5 y = 7 5 y = 10 y= 2

Solution: x = -3, y = 2

2 x  3 y  11 12.  3 x  7 y  41 Solution 6 x + 9 y = 33

2 x + 3 y = 11

Solution:

3 x - 7 y = -41  ´ (-2) -6 x + 14 y = 82

2 x + 3(5) = 11

x = -2, y = 5

23 y = 115

2 x = -4

y=

5

x = -2

2( x + y ) - x = 0  x + 2 y = 0  ´ (-3) 3( x + y ) + 2 y = 1  3 x + 5 y = 1 

- 3x - 6 y = 0 3x + 5 y = 0

2x + 3 y =

11  ´ (3)

2  x  y   x  0 13.  3  x  y   2 y  1

Solution

-y = 1 y = -1

x + 2y = 0 x + 2(-1) = 0 x=2

Solution: x= 2 y =- 1

8 x  12 y  24 14.  2 x  3 y  4 Solution 8 x + 12 y = 24

2 x + 3 y = 4  ´ (-4)

8 x + 12 y = 24 -8 x - 12 y =- 16 0

¹

8  Inconsistent system: No solution

3 x  y  4 15.  9 x  3 y  12

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1632


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution 3 x - y = 4  ´ (-3)

-9 x + 3 y =- 12

9 x - 3 y = 12

9x - 3 y =

12

3x - y = 4 3x - 4 = y

0=

0

Dependent equations, General Solution:

( x , 3 x - 4) Solve each system of linear equations by any method. 3 x  2 y  z  2  16.  x  y  z  0 2 x  3 y  z  1 

Solution (1) 3 x + 2 y - z = 2

Add (1) and (2):

Add equations (1) and - (3):

(2) x+ y -z =0 (3) 2 x + 3 y - z = 1

(1) 3 x + 2 y - z = 2 -(2) -x - y + z = 0

(1) 3x + 2 y - z = 2 -(3) -2 x - 3 y + z = -1

(4)

2x + y

(5)

=2

x- y

=

1

Solve the system of two equations and two unknowns formed by equations (4) and (5): 2x + y = 2 x- y=1 3x x

=3 =1

2x + y = 2 2(1) + y = 2 y =0

x+ y-z = 0 1+0- z = 0

Solution: x = 1, y = 0, z = 1

- z = -1 z= 1

5 x  y  z  3  17. 3 x  y  2z  2 x  y  2 

Solution (1) 5 x - y + z = 3

Add (1) and (2):

Add equations (1) and (3):

(2) 3 x + y + 2z = 2 =2 (3) x+y

(1) 5 x - y + z = 3 (2) 3 x + y + 2z = 2

(1) (3)

5x - y + z = 3 x+y =2

(4) 8 x

(5)

6x

+ 3z = 5

+z =5

Solve the system of two equations and two unknowns formed by equations (4) and (5): 8 x + 3z = 5 6 x + z = 5  ´ (-3)

8 x + 3z = 5 -18 x - 3z = -15 -10 x x

Solution:

= -10 = 1

6x + z = 5 6(1) + z = 5 z = -1

x+y =2 1+ y = 2 y=1

x = 1, y = 1, z = -1

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1633


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

2 x  y  z  1  18.  x  y  2z  3 x  y  z  1 

Solution (1) 2 x - y + z = 1

(2) (3)

x - y + 2z = 3 x-y + z =1

Add (1) and - (2):

(1) 2 x - y + z = 1 - (2) - x + y - 2z = -3 (4)

- z = -2

x

Add equation ( 1) and - (3) :

(1) 2x - y + z = 1 - (3) -x + y - z = -1 = 0 (5) x

Solve the system of two equations and two unknowns formed by equations (4) and (5):

x - z = -2 0 - z = -2 z=

2

x- y +z =

1

Solution:

x = 0, y = 1, z = 2

0- y +2 = 1 - y = -1 y= 1

19. Department store order The buyer for a large department store must order 40 coats, some faux fur and some leather. He is unsure of the expected sales. He can buy 25 fur coats and the rest leather for $9300, or 10 fur coats and the rest leather for $12,600. How much does he pay if he decides to split the order evenly?

Solution ìï (1) 25 x + 15 y = 9300 Let x = cost of fake fur and let y = cost of leather. Then ïí ïïî(2) 10 x + 30 y = 12600

25 x + 15 y = 9300  ´ (-2) 10 x + 30 y = 12600

50 x  30 y  18600 10 x  30 y  12600 40 x x

25 x  15 y  9300 25(150)  15 y  9300 15 y  5550 y  370

  6000 150 

The fake fur coats cost $150 while the leather coats cost $370. The cost will be $10,400. 20. Ticket sales Adult tickets for the championship game are usually $5, but on Seniors’ Day, seniors paid $4. Children’s tickets were $2.50. Sales of 1800 tickets totaled $7425, and children and seniors accounted for one-half of the tickets sold. How many of each were sold?

Solution Let x = # adult tickets, y = # senior tickets and z = # children tickets.

(1) x + y + z = 1800 (2) 5 x + 4 y + 2.5z = 7425 (3) y + z = 900

Add - 4 (1) and (2):

Add equation (1) and - (3):

-4(1) -4 x - 4 y - 4 z = -7200 5 x + 4 y + 2.5z = 7425 (2)

(1) -(3)

(4)

x

- 1.5z =

225

(5)

x + y + z = 1800 - y - z = - 900 x

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=

900

1634


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solve the system of two equations and two unknowns formed by equations (4) and (5): 225 225

y + z = 900 y + 450 = 900

- 1.5z = -675 z = 450

y = 450

x - 1.5z = 900 - 1.5z =

There were 900 adult tickets, 450 senior tickets, and 450 children's tickets sold.

Use matrices to solve each system of linear equations. Write the solution as a ordered pair, triple, or quadruple. If the system has no solution, write no solution; inconsistent system. If the system has an infinite number of solutions, write dependent equations and provide a general solution.

2 x  5 y  7 21.  3 x  y  2 Solution é2 5 7 ù é 1 -6 -5ù é ù é ù ê ú ê ú  ê 1 -6 -5ú  ê 1 -6 -5ú  ê 3 - 1 2ú  ê 3 - 1 ú ê ú ê 2û 1 1úû ë û ë ë0 17 17 û ë0 1 - R1 + R2  R1 - 3R1 + R2  R2 R  R2 17 2 é 1 0 1ù ê ú ê0 1 1ú Solution: ë û 6R2 + R1  R1

x = 1, y = 1

3 x  y  4 22.  6 x  2 y  8 Solution é 3 - 1 -4 ù é ù  3 x - y = -4 ê ú  ê 3 - 1 -4 ú ê-6 2 ê0 0 8ú 0ú 3x + 4 = y ë û ë û 2R1 + R2  R2

Dependent equations General Solution: (x, 3x + 4)

x  3 y  z  8  23. 2 x  y  2z  11  x  y  5 z  8 

Solution é 1 3 -1 8ùú ê ê2 1 -2 11ú  ê ú ê 1 -1 5 -8ú ë û

é1 3 -1 8ùú ê ê0 -5 0 -5ú  ê ú ê0 -4 6 -16ú ë û - 2R1 + R2  R2

é1 3 -1 8ùú ê ê0 1 0 1ú  ê ú ê0 -4 6 -16ú ë û - 51 R2  R2

- R1 + R3  R3

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1635


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é 1 0 -1 5ùú ê ê0 1 0 1ú  ê ú ê0 0 6 -12ú ë û - 3R2 + R1  R1

é1 0 0 3ùú ê ê0 1 0 1ú Solution: x = 3, y = 1, z = -2 ê ú ê0 0 1 -2ú ë û 1 R R R +  1 1 6 3

4R2 + R3  R3

1 6

R3  R3

x  3 y  z  3  24. 2 x  y  z  11 3 x  2 y  3z  2 

Solution é1 3 1 3ùú ê ê2 -1 1 -11ú  ê ú ê3 2 3 2ú ë û

é1 3 1 3ùú ê ê0 -7 -1 -17ú  ê ú ê0 -7 0 -7ú ë û - 2R1 + R2  R2

é1 3 1 3ùú ê ê0 1 0 1ú  ê ú ê0 -7 -1 -17ú ë û R2  - 71 R3

- 3R1 + R3  R3 é1 0 é 1 0 0 -10ù 1 0ùú ê ê ú ê0 1 0 1ú  ê0 1 0 1ú ê ú ê ú ê0 0 -1 -10ú ê0 0 1 10ú ë û ë û -3R2 + R1  R1 R3 + R1  R1

7R2 + R3  R3

Solution:

x = -10, y = 1, z = 10

- R3  R3

x  y  z  4  25. 3 x  2 y  2z  3 4 x  y  z  0 

Solution é1 1 1 4ùú ê ê 3 -2 -2 -3ú  ê ú ê4 -1 -1 0ú ë û

é1 é1 1 1 4ùú 1 1 4ùú ê ê ê0 -5 -5 -15ú  ê0 -5 -5 -15ú ê ú ê ú ê0 -5 -5 -16ú ê0 0 0 -1ú ë û ë û - 3R1 + R2  R2 - R2 + R3  R3 - 4R1 + R3  R3

The last row indicates 0x + 0y + 0z = –1. This is impossible.  no solution

w  x  y  z  4  2w  x  2 y  3z  8 26.  w  2 x  3 y  z  4 3w  x  2 y  3z  9 

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1636


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 1 1 1 -1 4ùú ê ê 2 -1 2 3 -8ú ê ú  ê-1 2 -3 1 4ú ê ú 1 2 -3 9ú êë 3 û

é1 1 1 -1 4 ùú ê ê0 -3 0 5 -16ú ê ú  ê0 3 -2 0 8 ú ê ú êë0 -2 -1 0 -3 úû - 2R1 + R2  R2 R1 + R3  R3

é1 1 1 -1 4 ùú ê ê0 1 -3 0 5 ú ê ú ê0 3 -2 0 8 ú ê ú 0 5 -16ú êë0 -3 û R3 + R4  R2 R2  R4

- 3R1 + R3  R4 é 1 0 4 -1 -1ù ê ú ê0 1 -3 0 5ú ê ú  ê0 0 7 0 -7ú ê ú êë0 0 -9 5 -1úû - R2 + R1  R1

é 1 0 4 -1 -1ù ê ú ê0 1 -3 0 5 ú ê ú  ê0 0 1 0 -1ú ê ú êë0 0 -9 5 -1úû 1 R  R3 7 3

- 3R2 + R3  R3 3R2 + R4  R4

é 1 0 0 -1 3 ù ê ú ê0 1 0 0 2 ú ê ú ê0 0 1 0 -1 ú ê ú êë0 0 0 5 -10úû - 4R3 + R1  R1 3R3 + R2  R2 9R3 + R4  R4

é 1 0 0 -1 3ùú ê ê0 1 0 0 2ú ú  ê ê0 0 1 0 -1ú ê ú êë0 0 0 1 -2úû 1 R  R4 5 4

é1 0 0 0 1ùú ê ê0 1 0 0 2ú ê ú ê0 0 1 0 -1ú ê ú êë0 0 0 1 -2úû R4 + R1  R1

Solution: (1, 2, - 1, - 2)

Solve the matrix equation for x and y. 1  27.  x 0 

4   1   2    4 x  7   x  4

x  2 y 

Solution -4 = x , x = -4, 0 = x + 4, x + 7 = y  x = -4, y = 3

Perform the matrix operations, if possible.

3 2 1  2 1 3 28.    3 2 1  1 2 1  Solution é3 2 1ù é-2 1 3ùú éê 1 3 4ùú ê ú+ê = ê3 2 1ú ê 1 -2 1ú ê4 0 2ú ë û ë û ë û  2 3 5  0  2 1      29.  1 2 4    3 4 2  2 1 2   6  4 1     

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1637


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é2 3 5ùú éê0 -2 1ùú éê 2 5 4ùú ê ê 1 -2 ú ê ú ê 4 - 3 4 -2 = -2 -6 6ú ê ú ê ú ê ú êë2 1 -2úû êë6 -4 1úû êë-4 5 -3úû

 1 2  2 3 30.     3 1   1 2 Solution é 1 -2ù é 2 3ù é 4 -1ù ê úê ú=ê ú ê-3 1úû êë-1 2úû êë-7 -7úû ë  2 1  2 3 5    31.    1 2    1 2 3    2 3   

Solution

é 2 1ù é-2 ú é-17 3 5ùú ê 19ùú ê ê -1 2ú = ê ê 1 -2 -3ú ê ú ú ê ë û ê-2 3ú ë 10 -12û ë û 2    32.  1 3 2  1  3   

Solution

é2ù ê ú é 1 -3 2ù ê 1 ú = é5ù êë úû ê ú êë úû êë3úû  1   2 33.   2 1 1 3  1   5

Solution é ù é2 ê 1ú ê ê2ú ê4 é ù ê ú ê2 -1 1 3ú = ê û ê2 ê 1ú ë ê ú ê êë5úû êë 10

-1 1 3 ùú -2 2 6 ú ú -1 1 3 ú ú -5 5 15úû

 2  1 5 3    1 1  1 34.   2    1 1    1 3   2 2  3  

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1638


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

é ù é ù ê 2ú é ùé ù é ùé ùé ù ê 1 -5 3ú ê-2ú ê 1 -1ú ê 1ú = ê 21 ú ê 1 -1ú ê 1ú  not possible ê2 1 -1úû ê ú êë-1 3úû êë-2úû êë-1úû êë-1 3úû êë-2úû ë êë 3úû  2 2    35.  1 3 2  1   1 3   5   5   

Solution

é 2ù é ù ê ú é 1 -3 2ù ê 1ú + é 1 -3ù ê2ú = é-11ù + é-13ù = é-24ù êë úû ê ú êë úû ê5ú êë úû êë úû êë úû ë û êë-5úû

  1 3  1 3   1  36.     3 1   1 1    5        Solution æ ö çç éê 1 -3ùú + éê-1 3ùú ÷÷ éê 1ùú = éê0 0ùú éê 1ùú = éê 0ùú çç ê3 ú ê ú 1û ë 1 1û ø÷÷ ëê-5ûú ëê4 2ûú ëê-5ûú ëê-6ûú èë  0 2   1 2      For Exercise 37 and 38, let A   3 3  and B   3 9 .  1 0  5 1   

37. Solve X + A = –B for X.

Solution X + A = -B

é 0 -2ù é 1 -2ù é -1 4ùú ê ú ê ú ê 3ú - ê3 9ú = ê 0 -12ú X = -A - B = - ê-3 ê ú ê ú ê ú êë -1 0úû êë5 1úû êë-4 -1úû 38. Solve 4X – A = B for X.

Solution 4X - A = B 4X = A + B é 0 -2ù é ù é 1 -1ù ú ú 1 ê 1 -2ú ê 4 1 1 1ê X = A + B = ê-3 3ú + ê3 9ú = ê0 3ú ê ú ê ú ê ú 4 4 4ê 4ê 0úû 1úû ê 1 41 ú ë -1 ë5 ë û

Find the multiplicative inverse of each matrix, if possible.

2 3 39.   3 5 

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1639


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 1 3 1 0ù é 1 3 1 0ù é2 3 1 0ù 2 2 2 2 ê ú ê ú ê ú ê 3 5 0 1 ú  ê 3 5 0 1 ú  ê0 1 - 3 1 ú  êë 2 2 ë û ûú ëê ûú 1 R R R R R 3  +  1 1 2 2 2 1 é 1 0 5 -3ù é 1 0 5 -3ù é 5 -3ù ê ú ê ú ê ú ê0 1 - 3 1 ú  ê0 1 -3 2 ú  Inverse: ê-3 2 ú êë úû 2 2 ë û ë û -3R2 + R1  R1 2R2  R2

 2 1 40.    6 4 Solution 1 1 1 1 1ù é ù é ù é ù é ê 2 -1 1 0ú  ê 1 - 2 2 0ú  ê 1 - 2 2 0ú  ê 1 0 2 2 ú ê-6 4 0 1 ú ê0 1 3 1 ú ê-6 4 0 1 ú ê0 1 3 1 ú êë úû êë úû ë û ë û 1 1  +  +  6 R R R R R R R R1 1 1 2 2 1 2 1 2 2 é2 1 ù 2ú Inverse: ê ê3 1ú ë û

 6 4  41.    3 2  Solution é-6 4 1 0ù é-6 4 1 0 ù ê ú ê ú ê-3 2 0 1 ú  ê 0 0 1 -2ú  No inverse exists. ë û ë û - 2R2 + R1  R2  1 0 0   42. 2 0 2  1 2 2  

Solution é 1 0 0 1 0 0ù é 1 0 0 1 0 0ù é1 0 0 1 0 0ùú ê ú ê ú ê ê2 0 -2 0 1 0ú  ê 1 2 2 0 0 1 ú  ê0 2 2 -1 0 1 ú  ê ú ê ú ê ú ê 1 2 2 0 0 1ú ê2 0 -2 0 1 0ú ê0 0 -2 -2 1 0ú ë û ë û ë û R2  R3 - R1 + R2  R2 - 2R1 + R3  R3 é1 0 0 é1 0 0 1 é 1 ù 1 0 0ùú 0 0ùú ê ê ê 3 01 01 ú 3 1 1 ê0 2 0 -3 1 1 ú  ê0 1 0 ú : Inverse = êú 2 2 2ú 2 2ú ê ú ê ê 2 ê0 0 -2 -2 1 0ú ê0 0 1 1 - 21 0ú ê 1 - 21 0ú ë û ë û ë û 1 R2 + R3  R2 R R  2 2 2 - 21 R3  R3

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1640


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

 1 0 8   43. 3 7 6   1 2 3  

Solution é 1 0 8 1 0 0ù é1 0 é1 0 8 8 1 0 0ùú 1 0 0ùú ê ú ê ê 18 ê3 7 6 0 1 0ú  ê0 7 -18 -3 1 0ú  ê0 1 - 73 71 0ú  7 ê ú ê ú ê ú ê 1 2 3 0 0 1ú ê0 0 -5 -1 0 1 ú ê 0 2 -5 -1 0 1 ú ë û ë û ë û 1 - 3R1 + R2  R2  R R 2 7 2

- R1 + R3  R3 é1 0 8 ù é 1 0 0 9 16 -56ù é 9 16 -56ù 1 0 0ú ê ê ú ê ú 1 ê0 1 - 18 - 3 ú 0ú  ê0 1 0 -3 -5 18 ú  Inverse: ê-3 -5 18 ú 7 7 7 ê ê ú ê ú 1 ê0 0 ê 0 0 1 1 -2 êë -1 -2 7 úû 7 ú - 71 - 72 1 ú 7 ë û ë û - 2R2 + R3  R3 - 56R3 + R1  R1 18R3 + R2  R2 7R3  R3  4 4 1   44.  1 1 1    1 1 0   

Solution é 4 4 1 1 0 0ù é 1 1 1 0 1 0ùú ê ú ê ê 1 ú ê 1 1 0 1 0  4 4 1 1 0 0ú  ê ú ê ú ê-1 -1 0 0 0 1 ú ê-1 -1 0 0 0 1 ú ë û ë û R1  R2 é 1 1 1 0 1 0ù é 1 1 1 0 1 0ù ê ú ê ú ê0 1 -3 1 -4 0ú  ê0 0 0 1 -1 3ú : No inverse exists. ê ú ê ú ê0 0 1 0 1 1 ú ê0 0 1 0 1 1 ú ë û ë û - 4R1 + R2  R2 3R3 + R2  R2 R1 + R3  R3 Use the multiplicative inverse of the coefficient matrix to solve each system of linear equations.

3 x  y  8 45.   x  2 y  5 Solution é3 -1ù é x ù é8ù ê úê ú = ê ú ê 1 2ú ê y ú ê5ú ë ûë û ë û -1 é ù é ù é ù é 2 ê x ú = ê3 -1ú ê8ú = ê 71 ê y ú ê 1 2ú ê5ú ê- 7 ë û ë û ë û ë

1 7 3 7

ùé ù é ù ú ê8ú = ê3ú ú ê 5ú ê 1 ú ûë û ë û

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1641


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

4 x  y  2z  0  46.  x  y  2z  1 x  z  0 

Solution é4 -1 2ù é x ù é0ù ê úê ú ê ú ê1 1 2ú ê y ú = ê 1 ú ê úê ú ê ú êë 1 0 1úû êë z úû êë0úû

é x ù é4 -1 2ù ê ú ê ú ê y ú = ê 1 1 2ú ê ú ê ú êë z úû êë 1 0 1 úû

-1

é0ù é 1 1 -4ùú éê0ùú éê 1ùú ê ú ê ê 1 ú = ê 1 2 -6ú ê 1 ú = ê 2ú ê ú ê úê ú ê ú êë0úû êë-1 -1 5úû êë0úû êë-1úû

w  3 x  y  3z  1  w  4 x  y  3z  2 47.  x  y  1 w  2 x  y  2z  1 

Solution é1 3 1 ê ê1 4 1 ê ê0 1 1 ê êë 1 2 -1

3ùú éêw ùú éê 1 ùú 3ú ê x ú ê2ú úê ú = ê ú 0ú ê y ú ê 1 ú úê ú ê ú 2úû êë z úû êë 1 úû

éw ù é 1 3 1 ê ú ê êxú ê 1 4 1 ê ú=ê ê y ú ê0 1 1 ê ú ê êë z úû êë 1 2 -1

3ùú 3ú ú 0ú ú 2úû

-1

é 1 ù é 3 -5 5 3ùú éê 1 ùú éê 1ùú ê ú ê ê2ú ê-1 1 0 0ú ê2ú ê 1ú ê ú=ê úê ú = ê ú ê 1 ú ê 1 -1 1 0ú ê 1 ú ê 0ú ê ú ê úê ú ê ú 1 -2 -1úû êë 1 úû êë-1úû êë 1 úû êë 0

Evaluate each determinant.

3 2 48.    1 3 Solution é3 -2ù ê ú = (3)(-3) - (-2)(1) = -9 + 2 = -7 ê 1 -3ú ë û  1 2 3    49. 2 1 3   1 1 0   

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1642


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 1 -2 3ù ê ú -1 3 2 3 2 -1 ê2 -1 3ú = 1 - (-2) +3 ê ú -1 0 1 0 1 -1 êë 1 -1 0úû

= 1(3) + 2(-3) + 3(-1) = 3 - 6 - 3 = -6  1 3 1   50.  1 2 1  1 0 2  

Solution é 1 3 -1ù ê ú 2 1 1 1 1 2 ê1 2 1ú = 1 -3 + (-1) ê ú 0 2 1 2 1 0 êë 1 0 2úû

= 1(4) - 3(1) - 1(-2) = 4 - 3 + 2 = 3  1  1 51.  0   3

2 3 4  3 3 2 0 0 1  3 4 0

Solution Expand along 3rd row

é 1 ê ê-1 ê ê 0 ê êë 3

2 3 4ùú 1 2 3 3 -3 2ú ú = 0(*) - 0(*) + 0(*) - (-1) -1 3 -3 0 0 -1ú 3 3 4 ú 3 4 3úû æ 3 -3 -1 -3 -1 3 ö÷ ÷ = 1ççç1 -2 +3 4 3 4 3 3 ÷÷ø èç 3 = 1(21) - 2(5) + 3(-12) = -25

Use Cramer’s Rule to solve each linear system.

 x  3 y  5 52.  2 x  y  4 Solution x=

-5 3 -4 1 1 3 -2 1

=

7 =1 7

y=

1 -5 - 2 -4 1 3 -2 1

=

-14 = -2 7

 x  y  z  1  53. 2 x  y  3z  4  x  3 y  z  1 

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1643


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

x=

-1 -1 1 -4 -1 3 -1 -3 1

=

1 -1 1 2 -1 3 1 -3 1

2 =1 2

y=

1 -1 1 2 -4 3 1 -1 1 1 -1 1 2 -1 3 1 -3 1

=

0 =0 2

z=

1 -1 -1 2 -1 -4 1 -3 - 1 1 -1 1 2 -1 3 1 -3 1

=

-4 = -2 2

x  3 y  z  7  54.  x  y  3z  9 x  y  z  3 

Solution

x=

7 -3 1 -9 1 -3 3 1 1

=

1 -3 1 1 1 -3 1 1 1

16 =1 16

y=

1 7 1 1 -9 -3 1 3 1 1 -3 1 1 1 -3 1 1 1

=

-16 = -1 16

z=

1 -3 7 1 1 -9 1 1 1 1 -3 1 1 1 -3 1 1 1

=

48 =3 16

w  x  y  z  4  2w  x  z  4 55.  x  2 y  z  0 w  y  z  2 

Solution

w=

y=

4 1 -1 4 1 0 0 1 2 2 0 1

1 1 1 1

1 1 -1 2 1 0 0 1 2 1 0 1

1 1 1 1

1 1 2 1 0 1 1 0

4 4 0 2

1 1 1 1

1 1 -1 2 1 0 0 1 2 1 0 1

1 1 1 1

=

=

-4 =1 -4

x=

4 -1 4 0 0 2 2 1

1 1 1 1

1 1 -1 2 1 0 0 1 2 1 0 1

1 1 1 1

1 1 -1 2 1 0 0 1 2 1 0 1

4 4 0 2

1 2 0 1

4 = -1 z = -4 1 1 -1 2 1 0 0 1 2 1 0 1

1 1 1 1

=

0 =0 -4

=

-8 =2 -4

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1644


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

a If d g

b c e f  7, evaluate each determinant. h i

3a 3b 3c 56. d e f g h i

Solution

3a 3b 3c a b c d e f = 3 d e f = 21 g h i g h i a b c 57. d  g e  h f  i g h i

Solution

a b c a b c d +g e+h f +i = d e f = 7 g h i g h i Decompose into partial fractions. 58.

7x  3 x2  x Solution A B 7x + 3 7x + 3 = = + 2 x( x + 1) x x+1 x +x A( x + 1) Bx = + x( x + 1) x( x + 1) Ax + A + Bx = x( x + 1) ( A + B) x + A = x( x + 1) ìïï A + B = 7 3 4 A = 3 7x + 3  = + í ïïî A =3 B = 4 x( x + 1) x x+1

59.

4 x 3  3x  x 2  2 x4  x2

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1645


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

A B Cx + D 4 x 3 + 3x + x 2 + 2 4 x 3 + x 2 + 3x + 2 = = + 2 + 2 4 2 2 2 x x x +x x ( x + 1) x +1 2 Ax( x + 1) B( x 2 + 1) (Cx + D) x 2 = 2 2 + 2 2 + 2 2 x ( x + 1) x ( x + 1) x ( x + 1) 3 2 3 Ax + Ax + Bx + B + Cx + Dx 2 = x 2 ( x 2 + 1) 3 ( A + C ) x + (B + D) x 2 + Ax + B = x 2 ( x 2 + 1) ìï A + C = 4 A=3 ïï ïï B + D = 1 B = 2 4 x 3 + x 2 + 3x + 2 3 x-1 2  = + 2+ 2 í 2 2 ïï A C=1 =3 x x x ( x + 1) x +1 ïï B D = 2 = 1 ïïî 60.

x2  5 x 3  x 2  5x Solution x2 + 5 x2 + 5 A Bx + C = = + 2 2 2 x x + x + 5x x ( x + x + 5) x + x +5 A( x 2 + x + 5) (Bx + C ) x = + 2 x ( x + x + 5) x ( x 2 + x + 5) 3

= =

Ax 2 + Ax + 5 A + Bx 2 + Cx x ( x 2 + x + 5) ( A + B) x 2 + ( A + C ) x + (5 A) x ( x 2 + x + 5)

ìï A + B =1 A= 1 ïï 1 1 x2 + 5 ïí A +C = 0  B = 0 = - 2 2 ïï x x + x +5 x( x + x + 5) =5 C = -1 ïïî5A 61.

x2  1

 x  1

3

Solution x2 + 1 A B C = + + 3 2 x + 1 ( x + 1) ( x + 1) ( x + 1)3

A( x + 1)2 B( x + 1) C + + 3 3 ( x + 1) ( x + 1) ( x + 1)3 2 Ax + 2 Ax + A + Bx + B + C = ( x + 1)3 Ax 2 + (2 A + B) x + ( A + B + C ) = ( x + 1)3

=

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1646


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

ìï A =1 A= 1 ïï x2 + 1 1 2 2 ïí2 A + B = 0  B = -2 = + 3 2 ïï x + 1 ( x + 1) ( x + 1) ( x + 1)3 C =2 ïïî A + B + C = 1

Solve the linear inequality in two variables by graphing. 62. Graph: y ≥ –2x – 1.

Solution y ≥ –2x – 1

63. Graph: x2 + y2 > 4.

Solution x 2 + y2 > 4

Solve each system of inequalities in two variables by graphing.

3 x  2 y  6 64.   x  y  3 Solution 3 x  2 y  6   x  y  3

 y  x 2  1 65.  2  y  x  1

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1647


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

 y  x 2  1  2  y  x  1

Maximize P subject to the given conditions. 66. P  2 x  y x  0  y  0 x  y  3 

Solution

Point

P = 2x + y

(0, 0)

= 2(0) + 0 = 0

(0, 3)

= 2(0) + 3 = 3

(3, 0)

= 2(3) + 0 = 6

Max: P = 6 at (3, 0) 67. P  3 x  y

y  1  y  2   y  3x  1 x  1  Solution

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1648


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

P = 3x – y

(0, 1)

= 3(0) – 1 = –1

(1, 1)

= 3(1) – 1 = 2

(1, 2)

= 3(1) – 2 = 1

( 31 , 2 )

= 3( 31 ) - 2 = - 1

Max: P = 2 at (1, 1) 68. P  2 x  3 y x  0  y  3 x  y  4 

Solution

Point

P = 2x + 3y

(0, 3)

= 2(0) – 3(3) = –9

(7, 3)

= 2(7) – 3(3) = 5

(0, –4)

= 2(0) – 3(–4) = 12

Max: P = 12 at (0, –4) 69. P  y  2 x

x  y  1   x  1  x y   2 2   x  y  2

Solution

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1649


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

P = y – 2x

(0, 2)

= 2 – 2(0) = 2

(1, 1)

= 1 – 2(1) = –1

(1, 0)

= 0 – 2(1) = –2

( - 23 , 53 )

= 53 - 2(- 23 ) = 3

Max: P = 3 at (- 23 , 53 ) 70. A company manufactures two fertilizers, x and y. Each 50-pound bag of fertilizer requires three ingredients, which are available in the limited quantities shown in the table. The profit on each bag of fertilizer x is $6 and on each bag of y is $5. How many bags of each product should be produced to maximize the profit?

Ingredient

Number of Pounds in Fertilizer x

Number of Pounds in Fertilizer y

Total Number of Pounds Available

Nitrogen

6

10

20,000

Phosphorus

8

6

16,400

Potash

6

4

12,000

Solution Let x = bags of Fertilizer x and y bags of Fertilizer y. Maximize P = 6x + 5y ìï6 x + 10 y £ 20000 ïï ïï8 x + 6 y £ 16400 subject to ïí ïï6 x + 4 y £ 12000 ïï ïïî x ³ 0, y ³ 0

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1650


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

P = 6x + 5y

(0, 0)

=0

(0, 2000)

= 10000

(1000, 1400)

= 13000

(1600, 600)

= 6(1600) + 5(600) = 12600

(2000, 0)

= 6(2000) + 5(0) = 12000

1000 bags of x and 1400 bags of y should be made, for a maximum profit of $13,000.

CHAPTER TEST SOLUTIONS Solve each system of linear equations in two variables equations by the graphing method. 1.

 x  3 y  5  2 x  y  0 Solution ìï x - 3 y = -5 ïí ïï2x - y = 0 î

solution: (1, 2) 2.

 x  2 y  5   y  2 x  4 Solution ìï x = 2 y + 5 ïí ïï y = 2 x - 4 î

solution: (1, –2)

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1651


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solve each system of linear equations by the substitution or elimination method. 3.

3 x  y  0  2 x  5 y  17 Solution 3 x + y = 0  ´ (5) 2 x - 5 y = 17

4.

3x + y = 0 3(1) + y = 0 y = -3

15 x + 5 y = 0 2 x - 5 y = 17 17 x

= 17

x

= 1

Solution: x = 1, y = -3

x  y x7  2   x  y  y  6  2 Solution x+y +x= 7 2 x- y - y = -6  2

3 x + y = 14

3 x + y = 14

x - 3 y = -12

x - 3 y = -12  -3 x + 9 y = 36

x - 3 (5) = -12

10 y = 50 y =5

Solution: x=3 y =5

x=3

5. Mixing solutions A chemist has two solutions; one has a 20% concentration and the other a 45% concentration. How many liters of each must she mix to obtain 10 liters of 30% concentration?

Solution Let x = liters of 20% solution and y = liters of 45% solution. The following system applies: x+ y = 10 0.2 x + 0.45 y = 3

 ´ (-2)  ´ (10)

-2 x - 2 y = -20 2 x + 4.5 y = 30

2.5 y = y=

She must use 4 liters of the 45% and 6 liters of the 20% solution.

10 4

6. Wholesale distribution Great Buy, Hi-Fi Electronics, and Sound World buy a total of 175 Bluetooth speakers used for large auditoriums from the same distributor each month. Because Sound World buys 25 more units than the other two stores combined, Sound World’s cost is only $160 per unit. The speakers cost Hi-Fi $165 each and Great Buy $170 each. How many speakers does each retailer buy each month if the distributor receives $28,500 each month from the sale of the speakers to the three stores?

Solution Let x = # from Ace, y = # from Hi-Fi and z = # from CD World.

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1652


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Add (1) and (2):

(1) x + y + z = 175 (2) -x - y + z = 25 (3) 170 x + 165 y + 160z = 28500

(1) x + y + z = 175 (2) -x - y + z = 25 (4)

2z = 200 z = 100

Add equations 170(2) and (3): -170 x - 170 y + 170z = 4250 170(2) (3) 170 x + 165 y + 160z = 28500 - 5 y + 330z = 32750

(5)

Solve the system of two equations and two unknowns formed by equations (4) and (5): -5 y + 330z = 32750 -5 y + 330(100) = 32750 -5 y = - 250

y = 50 x + y + z = 175 x + 50 + 100 = 175 x = 25 Ace buys 25 units per month. Hi-Fi buys 50 units per month. CD World buys 100 units per month.

Write each system of linear equations as a matrix and solve it by Gaussian elimination. 7.

3 x  2 y  4  2 x  3 y  7 Solution é 3 - 2 4ù é ù é ù ê ú  ê 1 -5 -3ú  ê 1 -5 -3ú  ê2 3 7 ú ê2 3 ê0 13 13 ú 7ú ë û ë û ë û - R2 + R1  R1 - 2R1 + R2  R2 From R2 : y = 1 From R1 : x - 5 y = -3

é 1 -5 -3ù ê ú ê0 1 1 ú ë û 1 R  R2 13 2

Solution: x = 2, y = 1

x - 5 (1) = -3 x=2

8.

x  3 y  z  6  2 x  y  2z  2 x  2 y  z  6 

Solution é 1 3 -1 6 ù é 1 3 -1 6 ù é 1 3 -1 6 ù ê ú ê ú ê ú ê2 -1 -2 -2ú  ê0 -7 0 -14ú  ê0 1 0 2 ú ê ú ê ú ê ú ê1 2 ê0 1 - 2 0 ú ê0 0 -14 -14ú 1 6ú ë û ë û ë û - 2R1 + R2  R2 - 71 R2  R2 - R3 + R1  R3

7R3 + R2  R3

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1653


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é 1 3 -1 6ù ê ú ê0 1 0 2ú ê ú ê0 0 1 1 ú ë û - 141 R3  R3

From (3): z = 1

From(1) :

From (2): y = 2

x + 3y - z = 6 x + 3(2) - (1) = 6 x=1

Solution: x = 1, y = 2, z = 1

Write each system of linear equations as a matrix and solve it by Gauss–Jordan elimination. If the system has infinitely many solutions, write a general solution.

9.

 x  2 y  3z  5  3 x  y  2z  7 y  z  2 

Solution é 1 2 3 -5ù é1 2 é1 2 3 -5ùú 3 -5ùú ê ú ê ê ê 3 1 -2 7 ú  ê0 -5 -11 22 ú  ê0 1 -1 2 ú  ê ú ê ú ê ú ê0 1 -1 2 ú ê0 1 ê0 -5 -11 22 ú -1 2 ú ë û ë û ë û R2  R3 - 3R1 + R2  R2 é 1 0 5 -9ù é 1 0 5 -9ù é1 0 0 1 ù ê ú ê ú ê ú ê0 1 - 1 ú ê ú 2  0 1 -1 2  ê0 1 0 0 ú Solution: ê ú ê ú ê ú ê0 0 -16 32 ú ê0 0 ê0 0 1 -2ú 1 - 2ú ë û ë û ë û -2R2 + R1  R1 - 161 R3  R3 - 5R3 + R1  R1 5R2 + R3  R3

x=1 y =0 z = -2

R2 + R3  R2

x  2 y  z  0  10. 3 x  2 y  2z  7 4 x  z  7 

Solution é1 é1 é1 2 1 2 1 0ùú 2 1 0ùú 0ùú ê ê ê ê 3 -2 -2 7 ú  ê0 -8 -5 7ú  ê0 1 5 - 7 ú  8 8ú ê ú ê ú ê ê4 ê0 -8 -5 7ú ê0 0 0 0 -1 7 ú 0ú ë û ë û ë û - 3R1 + R2  R2 - R2 + R3  R3 - 4R1 + R3  R3 - 81 R2  R2 7ù é1 0 -1 4 4ú ê 5 ê0 1 - 87 úú  Solution: x = 7 + 1 z 8 ê 4 4 ê0 0 0 0ú y = - 87 - 85 z ë û z = any real number -2R2 + R1  R1

Note: This answer is equivalent to the answer provided in the textbook.

Perform the operations.

 2 3 5  2 1 1 11. 3    5  0 3 1  0 3 2

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1654


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution é 2 -3 5ù é ù é ù é ù ú - 5 ê-2 1 -1ú = ê6 -9 15ú - ê-10 5 -5ú 3ê ê0 ú ê 0 3 2ú ê0 ú ê 0 15 10ú 3 1 9 3 ë û ë û ë û ë û é 16 -14 20ù ú =ê ê 0 -6 -13ú ë û  2 2    3 12.  1 2 3  2 2   2  1 0    

Solution é 2 -2ù é 3ù ê ú é 3ù é 1 2 3ù ê-2 2ú ê ú = éêë 1 2ùúû ê ú = éêë-1ùúû êë úû ê ê-2ú ú ê-2ú ë û êë 1 0úû ë û Find the multiplicative inverse of each matrix, if possible.

5 19 13.   2 7  Solution 19 1 19 1 é é ù é5 19 1 0ù 0ùú 5 5 ê ú  ê 1 5 5 0ú  ê 1 3 2 ê ú ê2 7 0 1 ú ê ú êë0 - 5 - 5 1úû ë2 7 0 1 û ëê ûú 1 - 2R1 + R2  R2 R  R1 5 1 19 ù 19 ù 7 19 1 é é é 7 0úù 3ú 3 ú ê1 5 5 ê 1 0 -3 ê- 3 ê0 1 2 - 5 ú  ê0 1 2 - 5 ú  Inverse: ê 2 - 5 ú êë 3 3 úû 3 3û 3û ë 3 ë - 53 R2  R2 - 195 R2 + R1  R1  1 3 2    14.  4 1 4   0 3 1  

Solution é-1 3 -2 1 0 0ù é 1 -3 2 -1 0 0ùú ê ú ê ê 4 1 ú ê 4 0 1 0  0 13 -4 4 1 0ú  ê ú ê ú ê 0 3 -1 0 0 1ú ê0 3 -1 0 0 1ú ë û ë û 4R1 + R2  R2 - R1  R1 é 1 -3 2 -1 0 ù é 0ú 11 3 -12ùú ê ê1 0 2 ê0 1 0 4 1 - 4 ú  ê0 1 0 4 1 -4 ú  ê ú ê ú ê0 ú ê 3 -1 0 0 1 0 0 -1 -12 -3 13ú ë û ë û 3R2 + R1  R1 - 4R3 + R2  R2 - 3R2 + R3  R3

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1655


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

é 1 0 0 -13 -3 é-13 -3 14ùú 14ùú ê ê ê0 1 0 ú ê 4 1 -4 Inverse: 4 1 - 4ú ê ú ê ú ê0 0 1 êë 12 12 3 -13ú 3 -13úû ë û 2R3 + R1  R1 - R3  R3

Use the multiplicative inverses found in Questions 13 and 14 to solve each system of linear equations.

5 x  19 y  3 15.  2 x  7 y  2 Solution é5 19ù é x ù é3ù ê úê ú = ê ú ê2 7 ú ê y ú ê2ú ë ûë û ë û -1 19 ù é ù é 17 ù é ù é ù é ù é- 7 3 3ú ê ú ê x ú = ê5 19ú ê3ú = ê 32 = ê 34 ú 5 ê y ú ê2 7ú ê2ú ê 3 - 3 ú ê2ú ê- 3 ú ë û ë û ë û ë ûë û ë û  x  3 y  2z  1  16. 4 x  y  4 z  3  3 y  z  1 

é-1 3 -2ù é x ù é 1ù ê úê ú ê ú ê 4 1 4ú ê y ú = ê 3ú ê úê ú ê ú êë 0 3 -1úû êë z úû êë-1úû é x ù é-1 3 -2ù é 1ù ê ú ê úê ú ê yú = ê 4 1 4ú ê 3ú ê ú ê úê ú êë z úû êë 0 3 -1úû êë-1úû é x ù é-13 -3 14ùú éê 1ùú éê-36ùú ê ú ê ê yú = ê 4 1 -4ú ê 3ú = ê 11ú ê ú ê úê ú ê ú êë z úû êë 12 3 -13úû êë-1úû êë 34úû

Evaluate each determinant. 17.

3 5 3 1 Solution

3 -5 = (3)(1) - (-5)(-3) = 3 - 15 = -12 -3 1 3 5 1 18. 2 3 2 1 5 3

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1656


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Solution

3 5 -1 -2 -2 -2 3 3 -2 -2 3 -2 = 3 -5 + (-1) 5 -3 1 -3 1 5 1 5 -3

= 3(1) - 5(8) - 1(-13) = 3 - 40 + 13 = -24 Use Cramer’s Rule to solve each system of linear equations for y.

3 x  5 y  3 19.   3 x  y  2 Solution y=

3 3 -3 2 3 -5 -3 1

=

15 5 =-12 4

3 x  5 y  z  2  20.  2 x  3 y  2z  1  x  5 y  3z  0 

Solution

y=

3 2 -1 -2 1 - 2 1 0 -3 3 5 -1 -2 3 -2 1 5 -3

=

-24 =1 -24

Decompose each fraction into partial fractions. 21.

5x 2x  x  3 2

Solution A B 5x 5x = = + 2 2 x - x - 3 (2 x - 3)( x + 3) 2 x - 3 x + 1 A( x + 1) B(2 x - 3) = + (2 x - 3)( x + 1) (2 x - 3)( x + 1) Ax + A + 2Bx - 3B = (2 x - 3)( x + 1) A( A + 2B) x + ( A - 3B) = (2 x - 3)( x + 1) ìïï A + 2B = 5 5x 3 1 A=3  = + í ïïî A - 3B = 0 B = 1 (2 x - 3)( x + 1) 2 x - 3 x + 1

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1657


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

22.

3x 2  x  2 x 3  2x Solution

A Bx + C 3x 2 + x + 2 3x 2 + x + 2 = = + 2 3 2 x x + 2x x( x + 2) x +2 A( x 2 + 2) (Bx + C ) x = + x( x 2 + 2) x( x 2 + 2) 2 Ax + 2 A + Bx 2 + Cx = x ( x 2 + 2)

= ìï A + B = 3 A=1 ïï C 1 B =  =2 í ïï C=1 =2 ïïî2 A

( A + B) x 2 + Cx + 2 A x( x 2 + 2)

3x 2 + x + 2 1 2x + 1 = + 2 2 x x ( x + 2) x +2

Graph the solution set of each system of inequalities.

 x  3 y  3 23.   x  3 y  3 Solution ìïï x - 3 y ³ 3 í ïïî x + 3 y £ 3

3 x  4 y  12  3 x  4 y  6 24.  x  0 y  0 

Solution ìï3 x + 4 y £ 12 ïï í3 x + 4 y ³ 6 ïï ïïî x ³ 0, y ³ 0

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1658


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

25. Maximize P = 3x + 2y subject to y  0  x  0  2 x  y  4 y  2 

Solution

Point

P = 3x + 2y

(0, 2)

= 3(0) + 2(2) = 4

(1, 2)

= 3(1) + 2(2) = 7

(2, 0)

= 3(2) + 2(0) = 6

(0, 0)

= 3(0) + 2(0) = 0

Max: P = 7 at (1, 2) 26. Minimize P = y – x subject to x  0  y  0  x  y  8 2 x  y  2 

Solution

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1659


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

Point

P=y–x

(0, 2)

=2–0=2

(0, 8)

=8–0=8

(8, 0)

= 0 – 8 = –8

(1, 0)

= 0 – 1 = –1

Min: P = –8 at (8, 0)

GROUP ACTIVITY SOLUTIONS Vacation Budget Real-World Example of Systems of Inequalities Planning a family vacation is exciting. Everyone loves getting away for some relaxation and adventure. Planning ahead, saving money monthly, and creating a travel budget can help families have an amazing vacation and keep expenses less than or equal to a specific dollar amount. Vacation expenses include, transportation, lodging, entertainment, food, and miscellaneous items.

Group Activity Your family plans a vacation to visit two national parks, Yellowstone and Grand Teton. The parks are only 31 miles apart and you can’t wait to take a hike in both Yellowstone and Grand Teton, go kayaking and rafting, see wildlife in their natural habitat, watch geysers erupt, see giant colorful hot springs, and immerse yourself in the nearby Native American culture. a. While visiting the parks, your family plans to stay part of the time in a rustic cabin and the rest of the time at a resort lodge. The average cost of a rustic cabin is $135 a night and the average cost of a resort lodge is $275 a night. Let x represent the number of night’s stay in a cabin and y represent the number of night’s stay in a resort lodge. If your family lodging budget allows for no more than $1500 to be spent on lodging, write an inequality that represents this limitation. b. If your family plans to stay at least seven nights, write an inequality that represents your length of stay. c. Your parents insist that your family stays at least two nights in a resort lodge. Write an inequality that represents your parent’s requirement. d. You insist that your family stays at least four nights in a rustic cabin. Write an inequality that represents your requirement. e. After completing parts a–d, write a system of inequalities that models all of the conditions stated in the problem that pertain to lodging. f.

Graph the solution set of the system of inequalities.

g.

Based on your graphical solution, compile a list of possible nights you could stay in a cabin and in a resort lodge and stay on budget. List these in a matrix format with column 1 representing the number of nights in a cabin, column 2 representing the number of nights in a resort lodge, and column 3 representing the total cost for lodging.

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1660


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 6: Systems of linear equations, matrices, and inequalities

h. What is the greatest number of nights that your family could stay at the resort lodge and stay within your set budget? If you stay this number of nights in the resort lodge, identify the maximum number of nights you could stay in a cabin and not exceed your budget?

Solution a. 135 x  275 y  1500 b.

xy 7

c.

y 2

d.

x4

e.

 135 x  275 y  1500  x  y  7  y  2 x  4 

f.

g.

answers may vary – one matrix is: é4 3 1365 ù ê ú ê5 2 1221 ú ê ú ê5 3 1500ú ê ú ê6 2 1360ú ê ú ëê7 2 1495ûú

h. 3 nights, 5 nights

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1661


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution and Answer Guide GUSTAFSON/HUGHES, COLLEGE ALGEBRA 2023, 9780357723654; CHAPTER 7: CONIC SECTIONS AND SYSTEMS OF NONLINEAR EQUATIONS

TABLE OF CONTENTS End of Section Exercise Solutions ................................................................................ 1662 Exercises 7.1 ........................................................................................................................... 1662 Exercises 7.2 ........................................................................................................................... 1701 Exercises 7.3 .......................................................................................................................... 1735 Exercises 7.4 .......................................................................................................................... 1767 Chapter Review Solutions............................................................................................... 1791 Chapter Test Solutions .................................................................................................. 1803 Cumulative Review Solutions ........................................................................................ 1808 Group Activity Solutions ................................................................................................ 1820

END OF SECTION EXERCISE SOLUTIONS EXERCISES 7.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Complete the square to make x 2  14 x a perfect-square trinomial. Solution x 2  14 x  49

2. Complete the square to make y 2  8 y a perfect-square trinomial. 3

Solution 8 16 y2  y  3 9 3. Factor the perfect-square trinomial y 2  18 y  81.

Solution ( y  9)2

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1662


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

4. Factor the perfect-square trinomial x 2  5 x  25 6

144

Solution

 5 x   12  

2

5. If y 2  4 px , write the equation that results if p  5.

Solution y 2  4( 5) x y 2  20 x

6. If ( x  h)2  4 p( y  k ), write the equation that results if ( h, k )  (3,  4) and p  4.

Solution ( x  3)2  4( 4)( y  4) ( x  3)2  16( y  4)

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7.

( x  2)2  ( y  5)2  9: center (_____, _____ ); radius _____

Solution (2,  5), 3 8.

x 2  y 2  36  0: center (_____, _____ ); radius _____

Solution (0, 0), 6 9.

x 2  y 2  5: center (_____, _____ ); radius _____

Solution

(0, 0),

5

10. 2( x  9)2  2 y 2  7 : center (_____, _____ ); radius _____

Solution (9, 0),

7 , or 2

14 2

Determine whether the graph of the parabola opens up-ward, downward, to the left, or to the right. 11.

y 2  4 x : opens __________

Solution to the left

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1663


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

12. y 2  10 x : opens __________

Solution to the right 13. x 2  4( y  3): opens __________

Solution downward 14. ( x  2)2  ( y  3): opens __________

Solution upward Fill in the blanks. 15. A parabola is the set of all points in a plane equidistant from a line, called the __________, and a fixed point not on the line, called the __________.

Solution directrix, focus 16. The general form of a second-degree equation in the variables x and y is Ax 2  ________________  0.

Solution Bxy  Cy 2  Dx  Ey  F Identify the conic as a circle or parabola. 17. x 2  5 x  y 2  12

Solution Two squared variables: circle 18. 3 x 2  3 y 2  18 x  6 y  24

Solution Two squared variables: circle 19. x 2  8 y  6 x  1

Solution One squared variable: parabola 20. 2 y 2  4 y  6 x  4

Solution One squared variable: parabola

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1664


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Practice Write an equation of each circle shown or described. Write your answer in standard form and then general form. 21.

Solution ( x  h)2  ( y  k )2  r 2 ( x  0)2  ( y  0)2  72 x 2  y 2  49 x 2  y 2  49  0

22.

Solution r  ( 3  0)2  (2  0)2 

13

( x  h)2  ( y  k )2  r 2 ( x  0)2  ( y  0)2 

 13 

2

x 2  y 2  13 x 2  y 2  13  0

23.

Solution r  (3  2)2  (2  ( 2))2 

17

( x  h)  ( y  k )  r 2

2

( x  2)2  ( y  ( 2))2 

2

 17 

2

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1665


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

( x  2)2  ( y  2)2  17 x 2  4 x  4  y 2  4 y  4  17 x2  y 2  4x  4 y  9  0

24.

Solution  5  ( 2) 4  ( 3)  3 1 , O   O ,  2 2   2 2 2

2

 3  1 r  5    4    2  2 

49 2

( x  h)2  ( y  k )2  r 2 2 2  49   3  1   x     y      2  2   2  2

2

2

 3  1 49 x    y    2 2 2     9 1 49  y2  y   4 4 2 2 2 x  y  3 x  y  22  0

x 2  3x 

25. Radius of 6; center at the intersection of 3 x  y  1 and 2 x  3 y  4

Solution 3x  y  1

  (3)

9x  3 y  3

2 x  3 y  4

2 x  3 y  4 7 1

7x x

3x  y  1

Center:

3(1)  y  1 (1,  2) y  2 ( x  h)2  ( y  k )2  r 2

 x  1   y  (2)   6  x  1   y  2  36 2

2

2

2

2

x 2  2 x  1  y 2  4 y  4  36 x 2  y 2  2 x  4 y  31  0

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1666


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

26. Radius of 8; center at the intersection of x  2 y  8 and 2 x  3 y  5

Solution x  2 y  8   (  2) 2 x  4 y  16

2 x  3 y  5

2x  3 y   5 7 y y

 21  3

x  2 y  8 Center: x  2(3)  8 (2, 3) x 2 ( x  h)2  ( y  k )2  r 2

 x  2   y  3  8  x  2   y  3  64 2

2

2

2

2

x 2  4 x  4  y 2  6 y  9  64 x 2  y 2  4 x  6 y  51  0

Graph each circle. 27. x 2  y 2  4

Solution x2  y 2  4 ( x  0)2  ( y  0)2  22 C(0, 0), r  2

28. x 2  2 x  y 2  15

Solution x 2  2 x  y 2  15

x 2  2 x  1  y 2  15  1 ( x  1)2  ( y  0)2  42

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1667


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

C(1, 0), r  4

29. 3 x 2  3 y 2  12 x  6 y  12

Solution 3 x 2  3 y 2  12 x  6 y  12

x2  4x  y 2  2 y  4 x2  4x  4  y 2  2 y  1  4  4  1 ( x  2)2  ( y  1)2  32 C(2, 1), r  3

30. 2 x 2  2 y 2  4 x  8 y  2  0

Solution 2x 2  2 y 2  4 x  8 y  2  0

x 2  2 x  y 2  4 y  1 x 2  2 x  1  y 2  4 y  4  1  1  4 ( x  1)2  ( y  2)2  22 C ( 1, 2), r  2

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1668


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Find the vertex, focus, and directrix of each parabola. 31. x 2  12 y

Solution

x 2  12 y ( x  0)2  4  3( y  0) p  3, opens up V(0, 0), F (0, 3), D: y  3

32. x 2  32 y

Solution x 2  32 y ( x  0)2  32( y  0) ( x  0)2  4  8( y  0)

p  8, opens down V (0, 0), F (0,  8), D : y  8

33. y 2  12 x

Solution

y 2  12x ( y  0)2  4  ( 3)( x  0) p  3, opens left V(0, 0), F ( 3, 0), D : x  3

34. y 2  36 x

Solution y 2  36 x ( y  0)2  36( x  0) ( y  0)2  4  9( x  0)

p  9, opens right V (0, 0), F (9, 0), D : x  9

35. ( y  3)2  20 x

Solution

( y  3)2  20 x ( y  3)2  4  5( x  0) p  5, opens right V(0, 3), F (5, 3), D: x  5

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1669


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

36. ( x  4)2  4 y

Solution ( x  4)2  4 y ( x  4)2  4( y  0) ( x  4)2  4  1( y  0)

p  1, opens down V ( 4, 0), F ( 4,  1), D : y  1

37. y 2  16( x  2)

Solution y 2  16( x  2) ( y  0)2  16( x  ( 2)) ( y  0)2  4  4( x  ( 2))

p  4, opens left V ( 2, 0), F ( 6, 0), D : x  2

1 38. x 2   ( y  5) 2

Solution 1 x 2   ( y  5) 2  1 2 ( x  0)  4     ( y  ( 5))  8

1 p   , opens down 8  1 7 V (0,  5), F  0,  5  , D : y   4 8 8  

39. ( x  2)2  24( y  1)

Solution

( x  2)2  24( y  1) ( x  ( 2))2  4  ( 6)( y  1) p  6, opens down V( 2, 1), F ( 2,  5), D : y  7

40. ( x  3)2  8( y  4)

Solution ( x  3)2  8( y  4) ( x  3)2  4  2( y  4)

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1670


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

p  2, opens up V (3,  4), F (3,  2), D : y  6

41. ( y  1)2  28( x  2)

Solution

( y  1)2  28( x  2) ( y  ( 1))2  4  7( x  2) p  7, opens right V(2,  1), F (9,  1), D: x  5

42. ( y  2)2  40( x  5)

Solution ( y  2)2  40( x  5) ( y  2)2  4  10( x  5)

p  10, opens left V ( 5, 2), F ( 15, 2), D : x  5

Find an equation in standard form of each parabola described. 43. Vertex at (0, 0); focus at (0, 3)

Solution Vertical (up), p = 3

( x  h)2  4 p( y  k ) ( x  0)2  4(3)( y  0) x 2  12 y 44. Vertex at (0, 0); focus at (0, –3)

Solution Vertical (down), p = –3

( x  h)2  4 p( y  k ) ( x  0)2  4( 3)( y  0) x 2  12 y 45. Vertex at (0, 0); focus at (–3, 0)

Solution Horizontal (left), p = –3

( y  k )2  4 p( x  h) ( y  0)2  4(3)( x  0) y 2  12 x

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1671


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

46. Vertex at (0, 0); focus at (3, 0)

Solution Horizontal (right), p = 3

( y  k )2  4 p( x  h) ( y  0)2  4(3)( x  0) y 2  12x 47. Vertex at (3, 5); focus at (3, 2)

Solution Vertical (down), p = –3

( x  h)2  4 p( y  k ) ( x  3)2  4( 3)( y  5) ( x  3)2  12( y  5) 48. Vertex at (3, 5); focus at (–3, 5)

Solution Horizontal (left), p = –6

( y  k )2  4 p( x  h) ( y  5)2  4(6)( x  3) ( y  5)2  24( x  3) 49. Vertex at (3, 5); focus at (3, –2)

Solution Vertical (down), p = –7

( x  h)2  4 p( y  k ) ( x  3)2  4( 7)( y  5) ( x  3)2  28( y  5) 50. Vertex at (3, 5); focus at (6, 5)

Solution Horizontal (right), p = 3

( y  k )2  4 p( x  h) ( y  5)2  4(3)( x  3) ( y  5)2  12( x  3) 51. Vertex at (0, 2); directrix at y = 3

Solution Vertical (down), p = –1

( x  h)2  4p( y  k ) ( x  0)2  4(1)( y  2) x 2  4( y  2)

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1672


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

52. Vertex at (–3, 4); directrix at y = 2

Solution Vertical (up), p = 2

( x  h)2  4 p( y  k ) ( x  ( 3))2  4(2)( y  4) ( x  3)2  8( y  4) 53. Vertex at (1, –5); directrix at x = –1

Solution Horizontal (right), p = 2

( y  k )2  4 p( x  h) ( y  ( 5))2  4(2)( x  1) ( y  5)2  8( x  1) 54. Vertex at (3, 5); directrix at x = 6

Solution Horizontal (left), p = –3

( y  k )2  4 p( x  h) ( y  5)2  4(3)( x  3) ( y  5)2  12( x  3) 55. Vertex at (2, 2); passes through (0, 0)

Solution

( x  2)2  4 p( y  2) OR ( y  2)2  4 p( x  2) (0  2)2  4 p(0  2) 4  8p 1  p 2 2  4 p

(0  2)2  4 p(0  2) 4  8p 1  p 2 2  4 p

( x  2)2  2( y  2)

( y  2)2  2( x  2)

56. Vertex at (–2, –2); passes through (0, 0)

Solution

( x  ( 2))2  4 p( y  ( 2)) OR ( y  (2))2  4 p( x  ( 2)) (0  2)2  4 p(0  2) 4  8p 1 p 2 2  4p

(0  2)2  4 p(0  2) 4  8p 1 p 2 2  4p

( x  2)2  2( y  2)

( y  2)2  2( x  2)

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1673


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

57. Vertex at (–4, 6); passes through (0, 3)

Solution ( x  ( 4))2  4 p( y  6)

2 OR ( y  6)  4 p( x  ( 4)) (0  4)2  4 p(3  6) (3  6)2  4 p(0  4) 16  12p 9  16p 4 9  p p 3 16 16 9   4p  4p 3 4 16 9 ( x  4)2   ( y  6) ( y  6)2  ( x  4) 3 4

58. Vertex at (–2, 3); passes through (0, –3)

Solution ( x  ( 2))2  4 p( y  3)

OR

(0  2)2  4 p(3  3) 4  24 p 1  p 6 2   4p 3 2 ( x  2)2   ( y  3) 3

( y  3)2  4 p( x  ( 2)) (3  3)2  4 p(0  2) 36  8p 9 p 2 18  4 p ( y  3)2  18( x  2)

59. Vertex at (6, 8); passes through (5, 10) and (5, 6)

Solution

( x  6)2  4 p( y  8)

OR

( y  8)2  4 p( x  6)

(5  6)2  4 p(10  8) (10  8)2  4 p(5  6) 1  8p 4  4 p 1 1  p p 8 4  4 p 1 ( y  8)2  4( x  6)  4p 2 1 ( x  6)2  ( y  8) 2 Check to see which equation is satisfied by (5, 6) as well. Answer: ( y  8)2  4( x  6)

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1674


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

60. Vertex at (2, 3); passes through 1,

13 4

 and  1,  21 4

Solution

( x  2)2  4 p( y  3)  13  (1  2)  4 p   3  4  1 p 4  4p 2

( x  2)2  4( y  3)

OR

( y  3)2  4 p( x  2) 2

 13    3   4 p(1  2) 4  1  4 p 16 1   4p 16 1 ( y  3)2   ( x  2) 16

Check to see which equation is satisfied by 1, 21 4

 as well. Answer: ( x  2)  4( y  3) 2

61. Vertex at (3, 1); passes through (4, 3) and (2, 3)

Solution

( x  3)2  4 p( y  1) OR ( y  1)2  4 p( x  3) (4  3)2  4 p(3  1) (3  1)2  4 p(4  3) 1  8p 4  4p 1 ( y  1)2  4( x  3) p 8 1  4p 2 1 ( x  3)2  ( y  1) 2 Check to see which equation is satisfied by (2, 3) as well. Answer: ( x  3)2  62. Vertex at (–4, –2); passes through (–3, 0) and

1 ( y  1) 2

 , 3 9 4

Solution ( x  ( 4))2  4 p( y  ( 2)) OR ( y  ( 2))2  4p( x  (4))

( 3  4)2  4 p(0  2) 1  8p 1  4p 2 1 ( x  4)2  ( y  2) 2

(0  2)2  4p( 3  4) 4  4p ( y  2)2  4( x  4)

Check to see which equation is satisfied by

 , 3 as well. Answer: ( y  2)  4( x  4) 9 4

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2

1675


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Write each equation of a parabola in standard form and identify the vertex. 63. y 2  4 x  8  0 Solution y 2  4x  8  0 y 2  4x  8 y 2  4( x  2) vertex (2, 0)

64. y 2  8 x  8  0

Solution y 2  8x  8  0 y 2  8 x  8 y 2  8( x  1) vertex (–1, 0)

65. x 2  12 y  12  0

Solution x 2  12 y  12  0 x 2  12 y  12 x 2  12( y  1) vertex (0, –1)

66. x 2  16 y  32  0

Solution x 2  16 y  32  0 x 2  16 y  32 x 2  16( y  2) vertex (0, 2)

67. x 2  4 x  4 y  0

Solution x2  4x  4 y  0 x2  4x  4 y x2  4x  4  4 y  4 ( x  2)2  4( y  1) vertex (2, –1)

68. x 2  2 x  4 y  19  0

Solution x 2  2 x  4 y  19  0 x 2  2 x  4 y  19 x 2  2 x  1  4 y  19  1 ( x  1)2  4( y  5) vertex (–1, 5)

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1676


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

69. y 2  8 x  6 y  33  0

Solution y 2  8 x  6 y  33  0 y 2  8 x  6 y  33 y 2  6 y  9  8 x  33  9 ( y  3)2  8( x  3) vertex (–3, –3)

70. y 2  12 x  8 y  56  0

Solution y 2  12 x  8 y  56  0 y 2  8 y  12 x  56 y 2  8 y  16  12 x  56  16 ( y  4)2  12( x  6) vertex (–6, 4)

Graph each parabola. 71. y 2  8 x

Solution y 2  8 x Vertex (0, 0) opens right

72. x 2  20 y

Solution x 2  20 y Vertex (0, 0) opens down

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1677


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

73. x 2  8( y  1)

Solution x 2  8( y  1) Vertex (0, –1) opens down

74. y 2  4( x  2)

Solution y 2  4( x  2) Vertex (2, 0) opens right

75. ( x  1)2  4( y  1)

Solution ( x  1)2  4( y  1) Vertex (1, –1) opens up

76. ( y  2)2  8( x  1)

Solution ( y  2)2  8( x  1) Vertex (1, –2) opens left

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1678


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Write each parabola in standard form and graph it. 77. y  x 2  4 x  5

Solution y  x2  4x  5

y  5  x2  4x y  5  4  x2  4x  4 y  1  ( x  2)2

78. 2 x 2  12 x  7 y  10

Solution 2 x 2  12x  7 y  10

7 y 5 2 7 x 2  6x  9  y  5  9 2 7 2 ( x  3)  y  14 2 7 ( x  3)2  ( y  4) 2 x 2  6x 

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1679


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

79. y 2  4 x  6 y  1

Solution

y 2  4 x  6 y  1 y 2  6 y  4 x  1 y 2  6 y  9  4 x  1  9 ( y  3)2  4 x  8 ( y  3)2  4( x  2)

80. x 2  2 y  2 x  7

Solution

x 2  2 y  2 x  7 x 2  2x  2 y  7 x 2  2x  1  2 y  7  1 ( x  1)2  2 y  6 ( x  1)2  2( y  3)

81. y 2  4 y  4 x  8

Solution y 2  4 y  4x  8

y 2  4 y  4  4x  8  4 ( y  2)2  4 x  4 ( y  2)2  4( x  1)

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1680


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

82. y 2  2 x  2 y  5

Solution

y 2  2x  2 y  5 y 2  2 y  2 x  5 y 2  2 y  1  2 x  5  1 ( y  1)2  2 x  6 ( y  1)2  2( x  3)

83. y 2  4 y  8 x  20

Solution y 2  4 y  8 x  20

y 2  4 y  4  8 x  20  4 ( y  2)2  8 x  24 ( y  2)2  8( x  3)

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1681


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

84. y 2  2 y  9 x  17

Solution y 2  2 y  9 x  17

y 2  2 y  1  9 x  17  1 ( y  1)2  9 x  18 ( y  1)2  9( x  2)

85. x 2  6 y  22  4 x

Solution

x 2  6 y  22  4 x x 2  4 x  6 y  22 x 2  4 x  4  6 y  22  4 ( x  2)2  6 y  18 ( x  2)2  6( y  3)

86. 4 y 2  4 y  16 x  7

Solution 4 y 2  4 y  16 x  7

7 4 1 7 1 2 y  y   4 x   4 4 4 2  1  y    4 x  2 2  y 2  y  4 x 

2

  1 1  y    4  x   2 2  

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1682


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

87. 4 x 2  4 x  32 y  47

Solution 4 x 2  4 x  32 y  47

47 4 1 47 1 x 2  x   8 y   4 4 4 2   1  x    8 y  12 2   x 2  x  8 y 

2

  1 3  x    8  y   2 2   

88. 4 y 2  16 x  17  20 y

Solution 4 y 2  16 x  17  20 y

y 2  5 y  4x  y2  5y  2

17 4

25 17 25  4x   4 4 4

 5  y    4x  2 2  2

  5 1  y    4 x   2 2    

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1683


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Find the vertex, focus and directrix of each parabola. Then graph the parabola, showing the focus and directrix on the graph. 89. y 2  12 x

Solution y 2  12 x ( y  0)2  4  3( x  0)

V (0, 0), F (3, 0), D : x  3

90. y 2  12 x

Solution y 2  12 x ( y  0)2  4  3( x  0)

V (0, 0), F ( 3, 0), D : x  3

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1684


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

91. x 2  12 y

Solution x 2  12 y ( x  0)2  4  3( y  0)

V (0, 0), F (0,  3), D : y  3

92. x 2  12 y

Solution x 2  12 y ( x  0)2  4  3( y  0)

V (0, 0), F (0, 3), D : y  3

93. ( y  1)2  8( x  2)

Solution ( y  1)2  8( x  2) ( y  1)2  4  2( x  2)

V ( 2, 1), F ( 4, 1), D : x  0

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1685


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

94. ( y  1)2  8( x  2)

Solution ( y  1)2  8( x  2) ( y  1)2  4  2( x  2)

V ( 2, 1), F (0, 1), D : x  4

95. x 2  2 x  8 y  15  0

Solution x 2  2 x  8 y  15  0 x 2  2 x  8 y  15 x 2  2 x  1  8 y  15  1 ( x  1)2  8( y  2)

V (1,  2), F (1, 0), D : y  4

96. x 2  2 x  8 y  17

Solution x 2  2 x  8 y  17 x 2  2 x  8 y  17 x 2  2 x  1  8 y  17  1 ( x  1)2  8( y  2)

V (1,  2), F (1,  4), D : y  0

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1686


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

97. Use a graphing calculator to graph the parabola y 2  4 x  12. Sketch the parabola by hand and compare the results.

Solution y 2  4 x  12

y   4 x  12

98. Use a graphing calculator to graph the parabola y 2  8 x  24  0. Sketch the parabola by hand and compare the results.

Solution y 2  8 x  24  0 y 2  8 x  24 y   8 x  24

Fix It In exercises 99 and 100, identify the step where the first error is made and fix it. 99. Write the parabola y 2  2 x  6 y  1 in standard form and identify its vertex.

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1687


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution Step 5 was incorrect. Step 5: vertex is ( 4, 3) 100. Graph the parabola y 2  4 y  8 x  12. To do so, write it in standard form, identify its vertex, determine the y-intercepts, and then draw its graph.

Solution Step 4 was incorrect. Step 4:

Applications 101. Broadcast range A television tower broadcasts a signal with a circular range, as shown in the illustration. Can a city 50 miles east and 70 miles north of the tower receive the signal?

Solution Check the coordinates:

x 2  y 2  502  702  2500  4900  7400 7400  8100, so the city can receive. 102. Warning sirens A tornado warning siren can be heard in the circular range shown in the illustration. Can a person 4 miles west and 5 miles south of the siren hear its sound?

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1688


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution Check the coordinates:

x 2  y 2  ( 4)2  ( 5)2  16  25  41 Since 41  36, the siren cannot be heard. 103. Radio translators Some radio stations extend their broadcast range by installing a translator—a remote device that receives the signal and retransmits it. A station with a broadcast range given by x 2  y 2  1600, where x and y are in miles, installs a translator with a broadcast area bounded by x 2  y 2  70 y  600  0. Find the greatest distance from the main transmitter that the signal can be received.

Solution 2 2  x  y  1600  Graph both circles: x 2  ( y  35)2  625 

The point farthest from the transmitter (0, 0) is the point (0, 60). The greatest distance is 60 miles. 104. Ripples in a pond When a stone is thrown into the center of a pond, the ripples spread out in a circular pattern, moving at a rate of 3 feet per second. If the stone is dropped at the point (0, 0) in the illustration, when will the ripple reach the seagull floating at the point (15, 36)?

Solution

distance  (15  0)2  (36  0)2  225  1296  1521  39 Since the distance is 39 feet, the ripples will reach the seagull in 13 seconds.

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1689


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

105. Writing equations of circles Find an equation in standard form of the circle whose outer rim is the circular arch shown in the illustration.

Solution C(4, 0), r  4

( x  h)2  ( y  k )2  r 2 ( x  4)2  ( y  0)2  42 ( x  4)2  y 2  16 106. Writing equations of circles The shape of the window shown is a combination of a rectangle and a semicircle. Find an equation in standard form of the circle of which the semicircle is a part.

Solution C(5, 14), r  5

( x  h)2  ( y  k )2  r 2 ( x  5)2  ( y  14)2  52 ( x  5)2  ( y  14)2  25 107. Meshing gears For design purposes, the large gear is described by the circle x 2  y 2  16. The smaller gear is a circle centered at (7, 0) and tangent to the larger circle. Find an equation in standard form of the smaller gear.

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1690


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution x 2  y 2  16: C(0, 0), r  4 Small gear: C(7, 0), r  3

( x  h)2  ( y  k )2  r 2 ( x  7)2  ( y  0)2  32 ( x  7)2  y 2  9 108. Walkways The walkway shown is bounded by the two circles x 2  y 2  2500 and ( x  10)2  y 2  900, measured in feet. Find the largest and the smallest width of the walkway.

Solution x 2  y 2  2500

C(0, 0), r  50

( x  10)2  y 2  900 C(10, 0), r  30

Endpoints on x: ( 50, 0), (50, 0)

Endpoints on x: ( 20, 0), (40, 0)

Smallest width  50  40  10 ft Largest width  20  ( 50)  30 ft

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1691


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

109. Solar furnaces A parabolic mirror collects rays of the sun and concentrates them at its focus. In the illustration, how far from the vertex of the parabolic mirror will it get the hottest? (All measurements are in feet.)

Solution Find the distance to the focus: 4p  8  p  2 It will be hottest 2 feet from the vertex. 110. Searchlight reflector A parabolic mirror reflects light in a beam when the light source is placed at its focus. In the illustration, how far from the vertex of the parabolic reflector should the light source be placed? (All measurements are in feet.)

Solution Find the distance to the focus: 4 p  12  p  3 The light should be 3 feet from the vertex. 111. Writing equations of parabolas Derive an equation of the parabolic arch shown.

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1692


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution The vertex is (0, 0), while (15, –10) is on the curve (vertical parabola): ( x  h)2  4 p( y  k ) (15  0)2  4 p( 10  0) 225  40 p 45 45   4p  x 2   y 2 2 112. Projectiles The cannonball in the illustration follows the parabolic trajectory y  30 x  x 2 . How far short of the castle does it land?

Solution y  30 x  x 2 0  30 x  x 2 0  x(30  x ) x  0 or 30  x  0 30  x Hits at (30, 0)  5 feet short

113. Satellite antennas The cross section of the satellite antenna in the illustration is a parabola given by the equation y  1 x 2 , with distances measured in feet. If the dish 16

is 8 feet wide, how deep is it?

Solution The depth is the y-coordinate when x  4. [4 feet on each side of the vertex] 1 2 x 16 1 y  (4)2  1 16 The depth is 1 foot. y 

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1693


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

114. Design of a satellite antenna The cross section of the satellite antenna shown is a parabola with the pickup at its focus. Find the distance d from the pickup to the center of the dish.

Solution The vertex is (0, 0), while (1, 3) is on the curve (horizontal parabola): ( y  k )2  4 p( x  h) (3  0)2  4 p(1  0) 9  4p 9 9  pd  feet 4 4 115. Toy rockets A toy rocket is s feet above the Earth at the end of t seconds, where s  16t 2  80 3t . Find the maximum height of the rocket.

Solution

s  16t 2  80 3t 1 s  t 2  5 3t 16 1 75 75  s  t 2  5 3t  16 4 4 

 1 5 3   ( s  300)   t   16 2  

2

The maximum height is 300 feet. 116. Operating a resort A resort owner plans to build and rent n cabins for d dollars per week. The price d that she can charge for each cabin depends on the number of cabins she builds, where d  45

   . Find the number of cabins she should build n 32

1 2

to maximize her weekly income.

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1694


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution

Income  Pr ice  # rented  n 1 y  45   n 32 2   45 2 45 y  n  n 32 2 

32 y  n2  16n 45

32 y  64  n2  16n  64 45 32  y  64  (n  8)2 45 She should build 8 cabins. 

117. Design of a parabolic reflector Find the outer diameter (the length AB ) of the parabolic reflector shown.

Solution Place the vertex at (0, 0), with the focus at (1, 0)  p  1, 4 p  4. ( y  k )2  4 p( x  h)

y 2  4x Let x = 10:

y 2  4x y 2  4(10)  y   40

The width = 2 40  12.6 cm. 118. Design of a suspension bridge The cable between the towers of the suspension bridge shown in the illustration has the shape of a parabola with vertex 15 feet above the roadway. Find an equation in standard form of the parabola.

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1695


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution The vertex is (0, 15), while (450, 120) is on the curve (vertical parabola): ( x  h)2  4 p( y  k ) (450  0)2  4 p(120  15) 202, 500  420 p 13, 500 13, 500  4p  x 2  ( y  15) 7 7 119. Gateway Arch The Gateway Arch in St. Louis has a shape that approximates a parabola. (See the illustration.) Find the width w of the arch 200 feet above the ground. Round to the nearest foot.

Solution The vertex is (0, 0), while (315, –630) is on the curve (vertical parabola): ( x  h)2  4 p( y  k ) (315  0)2  4 p( 630  0) 99, 225  2520 p 19845 19845   p  4p   504 126 Let y = –430: 19845 x2   y 126 19845 x2   ( 430) 126 19845 x  ( 430) 126 x  260 The width is about 520 feet. 120. Building tunnels A construction firm plans to build a tunnel whose arch is in the shape of a parabola. (See the illustration.) The tunnel will span a two-lane highway 8 meters wide. To allow safe passage for vehicles, the tunnel must be 5 meters high at a distance of 1 meter from the tunnel’s edge. Find the maximum height of the tunnel.

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1696


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution The equation is ( x  h)2  4 p( y  k ). From the figure, h = 0. Also, the points (4, 0) and (3, 5) are on the graph: x 2  4 p( y  k ) x 2  4 p( y  k ) 42  4 p(0  k )

32  4 p(5  k )

16  4 pk

9  20 p  4 pk

4  pk

Substituting:

9  20 p  4 pk 9  20 p  4( 4) 7 80  pk  20 7

The maximum height is

80 meters. 7

Discovery and Writing 121. Describe a parabola.

Solution Answers may vary. 122. How can you recognize the equation of a parabola when compared to other equations?

Solution Answers may vary. 123. Show that the standard form of the equation of a parabola ( y  2)2  8( x  1) is a special case of the general form of a second-degree equation in two variables.

Solution ( y  2)2  8( x  1) y 2  4 y  4  8x  8 y 2  8 x  4 y  12  0 0 x 2  0 xy  y 2  8 x  4 y  12  0

124. Show that the standard form of the equation of a circle ( x  2)2  ( y  5)2  36 is a special case of the general form of a second-degree equation in two variables.

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1697


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution ( x  2)2  ( y  5)2  36

x 2  4 x  4  y 2  10 y  25  36 x 2  0 xy  y 2  4 x  10 y  7  0 Find an equation, in the form ( x  h )2  ( y  k )2  r 2, of the circle passing through the given points. 125. (0, 8), (5, 3), and (4, 6)

Solution

( x  h)2  ( y  k )2  r 2

( x  h)2  ( y  k )2  r 2

(0  h)2  (8  k )2  r 2

(5  h)2  (3  k )2  r 2

h2  64  16k  k 2  r 2

25  10h  h2  9  6k  k 2  r 2

h2  k 2  r 2  16k  64

h2  k 2  r 2  10h  6k  34

( x  h)2  ( y  k )2  r 2 (4  h)2  (6  k )2  r 2 16  8h  h2  36  12k  k 2  r 2 h2  k 2  r 2  8h  12k  52  16k  64  10h  6k  34    16k  64  8h  12k  52 

10k  10h  30 4k  8h  12

 k  3, h  0

Substitute into one of the above equations to get r = 5.

Circle: x 2  ( y  3)2  25

126. ( 2, 0), (2, 8), and (5,  1)

Solution

( x  h)2  ( y  k )2  r 2

( x  h)2  ( y  k )2  r 2

( 2  h)2  (0  k )2  r 2

(2  h)2  (8  k )2  r 2

4  4h  h2  k 2  r 2

4  4h  h2  64  16k  k 2  r 2

h2  k 2  r 2  4h  4

h2  k 2  r 2  4h  16k  68

( x  h)2  ( y  k )2  r 2 (5  h)2  ( 1  k )2  r 2 25  10h  h2  1  2k  k 2  r 2 h2  k 2  r 2  10h  2k  26 4k  4  4 h  16k  68    4k  4  10h  2k  26 

8h  16k  64 14h  2k  22

 h  2, k  3

Substitute into one of the above equations to get r = 5.

Circle: ( x  2)2  ( y  3)2  25

Find an equation of the parabola passing through the given points. Give the equation in the form y  ax 2  bx  c . 127. (1, 8), ( 2,  1), and (2, 15)

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1698


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution y  ax 2  bx  c

y  ax 2  bx  c

8  a(1)2  b(1)  c

1  a( 2)2  b( 2)  c 15  a(2)2  b(2)  c

8 abc

1  4a  2b  c

y  ax 2  bx  c 15  4a  2b  c

 abc 8  2 4a  2b  c  1  a  1, b  4, c  3  y  x  4 x  3 4a  2b  c  15 

128. (1,  3), ( 2, 12), and (  1, 3)

Solution y  ax 2  bx  c

y  ax 2  bx  c

3  a(1)2  b(1)  c

12  a( 2)2  b( 2)  c 3  a( 1)2  b( 1)  c

3  a  b  c

12  4a  2b  c

y  ax 2  bx  c 3 abc

 a  b  c  3  2 4a  2b  c  12  a  2, b  3, c  2  y  2 x  3 x  2  abc  3 

129. Projectile motion A stone tossed upward is s feet above the Earth after t seconds, where s  16t 2  128t. Show that the stone’s height x seconds after it is thrown is equal to its height x seconds before it hits the ground.

Solution The stone hits the ground when s = 0: 0  16t 2  128t 0  16t (t  8) It hits the ground after 8 seconds. Find s when t  x :

Find s when t  8  x : s  16(8  x )2  128(8  x )  16(64  16 x  x 2 )  1024  128 x  1024  256 x  16 x 2  1024  128 x  16 x 2  128 x

s  16 x 2  128 x

130. Ballistics Show that the stone in Exercise 93 reaches its greatest height in one-half of the time it takes until it strikes the ground.

Solution The maximum height occurs at the vertex: s  16 x 2  128 x 1  s  t 2  8t 16 1  s  16  t 2  8t  16 16 1  s  16  (t  4)2 16

The maximum height occurs after 4 seconds, which is half the time of 8 seconds (found in #129) it takes the stone to hit the ground.

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1699


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Critical Thinking In Exercises 131–134, match the equation of the parabola with its graph. 131. x 2  9 y

Solution b 132. x 2  9 y

Solution d 133. y 2  9 x

Solution c 134. y 2  9 x

Solution a a.

b.

c.

d.

In Exercises 135–138, match the equation of the parabola with its graph. 135. ( y  2)2  8( x  2)

Solution b 136. ( y  2)2  8( x  2)

Solution a

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1700


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

137. ( x  2)2  8( y  2)

Solution d 138. ( x  2)2  8( y  2)

Solution c a.

b.

c.

d.

EXERCISES 7.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

2

Consider the equation x  a2

y2  1. b2

If a = 7 and b = 11, write the equation that results.

Solution x2 y2  1 49 121 2. If c 2  a 2  b2 , find b2 if c  4 3 and a = 8.

Solution c 2  a 2  b2

4 3   8  b 2

2

2

48  64  b2 16  b2 16  b2

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1701


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

2

3. Given the equation x  81

y2  1. 4

a. Identify the x-intercepts. b. Identify the y-intercepts.

Solution a. (9, 0), (–9, 0) b. (0, 2), (0, –2) 4. Divide each term of the equation 9 x 2  25 y 2  225 by 225 and simplify.

Solution 9 x 2 25 y 2   225 225 225

x2 y 2  1 25 9 5. Given the equation of the circle x 2  y 2  4 x  6 y  4  0. Complete the square on x and y. Then write the equation in standard form.

Solution x2  y 2  4x  6 y  4  0 x 2  4 x  y 2  6 y  4 ( x 2  4 x  4)  ( y 2  6 y  9)  4  13 ( x  2)2  ( y  3)2  9

6. Given the equation of the parabola 3 x 2  12 x  8 y  20  0. Complete the square on x and write the equation in standard form.

Solution 3 x 2  12 x  8 y  20  0 3 x 2  12 x  8 y  20 3( x 2  4 x  4)  8 y  20  12 3( x  2)2  8( y  1)

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. An ellipse is the set of all points in the plane such that the __________ of the distances from two fixed points is a positive __________.

Solution sum, constant

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1702


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

8. Each of the two fixed points in the definition of an ellipse is called a __________ of the ellipse.

Solution focus 9. The chord that joins the __________ is called the major axis of the ellipse.

Solution vertices 10. The chord through the center of an ellipse and perpendicular to the major axis is called the __________ axis.

Solution minor 2

11. In the ellipse x  a2

y2  1 (a  b  0), the vertices are V(_____, _____) and Vʹ(_____, b2

_____).

Solution (a, 0), ( a, 0) 12. In an ellipse, the relationship between a, b, and c is __________.

Solution b2  a 2  c 2 , or c2  a2  c 2

13. To draw an ellipse that is 26 inches wide and 10 inches tall, how long should the piece of string be, and how far apart should the two thumbtacks be?

Solution 2a  26  String: 26 inches long 2b  10  b  5 b2  a 2  c 2 52  132  c 2  c  12 Thumbtacks: 2c  24 inches apart

14. To draw an ellipse that is 20 centimeters wide and 12 centimeters tall, how long should the piece of string be, and how far apart should the two thumbtacks be?

Solution 2a  20  String: 20 cm long 2b  12  b  6 b2  a 2  c 2 62  102  c 2  c  8 Thumbtacks: 2c  16 cm apart

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1703


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Identify the conic as a circle, parabola, or an ellipse. 15. x 2  6 x  y 2  7

Solution Both variables squared with equal coefficients: circle 16. 5 x 2  5 y 2  10 y 

24 0 5

Solution Both variables squared with equal coefficients: circle 17. x 2  4 y  5  2 x

Solution One variable squared: parabola 18. y 2  8 y  2 x  20

Solution One variable squared: parabola 19. 7 x 2  5 y 2  35  0

Solution Both variables squared with unequal coefficients: ellipse 20. 5 x 2  2 y 2  10 x  4 y  13

Solution Both variables squared with unequal coefficients: ellipse Practice Write an equation in standard form of the ellipse described. The center of each ellipse is the origin. 21. Major axis of length 8 units located on the x-axis and minor axis of length 6 units

Solution a  4, b  3; horizontal C(0, 0) x2 y 2  1 16 9

22. Major axis of length 14 units located on the y-axis and minor axis of length 10 units

Solution a  7, b  5; vertical C(0, 0) x2 y 2  1 25 49

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1704


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

23. Focus at (3, 0); vertex at (5, 0)

Solution c  3, a  5; horizontal

b2  a2  c2  25  9  16 x2 y 2  1 25 16 24. Focus at (0, 4); vertex at (0, 7)

Solution c  4, a  7; vertical

b2  a2  c2  49  16  33 2 x y2  1 33 49 25. Focus at (0, 1); 4 is one-half the length of the minor axis 3

Solution c  1, b 

4 ; vertical 3

a 2  b2  c 2 16 25  1 9 9

x2 y2  1 16 9 25 9 9x 2 9 y 2  1 16 25 26. Focus at (1, 0); 4 is one-half the length of the minor axis 3

Solution c  1, b 

4 ; horizontal 3

a 2  b2  c 2 16 25  1 9 9

x2 y2  1 25 9 16 9 9x 2 9 y 2  1 25 16 27. Focus at (0, 3); major axis equal to 8

Solution c  3, a  4; vertical

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1705


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

b2  a2  c2  16  9  7 2 x y2  1 7 16 28. Focus at (5, 0); major axis equal to 12

Solution c  5, a  6; horizontal

b2  a2  c2  36  25  11 2 x y2  1 36 11 Write an equation in standard form of each ellipse described. 29. Center at (3, 4); a = 3, b = 2; major axis parallel to the y-axis

Solution vertical

( x  3)2 ( y  4)2  1 4 9 30. Center at (3, 4); passes through (3, 10) and (3, –2); b =2

Solution a  6, vertical

( x  3)2 ( y  4)2  1 4 36 31. Center at (3, 4); a = 3, b = 2; major axis parallel to the x-axis

Solution horizontal

( x  3)2 ( y  4)2  1 9 4 32. Center at (3, 4); passes through (8, 4) and (–2, 4); b = 2

Solution a  5, horizontal

( x  3)2 ( y  4)2  1 25 4 33. Foci at (–2, 4) and (8, 4); b = 4

Solution Center: (3, 4), b  4, c  5, horizontal a2  b2  c2  16  25  41

( x  3)2 ( y  4)2  1 41 16

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1706


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

34. Foci at (8, 5) and (4, 5); b = 3

Solution Center: (6, 5), b  3, c  2, horizontal a2  b2  c 2  9  4  13

( x  6)2 ( y  5)2  1 13 9 35. Vertex at (6, 4); foci at (–4, 4) and (4, 4)

Solution Center: (0, 4), c  4, a  6, horizontal b2  a2  c 2  36  16  20

x 2 ( y  4)2  1 36 20 36. Center at ( 4, 5); c  1 ; vertex at (–4, –1) a

3

Solution a  6, c  2, vertical b2  a2  c 2  36  4  32

( x  4)2 ( y  5)2  1 32 36 37. Foci at (6, 0) and (–6, 0); c  3 a

5

Solution Center: (0, 0), c  6, a  10, horizontal b2  a2  c2  100  36  64

x2 y2  1 100 64 2

38. Vertices at (2, 0) and (–2, 0); 2b  2 a

Solution Center: (0, 0), a  2, b2  2, horizontal x2 y 2  1 4 2

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1707


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Write the standard form of the equation of each ellipse. 39.

Solution x2 y 2  1 92 52 x2 y 2  1 81 25 40.

Solution x2 y 2  1 82 62 x2 y 2  1 64 36 41.

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1708


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution x2 y 2  1 42 7 2 x2 y 2  1 16 49 42.

Solution x2 y2  1 32 102 x2 y2  1 9 100 Graph each ellipse. 43.

x2 y 2  1 25 9 Solution x2 y 2  1 25 9 Center: (0, 0), a  5, b  3, horizontal

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1709


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

44.

x2 y 2  1 36 25 Solution x2 y 2  1 36 25 Center: (0, 0), a  6, b  5, horizontal

45.

x2 y 2  1 25 49 Solution x2 y 2  1 25 49 Center: (0, 0), a  7, b  5, vertical

46. 4 x 2  y 2  4

Solution 4x2  y 2  4

4x2 y 2 4   4 4 4 x2 y 2  1 1 4 Center: (0, 0), a  2, b  1, vertical

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1710


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

47.

x 2 ( y  2)2  1 16 36 Solution x 2 ( y  2)2  1 16 36 Center: (0,  2), a  6, b  4, vertical

48. ( x  1)2 

4y2 4 25

Solution 4y2 4 25 ( x  1)2 4y2 4   4 4(25) 4 ( x  1)2 y 2  1 4 25 Center: (1, 0), a  5, b  2, vertical ( x  1)2 

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1711


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

49.

( x  4)2 ( y  2)2  1 49 9 Solution ( x  4)2 ( y  2)2  1 49 9 Center: (4, 2), a  7, b  3, horizontal

50.

( x  1)2 y 2  1 25 4 Solution ( x  1)2 y 2  1 25 4 Center: (1, 0), a  5, b  2, horizontal

Graph each ellipse and identify the foci. 51.

x2 y 2  1 9 4 Solution x2 y 2  1 9 4 Center: (0, 0) a  3, b  2 c 2  32  22 c2  9  4 c 5

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1712


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Foci: ( 5, 0), (  5, 0)

52. x 2  4 y 2  36

Solution

x2 y 2  1 36 9 Center: (0, 0) a  6, b  3 x 2  4 y 2  36 

c 2  62  32 c 2  36  9 c   27  3 3

Foci: (3 3, 0), ( 3 3, 0)

53. 25 x 2  4 y 2  100

Solution

x2 y 2  1 4 25 Center: (0, 0) a  5, b  2 25 x 2  4 y 2  100 

c 2  25  4 c 2  21 c 2   21

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1713


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Foci: (0,

54.

21), (0,  21)

x2 y 2  1 4 16 Solution x2 y 2  1 4 16 Center: (0, 0) a  4, b  2 c2  16  4 c2  12 c  2 3

Foci: (0, 2 3), (0,  2 3)

55.

( x  4)2 ( y  5)2  1 16 4 Solution ( x  4)2 ( y  5)2  1 16 4 Center: (4, 5) a  4, b  2 c 2  16  4 c 2  12 c  2 3

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1714


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Foci: (4  2 3, 5), (4  2 3, 5)

56.

( x  1)2 ( y  3)2  1 25 4 Solution ( x  1)2 ( y  3)2  1 25 4 Center: ( 1,  3) a  5, b  2 c 2  25  4 c 2  21 c   21 Foci: ( 1  21,  3), ( 1  21,  3)

57. 9( x  2)2  4( y  4)2  36

Solution

9( x  2)2  4( y  4)2  36 

( x  2)2 ( y  4)2  1 4 9

Center: (2,  4) a  3, b  2

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1715


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

c2  9  4 c2  5 c 5

Foci: (2,  4  5), (2,  4  5)

2

58.

( x  3)2 ( y  2)  1 9 16

Solution 2

( x  3)2 ( y  2)  1 9 16 Center: ( 3, 2) a  4, b  3

c 2  16  9 c2  7 c 7 Foci: ( 3, 2  7), ( 3, 2  7)

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1716


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Write each ellipse in standard form. 59. 4 x 2  y 2  2 y  15

Solution 4 x 2  y 2  2 y  15 4 x 2  y 2  2 y  1  15  1 4 x 2  ( y  1)2  16 4 x 2 ( y  1)2 16   16 16 16

x 2 ( y  1)2  1 4 16 60. 4 x 2  25 y 2  50 y  75  0

Solution 4 x 2  25 y 2  50 y  75  0 4 x 2  25( y 2  2 y )  75 4 x 2  25( y 2  2 y  1)  75  25 4 x 2  25( y  1)2  100

x 2 ( y  1)2  1 25 4 61. 4 x 2  25 y 2  8 x  96  0

Solution

4 x 2  25 y 2  8 x  96  0 4( x 2  2 x )  25 y 2  96 4( x 2  2 x  1)  25 y 2  96  4 4( x  1)2  25 y 2  100 4( x  1)2 25 y 2 100   100 100 100 2 2 ( x  1) y  1 25 4 62. 9 x 2  49 y 2  54 x  360  0

Solution 9 x 2  49 y 2  54 x  360  0 9 x 2  54 x  49 y 2  360 9( x 2  6 x  9)  49 y 2  360  81 9( x  3)2  49 y 2  441

( x  3)2 y 2  1 49 9

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1717


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

63. 9 x 2  4 y 2  18 x  16 y  11  0

Solution

9 x 2  4 y 2  18 x  16 y  11  0 9( x 2  2 x )  4( y 2  4 y )  11 9( x 2  2 x  1)  4( y 2  4 y  4)  11  9  16 9( x  1)2  4( y  2)2  36 9( x  1)2 4( y  2)2 36   36 36 36 2 2 ( x  1) ( y  2)  1 4 9 64. x 2  4 y 2  10 x  8 y  13

Solution

x 2  4 y 2  10 x  8 y  13 x 2  10 x  4( y 2  2 y )  13 x  10 x  25  4( y 2  2 y  1)  13  25  4 2

( x  5)2  4( y  1)2  16 ( x  5)2 4( y  1)2 16   16 16 16 2 2 ( x  5) ( y  1)  1 16 4 65. 9 x 2  16 y 2  36 x  96 y  36

Solution 9 x 2  16 y 2  36 x  96 y  36 9( x 2  4 x )  16( y 2  6 y )  36 9( x 2  4 x  4)  16( y 2  6 y  9)  36  36  144 9( x  2)2  16( y  3)2  144

( x  2)2 ( y  3)2  1 16 9 66. 25 x 2  4 y 2  50 x  56 y  121  0

Solution

25 x 2  4 y 2  50 x  56 y  121  0 25 x 2  50 x  4 y 2  56 y  121 25( x 2  2 x )  4( y 2  14 y )  121 25( x 2  2 x  1)  4( y 2  14 y  49)  121  25  196 25( x  1)2  4( y  7)2  100

( x  1)2 ( y  7)2  1 4 25

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1718


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Graph each ellipse. 67. x 2  4 y 2  4 x  8 y  4  0

Solution x2  4 y 2  4x  8 y  4  0 x 2  4 x  4( y 2  2 y )  4 x 2  4 x  4  4( y 2  2 y  1)  4  4  4 ( x  2)2  4( y  1)2  4 ( x  2)2 ( y  1)2  1 4 1 Center: (2,  1), a  2, b  1, horizontal

68. x 2  4 y 2  2 x  16 y  13

Solution x 2  4 y 2  2 x  16 y  13 x 2  2 x  4( y 2  4 y )  13 x 2  2 x  1  4( y 2  4 y  4)  13  1  16 ( x  1)2  4( y  2)2  4 ( x  1)2 ( y  2)2  1 4 1 Center: (1, 2), a  2, b  1, horizontal

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1719


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

69. 4 x 2  y 2  24 x  2 y  36  0

Solution 4 x 2  y 2  24 x  2 y  36  0 4( x 2  6 x )  ( y 2  2 y )  36 4( x  6 x  9)  ( y 2  2 y  1)  36  36  1 2

( x  3)2 ( y  1)2  1 1 4

70. 9 x 2  y 2  18 x  10 y  33  0

Solution 9 x 2  y 2  18 x  10 y  25  0 9 x 2  18 x  y 2  10 y  25 9( x 2  2 x  1)  ( y 2  10 y  25)  25  9  25 9( x  1)2  ( y  5)2  9

( x  1)2 ( y  5)2  1 1 9

71. 16 x 2  25 y 2  160 x  200 y  400  0

Solution 16 x 2  25 y 2  160 x  200 y  400  0 16( x 2  10 x )  25( y 2  8 y )  400 16( x 2  10 x  25)  25( y 2  8 y  16)  400  400  400 16( x  5)2  25( y  4)2  400

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1720


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

( x  5)2 ( y  4)2  1 25 16 Center: (5, 4), a  5, b  4, horizontal

72. 3 x 2  2 y 2  7 x  6 y  1

Solution 3 x 2  2 y 2  7 x  6 y  1  7  3  x 2  x   2( y 2  3 y )  1 3     2 7 49  9 49 9 3  x2  x     2  y  3 y    1  3 36  4 12 2   2

2

  7 3 91 3 x    2 y    6 2 12   2

2

  7 3 x   y   6 2   1 91 36 91 24  7 3 Center:   ,  , a   6 2

91 ,b 36

91 , vertical 24

Graph each ellipse and identify the foci. 73. x 2  9 y 2  6 x  36 y  36  0

Solution x 2  9 y 2  6 x  36 y  36  0 x 2  6 x  9 y 2  36 y  36 ( x  6 x  9)  9( y 2  4 y  4)  36  9  36 2

( x  3)2  9( y  2)2  9

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1721


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

( x  3)2 ( y  2)2  1 9 1 Center: ( 3, 2) a  3, b  1 c2  9  1 c2  8 c  2 2 Foci: (  3  2 2, 2), ( 3  2 2, 2)

74. 4 x 2  y 2  8 x  2 y  1  0

Solution 4 x 2  y 2  8x  2 y  1  0 4 x 2  8 x  y 2  2 y  1 4( x 2  2 x  1)  ( y 2  2 y  1)  1  4  1 4( x  1)2  ( y  1)2  4 ( x  1)2 ( y  1)2  1 1 4 Center: (1,  1) a  4, b  1

c2  4  1 c 3 Foci: ( 1,  1  3), (1,  1  3)

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1722


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

75. 9 x 2  4 y 2  18 x  16 y  11  0

Solution

9 x 2  4 y 2  18 x  16 y  11  0 9 x 2  18 x  4 y 2  16 y  11 9( x 2  2 x )  4( y 2  4 y )  11 9( x  2 x  1)  4( y 2  4 y  4)  11  9  16 2

9( x  1)2  4( y  2)2  36

( x  1)2 ( y  2)2  1 4 9 Center: ( 1,  2) a  3, b  2 c2  9  4 c2  5 c 5

Foci: (  1,  2  5), ( 1,  2  5)

76. 25 x 2  16 y 2  150 x  128 y  81  0

Solution 25 x 2  16 y 2  150 x  128 y  81  0 25 x 2  150 x  16 y 2  128 y  81 25( x  6 x  9)  16( y 2  8 y  16)  81  225  256 2

25( x  3)2  16( y  4)2  400

( x  3)2 ( y  4)2  1 16 25 Center: (3, 4) a  5, b  4 c 2  25  16 c2  9 c  3

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1723


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Foci: (3, 7), (3, 1)

77. 9 x 2  25 y 2  90 x  200 y  400  0

Solution 9 x 2  25 y 2  90 x  200 y  400  0 9 x 2  90 x  25 y 2  200 y  400 9( x 2  10 x  25)  25( y 2  8 y  16)  400  225  400 9( x  5)2  25( y  4)2  225

( x  5)2 ( y  4)2  1 25 9 Center: (5,  4) a  5, b  3 c 2  25  9 c 2  16 c  4 Foci: (9,  4), (1,  4)

78. x 2  9 y 2  10 x  108 y  313  0

Solution

x 2  9 y 2  10 x  108 y  313  0 x 2  10 x  9 y 2  108 y  313 x 2  10 x  9( y 2  12 y )  313 ( x 2  10 x  25)  9( y 2  12 y  36)  313  25  324 ( x  5)2  9( y  6)2  36

( x  5)2 ( y  6)2  1 36 4 Center: ( 5, 6) a  6, b  2

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1724


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

c 2  36  4 c 2  32 c  4 2 Foci: (  5  4 2, 6), (  5  4 2, 6)

2

79. Use a graphing calculator to graph the ellipse x  4

y2  1. 36

Sketch the ellipse by hand

and compare the results.

Solution x2 y 2  1 4 36 9 x 2  4 y 2  36 4 y 2  36  9 x 2 36  9 x 2 4 36  9 x 2 y  4

y2 

80. Use a graphing calculator to graph the ellipse

( y  2)2 ( x  3)2   1. 4 25

Sketch the ellipse by

hand and compare the results.

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1725


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution ( x  3)2 ( y  2)2  1 4 25 25( x  3)2  4( y  2)2  100 4( y  2)2  100  25( x  3)2 4( y  2)2   100  25( x  3)2 2( y  2)   100  25( x  3)2 100  25( x  3)2 2 100  25( x  3)2 y  2 2

y 2  

Fix It In exercises 81 and 82, identify the step where the first error is made and fix it. 81. Write the ellipse 4 x 2  25 y 2  16 x  100 y  16  0 in standard form.

Solution Step 3 was incorrect: Step 3: 4( x 2  4 x  4)  25( y 2  4 y  4)  16  16  100 Step 4: 4( x  2)2  25( y  2)2  100 Step 5:

( x  2)2 ( y  2)2  1 25 4

82. Graph the ellipse 25 x 2  8 y 2  200. To do so, write it in standard form, identify its vertex, determine the x and y-intercepts, and then draw its graph.

Solution Step 5 was incorrect.

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1726


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Step 5:

Applications 83. Pool tables Find an equation in standard form of the outer edge of the elliptical pool table shown below.

Solution C(0, 0), a  30, b  20, horizontal

( x  0) 302

2

( y  0)

2

1 202 x2 y2  1 900 400

84. Equation of an arch An arch is a semiellipse 12 meters wide and 5 meters high. Write an equation in standard form of the ellipse if the ellipse is centered at the origin.

Solution a  6, b  5

x2 y 2  1 36 25 85. Design of a track A track is built in the shape of an ellipse with a maximum length of 100 meters and a maximum width of 60 meters. Write an equation in standard form of

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1727


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

the ellipse and find its focal width. That is, find the length of a chord that is perpendicular to the major axis and passes through either focus of the ellipse.

Solution a  50, b  30

c 2  a 2  b2  2500  900  1600 c  40

x2 y2  1 2500 900

402 y2  1 2500 900 900 y2  900 2500 30 y  30 50 y  18 The focal width is 36 meters.

86. Whispering galleries Any sound from one focus of an ellipse reflects off the ellipse directly back to the other focus. This property explains whispering galleries such as Statuary Hall in Washington, D.C. The ceiling of the whispering gallery shown has the shape of a semiellipse. Find the distance sound travels as it leaves focus F and returns to focus Fʹ.

Solution Using the Pythagorean Theorem:

The distance is 26 meters.

87. Finding the width of a mirror Many mirrors are oval shaped. The dimensions of a mirror are shown, and the mirror is in the shape of an ellipse. Find the width of the mirror 12 inches above its base.

Solution a  24, b  12

x2 y2  1 144 576 x2 ( 12)2  1 144 576 x2 432  144 576 x  10.4 The width is about 20.8 inches.

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1728


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

88. Finding the height of a window The window shown has the shape of an ellipse. Find the height of the window 20 inches from one end.

Solution a  36, b  30 x2 y2  1 1296 900 162 y2  1 1296 900 1040 y2  900 1296 y  13.45

The width is about 26.9 inches. 89. Astronomy The moon has an orbit that is an ellipse, with the Earth at one focus. If the major axis of the orbit is 378,000 miles and the ratio of c to a is approximately 11 , how far does the moon get from the Earth? (This farthest point in an orbit is 200

called the apogee.)

Solution The farthest distance = a + c:

378000  189000 2 11 c  a 200 11a  10395 c 200 distance = a + c = 199,395 miles

a

2

90. Area of an ellipse The area A of the ellipse x  a2

y2  1 is given by b2

A   ab. Find the

area of the ellipse 9 x 2  16 y 2  144.

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1729


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution 9 x 2  16 y 2  144

9 x 2 16 y 2 144   144 144 144 x2 y 2  1 16 9 a  4, b  3 A   ab   (4)(3)  12 square units,

or about 37.7 square units

Discovery and Writing 91. Describe an ellipse.

Solution Answers may vary. 92. Explain the difference between the foci and vertices of an ellipse.

Solution Answers may vary. 93. How do you distinguish among the equations of circles, parabolas, and ellipses?

Solution Answers may vary. 94. If F is a focus of the ellipse shown and B is an end-point of the minor axis, use the distance formula to prove that the length of segment FB is a. (Hint: In an ellipse, b2  a 2  c 2 . )

Solution FB 

(c  0)2  (0  b)2 

c 2  b2 

a 2  a  a(a  0)

95. If F is a focus of the ellipse shown and P is any point on the ellipse, use the distance formula to show that the length of FP is a  c x . (Hint: In an ellipse, c 2  a2  b2 . ) a

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1730


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution

x2 y 2   1. a2 b2 P is a point on the ellipse, so solve for y2: The equation of the ellipse is

2

2

x y  2 1 2 a b y2 x2  1 2 2 b a  x2  y 2  b2  1  2  a  

FP  (c  x )2  (0  y )2  (c  x )2  y 2  x2   (c  x )2  b2  1  2  a    x2   (c  x )2  (a2  c 2 )  1  2  a    c 2  2cx  x 2  a2  x 2  c2   a2  2cx 

c

2

a2

c2 a

2

x2

x2

2

 c  c  a  x   a  x a a   96. Finding the focal width In the ellipse shown, chord AAʹ passes through the focus F and is perpendicular to the major axis. Show that the length of AAʹ (called the focal 2

width) is 2b . a

Solution The equation of the ellipse is Let x = c and solve for y2: x2

 2

y2

1 a b2 c2 y 2  1 a 2 b2

 c2  y 2  b2  1  2  a    a 2  b2   b2  1   a2  

x2 y 2   1. a2 b2

Thus, y  

b2 a

2



b2 . a

 b2  The coordinates of Aʹ and A are  c,  a    b2  and  c,   . Therefore, the focal width is a   2b2 . a

 b2  b4  b2  1  1  2   2 a  a 

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1731


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

97. Prove that segment FV in Example 7 is the shortest distance between the Earth and the sun. (Hint: Refer to Exercise 95.)

Solution By the result of #95, the distance between a point P( x , y ) on the ellipse and a focus

c x. Since P is a point on the ellipse, x must take on values from –a a to a. To make D as small as possible, x must be positive and as large as possible. This occurs when x  a. If x  a, then point P is actually at point V. F (c, 0) is D  a 

98. Constructing an ellipse The ends of a piece of string 6 meters long are attached to two thumbtacks that are 2 meters apart. A pencil catches the loop and draws it tight. As the pencil is moved about the thumbtacks (always keeping the tension), an ellipse is produced, with the thumbtacks as foci. Write an equation in standard form of the ellipse. (Hint: You’ll have to establish a coordinate system.)

Solution Let the origin be at the midpoint of the line segment between the two thumbtacks and let the x-axis be parallel to that segment. Then 2a  6, so a  3. Also, 2c  2, so c  1. Find b: b2  a2  c 2  32  12  8. The equation is

x2 y 2 x2 y 2   1, or   1. 9 8 a2 b2

99. The distance between point P(x, y) and the point (0, 2) is 1 of the distance of point P 3

from the line y = 18. Find an equation in standard form of the ellipse on which point P lies.

Solution Consider the following diagram:

1 PN 3 1 ( x  0)2  ( y  2)2  ( x  x )2  ( y  18)2 3 1 x 2  ( y  2)2  [0  ( y  18)2 ] 9 1 x 2  y 2  4 y  4  ( y  18)2 9 PM 

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1732


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

9 x 2  9 y 2  36 y  36  y 2  36 y  324 9 x 2  8 y 2  288

x2 y 2  1 32 36 100. Prove that a > b in the development of the standard equation of an ellipse.

Solution In the standard development, b2 is defined as a2 – c2, where a and c are assumed to be greater than 0. Thus, a2  b2 . Since a  0, this implies a  b. 101. Show that the expansion of the standard equation of an ellipse is a special case of the general second-degree equation in two variables.

Solution

( x  h)2 ( y  k )2  1 a2 b2  ( x  h)2 ( y  k )2  2 2 a2 b2     a b (1) 2 2 b  a  b2 ( x  h)2  a2 ( y  k )2  a2 b2 b2 ( x 2  2hx  h2 )  a2 ( y 2  2ky  k 2 )  a2 b2 b2 x 2  2b2 hx  b2 h2  a2 y 2  2a2 ky  a2 k 2  a2 b2  0 b 2 x 2  0 xy  a 2 y 2  ( 2 b 2 h) x  ( 2a 2 k ) y  ( b 2 h2  a 2 y 2  a 2 b 2 )  0 102. The eccentricity of an ellipse provides a measure of how much the curve resembles a true circle. Specifically, the eccentricity of a true circle equals 0. Note that the semimajor axis is perpendicular to the axis containing the foci of an ellipse. When analyzing planetary orbits, astronomers plot the relationship between the length of the orbit’s semimajor axis (measured in Astronomical Units, where 1 AU = 149,598,000 km) and the eccentricity of the orbit. Use the given data plot to estimate how many of the 75 planets shown follow orbits that are true circles.

Source for right panel: Eccentricity vs. semimajor axis for extrasolar planets. The 75 planets shown were found in a Doppler survey of 1300 FGKM main sequence stars using the Lick, Keck, and AAT telescopes. The survey was carried out by the CaliforniaCarnegie planet search team. http://exoplanets.org/newsframe.html

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1733


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution There are 4 points with an Orbital Eccentricity coordinate of 0.0 on the graph, so there are 4 planets. Critical Thinking In Exercise 103–106, match the equation of the ellipse with its graph. 103.

x2 y 2  1 64 9 Solution d

104.

x2 y 2  1 9 64 Solution c

105.

4x2 y 2  1 49 81 Solution a

106.

x2 4 y 2  1 81 49 Solution b a.

b.

c.

d.

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1734


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

In Exercise 107–110, match the equation of the ellipse with its graph. 107.

( x  2)2 ( y  2)2  1 16 49 Solution b

108.

( x  2)2 ( y  2)2  1 16 49 Solution a

109.

( x  2)2 ( y  2)2  1 49 16 Solution c

110.

( x  2)2 ( y  2)2  1 49 16 Solution d a.

b.

c.

d.

EXERCISES 7.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Consider the equation

x2 a2

y2 b2

 1. If a  5 and b  12, write the equation that results.

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1735


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution x2 y2  1 25 144

2. If a2  b2  c2 , find b2 if c  2 13 and a  6.

Solution

a2  b2  c2

62  b2  2 13

2

36  b2  4  13 b2  16 3. Given the equation

x2 y2   1. 100 4

a. Identify the x-intercepts, if any exist. b. Identify the y-intercepts, if any exist.

Solution a.

0 x2  1 100 4 x 2  100 x  10

x-intercepts:  10, 0   10, 0  b.

0 y2  1 100 4 y2  4 y 2  4, so y is not a real number

There are no y-intercepts. 4. Given the equation

y2 x2   1. 4 100

a. Identify the x-intercepts, if any exist. b. Identify the y-intercepts, if any exist.

Solution a.

0 x2  1 4 100  x 2  100 x 2  100, so x is not a real number There are no x-intercepts

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1736


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

b.

y2 0  1 4 100 y2  4 y 2  2

x-intercepts: 0, 2, 0,  2 5. Divide each term of the equation 9x 2  25 y 2  225 by 225 and simplify.

Solution x2 y 2  1 25 9

6. Given the equation of the ellipse 4 x 2  9 y 2  56x  18 y  169  0. Complete the square on x and y and write the equation in standard form.

Solution

4 x 2  9 y 2  56 x  18 y  169  0

4 x 2  56 x  9 y 2  18 y  169

 

4 x 2  14 x  49  9 y 2  2 y  1  169  196  9 4  x  7   9  y  1  36 2

2

 x  7    y  1  1 2

9

2

4

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A hyperbola is the set of all points in the plane such that the absolute value of the __________ of the distances from two fixed points is a positive __________.

Solution difference, constant 8. Each of the two fixed points in the definition of a hyperbola is called a __________ of the hyperbola.

Solution focus 9. The vertices of the hyperbola

x2 a2

y2 b2

 1 are V( _____, _____) and V'(_____, _____).

Solution (a, 0), ( a, 0)

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1737


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

10. The vertices of the hyperbola

y2 a2

x2 b2

 1 are V( _____, _____) and V'(_____, _____)

Solution (0, a), (0,  a) 11. The chord that joins the vertices is called the __________ of the hyperbola.

Solution transverse axis 12. In a hyperbola, the relationship between a, b, and c is __________.

Solution a 2  b2  c 2

Identify the conic as a circle, parabola, ellipse, or hyperbola. 13. x 2  ( y  4)2  12

Solution Both variables squared with equal coefficients and same sign: circle 14. 7 x 2  7 y 2  70x  14 y  119

Solution Both variables squared with equal coefficients and same sign: circle 15. y 2  2x  23  5 y

Solution One variable squared: parabola 16. ( x  8)2  5( y  6)

Solution One variable squared: parabola 17.

x2 y 2  1 35 9

Solution Both variables squared with unequal coefficients and same sign: ellipse 18. 2x 2  32 y 2  4 x  30  0

Solution Both variables squared with unequal coefficients and same sign: ellipse 19. x 2  2 y 2  4 y  6  0

Solution Both variables squared with opposite signs: hyperbola

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1738


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

20.

( x  3)2 y 2  1 4 9

Solution Both variables squared with opposite signs: hyperbola Practice Write an equation in standard form of each hyperbola described. 21. Vertices (5, 0) and ( 5, 0); focus (7, 0)

Solution a  5, c  7; harizontal b2  c 2  a 2  49  25  24 2

x y2  1 25 24

22. Focus (3, 0); vertex (2, 0); center (0, 0)

Solution a  2, c  3; harizontal b2  c 2  a 2 94 5 2

x y2  1 4 5

23. Center (2, 4); a  2, b  3; transverse axis is horizontal

Solution a  2, b  3; harizontal ( x  2)2 ( y  4)2  1 4 9

24. Center ( 1, 3); vertex (1, 3); focus (2, 3)

Solution a  2, c  3; harizontal b2  c 2  a 2  94 5 ( x  1)2 ( y  3)2  1 4 5

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1739


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

25. Center (5, 3); vertex (5, 6); passes through (1, 8)

Solution a  3; vertical ( y  3)2 ( x  5)2  9 b2

1

(8  3)2 (1  5)2  1 9 b2 25 16  1 9 b2 16 16   2 9 b b2  9 ( y  3)2 ( x  5)2  1 9 9

26. Foci (0, 10) and (0,  10);

c 5  a 4

Solution Center: (0, 0), c  10, a  8; vertical b2  c 2  a 2  100  64  36 2

y x2  1 64 36

27. Vertices (0, 3) and (0,  3);

c 5  a 4

Solution Center: (0, 0), a  3, c  5; vertical b2  c 2  a 2  25  9  16 2

y x2  1 9 16

28. Focus (4, 0); vertex (2, 0); center (0, 0)

Solution c  4, a  2; horizontal b2  c 2  a 2  16  4  12 2

x y2  1 4 12

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1740


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

29. Center (1, 4); focus (7, 4); vertex (3, 4)

Solution c  6, a  2; horizontal b2  c 2  a 2  36  4  32 ( x  1)2 ( y  4)2  1 4 32

30. Center (1,  3); a2  4; b2  16

Solution ( x  1)2 ( y  3)2  1 4 16 OR ( y  3)2 ( x  1)2  1 4 16

31. Center at the origin; passes through (4, 2) and (8, 6)

Solution

x2 a2 42 a2 16 a

2

  

y2 b2 22 b2 4 2

b 16 2

a 64 a2

x2

1

a2 82

1

a2

1  1 4

4 b2 16

y2

b2 ( 6)2

64

1

1 b2 64 36  1 a2 b2

4

a2 16 b2

 

36 b2 36 b2

1

x2 3 y 2  1 10 20

1

3

20

b2 20 b2  3 2 a  10

b2

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1741


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

  3 5 , 0 32. Center (3,  1); y-intercept (0,  1); x-intercept  3    2   Solution ( x  3)2 a2 (0  3)2 a

2

 

( y  1)2 b2 ( 1  1)2 2

b

( x  3)2 ( y  1)2  1 9 b2

1

(3  3 25  3)2

1

9

9

(0  1)2 b2

( x  3)2 ( y  1)2  1 9 4

1

5 1  1 4 b2 1 1  4 b2

1 a2 9  a2

4  b2

Find the area of the fundamental rectangle of each hyperbola.

33. 4( x  1)2  9 y  2

  36 2

Solution 4( x  1)2  9  y  2   36 2

4( x  1)2

9  y  2

2

36 36

36 36 ( x  1)2 ( y  2)2  1 9 4 a  3, b  2 Area  (2a)(2b)  (6)(4)  24 sq. units

34. x2  y 2  4x  6 y  6

Solution x2  y 2  4x  6 y  6 x2  4x  ( y 2  6 y )  6 x 2  4 x  4  ( y 2  6 y  9)  6  4  9 ( x  2)2  ( y  3)2  1 ( x  2)2 ( y  3)2  1 1 1 a  1, b  1 Area  (2a)(2b)  (2)(2)  4 sq. units

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1742


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

35. x 2  6x  y 2  2 y  11

Solution x 2  6 x  y 2  2 y  11 x 2  6 x  ( y 2  2 y )  11 x 2  6 x  9  ( y 2  2 y  1)  11  9  11 ( x  3)2  ( y  1)2  3 ( x  3)2 ( y  1)2  1 3 3 ( y  1)2 ( x  3)2   1a  3 3

3, b 

3;

  

Area  (2a)(2b)  2 3 2 3  12 sq. units

36. 9 x 2  4 y 2  18 x  24 y  63

Solution

9 x 2  4 y 2  18 x  24 y  63 9( x 2  2 x )  4( y 2  6 y )  63 9( x 2  2 x  1)  4( y 2  6 y  9)  63  9  36 9( x  1)2  4( y  3)2  36 9( x  1)2 4( y  3)2 36   36 36 36 ( x  1)2 ( y  3)2   1  a  2, b  3; 4 9 Area  (2a)(2b)  (4)(6)  24 sq. units Write an equation in standard form of each hyperbola described. 37. Center ( 2,  4); a  2; area of fundamental rectangle is 36 square units

Solution

(2a)(2b)  36 4(2b)  36 9 2 ( x  2)2 4( y  4)2  1 4 81 OR b

( y  4)2 4( x  2)2  1 4 81

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1743


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

38. Center (3,  5); b  6; area of fundamental rectangle is 24 square units

Solution (2a)(2b)  24

(2a)(12)  24 a1 2

( x  3) ( y  5)2  1 1 36 OR ( y  5)2 ( x  3)2  1 1 36  5 39. Vertex (6, 0); one end of conjugate axis at  0,   4

Solution Center: (0, 0), a  6, b  45

x2 62

y2

  5 4

2

1

x 2 16 y 2  1 36 25 40. Vertex (3, 0); focus ( 5, 0); center (0, 0)

Solution a  3, c  5; horizontal b2  c 2  a 2  25  9  16 2

x y2  1 9 16

Write the standard form of the equation of the hyperbola. 41.

Solution a  2, b  3

x2 y 2  1 4 9

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1744


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

42.

Solution a  3, b  5

y 2 x2  1 9 25 Write each hyperbola in standard form. 43. x 2  9 y 2  8x  36 y  29  0

Solution x 2  9 y 2  8 x  36 y  29  0

 x  8x  16  9  y  4 y  4  29  16  36 2

2

 x  4   9  y  2  9 2  x  4  y  2 2  1   9 2

2

44. 36x 2  y 2  72x  10 y  25  0

Solution

36 x 2  y 2  72x  10 y  25  0

 

36 x 2  2 x  1  y 2  10 y  25  25  36  25 36  x  1   y  5   36 2

2

 y  5  x  1  36  1 2

2

45. 4 x 2  9 y 2  64x  54 y  211  0

Solution 4 x 2  9 y 2  64 x  54 y  211  0

 

4 x 2  16 x  64  9 y 2  6 y  9  211  256  81 9  y  3   4  x  4   36 2

2

 y  3   x  8  1 2

4

2

9

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1745


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

46. 25x 2  4 y 2  50x  48 y  19  0

Solution

25 x 2  4 y 2  50 x  48 y  19  0 4 y 2  48 y  25 x 2  50 x  19

4 y 2  12 y  36  25 x 2  2 x  1  19  144  25 4  y  6   25  x  1  100 2

2

 y  6   x  1  1 2

25

2

4

Graph each hyperbola. 47.

x2 y 2  1 9 4

Solution x2 y 2  1 9 4 Center: (0, 0), a  3, b  2, horizontal

48.

y 2 x2  1 4 9

Solution y 2 x2  1 4 9 Center: (0, 0), a  2, b  3, vertical

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1746


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

49. 4 x 2  3 y 2  36

Solution 4 x 2  3 y 2  36 4 x 2 3 y 2 36   36 36 36 x2 y 2  1 9 12 Center: (0, 0), a  3, b  2 3, horizontal

50. 3x 2  4 y 2  36

Solution

3 x 2  4 y 2  36 3 x 2 4 y 2 36   36 36 36 x2 y 2  1 12 9 Center: (0, 0), a  2 3, b  3, horizontal

51. y 2  x 2  1

Solution y 2  x2  1 y 2 x2  1 1 1

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1747


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Center: (0, 0), a  1, b  1, vertical

52. x 2 

y2 1 4

Solution y2 1 4 x2 y 2  1 1 4 Center: (0, 0), a  1, b  2, horizontal x2 

53.

( x  2)2 9

y2 1 4

Solution ( x  2)2

y2 1 9 4 Center: (  2, 0), a  3, b  2, horizontal 

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1748


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

54.

y 2 ( x  2)2  1 9 36

Solution y 2 ( x  2)2  1 9 36 Center: ( 2, 0), a  3, b  6, vertical

55. 4( y  2)2  9( x  1)2  36

Solution

4( y  2)2  9( x  1)2  36 4( y  2)2 9( x  1)2 36   36 36 36 ( y  2)2 ( x  1)2  1 9 4 Center: (  1, 2), a  3, b  2, vertical

56. 9( y  2)2  4( x  1)2  36

Solution 9( y  2)2  4( x  1)2  36 2

9( y  2)2 4( x  1) 36   36 36 36 2 2 ( y  2) ( x  1)  1 4 9

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1749


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Center: ( 1,  2), a  2, b  3, vertical

57. 4 x 2  2 y 2  8x  8 y  8

Solution 4 x 2  2 y 2  8x  8 y  8 4( x 2  2 x )  2( y 2  4 y )  8 4( x 2  2 x  1)  2( y 2  4 y  4)  8  4  8 4( x  1)2  2( y  2)2  4 4( x  1)2 2( y  2)2 4   4 4 4 ( x  1)2 ( y  2)2  1 1 2 Center: (  1,  2), a  1, b 

2, horizontal

58. x2  y 2  4x  6 y  6

Solution x2  y 2  4x  6 y  6 x2  4x  ( y 2  6 y )  6 x 2  4 x  4  ( y 2  6 y  9)  6  4  9 ( x  2)2  ( y  3)2  1 ( x  2)2 ( y  3)2  1 1 1

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1750


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Center: ( 2,  3), a  1, b  1, horizontal

59. y 2  4 x 2  6 y  32x  59

Solution y 2  4 x 2  6 y  32 x  59 y 2  6 y  4( x 2  8 x )  59 y 2  6 y  9  4( x 2  8 x  16)  59  9  64 ( y  3)2  4( x  4)2  4 ( y  3)2 ( x  4)2  1 4 1 Center: ( 4,  3), a  2, b  1, vertical

60. x 2  6x  y 2  2 y  11

Solution x 2  6 x  y 2  2 y  11 x 2  6 x  ( y 2  2 y )  11 x 2  6 x  9  ( y 2  2 y  1)  11  9  1 ( x  3)2  ( y  1)2  3 ( x  3)2 ( y  1)2  1 3 3 ( y  1)2 ( x  3)2  1 3 3

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1751


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Center: (  3, 1), a 

3, b 

3, vertical

61.  xy  6

Solution  xy  6

62. xy  20

Solution xy  20

Graph each hyperbola. Identify the center, vertices, and foci. 63.

x2 y 2  1 9 36

Solution x2 y 2  1 9 36

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1752


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

a=

3

b=

6

c=

c2  9  36 c2  45 c  3 5

Center: Vertices: Foci:

64.

0, 0  3, 0  ,  3, 0 

3 5, 0 ,  3 5, 0

y 2 x2  1 16 4

Solution y 2 x2  1 16 4

a=

4

b=

2

c=

c2  16  4 c2  20 c  2 5

Center:

(0, 0)

Vertices:

(0, 4), (0,  4)

Foci:

(0, 2 5), (0,  2 5)

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1753


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

65.

( x  1)2 ( y  3)2  1 4 9

Solution ( x  1)2 ( y  3)2  1 4 9

a=

2

b=

3

c=

c2  4  9 c 2  13 c   13

66.

Center:

(1, 3)

Vertices:

(3, 3), ( 1, 3)

Foci:

(1  13, 3), (1  3, 3)

( y  1)2 ( x  2)2  1 25 4

Solution ( y  1)2 ( x  2)2  1 25 4

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1754


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

a=

5

b=

2

c=

c2  25  4 c2  29 c   29

Center:

( 2,  1)

Vertices:

( 2, 4), ( 2,  6)

Foci:

( 2,  1  29), ( 2,  1  29)

67. 4x 2  9 y 2  8x  54 y  113  0

Solution

4 x 2  9 y 2  8 x  54 y  113  0

 x  1   y  3  1 2

2

9

4

a=

3

b=

2

c=

c2  9  4 c 2  13

Center:

c   13 (1,  3)

Vertices:

(4,  3), (  2,  3)

Foci:

(1 

13,  3), (1 

13,  3)

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1755


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

68. 16x 2  25 y 2  96x  200 y  144  0

Solution

16 x 2  25 y 2  96 x  200 y  144  0

 y  4    x  3  1 2

16

2

25

a=

4

b=

5

c=

c2  16  25 c2  41

Center:

c   41 (3,  4)

Vertices:

(3, 0), (3,  8)

Foci:

(3,  4  41), (3,  4  41)

Find an equation in standard form of the hyperbola on which point P lies. 69. The difference of the distances between P ( x , y ) and the points ( 2, 1) and (8, 1) is 6.

Solution Foci: ( 2, 1), (8, 1)

Center: (3, 1), c  5 2a  6  a  3, b  4 ( x  3)2 ( y  1)2  1 9 16

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1756


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

70. The difference of the distances between P ( x , y ) and the points (3,  1) and (3, 5) is 5.

Solution Foci: (3,  1), (3, 5)

Center: (3, 1), c  5 5 11 , b 2 2 ( y  2)2 ( x  3)2  1

2a  5  a  25 4

11 4

71. The distance between point P ( x , y ) and the point (0, 3) is 32 of the distance between P and the line y  2.

Solution The distance between the point ( x , y ) and the line y  2 is the difference between the y-coordinates, or y  ( 2)  y  2.

3 ( y  2) 2 9 x 2  ( y  3)2  ( y  2)2 4 4 x 2  4( y  3)2  9( y  2)2

( x  0)2  ( y  3)2 

4 x 2  4( y 2  6 y  9)  9( y 2  4 y  4) 4 x 2  5 y 2  60 y  0 4 x 2  5( y 2  12 y )  0 4 x 2  5( y 2  12 y  36  0  5(36) 4 x 2  5( y  6)2  180 4 x 2 5( y  6)2 180   180 180 180 2 2 x ( y  6)  1 36 45 72. The distance between point P ( x , y ) and the point (5, 4) is 53 of the distance between P and the line x  3.

Solution The distance between the point ( x , y ) and the line y  3 is the difference between the x-coordinates, or x  ( 3)  x  3.

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1757


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

5 ( x  3) 3 25 ( x  5)2  ( y  4)2  ( x  3)2 9 9( x  5)2  9( y  4)2  25( x  3)2 ( x  5)2  ( y  4)2 

9( x 2  10 x  25)  9( y 2  8 y  16)  25( x 2  6 x  9) 9 x 2  90 x  225  9 y 2  72 y  144  25 x 2  150 x  225 9 y 2  72 y  16 x 2  240 x  144  0 9( y 2  8 y )  16( x 2  15 x )  144  225  9( y 2  8 y  16)  16  x 2  15 x    144  144  900 4  

9( y  4)2  16 x  15 2

  900 2

   9( y  4)  900  x    ( y  4)  1

16 x  15 2

15 2 225 4

2

2

2

2

100

Use a graphing calculator to graph each hyperbola. Then sketch the hyperbola by hand and compare the results. 73. x 2 

y2 1 4

Solution

y2 1 4 y2    x2  1 4 y 2  4x2  4

x2 

y   4x2  4

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1758


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

74.

( x  3)2 ( y  2)2  1 4 25

Solution ( x  3)2 ( y  2)2  1 4 25 ( y  2)2 ( x  3)2   1 25 4 ( y  2)2 ( x  3)2   1 25 4 25( x  3)2 ( y  2)2   25 4 y 2  

25( x  3)2  25 4

y  2

25( x  3)2  25 4

Fix It In exercises 75 and 76, identify the step where the first error is made and fix it. 75. Write the hyperbola 9x 2  4 y 2  90x  64 y  67  0 in standard form.

Solution Step 5 was incorrect. Step 5:

( x  5)2 ( y  8)2  1 4 9

76. Given the hyperbola 25 y 2  4x 2  100. Write its equation in standard form, identify the vertices, identify the foci, and then draw its graph.

Solution Step 3 was incorrect. Step 3: The foci are (0,

29), (0, 

29)

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1759


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Step 4: The graph is:

Applications 77. Fluids See the illustration below. Two glass plates in contact at the left, and separated by about 5 millimeters on the right, are dipped in beet juice, which rises by capillary action to form a hyperbola. The hyperbola is modeled by an equation of the form xy  k . If the curve passes through the point (12, 2), what is k?

Solution xy  k (12)(2)  k 24  k 78. Astronomy Some comets have a hyperbolic orbit, with the sun as one focus. When the comet shown in the illustration is far away from Earth, it appears to be approaching Earth along the line y  2 x. Find the equation in standard form of its orbit if the comet comes within 100 million miles of Earth.

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1760


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution a  100, 000, 000 c  200, 000, 000 x2 100, 000, 0002

y2 200, 000, 0002

1

79. Alpha particles The particle in the illustration approaches the nucleus at the origin along the path 9 y 2  x2  81 in the coordinate system shown. How close does the particle come to the nucleus?

Solution

9 y 2  x 2  81 9 y 2 x 2 81   81 81 81 y 2 x2  1 9 81 a  3  3 units 80. Physics Parallel beams of similarly charged particles are shot from two atomic accelerators 20 meters apart, as shown in the illustration. If the particles were not deflected, the beams would be 2.0  104 meters apart. However, because the charged particles repel each other, the beams follow the hyperbolic path y  kx , for some k. Find k.

Solution The point (10, 1  104 ) is on the graph.

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1761


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

k x k 4 1  10  10 1  103  k y 

81. Navigation The LORAN (LOng RAnge Navigation) system in the illustration uses two radio transmitters 26 miles apart to send simultaneous signals. The navigator on a ship at P ( x , y ) receives the closer signal first and determines that the difference of the distances between the ship and each transmitter is 24 miles. That places the ship on a certain curve. Identify the curve and find an equation in standard form.

Solution 2a  24  a  12

c  13  b  5 x2 y2  1 144 25 82. Navigation By determining the difference of the distances between the ship in the illustration and two radio transmitters, the LORAN navigation system places the ship on the hyperbola x 2  4 y 2  576 in the coordinate system shown. If the ship is 5 miles out to sea, find its coordinates.

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1762


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution x 2  4 y 2  576 x 2  4(5)2  576 x 2  100  576 x 2  676 x   676  26 (26, 5) 83. Wave propagation Stones dropped into a calm pond at points A and B create ripples that propagate in widening circles. In the illustration, points A and B are 20 feet apart, and the radii of the circles differ by 12 feet. The point P ( x , y ) where the circles intersect moves along a curve. Identify the curve and find an equation in standard form.

Solution 2a  12  a  6

c  10  b  8 x2 y 2  1 36 64 84. Sonic boom The position of a sonic boom caused by the faster-than-sound aircraft is one branch of the hyperbola y 2  x 2  25 in the coordinate system shown. How wide is the hyperbola 5 miles from its vertex?

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1763


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution y 2  x 2  25 y 2 x2  1 25 25 C(0, 0), V (0, 5) y 2  x 2  25 102  x 2  25 75  x 2 x  75  5 3 2(5 3)  10 3 miles

Discovery and Writing 85. Describe a hyperbola.

Solution Answers may vary. 86. Explain a strategy you would use to graph a hyperbola.

Solution Answers may vary. 87. Explain how to determine the dimensions of the fundamental rectangle.

Solution Answers may vary. 88. How do you distinguish among the equations of circles, parabolas, ellipses, and hyperbolas?

Solution Answers may vary. 89. Prove that c > a for a hyperbola with center at (0, 0) and line segment FFʹ on the x-axis.

Solution Answers may vary. 90. Show that the extended diagonals of the fundamental rectangle of the hyperbola y2 x2  2 1 a2 b

are y  ab x and y   ab x .

Solution Answers may vary. 91. Show that the expansion of the standard equation of a hyperbola is a special case of the general equation of second degree with B  0.

Solution Answers may vary.

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1764


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

92. Write a paragraph describing how you can tell from the equation of a hyperbola whether the transverse axis is vertical or horizontal.

Solution Answers may vary. Critical Thinking In Exercises 93–96, match the equation of the hyperbola with its graph. 93.

x2 y 2  1 25 9

Solution b 94.

x2 y 2  1 9 25

Solution d 95.

y 2 x2  1 9 25

Solution c 96.

y 2 x2  1 25 9

Solution a a.

b.

c.

d.

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1765


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

In Exercises 97–100, match the equation of the hyperbola with its graph. 97.

( y  3)2 ( x  3)2  1 9 16

Solution b 98.

( y  3)2 ( x  3)2  1 9 16

Solution d 99.

( x  3)2 ( y  3)2  1 16 9

Solution a 100.

( x  3)2 ( y  3)2  1 16 9

Solution c a.

b.

c.

d.

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1766


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

EXERCISES 7.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Graph the parabola x 2  2x  y  1.

Solution x 2  2 x  y  1. x 2  2x  1  y  1  1 ( x  1)2  y  2 Vertex (1,  2)

2. Graph the ellipse 4 x 2  25 y 2  100.

Solution 16 x 2  25 y 2  400 x2 y 2  1 25 16

3 x  y  9 3. Solve the system of linear equations in two variables by graphing.  5 x  2 y  4

Solution 3 x  y  9  5 x  2 y  4

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1767


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

3 x  y  17 4. Solve the system of linear equations in two variables by substitution.  2 x  3 y  4

Solution 3 x  y  17  2 x  3 y  4 From (1) y  3 x  17. Substitute into (2).

2x  3(3 x  17)  4 2x  9 x  51  4 11x  55 x 5 Back substitute into (1). 3(5)  y  17 15  y  17 y  2 y  2 Solution: (5,  2) 3 x  4 y  4 5. Solve the system of linear equations in two variables by elimination.  7 x  6 y  17

Solution 3 x  4 y  4 ( 7)  3 7 x  6 y  17 21x  28 y  28 21x  18 y  51 46 y  23 y 

1 2

Back substitute into (1).  1 3 x  4    4 2 3 x  2  4

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1768


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

3 x  6 x  2

 1 Solution:  2,  2   6. Solve the quadratic equation 9 x 2  8.

Solution 9x 2  8 x2 

8 9

x

8 2 2  9 3

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. Solutions of nonlinear systems of equations are the points of intersection of the __________ of conic sections.

Solution graphs 8. Approximate solutions of nonlinear systems can be found __________, and exact solutions can be found algebraically using the methods of __________ or __________.

Solution graphically, substitution, elimination Practice Solve each system of nonlinear equations in two variables by graphing. 9.

2 2 8 x  32 y  256   x  2 y

Solution 8 x 2  32 y 2  256   x  2 y

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1769


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

(4, 2), ( 4,  2)

 x 2  y 2  2 10.   x  y  2

Solution  x 2  y 2  2   x  y  2

(1, 1)

 x 2  y 2  90 11.  2  y  x

Solution  x 2  y 2  90  2  y  x

(3, 9), ( 3, 9)

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1770


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

 x 2  y 2  5 12.   x  y  3

Solution  x 2  y 2  5   x  y  3

(1, 2), (2, 1)

 x 2  y 2  25 13.  2 2  12 x  64 y  768

Solution  x 2  y 2  25  2 2  12 x  64 y  768

( 4, 3), (4, 3) ( 4,  3), (4,  3)

 x 2  y 2  13 14.  2  y  x  1

Solution  x 2  y 2  13  2  y  x  1

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1771


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

( 2, 3), (2, 3)

 x 2  13   y 2 15.   y  2 x  4

Solution  x 2  13   y 2   y  2 x  4

( 51 ,  18 ), (3, 2) 5

 x 2  y 2  20 16.  2  y  x

Solution  x 2  y 2  20  2  y  x

(2, 4), ( 2, 4)

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1772


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

 x 2  6 x  y  5 17.  2  x  6 x  y  5

Solution 2  x  6 x  y  5  2  x  6 x  y  5

(1, 0), (5, 0)

 x 2  y 2  5 18.  2 2 3 x  2 y  30

Solution  x 2  y 2  5  2 2 3 x  2 y  30

( 2, 3), (2, 3) ( 2,  3), (2,  3)

Use a graphing calculator to solve each system of nonlinear equations in two variables.  y  x  1 19.  2  y  x  x Solution  y  x  1  2  y  x  x

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1773


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

(1, 2), ( 1, 0)

 y  6  x 2 20.  2  y  x  x

Solution  y  6  x 2  2  y  x  x

(2, 2), ( 1.5, 3.75) 2 2 6 x  9 y  10 21.  3 y  2 x  0

Solution  2 2 6 x  9 y  10  3 y  2 x  0

 y 

10 6 x 2 9

 y  23 x

(1, 0.67), ( 1, 0.67)

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1774


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

 x 2  y 2  68 22.  2 2  y  3 x  4

Solution  x 2  y 2  68  y   68  x 2    y 2  3 x 2  4  y   4  3 x 2

(4, 7.2), (4,  7.2), ( 4, 7.2), ( 4,  7.2)

Solve each system of nonlinear equations in two variables using substitution or elimination for real values of x and y. 2 x 2  9 y 2  225 23.  5 x  3 y  15

Solution

5 x  3 y  15  y 

15  5 x 3

25 x 2  9 y 2  225 2

 15  5 x  25 x 2  9    225  3  25 x 2  (15  5 x )2  225 25 x 2  225  150 x  25 x 2  225 50 x 2  150 x  0 50 x ( x  3)  0 x 0 x3 15  5(0) 15  5(3) 5 y  0 y  3 3 (0, 5) (3, 0)  x 2  y 2  20 24.  2  y  x

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1775


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution

y  x2  x2  y x 2  y 2  20 y  y 2  20 y 2  y  20  0 ( y  4)( y  5)  0 x 0 y  5 4  x 2  x  2 (2, 4), (  2, 4)

5  x 2 no real solution

2 2  x  y  2 25.   x  y  2

Solution x  y  2 y  2 x x2  y 2  2 x 2  (2  x )2  2 x2  4  4x  x2  2 2x 2  4 x  2  0 2( x  1)( x  1)  0 x1 y  21 1 (1, 1)  x 2  y 2  36 26.  2 2 49 x  36 y  1764

Solution x 2  y 2  36  x 2  36  y 2 49 x 2  36 y 2  1764 49(36  y 2 )  36 y 2  1764 1764  49 y 2  36 y 2  1764 13 y 2  0 y 0 x 2  36  02  x  6 (6, 0), ( 6, 0)  x 2  y 2  5 27.   x  y  3

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1776


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution x  y  3 y  3 x x2  y 2  5 x 2  (3  x )2  5 x2  9  6x  x2  5 2x 2  6x  4  0 2( x  1)( x  2)  0 x1 x2 y  31 2 y  32  1 (1, 2)

(2, 1)

 x 2  x  y  2 28.  4 x  3 y  0

Solution x2  x  y  2  y  x2  x  2 4x  3 y  0 4 x  3( x 2  x  2)  0 4 x  3x 2  3x  6  0 3x 2  7 x  6  0 (3 x  2)( x  3)  0 x   23

x3

      2 2

y   23

2 3

  89

y  32  3  2 4

 ,   8 9

2 3

(3, 4)

 x 2  y 2  13 29.  2  y  x  1

Solution y  x2  1  x2  y  1 x 2  y 2  13 y  1  y 2  13

y 3 2

3  1  x  x  2 (2, 3), (  2, 3)

y  4 4  1  x 2 no real solutions

y 2  y  12  0 ( y  3)( y  4)  0

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1777


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

 x 2  y 2  25 30.  2 2 2 x  3 y  5

Solution y 3

x 2  y 2  25  x 2  25  y 2

2

2x 2  3 y 2  5 2(25  y 2 )  3 y 2  5

y  3

2

2

x  25  3  x  4

x  25  ( 3)2  x  4

(4, 3), (  4, 3)

(4,  3), (  4,  3)

50  2 y 2  3 y 2  5 45  5 y 2 9  y2  x 2  y 2  30 31.  2  y  x

Solution

y  x2  x2  y x 2  y 2  30 y  y 2  30 y 2  y  30  0 ( y  5)( y  6)  0 y 5 y  6 6  x 2

5  x2  x   5 ( 5, 5), (  5 , 5)

no real solutions

9 x 2  7 y 2  81 32.  2 2  x  y  9

Solution x2  y 2  9  x2  9  y 2 9 x 2  7 y 2  81 9(9  y 2 )  7 y 2  81 81  9 y 2  7 y 2  81 16 y 2  0 y 0 x 2  9  02  x  3 (3, 0), (  3, 0) 2 x 2  y 2  6 33.  2 2  x  y  3

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1778


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution 2x 2  y 2 

6

2x 2  y 2  6

x2  y 2 

3

2(3)  y 2  6

3x2

9

y2  0

x2

3

y 0

x

 3

( 3, 0), (  3, 0)

 x 2  y 2  13 34.  2 2  x  y  5

Solution x 2  y 2  13

x 2  y 2  13

x2  y 2  5

9  y 2  13

2x 2

 18

y2  4

x2

 9

y  2

x

 3

(3, 2), ( 3, 2), (3,  2), ( 3,  2)

 x 2  y 2  20 35.  2 2  x  y  12

Solution x2  y 2 

20

x 2  y 2  20

x 2  y 2  12

4  y 2  20

2x 2

8

y 2  16

x2

4

y  4

x

2

(2, 4), ( 2, 4), (2,  4), ( 2,  4)

 9  xy   36.  2 3 x  2 y  6 

Solution 9 9 xy    y   2 2x 3x  2 y  6

 9  3x  2   6  2x 

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1779


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

9 6 x 3x 2  6x  9  0 3( x  3)( x  1)  0 x3 x  1 3x 

9  3 y   2(3) 2

y   2(91)  92

(3,  32 )

9 ) 2

( 1,

 y 2  40  x 2 37.  2  y  x  10

Solution y  x 2  10  x 2  y  10 y 2  40  x 2 y 2  40  y  10

y 5

y  6

2

x  5  10  x   15 ( 5, 5), (  5, 5)

2

x  6  10  x  2 (2,  6), ( 2,  6)

y 2  y  30  0 ( y  5)( y  6)  0

 x 2  6 x  y  5 38.  2  x  6 x  y  5

Solution x2  6x  y

 5

x2  6x  y

 5

2 x 2  12 x 2( x  1)( x  5)

 10  0

x1 2

y  x  6x  5  0 (1, 0)

x 5 2

y  x  6x  5  0 (5, 0)

 y  x 2  4 39.  2 2  x  y  16

Solution y  x2  4  x2  y  4 x 2  y 2  16 y  4  y 2  16

y 5 2

x  5  4  x  3 (3, 5), (  3, 5)

y  4 2

x  4  4  x  0 (0,  4)

y 2  y  20  0 ( y  5)( y  4)  0

6 x 2  8 y 2  182 40.  2 2 8 x  3 y  24

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1780


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution

6 x 2  8 y 2  182

6 x 2  8 y 2  182

8x2  3 y 2 

6(9)  8 y 2  182

24

8 y 2  128  546

y 2  16

64 x 2  24 y 2  192

y  4

18 x 2  24 y 2 82 x 2 x

 738

2

9

x  3 (3, 4), (  3, 4), ( 3,  4), (  3,  4)

 x 2  y 2  5 41.  2 2 3 x  2 y  30

Solution

x2  y 2

 5

3 x 2  2 y 2  30

3x 2  2 y 2

 30

3(4)  2 y 2  30 2 y 2  18

2 x 2  2 y 2  10

y2  9

3 x 2  2 y 2  30

y  3

5x 2 x

 20

2

4

x  2 (2, 3), (  2, 3), (2,  3), (  2,  3) 1 1   5 x y 42.   1  1  3  x y

Solution 1  y1  x

1  y1 x 1  y1 1

5

5

1  y1 x

 3

2 x

2

1 y

1

y  41

x

 1, 

5 4

1 y

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1781


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

1 2   1 x y 43.  2  1  1  x y 3

Solution 1  2y  1 x 2  y1  31 x

1  2y  1 x 4  2y  32 x 5 x

1  2y  1 x 1  2y  1 3

 53

x

2  23 y

3

y 3

(3, 3) 1 3 4   x y 44.  2  1  7  x y

Solution 1  3y  4 x 2  y1  7 x

1  3y  4 x 6  3y  21 x 7 x

 25

x

7  25

1  3y  4 x 1  3y  4 7 25 3  73 y

y 3

 , 7 7 25

3 y 2  xy 45.  2 2 x  xy  84  0

Solution 3 y 2  xy 3 y 2  xy  0 y (3 y  x )  0 y  0 or x  3 y

y 0 2

x  3y 2

2 x  xy  84  0

2 x  xy  84  0

2

2

2 x  x (0)  84  0 2

2(3 y )  (3 y ) y  84  0

2 x  84

18 y 2  3 y 2  84  0

x 2  42

21 y 2  84

x   42 ( 42, 0), (  42, 0)

y 2  4  y  2 (6, 2), (  6,  2)

 x 2  y 2  10 46.  2 2 2 x  3 y  5

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1782


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution y  3

x 2  y 2  10  x 2  10  y 2

2

2x 2  3 y 2  5 2(10  y 2 )  3 y 2  5 20  2 y 2  3 y 2  5

x  10  ( 3)

x  10  (  3)2

x2  7

x2  7

x 7

x 7

( 7,

15  5 y 2

y  3 2

3), (  7,

2

3) ( 7,  3), (  7,  3)

3  y2

 1  xy  47.  6  y  x  5 xy 

Solution xy  61  y  61x

y  x  5 xy 1 5x x 6x 6x 1  6x2  5x

6x2  5x  1  0 (2 x  1)(3 x  1)  0 x  31

x  21 y 

1 1  6(1 2) 3 ( 21 ,

y 

1) 3

1 1  6(1 3) 2 ( 31 ,

1) 2

 1  xy  48.  12  y  x  7 xy 

Solution

xy  121  y  121x

y  x  7 xy 1 7x x 12 x 12 x 1  12 x 2  7 x

12 x 2  7 x  1  0 (4 x  1)(3 x  1)  0

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1783


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

x  31

x  41 y 

1 1  12(1 4) 3 ( 41 ,

y 

1) 3

1 1  12(1 3) 4 ( 31 ,

1) 4

Fix It In exercises 49 and 50, identify the step where the first error is made and fix it. 2 2 4 x  y  5 49. Solve the nonlinear system of equations in two variables by substitution.  2  y  x

Solution Step 4 was incorrect. Step 4: x  1 or x  1 Step 5: Solutions are (1, 1), ( 1, 1) 50. Solve the nonlinear system of equations in two variables by substitution. 4 x 2  y 2  20  2 2 3 x  y  8

Solution Step 5 was incorrect. Step 5: Solutions are (2, 2), (2,  2), ( 2, 2), ( 2,  2)

Applications 51. Geometry The area of a rectangle is 63 square centimeters, and its perimeter is 32 centimeters. Find the dimensions of the rectangle.

Solution

Let x  width and y  length.  xy  63  2 x  2 y  32 xy  63  y 

63 x

2 x  2 y  32

x 9 7 y  63 9

x7 9 y  63 7

 63  2 x  2    32 The dimensions are 9 cm by 7 cm.  x  2x 2  126  32 x 2 x 2  32 x  126  0 2( x  9)( x  7)  0

52. Dimensions of a whiteboard The area of a SMART Board interactive whiteboard is 2880 square inches, and its perimeter is 216 inches. Find the dimensions of the whiteboard.

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1784


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution

Let x  width and y  length.  xy  2880  2 x  2 y  216 xy  2880  y 

2 x  2 y  216

x  60 y  2880  48 60

x  48 y  2880  60 48

 2880  2x  2  The dimensions are 48 in. by   216  x  60 in. 2 x 2  5760  216 x

2880 x

2 x 2  216 x  5760  0 2( x  60)( x  48)  0

53. Fencing pastures The rectangular pasture shown below is to be fenced in along a riverbank. If 260 feet of fencing is to enclose an area of 8000 square feet, find the dimensions of the pasture.

Solution

Let x  width and y  length.

2 x  y  260

 xy  8000  2 x  y  260 xy  8000  y 

8000 x

x  50 x  80 8000 y  50  160 y  8000  100 80

 8000  2x   The dimensions are 50 ft by 160 ft   260  x  or 80 ft by 100 ft. 2 x 2  8000  260 x 2 x 2  260 x  8000  0 2( x  50)( x  80)  0

54. Investments Grant receives $225 annual income from one investment. Sasha invested $500 more than Grant, but at an annual rate of 1% less. Sasha’s annual income is $240. Find the amount and rate of Grant’s investment.

Solution Let x  Grant's principal.

Jeff's Grant's   0.01 rate rate 240 225   0.01 x  500 x I

P

r

Grant

225

x

225 x

Jeff

240

x + 500

240 x  500

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1785


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

240 x  225( x  500)  0.01x( x  500) 0.01x  20 x  112,500  0 2

x  2000 x  11, 250,000  0 ( x  2500)( x  4500)  0 2

x  2500  0 or x  2500

x  4500  0 x  4500  Grant invested $2500 at 9% interest.

55. Investments Carol receives $67.50 annual income from one investment. Francisco invested $150 more than Carol at an annual rate of 1 21 % more. Francisco’s annual income is $94.50. Find the amount and rate of Carol’s investment. (Hint: There are two answers.)

Solution Let x  Carol's principal.

John's rate

Carol's rate

 0.015

94.50 67.50   0.015 x  150 x I

P

r

Carol

67.50

x

67.50 x

John

94.50

x  150

9450 x  150

94.5 x  67.5( x  150)  0.015 x( x  150) 0.015 x  24.75 x  10, 125  0 2

x 2  1650 x  675,000  0 ( x  750)( x  900)  0 x  750  0 or x  750

x  900  0 Carol invested either $750 at 9% or x  900  she invested $900 at 7.5% interest.

56. Finding the rate and time Franco drove 306 miles. Franco’s brother made the same trip at a speed 17 miles per hour slower than Franco did and required an extra 1 21 hours. Find Franco’s rate and time.

Solution

Let r  Jim's rate. Then his brother's rate is r  17. Jim's time  Brother's time  1.5 306 306   1.5 r r  17 Rate

Time

Dist.

Jim's trip

r

306 r

306

Brother's trip

r  17

306 r  17

306

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1786


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

306(r  17)  306r  1.5r (r  17) 1.5r  25.5r  5202  0 2

r 2  17r  3468  0 (r  68)(r  51)  0 r  68  0 or r  51  0 r  68 r  51

Jim drove 68 miles per hour for 4.5 hours.

57. Paintball See the illustration. A liquid-filled paintball is shot from the base of an 1 incline and follows the parabolic path y   300 x 2  51 x , with distances measured in

feet. The incline has a slope of 101 . Find the coordinates of the point of intersection of the paintball and the ground at impact.

Solution  y  1 x 10  2 1 1  y   300 x  5 x 1 1 x   300 x 2  51 x 10 30 x   x 2  60 x x 2  30 x  0 x( x  30)  0 x  0 or x  30  0 x  30 x  30  y  101 (30)  3

(30, 3)

58. Artillery See the illustration for Exercise 57. A shell fired from the base of a hill follows the parabolic path y   61 x   2 x , with distances measured in miles. The hill has a slope of 31 . How far from the base of the hill is the point of impact? (Hint: Find the coordinates of the point and then the distance.) Solution  y  1 x 3  1  y x  2x    6

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1787


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

1 x   61 x 2  2 x 3 2

2 x   x  12 x

2

x  10 x  0 x ( x  10)  0 x  0 or x  10  0 x  10

x  10  y  31 (10)  10 3 (10  0)2  ( 10  0)2  100  3

100 9

1000 9 10 10  mi 3 

59. Air-traffic control A plane is flying over an airport on a path whose equation is

y  x  . If a second plane, flying at the same altitude, is traveling on a path whose equation is x  y  2, is there any danger of a mid-air collision? Solution  y  x 2   x  y  2 xy 2 x  x2  2 x2  x  2  0 ( x  2)( x  1)  0 x  2  0 or x  2

x10 x1

x  2  y  ( 2)2  4 x  1  y  12  1 There are potential collision points at (  2, 4) and (1, 1). 60. Ship traffic One ship is steaming on a path whose equation is y  x2  1 and another is steaming on a path whose equation is x  y  4. Is there any danger of collision? Solution  y  x 2  1   x  y  4 x  y  4 x  x 2  1  4 x2  x  5  0

The two solutions to this equation are not real, so there is no danger of collision.

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1788


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

61. Radio reception A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Collinsville with Harmony, a town 200 miles to the east and 100 miles north. See the illustration. If a driver leaves Collinsville and heads toward Harmony, how far from Collinsville will the driver pick up the station?

Solution  x  2 y  2 2 2 ( x  120)  y  100

The x-coordinate of the point where the line crosses the circle closest to the origin has the approximate coordinates (20.5, 10.25). The distance to the origin is about 23 miles. 62. Listening ranges For how many miles will a driver in Exercise 61 continue to receive the signal? Solution  x  2 y  2 2 2 ( x  120)  y  100

The x-coordinate of the point where the line crosses the circle farthest from the origin has the approximate coordinates (171.5, 85.7). The distance the point found in #61 is about 169 miles.

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1789


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Discovery and Writing 63. What is a system of nonlinear equations? Give an example to support your answer. Solution Answers may vary. 64. Describe three methods that can be used to solve a nonlinear system of equations. Solution Answers may vary. 65. Describe any disadvantages in using the graphing method to solve nonlinear systems of equations. Solution Answers may vary. 66. Explain why the elimination method, not the substitution method, is the better  x 2  4 y 2  16 . method to solve the system  2 2  x  y  1 Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 67. The graphing method is the most precise method for solving nonlinear systems of equations Solution False. The substitution and elimination methods are more precise. 68. The substitution method is the best method for solving systems consisting of a firstdegree equation and a second-degree equation. Solution True. 69. If both equations of the system are of the form Ax 2  By 2  C, the substitution method is the best method to use to solve the system. Solution False. Use the elimination method. 70. It is possible for a system of two equations and two variables whose graphs are a line and a parabola to have no solution. Solution True.

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1790


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

71. It is possible for a system of two equations and two variables whose graphs are a parabola and a circle to have exactly one real ordered-pair solution. Solution True. 72. It is possible for a system of two equations and two variables whose graphs are a hyperbola and an ellipse to have exactly five real ordered-pair solutions. Solution False. The greatest number of possible solutions is 4. 73. The product of two integers is 32 and their sum is 12. If you find the two numbers and subtract the smaller one from the larger one, you will obtain 2. Solution False. The difference is 4. 74. The sum of the squares of two numbers is 221, and the sum of the numbers is 9. If you find the two numbers and subtract the smaller one from the larger one, you will obtain 19. Solution True.

CHAPTER REVIEW SOLUTIONS Exercises Write an equation in standard form of each circle described. 1.

Center (0, 0); radius 4 Solution ( x  h)2  ( y  k )2  r 2 ( x  0)2  ( y  0)2  42 x 2  y 2  16

2. Center (0, 0); passes through (6, 8) Solution

r  (6  0)2  (8  0)2  10 ( x  h)2  ( y  k )2  r 2 ( x  0)2  ( y  0)2  102 x 2  y 2  100 3. Center (3,  2); radius 5

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1791


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution ( x  h)2  ( y  k )2  r 2 ( x  3)2  ( y  ( 2))2  52 ( x  3)2  ( y  2)  25 4. Center ( 2, 4); passes through (1, 0) Solution

r  ( 2  1)2  (4  0)2  5 ( x  h)2  ( y  k )2  r 2 ( x  ( 2))2  ( y  4)2  52 ( x  2)2  ( y  4)2  25 5. Endpoints of diameter ( 2, 4) and (12, 16) Solution

 2  12 4  16  , C   C(5, 10) 2   2 r  (12  5)2  (16  10)2  85 ( x  h)2  ( y  k )2  r 2 ( x  5)2  ( y  10)2  ( 85)2 ( x  5)2  ( y  10)2  85 6. Endpoints of diameter ( 3, 6) and (7, 10) Solution

 3  7 6  10  , C   C(2, 2) 2   2 r  (7  2)2  (10  2)2  89 ( x  h)2  ( y  k )2  r 2 ( x  2)2  ( y  2)2  ( 89)2 ( x  2)2  ( y  2)2  89 Write an equation of each circle in standard form and graph the circle. 7.

x 2  y 2  6x  4 y  3 Solution

x2  y 2  6x  4 y  3 x2  6x  y 2  4 y  3 x2  6x  9  y 2  4 y  4  3  9  4 ( x  3)2  ( y  2)2  16

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1792


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

8.

x 2  4 x  y 2  10 y  13 Solution x 2  4 x  y 2  10 y  13 x 2  4 x  4  y 2  10 y  25  13  4  25 ( x  2)2  ( y  5)2  16

Write an equation in standard form of each parabola described. 9. Vertex (0, 0); passes through ( 8, 4) and ( 8,  4) Solution Horizontal

( y  0)2  4 p( x  0) (4  0)2  4 p( 8  0) 16  32p 2  4 p y 2  2 x 10. Vertex (0, 0); passes through ( 8, 4) and (8, 4) Solution Vertical

( x  0)2  4 p( y  0) ( 8  0)2  4 p(4  0) 64  16p 16  4 p x 2  16 y

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1793


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

11. Find an equation in standard form of the parabola with vertex at ( 2, 3), curve passing through point ( 4,  8), and opening down. Solution Vertical

( x  2)2  4 p( y  3) ( 4  2)2  4 p( 8  3) 4  4 p( 11) 4   4p 11 4 ( x  2)2   ( y  3) 11 Graph each parabola. Identify the focus and directrix. 12. ( x  1)2  8( y  2) Solution

( x  1)2  8( y  2) V : (  1,  2) F : (  1,  4), D: y  0

13. ( y  4)2  12( x  1) Solution

( y  4)2  12( x  1) V : (  1, 4) F : (  4, 4), D: x  2

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1794


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Graph each parabola. 14. x 2  4 y  2x  9  0 Solution

x 2  4 y  2x  9  0 x 2  2x  4 y  9 x 2  2x  1  4 y  9  1 ( x  1)2  4( y  2)

15. y 2  6 y  4 x  13 Solution y 2  6 y  4 x  13 y 2  6 y  9  4 x  13  9 ( y  3)2  4( x  1)

16. Write an equation of the ellipse in standard form with center at the origin, major axis that is horizontal and 12 units long, and minor axis 8 units long. Solution a  6, b  4, horizontal

x2 y 2  1 36 16 17. Write an equation of the ellipse in standard form with center at the origin, major axis that is vertical and 10 units long, and minor axis 4 units long.

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1795


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution a  5, b  2, vertical

x2 y 2  1 4 25 18. Write an equation of the ellipse in standard form with center at point ( 2, 3) and curve passing through points ( 2, 0) and (2, 3). Solution

a  4, b  3, horizontal 

( x  2)2 ( y  3)2  1 16 9

19. Write the equation of the ellipse in standard form and graph it.

4x 2  y 2  16x  2 y  13 Solution

4 x 2  y 2  16 x  2 y  13 4( x 2  4 x )  y 2  2 y  13 4( x 2  4 x  4)  y 2  2 y  1  13  16  1 4( x  2)2  ( y  1)2  4 ( x  2)2 ( y  1)2  1 1 4 Center: (2,  1), a  2, b  1, vertical

Graph each ellipse. Identify the foci. 20.

x2 y 2  1 16 49 Solution

x2 y 2  1 16 49 Center: (0, 0)

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1796


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

c2  49  16 c2  33 c   33 Foci: (0,

21.

33), (0, 

33)

( x  2)2 ( y  3)2  1 64 25 Solution

( x  2)2 ( y  3)2  1 64 25 Center: ( 2, 3) c2  64  25 c2  39 c   39 Foci: ( 2 

39, 3), (  2 

39, 3)

22. Write an equation of the hyperbola in standard form with center at the origin, passing through points ( 2, 0) and (2, 0), and having a focus at (4, 0).

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1797


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution a  2, c  4; horizontal

b2  c2  a2  16  4  12 x2 y 2  1 4 12 23. Write an equation of the hyperbola in standard form with center at the origin, one focus at (0, 5), and one vertex at (0, 3). Solution a  3, c  5; vertical

b2  c2  a2  25  9  16 y 2 x2  1 9 16 24. Write an equation of the hyperbola in standard form with vertices at points ( 3, 3) and (3, 3) and a focus at point (5, 3).

Solution C(0, 3), a  3, c  5; vertical

b2  c2  a2  25  9  16 x 2 ( y  3)2  1 9 16 25. Write an equation of the hyperbola in standard form with vertices at points (3, 3) and (3, 3) and a focus at point (3, 5).

Solution C(0, 3), a  3, c  5; vertical

b2  c2  a2  25  9  16 x 2 ( y  3)2  1 9 16 2

y2

x  16  1. 26. Write equations of the asymptotes of the hyperbola 25

Solution

y

b 4 xy x a 5

27. Write the equation in standard form and graph it. 9x 2  4 y 2  16 y  18x  43

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1798


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution

9 x 2  4 y 2  16 y  18 x  43 9( x 2  2 x )  4( y 2  4 y )  43 9( x 2  2 x  1)  4( y 2  4 y  4)  43  9  16 9( x  1)2  4( y  2)2  36 ( x  1)2 ( y  2)2  1 4 9 Center: (1,  2), a  2, b  3, horizontal

28. Graph: 4 xy  1. Solution 4 xy  1.

Graph each hyperbola. Identify the foci. 29.

x2 y 2  1 9 25 Solution

x2 y 2  1 9 25 c2  9  25 c2  34 c   34

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1799


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Center: (0, 0) Foci: ( 34, 0), ( 

30.

34, 0)

( y  2)2 ( x  2)2  1 16 25 Solution

( y  2)2 ( x  2)2  1 16 25 c2  16  25 c2  41 c   41 Center: ( 2, 2) Foci: ( 2, 2 

41), (  2, 2 

41)

2 2  x  y  16 . 31. Solve the system of nonlinear equations in two variables by graphing:   y  x  4 Solution  x 2  y 2  16   y  x  4

( 4, 0), (0, 4)

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1800


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

3 x 2  y 2  52 . 32. Solve the system of nonlinear equations in two variables by graphing:  2 2  x  y  12 Solution

3 x 2  y 2  52  2 2  x  y  12

( 4, 2), (4, 2), (  4,  2), (4,  2)

 x2 y 2  1   . 33. Solve the system of nonlinear equations in two variables by graphing:  16 12 2  x2  y  1  3 Solution

 x2 y 2  1   16 12  2 x2  y  1  3

( 2, 3), (2, 3), (  2,  3), (2,  3)

34. Solve the system of nonlinear equations in two variables by substitution or elimination:

3 x 2  y 2  52 .  2 2  x  y  12

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1801


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution

3x 2  y 2

 52

3 x 2  y 2  52

x 2  y 2  12

3(16)  y 2  52

4x2

 64

x2 x

 16  4

y2  4 y  2

(4, 2), ( 4, 2), (4,  2), ( 4,  2) 35. Solve the system of nonlinear equations in two variables by substitution or elimination:  x 2  y 2  16 .   3 y  4 3  3 x Solution

 3y  4 3 3 x 2  y 2  16

y 4

 3 y  4 3  3x  x 

  3y  4 3     y 2  16   3   3 y 2  24 y  48  9 y 2  144

x

x

 3(4)  4 3  0  (0, 4) 3 y  2  3( 2)  4 3  2 3  (2 3,  2) 3

12 y 2  24 y  96  0 12( y  4)( y  2)  0 36. Solve the system of nonlinear equations in two variables by substitution or elimination:

 x2 y 2  1   16 12 .  2 x2  y  1  3 Solution 2 x2  y  1 16 12 2 x 2  y3  1

2 2 5 y 2  45 3 x  y  3

y2  9

3x 2  9  3

y  3

3 x 2  12

3 x 2  4 y 2  48

x2  4

3x 2  y 2  3

x  2  (2, 3), (  2, 3), (2,  3), (  2,  3)

5 y 2  45

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1802


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

CHAPTER TEST SOLUTIONS Write an equation in standard form of each circle described. 1.

Center (2, 3); r = 3 Solution ( x  h)2  ( y  k )2  r 2 ( x  2)2  ( y  3)2  32 ( x  2)2  ( y  3)2  9

2. Ends of diameter at (–2, –2) and (6, 8) Solution  2  6 2  8  , C   C(2, 3) 2   2 r  (6  2)2  (8  3)2  41 ( x  h)2  ( y  k )2  r 2 ( x  2)2  ( y  3)2 

 41 

2

( x  2)2  ( y  3)2  41 3. Center (2, –5), passes through (7, 7) Solution r  (7  2)2  (7  ( 5))2  13 ( x  h)2  ( y  k )2  r 2 ( x  2)2  ( y  ( 5))2  132 ( x  2)2  ( y  5)2  169

4. Change the equation of the circle x 2  y 2  4 x  6 y  4  0 to standard form and graph it. Solution x2  y 2  4x  6 y  4  0 x 2  4 x  y 2  6 y  4 x 2  4 x  4  y 2  6 y  9  4  4  9 ( x  2)2  ( y  3)2  9 C(2,  3), r  3

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1803


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Find an equation in standard form of each parabola described. 5. Vertex (3, 2); focus at (3, 6) Solution Vertical (up), p = 4

( x  h)2  4 p( y  k ) ( x  3)2  4(4)( y  2) ( x  3)2  16( y  2) 6. Vertex (4, –6); passes through (3, –8) and (3, –4) Solution Horizontal ( y  6)2  4 p( x  4) ( 4  6)2  4 p(3  4) 4  4 p 4  4 p ( y  6)2  4( x  4)

7. Vertex (2, –3); passes through (0, 0) Solution ( x  2)2  4 p( y  3) OR ( y  3)2  4 p( x  2)

(0  2)2  4 p(0  3) 4  4 p(3)

(0  3)2  4 p(0  2) 9  4 p( 2)

4  4p 3 4 ( x  2)2  ( y  3) 3

9  4p 2 9 ( y  3)2   ( x  2) 2 

8. Change the equation of the parabola x 2  6 x  8 y  7 into standard form and graph it. Solution x2  6x  8 y  7 x 2  6x  8 y  7 x2  6x  9  8 y  7  9 ( x  3)2  8( y  2) Vertex: (3,  2), vertical

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1804


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Find an equation in standard form of each ellipse described. 9. Vertex (10, 0), center at the origin, focus at (6, 0) Solution a  10, c  6, horizontal

b2  a2  c2  100  36  64 2 x y2  1 100 64 10. Minor axis 24, center at the origin, focus at (5, 0) Solution a  12, c  5, horizontal

a2  b2  c2  144  25  169 2 x y2  1 169 144 11. Center (2, 3); passes through (2, 9) and (0, 3) Solution a  6, b  2, Vertical

( x  2)2 ( y  3)2  1 4 36 12. Change the equation of the ellipse 9 x 2  4 y 2  18 x  16 y  11  0 into standard form and graph it. Solution 9 x 2  4 y 2  18 x  16 y  11  0 9( x 2  2 x )  4( y 2  4 y )  11 9( x 2  2 x  1)  4( y 2  4 y  4)  11  9  16 9( x  1)2  4( y  2)2  36

( x  1)2 ( y  2)2  1 4 9 Center: (1, 2), a  3, b  2, vertical

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1805


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Find an equation in standard form of each hyperbola described. 13. Center at the origin, focus at (13, 0), vertex at (5, 0) Solution a  5, c  13; horizontal

b2  c2  a2  169  25  144 2 x y2  1 25 144 14. Vertices (6, 0) and (–6, 0); c  13 a

12

Solution C(0, 0), a  6, c 

13 horizontal 2

b2  c 2  a 2 169 25   36  4 4 2 2 x y  1 36 25 4

15. Center (2, –1), major axis horizontal and of length 16, distance of 20 between foci Solution a  8, c  10; horizontal

b2  c2  a2  100  64  36 ( x  2)2 ( y  1)2  1 64 36 16. Change the equation of the hyperbola x 2  4 y 2  16 y  8 into standard form and graph it. Solution

x 2  4 y 2  16 y  8 x 2  4( y 2  4 y )  8 x 2  4( y 2  4 y  4)  8  16 x 2  4( y  2)2  8 ( y  2)2 x 2  1 2 8

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1806


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Center: (0, 2), a  2, b  8, vertical

Solve each system of nonlinear equations in two variables using substitution or elimination. 2 2  x  y  23 17.  2  y  x  3

Solution y 4

y  5

4  3  x2  x   7

 5  3  x2

y  x2  3  x2  y  3 x 2  y 2  23 y  3  y 2  23 y  y  20  0 2

 7, 4 ,   7, 4

no real solutions

( y  4)( y  5)  0

2 x 2  3 y 2  9 18.  2 2  x  y  27 Solution y 3

x 2  y 2  27  x 2  27  y 2 2x  3 y  9 2

2

2(27  y 2 )  3 y 2  9 54  2 y 2  3 y 2  9

x  27  32  x  3 2 2

3 2, 3 ,  3 2, 3

45  5 y 2 9  y2

y  3 x 2  27  ( 3)2  x  3 2

3 2,  3 ,  3 2,  3

Complete the square to write each equation in standard form, and identify the curve. 19. y 2  4 y  6 x  14  0 Solution y 2  4 y  6 x  14  0 y 2  4 y  6 x  14 y 2  4 y  4  6 x  14  4 ( y  2)2  6( x  3)  Parabola

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1807


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

20. 2 x 2  3 y 2  4 x  12 y  8  0 Solution

2 x 2  3 y 2  4 x  12 y  8  0 2( x 2  2 x )  3( y 2  4 y )  8 2( x 2  2 x  1)  3( y 2  4 y  4)  8  2  12 2( x  1)2  3( y  2)2  6 ( x  1)2 ( y  2)2   1  ellipse 3 2

CUMULATIVE REVIEW SOLUTIONS Simplify each expression. Assume that all variables represent positive numbers, and write answers without using negative exponents. 1.

23

64

Solution 64 2.

8

23

 64

13

  4  16 2

2

1 3

Solution 1 1 1 3  13  8 2 8 3.

y

23

y

53

y1 3 Solution y

23

y

53

y1 3 4.

(x

53

y

7 3

63

 y2

x

34 24

 y

y1 3

12

)( x )

x

34

Solution (x

53

x

5.

(x

2 3

12

)( x ) 34

x

13 6

x

34

)( x

23

x )

 x )( x

23

x )x

x

13

x

17 12

13

Solution (x

23

13

13

4 3

x

33

x

33

x

23

 x

4 3

x

23

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1808


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

6.

(x

1 2

12

 x )2

Solution (x

7.

3

1 2

12

 x )2  ( x

1 2

12

 x )( x

1 2

12

x )x

2 2

 x0  x0  x

22

1 2 x x

27 x 3

Solution 3

27 x 3  3 ( 3 x )3  3 x 48t 3

8.

Solution 48t 3 

9.

3

16t 2 3t  4t 3t

128 x 4 2x

Solution 3

10.

128 x 4  3 64 x 3  4 x 2x x2  6x  9

Solution x 2  6 x  9  ( x  3)2  x  3

11.

50  8  32

Solution 50  8  32  5 2  2 2  4 2  7 2

12. 3 4 32  2 4 162  5 4 48 Solution 3 4 32  2 4 162  5 4 48  3  2 4 2  2  3 4 2  5  2 4 3  12 4 2  10 4 3

13. 3 2 2 3  4 12 Solution

 

3 2 2 3  4 12  6 6  12 24  6 6  12  2 6  18 6

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1809


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

14.

5 3

x

Solution

5 3

x

3

5 x2 3

3

x x2

3

5 x2 x

x 2

15.

x 1 Solution

 x  2 x  1 x  1  x  1 x  1

x 2

16.

6

x3 x 2 x1

x3 y 3

Solution 6

x3 y 3  ( x3 y 3 )

16

x

36

y

36

x

12

y

12

 ( xy )

12

xy

Solve each equation. 17. 5 x  2  x  8 Solution 5 x 2  x 8

5 x  2   ( x  8) 2

2

25( x  2)  x 2  16 x  64 25 x  50  x 2  16 x  64 0  x 2  9 x  14 0  ( x  2)( x  7) x  2 or x  7 (both check) 18.

x 

x2 2

Solution

x  x2 2 x 2  2 x

 x  2   2  x  2

2

x 2  44 x  x 4 x 2

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1810


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

4 x   2 2

2

16 x  4 1 x 4 19. Use the method of completing the square to solve the equation 2 x 2  x  3  0. Solution

2x 2  x  3  0 1 3 x2  x  2 2 1 1 24 1 2 x  x   2 16 16 16 2  1 25 x    4 16  1 5 x  4 4 1 5 x  4 4 x  1 or x  

3 2

20. Use the quadratic formula to solve the equation 3 x 2  4 x  1  0. Solution 3 x 2  4 x  1  0  a  3, b  4, c  1

b  b2  4ac 2a 4  42  4(3)( 1)  2(3)

x

4  16  12 6 4  28 2  7   6 3 

Perform each operation and write each complex number in a  bi form. 21. (3  5i )  (4  3i ) Solution (3  5i )  (4  3i )  3  5i  4  3i  7  2i 22. (7  4i )  (12  3i ) Solution (7  4i )  (12  3i )  7  4i  12  3i  5  7i

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1811


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

23. (2  3i )(2  3i ) Solution (2  3i )(2  3i )  4  6i  6i  9i 2  4  9( 1)  4  9  13  0i 24. (3  i )(3  3i ) Solution (3  i )(3  3i )  9  9i  3i  3i 2  9  6i  3( 1)  9  6i  3  12  6i 25. (3  2i )  (4  i )2 Solution (3  2i )  (4  i )2  3  2i  (16  8i  i 2 )  3  2i  (15  8i )  3  2i  15  8i  12  10i 26.

5 3i Solution 5 5(3  i ) 5(3  i ) 5(3  i ) 3  i 3 1       i 3  i (3  i )(3  i ) 10 2 2 2 9  i2

Find each value. 27. 3  2i Solution 3  2i 

3 2  22 

13

28. 5  6i Solution 5  6i 

52  ( 6)2 

61

29. For what values of k will the solutions of 2x 2  4 x  k be equal? Solution 2x 2  4 x  k  2x 2  4 x  k  0 a  2, b  4, c  k : Set b2  4ac  0. b2  4ac  0 42  4(2)( k )  0 16  8k  0 k  2

30. Find the coordinates of the vertex of the graph of the equation y  1 x 2  x  1. 2

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1812


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution y  1 x 2  x  1: a  2

x y 

1 , b  1, c  1 2

b 1  1 2a 2 1

2

1 2 1 (1)  1  1  2 2

Solve each inequality and give the result in interval notation. 31. Solve: x 2  x  6  0. Solution x2  x  6  0 ( x  2)( x  3)  0 factors = 0: x  2, x  3 intervals: ( ,  2), ( 2, 3), (3,  ) interval

test number

value of x 2  x  6

( ,  2)

–3

+6

( 2, 3)

0

–6

(3,  )

4

+6

Solution: ( ,  2)  (3,  ) 32. Solve: x 2  x  6  0. Solution x2  x  6  0 ( x  2)( x  3)  0 factors = 0: x  2, x  3 intervals: ( ,  2), ( 2, 3), (3,  ) interval

test number

value of x 2  x  6

( ,  2)

–3

+6

( 2, 3)

0

–6

(3,  )

4

+6

Solution: [ 2, 3]

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1813


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Let f ( x )  3 x 2  2 and g( x )  2 x  1. Find each value or function. 33. f ( 1)

Solution f (1)  3( 1)2  2  3  2  5 34. ( g  f )(2)

Solution ( g  f )(2)  g(f (2))  g(3(2)2  2)  g(14)  2(14)  1  27 35. (f  g)( x )

Solution (f  g)( x )  f ( g( x ))  f (2 x  1)  3(2 x  1)2  2  3(4 x 2  4 x  1)  2  12 x 2  12 x  3  2  12 x 2  12 x  5

36. ( g  f )( x )

Solution ( g  f )( x )  g(f ( x ))  g(3 x 2  2)  2(3 x 2  2)  1  6x 2  4  1  6x 2  3 37. Write y  log 2 x in exponential notation.

Solution y  log 2 x  2 y  x 38. Write 3b  a in logarithmic notation.

Solution 3b  a  log 3 a  b Find x. 39. log x 25  2

Solution log x 25  2  x 2  25  x  5

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1814


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

40. log 5 125  x

Solution log 5 125  x  5x  125  x  3 41. log 3 x  3

Solution

log 3 x  3  33  x  x 

1 27

42. log 5 x  0

Solution log 5 x  0  50  x  x  1 43. Find the inverse of y  log 2 x.

Solution y  log 2 x; inverse: y  2x 44. If log 10 10 x  y , then y equals what quantity?

Solution log 10 10 x  x, so y  x. log 7  0.8451 and log 14  1.1461 Approximate each expression without using a calculator or tables. 45. log 98

Solution log 98  log(14  7)  log 14  log 7  1.1461  0.8451  1.9912 46. log 2

Solution log 2  log

14  log 14  log 7  1.1461  0.8451  0.3010 7

47. log 49

Solution log 49  log 72  2log 7  2(0.8451)  1.6902 48. log

7 (Hint: log 10 = 1.) 5

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1815


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution 7 14 log  log  log 14  log 10  1.14641  1  0.1461 5 10 49. Solve: 2x  2  3x .

Solution

2x  2  3x log 2x  2  log 3x ( x  2)log 2  x log 3 x log 2  2log 2  x log 3 2log 2  x log 3  x log 2 2log 2  x(log 3  log 2) 2log 2 x log 3  log 2 50. Solve: 2log 5  log x  log 4  2.

Solution 2log 5  log x  log 4  2 log 52  log x  log 4  2 log

25 x 2 4 25 x 102  4 400  25 x 16  x

Use a calculator for Exercises 51 and 52. 51. Boat depreciation How much will a $9000 boat be worth after 9 years if it depreciates 12% per year?

Solution

 r A  A0  1   k 

kt

 0.12   9000  1   1    $2848.31

1(9)

52. Find log 6 8.

Solution log 8 log 6 8   1.16056 log 6

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1816


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

2 x  y  5 53. Use graphing to solve  .  x  2 y  0

Solution 2 x  y  5   x  2 y  0

solution: (2, 1) 3 x  y  4 54. Use substitution to solve  . 2 x  3 y  1

Solution (1) 3 x  y  4  (2) 2 x  3 y  1 Substitute y  3 x  4 from (1) into (2):

2 x  3 y  1 2 x  3( 3 x  4)  1 2 x  9 x  12  1 11x  11 x1 Substitute and solve for y: y  3 x  4  3(1)  4  1 x  1, y  1  x  2 y  2 55. Use elimination to solve  . 2 x  y  6

Solution x  2 y  2 x  2 y  2 2 x  y  6 2 x  y  6   2 4 x  2 y  12 2(2)  y  6 5x

 10

y  2

x

 2

y  2

Solution: x  2, y  2

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1817


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

x  y  1  56. Use any method to solve  10 5 2 . y 13  2x  5  10

Solution x  y  1   10 10 5 2 x  y  13   10 2 5 10

57. Evaluate:

3 2 1

1

x  2y  5

x  2y  5

Solution:

5 x  2 y  13

3  2y  5

x  3, y  1

6x

 18

2y  2

x

 3

y 1

.

Solution 3 2  3( 1)  ( 2)1 1 1  3  2  1 4 x  3 y  1 58. Use Cramer’s Rule and solve for y only:  . 3 x  4 y  7

Solution 4 1

y 

3 7 4 3 3 4

25  1 25

x  y  z  1  59. Solve: 2 x  y  z  4 . x  2 y  z  4 

Solution 1 1  x  1 1 1  1  3       1  y   2 1 1  4    3  z   1 2 1  4   1       3

0   1  1     0  1   4    1 3   1   4 1    3 3 3 1 3

 x  2 y  3z  6  60. Solve: 3 x  2 y  z  6 . 2 x  3 y  z  6 

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1818


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution 1 1  x   1 2 3 6   12       1  y   3 2 1 6    12       5  z  2 3 1 6  12

7 12 5 12 1 12

 1  6   1 3 2  6    1   3       1 6      1 3

61. Identify the vertex of the parabola ( y  3)2  8( x  3).

Solution ( y  3)2  8( x  3) Vertex: (–3, 3) 62. Graph x 2  8 y .

Solution x 2  8 y ; Vertex (0, 0) 4 p  8, p  2, opens down

63. Graph

( x  1)2 ( y  3)2   1. 9 25

Solution ( x  1)2 ( y  3)2  1 9 25 Center: (1, 3), a  5, b  3, vertical

64. Graph

( y  3)2 ( x  2)2  1 9 16

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1819


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution ( y  3)2 ( x  2)2  1 9 16 Center: (2, 3), a  3, b  4, vertical

GROUP ACTIVITY SOLUTIONS Suspension Bridges Real-World Example of a Conic Suspension bridges, like the Brooklyn Bridge and the Golden Gate Bridge, suspend the roadway by cables, ropes, or chains from two tall towers. These towers support the majority of the weight as compression pushes down on the suspension bridge's deck and then travels up the cables, ropes, or chains to transfer compression to the towers. The towers then dissipate the compression directly into the earth. Their parabolic shape helps ensure that the bridge stays up and that the cables can sustain the weight of hundreds of cars and trucks each hour. The largest suspension bridge in the world is the Akashi Kaikyō Bridge in Japan. It is 1191 feet long.

Group Activity A suspension bridge cable is suspended in the shape of a parabola between two tall towers that are 75 feet above the road. The tops of the towers are 2500 feet apart. The cables are 4.75 feet above the road midway between the towers. Find the height of the cable 625 feet from the center of the bridge. a. Draw a figure of the suspension bridge and label the information given in the problem. Assume the origin to be midway between the towers on the road or deck. b. Find an equation in standard form of the equation of the parabola. State the vertex of the parabola and determine p. Round p to three decimal places. c. Substitute into the equation of the parabola and determine the height of the cable 625 feet from the center of the bridge. Round the height to three decimal places.

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1820


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 7: Conic Sections and Systems of Nonlinear Equations

Solution a.

x 2  4 p( y  4.75) b.

12502  4 p(75  4.75) p  5560.498 So, x 2  22241.992( y  4.75)

Vertex (0, 4.75)

c. 6252  2241.992( y  4.75) y  22.313 feet

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1821


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution and Answer Guide GUSTAFSON/HUGHES, C OLLEGE ALGEBRA 2023, 9780357723654; C HAPTER 8: SEQUENCES, SERIES, INDUCTION, AND PROBABILITY

TABLE OF CONTENTS End of Section Exercise Solutions ................................................................................ 1822 Exercises 8.1 ........................................................................................................................... 1822 Exercises 8.2 .......................................................................................................................... 1836 Exercises 8.3 .......................................................................................................................... 1854 Exercises 8.4 .......................................................................................................................... 1868 Exercises 8.5 .......................................................................................................................... 1885 Exercises 8.6 .......................................................................................................................... 1903 Exercises 8.7 .......................................................................................................................... 1920 Chapter Review Solutions.............................................................................................. 1933 Chapter Test Solutions .................................................................................................. 1948 Group Activity Solutions ................................................................................................ 1954

END OF SECTION EXERCISE SOLUTIONS EXERCISES 8.1 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Square each binomial. a.  2 x  5 y 

2

b.  2 x  5 y 

2

Solution a. 4 x 2  20 xy  25 y 2 b. 4 x 2  20 xy  25 y 2

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1822


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

2. Cube each binomial. a.  x  4 y 

3

b.  x  4 y 

3

Solution a. x 3  12 x 2 y  48 xy 2  64 y 3 b. x 3  12 x 2 y  48 xy 2  64 y 3 3. Given  a  b   a4  4a3 b  6a2 b2  4ab3  b4 . 4

a. What is the sum of the exponents on the variables of each term? b. The pattern of the exponents on a is 4, 3, 2, 1, 0. What is the pattern of the exponents on b?

Solution a. 4 b. 0, 1, 2, 3, 4 4. Given  a  b   a5  5a4 b  10a3 b2  10a2 b3  5ab4  b5 . 5

a. What is the sum of the exponents on the variables of each term? b. The pattern of the exponents on b is 0, 1, 2, 3, 4, 5. What is the pattern of the exponents on a?

Solution a. 4 b. 5, 4, 3, 2, 1, 0 5. Evaluate: 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1.

Solution 5040 6. Evaluate:

654321 without a calculator. 4  3  2  1 2  1

Solution

65 4  3  2 1 4  3  2  1 2  1

30  15 2

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. In the expansion of a binomial, there will be one more term than the __________ of the binomial.

Solution power

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1823


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

8. The __________ of each term in a binomial expansion is the same as the exponent of the binomial.

Solution degree 9. The __________ term in a binomial expansion is the first term raised to the power of the binomial.

Solution first 10. In the expansion of  a  b  , the __________ on a decrease by 1 in each successive term. n

Solution exponents 11. Expand 7!: __________

Solution 7∙6∙5∙4∙3∙2∙1 12. 0! = _____

Solution 1 13. n ∙ _________ = n!

Solution

 n  1 !

14. In the seventh term of  a  b  , the exponent on a is _____. 11

Solution 11 – 7 + 1 = 5 Practice Evaluate each expression. 15. 5!

Solution 5!  5  4  3  2  1  120 16. –5!

Solution

5!    5  4  3  2  1  120

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1824


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

17. 3! ∙ 6!

Solution 3!  6!  6  720  4320 18. 0 !  7 !

Solution 0!  7 !  1  5040  5040 19. 6!  6!

Solution 6!  6!  720  720  1440 20. 5 !  2 !

Solution 5!  2!  120  2  118 21.

9! 12!

Solution 9! 9! 1   12! 12  11  10  9! 1320 22.

8! 5!

Solution 8! 8  7  6  5!   8  7  6  336 5! 5! 23.

5!  7 ! 9!

Solution 5!  7 ! 5!  7 ! 120 5    9! 9 8  7! 72 3 24.

3!  5!  7 ! 1!8!

Solution 3!  5!  7 ! 3!  5 !  7 ! 720    90 1! 8! 8  7! 8

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1825


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

25.

18!

6!  18  6  !

Solution

18!

6!  18  6  ! 26.

18  17  16  15  14  13  12! 13, 366,080 18!    18,564 6!  12! 6!  12! 720

15  14  13  12  11  10  9! 3,603,600 15!    5005 9!  6! 9!  6! 720

15!

9!  15  9 !

Solution

15!

9!  15  9 !

Use Pascal’s Triangle to expand each binomial. 27.  x  y 

3

Solution

 x  y   x  3 x y  3 xy  y 3

28.  x  y 

3

2

2

3

3

Solution Row 3 of Pascal's triangle: 1 3 3 1

 x  y   x  3 x   y   3 x   y     y   x  3 x y  3 xy  y 3

3

2

2

3

3

2

2

3

29.  a  b 

5

Solution Row 5 of Pascal's triangle: 1 5 10 10 5 1

a  b   a  5a b  10a b  10a b  5ab  b 5

5

4

3

2

2

3

4

5

30.  a  b 

5

Solution

a  b   a  5a b  10a b  10a b  5ab  b 5

31.

x  y

5

4

3

2

2

3

2

5

7

Solution Row 7 of Pascal's triangle: 1 7 21 35 35 21 7 1

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1826


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 x  y   x  7 x   y   21x   y   35 x   y   35 x   y   21x   y   7 x   y     y  7

7

6

2

5

3

4

4

3

5

2

6

7

 x 7  7 x 6 y  21x 5 y 2  35 x 4 y 3  35 x 3 y 4  21x 2 y 5  7 xy 6  y 7

32.  a  b 

7

Solution Row 7 of Pascal's triangle: 1 7 21 35 35 21 7 1

a  b   a  7a b  21a b  35a b  35a b  21a b  7ab  b 7

7

6

5

2

4

3

3

4

2

5

6

7

Use the Binomial Theorem to expand each binomial. 33.  a  b 

3

Solution

a  b   a  1!3!2! a b  2!3!1! ab  b  a  3a b  3ab  b 3

34.  a  b 

3

2

2

3

3

2

2

3

4

Solution

a  b   a  1!43!! a b  2!42!! a b  3!4 1!! ab  b  a  4a b  6a b  4ab  b 4

4

3

2

2

3

4

4

3

2

2

3

4

35.  a  b 

5

Solution

a  b   a  1!5!4 ! a  b   2!5!3! a  b   3!5!2! a  b   45!! 1! a  b    b  5

5

4

2

3

2

3

4

5

 a5  5a4 b  10a3 b2  10a2 b3  5ab4  b5

36.  x  y 

4

Solution

 x  y   x  1!4!3! x   y   2!4!2! x   y   3!4!1! x   y     y  4

4

3

2

2

3

4

 x 4  4 x 3 y  6 x 2 y 2  4 xy 3  y 4

37.  2 x  y 

3

Solution

 2 x  y    2 x   1!3!2!  2 x  y  2!3!1!  2 x  y  y  8 x  12 x y  6 xy  y 3

3

2

2

3

3

2

2

3

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1827


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

38.  x  2 y 

3

Solution

 x  2 y   x  1!3!2! x  2 y   2!3!1! x  2 y    2 y   x  6 x y  12 xy  8 y 3

39.  x  2 y 

3

2

2

3

3

2

2

3

3

Solution

 x  2 y   x  1!3!2! x  2 y   2!3!1! x  2 y    2 y   x  6 x y  12 xy  8 y 3

40.  2x  y 

3

2

2

3

3

2

2

3

3

Solution

 2 x  y    2 x   1!3!2!  2 x    y   2!3!1!  2 x   y     y   8 x  12 x y  6 xy  y 3

3

41.  2 x  3 y 

2

2

3

3

2

2

3

4

Solution

 2x  3 y    2x   1!4!3!  2x   3 y   2!4!2!  2x   3 y   3!4!1!  2x  3 y    3 y  4

4

3

2

2

3

4

 16 x 4  96 x 3 y  216 x 2 y 2  216 xy 3  81 y 4

42.  2 x  3 y 

4

Solution

 2x  3 y    2x   1!4!3!  2x   3 y   2!4!2!  2x   3 y   3!4!1!  2x  3 y    3 y  4

4

3

2

2

3

4

 16 x 4  96 x 3 y  216 x 2 y 2  216 xy 3  81 y 4

43.  x  2 y 

4

Solution

 x  2 y   x  1!4!3! x  2 y   2!4!2! x  2 y   3!4!1! x  2 y    2 y  4

4

3

2

2

3

4

 x 4  8 x 3 y  24 x 2 y 2  32 xy 3  16 y 4

44.  x  2 y 

4

Solution

 x  2 y   x  1!4!3! x  2 y   2!4!2! x  2 y   3!4!1! x  2 y    2 y  4

4

3

2

2

3

4

 x 4  8 x 3 y  24 x 2 y 2  32 xy 3  16 y 4

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1828


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

45.  x  3 y 

5

Solution

 x  3 y   x  1!5!4! x  3 y   2!5!3! x  3 y   3!5!2! x  3 y   4!5!1! x  3 y    3 y  5

5

4

2

3

2

3

4

5

 x 5  15 x 4 y  90 x 3 y 2  270 x 2 y 3  405 xy 4  243 y 5

46.  3x  y 

5

Solution

 3x  y    3x   1!5!4!  3x    y   2!5!3!  3x    y   3!5!2!  3x    y   4!5!1!  3x   y     y  5

5

4

3

2

2

3

4

5

 243 x 5  405 x 4 y  270 x 3 y 2  90 x 2 y 3  15 xy 4  y 5

x  47.   y  2  

4

Solution 4

4

3

2

x  x 4!  x  4!  x  2 4!  x  3 4   y       y   y   y  y 1! 3!  2  2! 2!  2  3! 1!  2  2  y 1 4 1 3 3  x  x y  x 2 y 2  2 xy 3  y 4 16 2 2

 y 48.  x   2 

4

Solution 4

2

3

 y 4! 3  y  4! 2  y  4!  y   y  4 x   x    x    x    x  2 1! 3! 2 2! 2! 2 3! 1!  2   2        3 1 1 4  x 4  2 x 3 y  x 2 y 2  xy 3  y 2 2 16

4

Find the required term in each binomial expansion. 49.  a  b  ; 3rd term 4

Solution The 3rd term will involve b2. 4! 2 2 a b  6a2 b2 2! 2!

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1829


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

50.  a  b  ; 2nd term 4

Solution The 2nd term will involve  b  . 1

4! 3 a  b   4a3 b 3! 1!

51.

a  b  ; 5th term 7

Solution The 5th term will involve b4. 7! 3 4 a b  35a3 b4 3! 4 ! 52.  a  b  ; 4th term 5

Solution The 4th term will involve b3. 5! 2 3 a b  10a2 b3 2! 3! 53.  a  b  ; 6th term 5

Solution The 6th term will involve (–b)5. 5 5! 0 a  b   b5 0!5! 54.  a  b  ; 7th term 8

Solution The 7th term will involve (–b)6. 6 8! 2 a   b   28a2 b6 2! 6! 55.  a  b  ; 5th term 17

Solution The 5th term will involve b4. 17 ! 13 4 a b  2380a 13 b4 13! 4 !

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1830


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

56.  a  b  ; 3rd term 12

Solution The 3rd term will involve (–b)2. 2 12! 10 a  b   66a 10 b2 10 ! 2!

57. a  2

 ; 2nd term 4

Solution

 

1

The 2nd term will involve  2 .

 

1 4! 3 a  2  4 2 a3 3! 1!

8

58. a  3 ; 3rd term

Solution

 .

The 3rd term will involve  3

2

 

2 8! 6 a  3  84 a6 6! 2!

9

59. a  3b ; 5th term

Solution

 3b . 4

The 5th term will involve 9! 5 a 5! 4 !

60.

 3b  1134 a b 4

5

4

 2a  b ; 4th term 7

Solution The 4th term will involve  b  . 3

7! 4 ! 3!

 2a   b  140 a b 4

3

4

3

4

x  61.   y  ; 3rd term 2  Solution The 3rd term will involve y 2 .

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1831


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability 2

4!  x  2 3 2 2   y  x y 2! 2!  2  2 8

 n 62.  m   ; 3rd term 2  Solution 2

 n The 3rd term will involve   . 2 2

 n 8! m6    7m6 n2 6! 2! 2 11

r s 63.    ; 10th term 2 2 Solution 9

 s The 10th term will involve    .  2 2

9

11!  r   s  55 2 9 r s       2!9!  2   2  2048 9

 p q 64.    ; 6th term 2 2 Solution 5

 q The 6th term will involve    .  2 4

5

9!  p   q  63 4 5 pq       4!5!  2   2  256 65.  a  b  ; 4th term n

Solution The 4th term will involve b3. n!

 n  3 ! 3!

an3 b3

66.  a  b  ; 5th term n

Solution The 5th term will involve (–b)4.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1832


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

n!

 n  4  ! 4!

a n  4  b   4

n!

 n  4  ! 4!

a n  4 b4

67.  a  b  ; rth term n

Solution The rth term will involve br – 1. n!

 n  r  1 !  r  1 !

a n  r  1 br  1

68.  a  b  ;  r  1 th term n

Solution The (r + 1)th term will involve br. n!

n  r  ! r  !

a n  r br

Fix It In exercises 69 and 70, identify the step where the first error is made and fix it. 69. Evaluate:

8!

6!  8  6  !

Solution Step 2 was incorrect. Step 2:

8  7  6! 6! 2!

Step 3:

87 2

Step 4:

56 2

Step 5: 28 70. Use Pascal’s Triangle to expand ( x  y )4 . To do so, first write down the appropriate row of Pascal’s Triangle, then determine ( x  y )4 , and then substitute –y in for y.

Solution Step 4 was incorrect. Step 4:  x  y   x 4  4 x 3 y  6 x 2 y 2  4 xy 3  y 4 4

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1833


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Discovery and Writing 71. Describe how to construct Pascal’s Triangle.

Solution Answers may vary. 72. What is binomial expansion?

Solution Answers may vary. 73. Explain why the terms alternate in the binomial expansion of  x  y  . 8

Solution Answers may vary. 74. Define factorial notation and explain how to evaluate 10!.

Solution Answers may vary. 75. With a calculator, evaluate 69!. Explain why we cannot find 70! with a calculator.

Solution Answers may vary. 76. Find the sum of the numbers in each row of the first ten rows of Pascal’s Triangle. Do you see a pattern?

Solution Answers may vary. 77. Show that the sum of the coefficients in the binomial expansion of (x + y)n is 2n. (Hint: Let x = y = 1.)

Solution Answers may vary. 78. Explain how the rth term of a binomial expansion is constructed.

Solution Answers may vary. 79. If we applied the pattern of coefficients to the coefficient of the first term in the Binomial Theorem, it would be

n!

0!  n  0  !

. Show that this expression equals 1.

Solution n!

0!  n  0  !

n! n!  1 0! n ! 1  n !

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1834


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

80. If we applied the pattern of coefficients to the coefficient of the last term in the Binomial Theorem, it would be

n!

n !  n  n !

. Show that this expression equals 1.

Solution n!

n !  n  n !

n! n!  1 n ! 0! n !  1

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 81. 0!  0

Solution

False. 0!  1 82. The first term in the expansion of  a  b 

999

is a999 .

888

is b888 .

Solution True. 83. The last term in the expansion of  a  b 

Solution True. 84. For the expansion of  a  b 

777

, the exponents on a increase by 1 in each successive term.

Solution False. The exponents on a decrease by 1 in each successive term. 85. For the expansion of  a  b 

666

, the exponents on b decrease by 1 in each successive term.

Solution False. The exponents on b increase by 1 in each successive term. 86. The sum of the exponents on the variables in any term in the expansion of  a  b 

555

is

555.

Solution True. 87. The number of terms in the binomial expansion of  a  b 

444

is 444.

Solution False. The number of terms is 445.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1835


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

88. To find the binomial expansion of x 333  y 222

 , it is helpful to rewrite the expression 111

inside the parentheses as x 333   y 222 .

Solution True.

 1 89. The constant term in the expansion of  a   a 

10

is 252.

Solution False. The constant terms is –252. 9

 1 90. The coefficient of x5 in the expansion of  x   is 36. x   Solution True.

EXERCISES 8.2 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

 

Given the function f n  n2  3. Determine f(1), f(2), f(3), f(4), f(5), and f(6).

Solution

f  1  12  3  4

f  2   22  3  7

f  3   32  3  12

f  4   42  3  19

f  5   52  3  28

f  6   62  3  39 2. Evaluate

n1 for n = 1, 2, 3, 4, and 5. 4n

Solution 1 1 2 1   4 2 4 1



21

4  2

3 8

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1836


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

31

4  3 41

4 4 51

4 5

4 1  12 3

5 16

6 3  20 10

3. Evaluate  n 2  7 for n = 1, 2, 3, 4, and 5.

Solution 12  7  6 22  7  3 32  7  2 42  7  9 52  7  18

4. Evaluate:

  1 b.  1

8

a.

9

Solution a. 1 b. –1

 1 5. Evaluate 2n

n

for n = 1, 2, 3, 4, and 5.

Solution

 1  1 1

21

2

 1  1 22

2

4

 1  1 23

3

8

 1   24

4

16

 1  1 25

5

32

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1837


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 1 6. Evaluate

n 1

for n = 1, 2, 3, 4, and 5. Then find the sum of the five evaluations.

2n

Solution

 1 21

 1

1 1

 1

2

 1  1

 1  1

 1  1

 1

 1  1

2

4

8

8

5

5 1

5

4

4

4 1

24

2

3

3 1

23

 1

2

2 1

2

2

 1  1 

16

16

6

32

32

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. An infinite sequence is a function whose __________ is the set of natural numbers.

Solution domain 8. A finite sequence is a function whose __________ is the set of the first n natural numbers.

Solution domain 9. A __________ is formed when we add the terms of a sequence.

Solution series 10. A series associated with a finite sequence is a __________ series.

Solution finite 11. A series associated with an infinite sequence is an __________ series.

Solution infinite

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1838


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

12. If the signs between successive terms of a series alternate, the series is called an __________ series.

Solution alternating 13. __________ is a shorthand way to indicate the sum of the first n terms of a sequence.

Solution Summation notation 5

14. The symbol  k 2  3 indicates the __________ of the five terms obtained when we k 1

successively substitute 1, 2, 3, 4, and 5 for k.

Solution sum 15.

5

5

 6k  ______  k 2

k 1

2

k 1

Solution 6 16.

  k  3k    k  ________ 5

5

2

k 1

2

k 1

Solution 5

5

k 1

k 1

  3k   3 k 17.

5

 c, where c is a constant, equals ______. k 1

Solution 5c 18. The summation of a sum is equal to the ______ of the summations.

Solution sum Practice Write the first six terms of the sequence defined by each function.

 

19. f n  5n n  1

Solution

f  1  5  1 1  1  0

f  2   5  2  2  1  10

f  3   5  3  3  1  30

f  4   5  4  4  1  0 f  5   5  5  5  1  100 f  6   5  6  6  1  150

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1839


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 n  1 n  2 20. f  n  n     2  3  Solution  1  1 1  2  f  1  1   0  2  3 

 2  1 2  2  3  1 3  2  f  2  2     0 f  3  3   1 2 3     2  3   4  1 4  2   5  1 5  2  6  1 6  2  f 4  4     4 f 5  5     10 f  6   6     20  2  2   2  3   2  3 

 

21. f n  n3  1

Solution a1  13  1  2 a2  23  1  9 a3  33  1  28 a4  43  1  65 a5  53  1  126 a6  63  1  217

 

22. f n  n4  2

Solution a1  14  2  1 a2  24  2  14 a3  34  2  79 a4  44  2  254 a5  54  2  623 a6  64  2  1294

 1 23. f  n   n

n

3

Solution

 1  1 a  1

1

13

 1  1 a  2

23

2

8

a3 

 1  1

a4 

 1  1

3

27

33

4

43

64

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1840


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 1  1 a  5

5

125

53

 1  1 6

a6 

216

63

 1 24. f  n  3   2

n

Solution 1

 1 3 a1  3    2 2 2

 1 3 a2  3    4 2 3

 1 3 a3  3    8 2 4

 1 3 a4  3    16 2 5

 1 3 a5  3    32 2 6

 1 3 a6  3    64 2

Find the next term of each sequence. 25. 1, 6, 11, 16, ⋯

Solution 1, 6, 11, 16, ⋯ Add 5 to get the next term. The next term is 21. 26. 1, 8, 27, 64, ⋯

Solution 1, 8, 27, 64, ⋯ Each term is a perfect cube. The next term is 53 = 125. 27. a, a + d, a + 2d, a + 3d, ⋯

Solution

a, a  d, a  2d, a  3d,  Add d to get the next term. The next term is a + 4d. 28. a, ar , ar 2 , ar 3 , 

Solution a, ar , ar 2 , ar 3 ,  Multiply by r to get the next term. The next term is ar4.

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1841


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

29. 1, 3, 6, 10, ⋯

Solution 1, 3, 6, 10, ⋯ The difference between terms increases by 1 each time. The next term is 10 + 5 = 15. 30. 20, 17, 13, 8, ⋯

Solution 20, 17, 13, 8, ⋯ The difference between terms increases by 1 each time. The next term is 8 – 6 = 2. Write the first five terms of each sequence and then find the specified term. 31. an  9n  1; a30

Solution

a1  9  1  1  8

a2  9  2   1  17

a3  9  3   1  26

a4  9  4   1  35 a5  9  5   1  44

a30  9  30   1  269 32. an  7n  3; a14

Solution

a1  7  1  3  10

a2  7  2   3  17

a3  7  3   3  24

a4  7  4   3  31

a5  7  5   3  38

a14  7  14   3  101 33. an   n2  5; a20

Solution a1    1  5  4 2

a2    2   5  1 2

a3    3   5  4 2

a4    4   5  11 2

a5    5   5  20 2

a20    20   5  395 2

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1842


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

34. an   2n2  n; a13

Solution a1  2  1   1  1 2

a2  2  2    2   6 2

a3  2  3    3   15 2

a4  2  4    4   28 2

a5  2  5    5   45 2

a13  2  13    13   325 2

35. an  n3  6; a10

Solution

a1   1  6  7 3

a2   2  6  14 3

a3   3  6  33 3

a4   4   6  70 3

a5   5  6  131 3

a6   10  6  1006 3

36. an   n3  7; a15

Solution a1    1  7  8 3

a2    2   7  15 3

a3    3   7  34 3

a4    4   7  71 3

a5    5  7  132 3

a15    15   7  3382 3

37. an 

n1 ; a30 n

Solution

 1  1  0  1  2  1  1 a  2 2 a1 

2

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1843


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 3  1  2  3 3 4  1  3 a  4 4 5  1  4 a  5 5  30  1  29 a   30 30 a3 

4

5

30

38. an 

n1 ; a15 3n

Solution

 1  1  2 3 3  1  2  1  3  1 a  6 2 3  2  3  1  4 a  9 3  3 4  1  5 a  12 3 4 5  1  6  2 a  15 5 3 5  15  1  16 a  45 3  15  a1 

2

3

4

5

15

 1 ; a 39. a  n

n

8

4n

Solution

 1   1 a  1

1

41

4

a2 

 1  1

a3 

 1   1

42 43

2

16

3

64

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1844


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 1  a  4

4

4

1 256

4

 1   a  5

5

 1  a  8

6

1 1024

45

48

1 65, 536

 1 ; a 40. a  n

n

9

5n

Solution

 1   1 a  1

1

5

51

a2 

 1  1

a3 

 1   1

a4 

 1  1

a5 

 1  

2

25

52

3

125

53

4

625

54

5

1 3125

5

5

 1

9

a9 



9

5

 1 41. a  n

1 1,953, 125

n 1

n2

; a16

Solution

 1 a 

1 1

 1 a 

2 1

 1 a 

3 1

 1 a 

4 1

1

2

3

4

2

1

2

2

2

3

42

1 1 1

 

1 4

1 9



1 16

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1845


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 1 a 

5 1

5

2

5

 1 a  16

16  1



162

 1 42. a  n

1 25 1 256

n 1

; a11

n3

Solution

 1 a 

1 1

 1 a 

2 1

 1 a 

3 1

 1 a 

4 1

1

3

1

2

3

2

3

3

3

4

4

3

 1 a  5

5

 1 a  11

 

11

1 64

1 125

1 1331

11 1

3

1 8

1 27



5 1

3

1 1 1

Find the sum of the first five terms of the sequence with the given general term. 43. an  n

Solution 1  2  3  4  5  15 44. an  2n

Solution

2  1  2  2  2  3  2  4   2  5  30

45. an  3

Solution 3  3  3  3  3  15 46. an  4n2

Solution 4  1  4  2   4  3   4  4   4  5   220 2

2

2

2

2

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1846


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 1 47. an  2   3

n

Solution 1

2

3

4

5

 1  1  1  1  1 2 2 2 2 2 242 2   2   2   2   2        3 9 27 81 243 242 3 3 3 3 3 48. an   1

n

Solution

 1  1  1   1  1   1  1 n

49. an  3n  2

Solution 3  1  2    3  2   2   3  3   2   3  4   2   3  5   2   1  4  7  10  12  35           50. an  2n  1

Solution 2  1  1  2  2   1  2  3   1  2  4   1  2  5   1  3  5  7  9  11  35           Assume that each sequence is defined recursively. Find the first four terms of each sequence. 51. a1  3 and an  1  2an  1

Solution

a1  3

a2  2a2  1  2  3  1  7

a3  2a2  1  2  7   1  15

a4  2a3  1  2  15  1  31 52. a1  5 and an 1  an  3

Solution

a1  5

a2  a2  3    5   3  2 a3  a2  3    2   3  5

a4  a3  3    5  3  2

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1847


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

an

53. a1  4 and an  1 

2

Solution a1  4 a 4 a2  1   2 2 2 a 2 a3  2   1 2 2 a 1 1 a4  3   2 2 2 54. a1  0 and an  1  2an2

Solution

a1  0

a2  2a12  2  0   0 2

a3  2a22  2  0   0 2

a4  2a32  2  0   0 2

55. a1  k and an  1  2an2

Solution a1  k a2  a12  k 2

  k a  a  k   k 2

a3  a22  k 2 4

2 3

2

4

4

8

56. a1  3 and an  1  kan

Solution a1  3

a2  ka1  3k

a3  ka2  k  3k   3k 2

 

a4  ka3  k 3k 2  3k 3 57. a1  8 and an  1 

2an k

Solution

a1  8 a2 

2an k

2  8 k

16 k

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1848


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

a3  a4 

2a2

k 2a3

k

2  16k  k

32 k2

   64

2 32 k2 k

58. a1  m and an  1 

k3

an2 m

Solution a1  m a12

m2 m m m a2 m2 m a3  2  m m a2 m2 a4  3  m m m a2 

Determine whether each series is an alternating infinite series. 59. 1  2  3     1 n   n

Solution alternating 60. 1  4  9     1

n 1

n2  

Solution alternating 61. a  a  a    a  ; a  3 2

3

n

Solution not alternating 62. a  a  a    a  ; a  2 2

3

n

Solution alternating Evaluate each sum. 5

63.  2k k 1

Solution 5

5

k 1

k 1

 2k  2 k  2  1  2  3  4  5   2  15   30

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1849


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

6

64.  3k k 3

Solution 6

6

k 3

k 3

 3k  3 k  3  3  4  5  6   3  18   54 4

65.  2k 2 k 3

Solution

  2k   2 k  2  3  4   1  25   50 4

4

2

k 3

2

2

2

k 3

100

66.  5 k 1

Solution 100

 5  100 5   500 k 1

5

67.   3k  1 k 1

Solution 5

5

5

k 1

k 1

k 1

  3k  1  3 k   1  3  1  2  3  4  5  5  1  3  15   5  40 5

68.  k 2  3k k 1

Solution

  k  3k    k  3 k   2  3  4  5   3  2  3  4  5 5

2

k 2

5

k 2

5

2

k 2

2

2

2

2

  4  9  16  25  3  14   54  42  96

1 2 k 1

1000

69. 

Solution 1000

1

 1

 2  1000  2   500 k 1

 

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1850


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

2 k 4 k 5

70. 

Solution 5 2 2 2 1 2 5 4 9         k 4 5 2 5 10 10 10 k 4 71.

4

1

k k 3

Solution 4 1 1 1 4 3 7       k 3 4 12 12 12 k 3 6

6

72.  3k 2  2k  3 k 2 k 2

k 2

Solution

  3k  2k   3 k    3k  2k    3k 6

6

2

k 2

6

2

k 2

6

2

k 2 6

6

k 2

k 2

2

k 2

  2k  2 k  2  2  3  4  5  6   2  20   40 4

4

73.   4k  1    4k  1 2

k 1

2

k 1

Solution

  4k  1    4k  1    16k  8k  1    16k  8k  1 4

2

k 1

4

2

k 1

4

4

2

k 1 4

4

k 1

k 1

2

k 1

  16k  16 k  16  1  2  3  4   16  10   160 10

10

74.   2k  1  4  k  1  k  2

k 0

k 0

Solution

  2k  1  4 k  1  k     4k  4k  1  4  k  k  10

k 0

2

10

10

k 0

k 0 10

10

2

k 0 10

2

10

  4k  4k  1   4k  4k 2   1  11  1  11 k 0

2

k 0

k 0

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1851


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

8

8

75.   5k  1    10k  1 2

k 6

k 6

Solution

 5k  1    10k  1    25k  10k  1    10k  1 8

2

k 6

8

8

k 6

k 6 8

k 6

  25k k 6

7

8

2

2

  25 k  25 6  7  8   3725 8

2

2

2

2

k 6

7

76.   3k  1  3 k  3k  2  2

k 2

k 2

Solution

  3k  1  3 k  3k  2   9k  6k  1  3  3k  2k  7

7

7

k 2

k 2 7

2

k 2

7

2

k 2 7

7

  9k  6k  1   9k 2  6k   1  1  6   6 k 2

2

k 2

k 2

Fix It In exercises 77 and 78, identify the step where the first error is made and fix it. 77. If an   1

n 1

2 , determine a4. 5n  2

Solution Step 3 was incorrect. 2 Step 3: a4   1 15, 625 Step 4: a4  

2 15, 625

4

78. Evaluate   2k  5  using the summation properties. 2

k 1

Solution Step 4 was incorrect.

 

 

 

Step 4: 4 30  20 10  25 4 Step 5: 20

Discovery and Writing 79. What is the difference between a sequence and a series?

Solution Answers may vary.

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1852


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

80. What is the symbol  and how is it used in this section?

Solution Answers may vary. 81. Find a counterexample to disprove the proposition that the summation of a product is the product of the summations. In other words, prove that n

n

n

k 1

k 1

l1

 f  k  g  k    f  k  g  k  Solution Answers may vary. 82. Find a counterexample to disprove the proposition that the summation of a quotient is the quotient of the summations. In other words, prove that n

f k 

g k  k 1

n

 f k  k 1 n

   g k  k 1

Solution Answers may vary. 83. Explain what it means to define a sequence recursively.

Solution Answers may vary. 84. Explain why the Binomial Theorem can be stated as n

n!

 r! n r !a

k 0

n r

br

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 85. The next term in the sequence 1, –8, 27, –67, 125, ⋯ is 216.

Solution False. The next term is –216. 86. The next term in the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ⋯ is 243.

Solution False. The next term is 233.

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1853


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

  1 87. If a  n

n 1

n

, then a324 

1 . 18

Solution True. 88. As n increases without bound, the terms of the sequence an 

2n approach 2. n1

Solution True. 1000

89.  5  5000 2

Solution 100

False.  5  999  5   4995. k 2

90. The graph of a sequence is a set of discrete points.

Solution True. 91.

999

 9k

99

k 1

 999   999     9    k 99   k 1   k 1 

Solution 999

999

k 1

k 1

False.  9k 99  9 k 99 .

92.

 k

8888

88

k 8

8888

8888

k 8

k 8

 k 888   k 888   k 88

Solution 8888

8888

8888

k 8

k 8

False.  k 88  k 888   k 88   k 888 . k 8

EXERCISES 8.3 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

If a = 2 and d = 5, write the five terms represented by the sequence of the form a, a + d, a + 2d, a + 3d, a + 4d.

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1854


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution a2 ad  25  7

a  2d  2  2  5   12 a  3d  2  3  5   17

a  4d  2  4  5   22

2. If a = 2 and d = –5, write the five terms represented by the sequence of the form a, a + d, a + 2d, a + 3d, a + 4d.

Solution a2 ad  25  73

a  2d  2  2  5   8

a  3d  2  3  5   13

a  4d  2  4  5   18 3. Consider the sequence 3, 9, 15, 21, 27, ⋯ . Determine each difference and state what you notice. a. a2  a1 b. a3  a2 c. a4  a3 d. a5  a4

Solution a. 9  3  6 b. 15  9  6 c. 21  15  6 d. 27  21  6 There is a common difference of 6. 4. Consider the sequence 3, –3, –9, –15, –21, –27, ⋯ . Determine each difference and state what you notice. a. a2  a1 b. a3  a2 c. a4  a3 d. a5  a4

Solution a. 3  3  6

  c. 15   9  6 d. 21   15  6

b. 9  3  6

5. Given the formula an = a + (n – 1)d. If an = 12, a = –4, and n = 4, what is d?

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1855


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

an  a   n  1 d 12  4  3d 16  3d

16 d 3

6. Given the formula an = a + (n – 1)d. If an = –12, a = –20, and n = 5, what is d?

Solution

an  a   n  1 d 12  20  4d 8  4d 2d

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. An arithmetic sequence is a sequence of the form a, a + d, a + 2d, a + 3d, . . . , a + __________ d, . . .

Solution

 n  1 d

8. An arithmetic series is a series of the form a + (a + d) + (a + 2d) + (a + 3d) + ⋯ + [a + (n – 1) _____ ] + ⋯

Solution d 9. If an arithmetic series has infinitely many terms, it is called an __________ arithmetic series.

Solution infinite 10. In an arithmetic sequence, a is the __________ term, d is the common __________, and n is the __________ of terms.

Solution first, difference, number 11. The last term of an arithmetic sequence is given by the formula __________.

Solution

an  a   n  1 d

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1856


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

12. The formula for the sum of the first n terms of an arithmetic series is given by the formula __________.

Solution

Sn 

n  a  an  2

13. __________ are numbers inserted between a first and last term of a sequence to form an arithmetic sequence.

Solution Arithmetic means 14. The formula __________ gives the distance (in feet) that an object will fall in t seconds.

Solution s  16t 2

Practice Write the first five terms of the arithmetic sequences with the given properties. 15. a  1; d  2

Solution 1, 3, 5, 7, 9 16. a  7; d  10

Solution –7, 3, 13, 23, 33 17. a  11; d  9

Solution 11, 2, –7, –16, –25 18. a  12; d  5

Solution –12, –17, –22, –27, –32 19. a = 5; 3rd term is 2

Solution

an  a   n  1 d a3  5   3  1 d 2  5  2d

3  2d 3 7 1 5   d  5, , 2, ,  1,  2 2 2 2

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1857


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

20. a = 4; 5th term is 12

Solution

an  a   n  1 d a5  4   5  1 d

12  4  2d 8  4d 2  d  4, 6, 8, 10, 12, 14

21. 7th term is 24; common difference is 52

Solution

an  a   n  1 d a7  a   7  1

5 2

5 24  a  6   2 24  a  15 23 33 43 9  a  9, , 14, , 19, 2 2 2 22. 20th term is –49; common difference is –3

Solution

an  a   n  1 d

a20  a   20  1 3 

49  a  19  3  49  a  57

8  a  8, 5, 2,  1,  4,  7

Find the missing term in each arithmetic sequence. 23. Find the 40th term of an arithmetic sequence with a first term of 6 and a common difference of 8.

Solution

an  a   n  1 d

a40  6   40  1 8

 6  39  8   6  312  318

24. Find the 35th term of an arithmetic sequence with a first term of 50 and a common difference of –6.

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1858


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

an  a   n  1 d

a35  50   35  1 6 

 50  34  6   50  204  154

25. The 6th term of an arithmetic sequence is 28, and the first term is –2. Find the common difference.

Solution

an  a   n  1 d

a6  2   6  1 d

28  2  5d 30  5d  d  6

26. The 7th term of an arithmetic sequence is –42, and the common difference is –6. Find the first term.

Solution

an  a   n  1 d

a7  a   7  1 6 

42  a  6  6 

42  a  36  a  6 27. Find the 55th term of an arithmetic sequence whose first three terms are –8, –1, and 6.

Solution

an  a   n  1 d

a55  8   55  1 7

 8  54  7   8  378  370

28. Find the 37th term of an arithmetic sequence whose second and third terms are –4 and 6.

Solution 2nd term 4  a  d

3rd term 6  a  2d

a  d  4 Solve the system:  a  2d  6 a  14, d  10 a37  14  36  10   346

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1859


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

29. If the fifth term of an arithmetic sequence is 14 and the second term is 5, find the 15th term.

Solution 5th term 14  a  4d

2nd term 5 ad

a  4d  14 Solve the system:  a  d  5 a  2, d  3 a15  2  14  3   44

30. If the fourth term of an arithmetic sequence is 13 and the second term is 3, find the 24th term.

Solution 4th term 13  a  3d

2nd term 3 ad

a  3d  13 Solve the system:  a  d  3 a  2, d  5 a24  2  23  5   113

Find the required means. 31. Insert three arithmetic means between 10 and 20.

Solution a  10, a5  20

20  10  4d 10  4d 5 25 35  d  10, , 15, , 20 2 2 2 32. Insert five arithmetic means between 5 and 15.

Solution a  5, a7  15

15  5  6d 10  6d 5 20 25 35 40  d  5, , , 10, , , 15 3 3 3 2 3

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1860


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

33. Insert four arithmetic means between –7 and 23 .

Solution a  7, a6 

2 3

2  7  5d 3 23  5d 3 23 d 15 82 59 12 13 2 7,  , , , , 15 15 5 15 3 34. Insert three arithmetic means between –11 and –2.

Solution a  11, a5  2

2  11  4d 9  4d 9 35 13 17  d  11,  , , , 2 4 4 2 4 Find the sum of the first n terms of each arithmetic series. 35. 5 + 7 + 9 + ⋯ (to 15 terms)

Solution a  5, d  2

a15  a   n  1 d  5  14  2   33 S15 

n  a  a15  2

15  5  33  2

 285

36. –3 + (–4) + (–5) + ⋯ (to 10 terms)

Solution a  3, d  1

a10  a   n  1 d  3  9  1  12 S10 

n  a  a10  2

10 3   12  2

  75

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1861


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

20 3  37.   n  12  n 1  2 

Solution 27 3 a ,d  2 2 a20  a   n  1 d  S20 

n  a  a20  2

3 27  19    42 2 2

20  272  42  2

 555

10 2 1 38.   n   3 n 1  3

Solution

a  1, d 

2 3

2 a10  a   n  1 d  1  9    7 3 n  a  a10  10  1  7  S10    40 2 2 Solve each problem. 39. Find the sum of the first 30 terms of an arithmetic sequence with 25th term of 10 and a common difference of 21 .

Solution 1 , a  10 2 25  1 10  a  24   2 10  a  12  a  2 d

 1  25 a30  a   n  1 d  2  29    2 2 25 n  a  a30  30  2  2  1 S30    157 2 2 2

40. Find the sum of the first 100 terms of an arithmetic sequence with 15th term of 86 and first term of 2.

Solution a  2, a15  86 86  a  14d 84  14d  d  6

a100  a   n  1 d  2  99  6   596 S100 

n  a  a100  2

100  2  596  2

 29, 900

41. Find the sum of the first 200 natural numbers.

Solution

a  1, d  1, n  a200  200; S200 

n  a  a200  2

200  1  200 2

 20, 100

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1862


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

42. Find the sum of the first 1,000 natural numbers.

Solution

a  1, d  1, n  a1000  1000; S1000 

n  a  a1000  2

1000  1  1000 2

 500,500

Fix It In exercises 43 and 44, identify the step where the first error is made and fix it. 43. The first term of an arithmetic sequence is 10 and the 50th term is 2,705. Determine the common difference d and then write the first six terms of the sequence.

Solution Step 4 was incorrect. Step 4: d = 55 Step 5: 10, 66, 120, 175, 230, 285 44. Find the sum of the first 40 terms of the arithmetic series given by 4 + 10 + 16 + 22 + 28 +∙∙∙. To do so, first find the common difference d, next determine the 40th term a40, and then substitute into the formula Sn 

n  a  an  2

.

Solution Step 5 was incorrect. Step 5: S40  4480

Applications 45. Interior angles The sums of the angles of several polygons are given in the table. Assuming that the pattern continues, complete the table.

Figure

Number of Sides

Sum of Angles

Triangle

3

180°

Quadrilateral

4

360°

Pentagon

5

540°

Hexagon

6

720°

Octagon

8

1080°

Dodecagon

12

1800°

Solution a  180, d  180, n  8  2  6

a  180, d  180, n  12  2  10

a6  a   n  1 d  180  5  180  a10  a   n  1 d  180  9  180   1080

 1800

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1863


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

46. Borrowing money To pay for college, a student borrows $5000 interest-free from his father. If he pays his father back at the rate of $200 per month, how much will he still owe after 12 months?

Solution

a  5000, d  200, n  13 Note: n = 13 occurs at the beginning of the 13th month, right after the 12th payment has been made 13th term  a   n  1 d

 5000  12  200   $2600

47. Borrowing money If Ellie borrows $5500 interest-free from her mother to buy a new car and agrees to pay her mother back at the rate of $105 per month, how much will she still owe after 4 years?

Solution

a  5500, d  105, n  49 Note: n = 49 occurs at the beginning of the 49th month, right after the 48th payment has been made 49th term  a   n  1 d

 5500  48  105   $460

48. Jogging

One day, two students jogged 21 mile. Because it was fun, they decided to

increase the jogging distance each day by a certain amount. If they jogged 6 43 miles on the 51st day, how much was the distance increased each day?

Solution 1 3 27 a  , a51  6  2 4 4 a51  a   n  1 d 27 1   50d 4 2 25 1  50d  d  4 8 The distance increased 81 mile per day.

49. Sales The year it incorporated, a company had sales of $237,500. Its sales were expected to increase by $150,000 annually for the next several years. If the forecast was correct, what will sales be in 10 years?

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1864


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution a  237, 500; d  150, 000; a10  a   n  1 d

 237, 500  9  150, 000   $1, 587, 500

50. Falling objects

Find how many feet a brick will travel during the 10th second of its fall.

Solution

a10  a   n  1 d

 16  9  32   304 feet

51. Falling objects If a rock is dropped from the Golden Gate Bridge, how far will it fall in the third second? Us the formula s = t2.

Solution

a3  a   n  1 d

 16  2  32   80 feet

52. Designing patios Each row of bricks in the following triangular patio is to have one more brick than the previous row, ending with the longest row of 150 bricks. How many bricks will be needed?

Solution a  1, d  1, n  150, a150  150

S150 

n  a  a150  2

150  1  150 

2  11, 325 bricks

53. Pile of logs Several logs are stored in a pile with 20 logs on the bottom layer, 19 on the second layer, 18 on the third layer, and so on. If the top layer has one log, how many logs are in the pile?

Solution

a  1, d  1, n  20, a20  20 S

n  a  a20  2

20  1  20  2

 210 logs

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1865


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

54. Theater seating The first row in a movie theater contains 24 seats. As you move toward the back, each row has 1 additional seat. If there are 30 rows, what is the capacity of the theater?

Solution 24  25  26    a  24, d  1, n  30

a30  a   n  1 d  24  29  1  53; S30 

n  a  a30  2

30  24  53 2

 1155 seats

Discovery and Writing 55. Define arithmetic sequence and provide two examples.

Solution Answers may vary. 56. Explain what the common distance d is in an arithmetic sequence.

Solution Answers may vary. 57. Describe how to determine a specific term of an arithmetic sequence.

Solution Answers may vary. 58. What formula is used to determine the sum of the first n terms of an arithmetic series? Explain how it is used.

Solution Answers may vary. 59. In an arithmetic sequence, can a and d be negative, but an positive?

Solution Answers may vary. 60. Can an arithmetic sequence be an alternating sequence? Explain.

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 61. 1, 4, 8, 13, 19, 26, ⋯ is not an arithmetic sequence.

Solution True. 62. The common difference for the arithmetic sequence 14, 9, 4, –1, –6, ⋯ is d = 5.

Solution False. d = –5.

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1866


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

63. An arithmetic sequence can have a first term of 4, a 25th term of 106, and a common difference of 4 41 .

Solution True. 64. Between 5 and 10 31 are three arithmetic means. One of them is 9 and the other two are 6 31 and 7 23 .

Solution True. 65. If we know the first term, last term, and number of terms of an arithmetic series, then we can find the sum of the terms of the series.

Solution True. 66. The formula Sn 

n  a  an  2

Solution False. The formula is Sn 

gives the sum of the first n terms of an arithmetic sequence.

n  a  an  2

.

67. The sum of the first 200 terms of the arithmetic sequence 1, 3, 5, 7, 9, ⋯ is 40,000.

Solution True. 68. The sum of the first 200 terms of the arithmetic sequence 2, 4, 6, 8, 20, ⋯ is 40,000.

Solution False. The sum is Sn 

n  a  an  2

200  2  400 2

 40, 200.

69. Each row of a formation of the members of a college marching band has one more person in it than the previous row. If 5 people are in the front row and 24 are in the 20th (and last) row, then there are 300 band members.

Solution False. There are Sn 

n  a  an  2

20  5  24  2

 290.

70. The discrete points on the graph of an arithmetic sequence are collinear.

Solution True.

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1867


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

EXERCISES 8.4 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Consider the infinite sequence of terms a, ar, ar2, ar3, ar4, … . If a = 3 and r = 5, write the first four terms of the sequence.

Solution a3

ar  3  5   15 ar 2  3  5   75 2

ar 3  3  5   375 3

2. Consider the infinite sequence of terms a, ar, ar2, ar3, ar4, … . If a = 5 and r  21 , write the first four terms of the sequence.

Solution a5  1 5 ar  5    2 2 2

 1 5 ar 2  5    2 4   3

 1 5 ar  5    8 2 3

3. Consider the infinite series a + ar + ar2 + ar3 + ar4 + … . If a = 10 and r = 2, write the series.

Solution 10  20  40  80   4. Consider the infinite series a + ar + ar2 + ar3 + ar4 + … . If a = –2 and r = 31 , write the series.

Solution 2 2 2 2     3 9 27 5. Simplify the complex fraction:

3 10

1  101

.

Solution  103   10  3 1     1   1  10   10  10  1 3

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1868


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

6  6  21 

5

6. Simplify the complex fraction:

1  21

.

Solution 6  6  21 

5

1  21

6   163   16  96  3 93    1 8 8  16  2

Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. A geometric sequence is a sequence of the form a, ar, ar2, ar3, . . . . The nth term is a (______).

Solution r n 1

8. In a geometric sequence, a is the _________ term, r is the common __________, and n is the _________ of terms.

Solution first, ratio, number 9. The last term of a geometric sequence is given by the formula an = ________.

Solution ar n  1

10. A geometric __________ is the sum of the terms of a geometric sequence.

Solution series 11. A geometric series with infinitely many terms is called an __________ geometric series.

Solution infinite 12. The formula for the sum of the first n terms of a geometric series is given by ____________.

Solution

Sn 

a  ar n ,r 1 1 r

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1869


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

13. __________ are numbers inserted between a first and a last term to form a geometric sequence.

Solution Geometric means 14. If r  1, the formula __________ gives the sum of the terms of an infinite geometric series.

Solution

S 

a 1 r

Practice Write the first four terms of each geometric sequence with the given properties. 15. a  10; r  2

Solution 10, 20, 40, 80 16. a  3; r  2

Solution –3, –6, –12, –24 17. a  2 and r  3

Solution –2, –6, –18, –54 18. a  64; r 

1 2

Solution 64, 32, 16, 8 19. a  3; r  2

Solution

3, 3 2, 6, 6 2 20. a  2; r  3

Solution

2, 2 3, 6, 6 3

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1870


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

21. a  3; r  7

Solution 3, –21, 147, –1029 22. a  2; r  

1 4

Solution 1 1 1 2,  , ,  2 8 32 23. a  5; r  

2 9

Solution 10 20 40 5, , , 9 81 729 1 24. a   ; r  6 3

Solution 1  , 2,  12, 72 3 25. a = 2; 4th term is 54

Solution

a4  ar 4  1 54  2r 3 27  r 3 3  r  2, 6, 18, 54 26. 3rd term is 4; r 

1 2

Solution

a3  ar 3 1 2

 1 4  a  2 1 4 a 4 16  a  16, 8, 4, 2

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1871


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Find the requested term of each geometric sequence. 27. Find the sixth term of the geometric sequence whose first three terms are 41 , 1, and 4.

Solution

a

 1 1 , r  4; a6  ar 6 1    45  256 4 4

28. Find the eighth term of the geometric sequence whose second and fourth terms are 0.2 and 5.

Solution a2  ar 1 0.2  ar 1 a4  ar 3 5  ar 3

0.2 0.2 1   5 25 r ar  r 3  5 0.2 0.2 1   r  5: a  0.2r 2  5 5 r 25 7 a8  ar 7 r 2  25 a8  ar 7 7 1 1 r  5   5  5  25 25  3125  3125 ar 3  5

r  5: a 

29. Find the fifth term of a geometric sequence whose second term is 6 and whose third term is –18.

Solution a2  ar 1 6  ar 1 a3  ar 2 18  ar 2

6 6  2 ar 2  18 r  3: a   3 r ar  r  18 4 a  ar 4 :  2  3   162 6r  18 5 r  3

30. Find the sixth term of a geometric sequence whose second term is 3 and whose fourth term is 31 .

Solution

a2  ar 1 3  ar 1 a4  ar 3 1  ar 3 3

1 3 1 2 ar  r  3 1 2 3r  3

1 3 3 9 : a  r 1/3 3 1 3 3 r : a   9 3 r 1/3 a6  ar 5 a6  ar 5

ar 3 

r

r

1 3

 1  9  3 1  27

5

 1  9     3 1  27

5

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1872


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solve each problem. 31. Insert three positive geometric means between 10 and 20.

Solution a5  ar 4 20  10r 4 2  r4 4

2  r problem specifies positive 

10, 10 4 2, 10 2, 10 4 8 , 20

32. Insert five geometric means between –5 and 5, if possible.

Solution

a7  ar 6 5  5r 6 1  r 6 r is not a real number  not possible 33. Insert four geometric means between 2 and 2048.

Solution

a6  ar 5 2048  2r 5 1024  r 5 4r 2, 8, 32, 128, 512 , 2048 34. Insert three geometric means between 162 and 2. (There are two possibilities.)

Solution

a5  ar 4 2  162r 4 1  r4 81 1  r 3 162, 54, 18,  6 , 2 or 162, 54, 18, 6 , 2 Find the sum of the indicated terms of each geometric series. 35. 4 + 8 + 16 + ⋯ (to 5 terms)

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1873


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution a  4, r  2, n  5

a  ar n 4  4  2  S5  1 r 12 124   124 1 5

36. 9 + 27 + 81 + ⋯ (to 6 terms)

Solution a  9, r  3, n  6

a  ar n 9  9  3  S6  1r 13 6552   3276 2 6

37. 2 + (–6) + 18 + ⋯ (to 10 terms)

Solution a  2, r  3, n  10 a  ar n 2  2  3  S10   1r 1   3  

38.

10

118, 096  29, 524 4

1 1 1      to 12 terms  8 4 2

Solution 1 a  , r  2, n  12 8 12 1 1  2 a  ar n 8 8   S12   1 r 12 4095  8  4095  8 1 6 3 39.  3   n 1  2 

n 1

Solution a  3, r 

3 ,n6 2

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1874


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

a  ar n 3  3  2  S6   1 r 1  32

6

3

6  1 40.  12    n 1  2

 1995 64 

1 2

1995 32

n 1

Solution

1 a  12, r   , n  6 2

a  ar n 12  12   2  S6   1 r 1    21  1

189

 163  2

6

63 8

Find the sum of each infinite geometric series. 41. 6  4 

8  4

Solution a  6, r  S 

2 3

a 6 6   1  18 2 1 r 1 3 3

42. 8 + 4 + 2 + 1 + ⋯

Solution a  6, r  S 

1 2

a 8 8   1  16 1 1 r 1 2 2

 1 43.  12    n 1  2 

n 1

Solution a  12, r   S 

1 2 12

a 12   3 8 1  r 1    21  2

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1875


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 1 44.    n 1  3  

n 1

Solution a  1, r  S 

1 3

a 1 1 3   2  1  r 1  31 2 3

Change each repeating decimal to a common fraction. 45. 0.5

Solution

a

5 1 1  ,r  10 2 10

S 

1 1 a 5  2 1  92  1  r 1  10 9 10

46. 0.6

Solution 6 3 1 a  ,r  10 5 10 3 3 a 2 S   5 1  95  1  r 1  10 3 10 47. 0.25

Solution 25 1 1 a  ,r  100 4 100 1 1 a 25 4 4 S    99  1 1  r 1  100 99 100 48. 0.37

Solution 37 1 a ,r  100 100 37 37 a 37 S   100 1  100  99 1  r 1  100 99 100

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1876


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Fix It In exercises 49 and 50, identify the step where the first error is made and fix it. 49. Determine the sixth term of a geometric sequence whose first three terms are 12, 4, and 43 . First, identify a, r, and n. Then, substitute into the formula nth term  ar n  1 .

Solution Step 4 was incorrect.

 1 Step 4: sixth term = 12   3 Step 5: sixth term =

5

12 4  243 81

50. Find the sum of the first six terms of the geometric series 10  5  52   . First, identify a, r, and n. Then, substitute into the formula Sn 

a  ar n 1 r

 r  1 .

Solution Step 3 was incorrect.

Step 3: S6  Step 4: S6  Step 5: S6 

10  10  641  10 

1 2 5 32

1 2

315 16

Applications Use a calculator to help solve each problem. 51. Staffing a department The number of students studying algebra at State College is 623. The department chair expects enrollment to increase 10% each year. How many professors will be needed in eight years to teach algebra if one professor can handle 60 students?

Solution a  623, r  1.10

a9  ar 8  623  1.1

8

 1335 students 1335 # professors   22.25 60 23 professors will be needed. 52. Bouncing balls On each bounce, the rubber ball in the illustration rebounds to a height one-half of that from which it fell. Find the total vertical distance the ball travels.

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1877


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution Down a  10, r  a 1 r 10  1  21

Up 1 2

S 

a  5, r  a 1r 5  1  21

1 2

S 

 20  10 Total vertical distance: 30 m 53. Bungee jumping A bungee jumper is attached to a cord that stretches to a length of 100 feet. If he rebounds to 60% of the height jumped, how far will he fall on his fifth descent? How far will he travel when he comes to rest?

Solution

a  100, r  0.6, a5  100  0.6  Down a  100, r  0.6

4

 12.96 ft Up a  60, r  0.6

a a S  1 r 1 r 100 60   1  0.6 1  0.6  250  150 Total vertical distance: 400 ft

S 

54. Bungee jumping A bungee jumper is attached to a cord that stretches to a length of 100 feet. If she rebounds to 70% of the height jumped, how far will she travel upward on the fifth rebound? How far will she have traveled when she comes to rest?

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1878


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution a  70, r  0.7, a5  70  0.7 

4

 16.807 ft Down Up a  100, r  0.7 a  70, r  0.7 a a S  S  1 r 1 r 100 70   1  0.7 1  0.7 1 1  333  233 3 3 2 Total vertical distance: 566 ft 3 55. Bouncing balls A SuperBall rebounds to approximately 95% of the height from which it is dropped. If the ball is dropped from a height of 10 meters, how high will it rebound after the 13th bounce? Round to two decimal places.

Solution a  10, r  0.95

a14  ar 13  10  0.95

13

 5.13 meters 56. Genealogy The following family tree spans three generations and lists seven people. How many names would be listed in a family tree that spans ten generations?

Solution a  1, r  2, n  10 S10 

a  ar n 1  1  210   1023 names 1 r 12

57. Investing money If a married couple invests $1000 in a 1-year certificate of deposit at 6 43 % annual interest, compounded daily, how much interest will be earned during the year? Round to two decimal places.

Solution

0.0675 , n  365 365 365  0.0675  365 ar  1000  1   365    $1069.82 The interent will be $69.82 a  1000, r  1 

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1879


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

58. Biology If a single cell divides into two cells every 30 minutes, how many cells will there be at the end of 10 hours?

Solution a  1, r  2, n  10  2  20

ar 20  1  2

20

 1,048,576

59. Depreciation A lawn tractor, costing C dollars when new, depreciates 20% of the previous year’s value each year. How much is the lawn tractor worth after five years? Round to two decimal places.

Solution a  c, r  0.80, n  5  ar 5  c  0.80   0.32768c, or about 0.33c 5

60. Financial planning

Enrique can invest $1000 at 7 21 %, compounded annually, or at

7 41 %, compounded daily. If he invests the money for a year, which is the better investment?

Solution 7 21 % compounded annually a  1000, r  1 

0.075 ,n1 1

ar 1  1000  1.075 

1

 $1075

7 41 % compounded daily .0725 , n  365 365 365  0.0725  ar 1  1000  1   365   a  1000, r  1 

 $1075.19  Better investments

61. Population study If the population of the Earth were to double every 30 years, approximately how many people would there be in the year 3020? (Consider the population in 2000 to be five billion and use 2000 as the base year.)

Solution a  5  109 , r  2

n   3020  2000  /30  34

ar 34  5  109  2 

34

 8.6  1019

62. Investing money If Elsa deposits $1300 in a bank at 7% interest, compounded annually, how much will be in the bank 17 years later? Round to two decimal places. (Assume that there are no other transactions on the account.)

Solution a  1300, r  1 

0.07 , n  17 1

ar 17  1300  1.07 

17

 $4106.46

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1880


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

63. Real estate appreciation If a house purchased for $50,000 in 1998 appreciates in value by 6% each year, how much will the house be worth in the year 2020? Round to two decimal places.

Solution a  50,000; r  1.06, n  22

ar 22  50,000  1.06 

22

 $180, 176.87 64. Compound interest Find the value of $1000 left on deposit for 10 years at an annual rate of 7%, compounded annually. Round to two decimal places.

Solution a  1000, r  1 

0.07 , n  10 1

ar 10  1000  1.07 

10

 $1967.15

65. Compound interest Find the value of $1000 left on deposit for 10 years at an annual rate of 7%, compounded quarterly. Round to two decimal places.

Solution 0.07 , n  40 4 40  0.07  ar 40  1000  1   4    $2001.60 a  1000, r  1 

66. Compound interest Find the value of $1000 left on deposit for 10 years at an annual rate of 7%, compounded monthly. Round to two decimal places.

Solution 0.07 , n  120 12 120  0.07  120 ar  1000  1   12    $2009.66 a  1000, r  1 

67. Compound interest Find the value of $1000 left on deposit for 10 years at an annual rate of 7%, compounded daily. Round to two decimal places.

Solution 0.07 , n  3650 365 3650  0.07  ar 3650  1000  1   365    $2013.62 a  1000, r  1 

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1881


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

68. Compound interest Find the value of $1000 left on deposit for 10 years at an annual rate of 7%, compounded hourly. Round to two decimal places.

Solution 0.07 , n  87600 8760 87600  0.07  ar 87600  1000  1   8760    $2013.75 a  1000, r  1 

69. Saving for retirement When Grayson was 20 years old, he opened an individual retirement account by investing $2000 at 11% interest, compounded quarterly. How much will his investment be worth when he is 65 years old? Round to two decimal places.

Solution 0.11 , n  180 4 180  0.1  ar 180  2000  1   4    $264, 094.58 a  2000, r  1 

70. Biology One bacterium divides into two bacteria every five minutes. If two bacteria multiply enough to completely fill a petri dish in two hours, how long will it take one bacterium to fill the dish?

Solution Since the bacteria double in 5 minutes, it will take 5 minutes for one bacterium to become two bacteria. It will then take 2 hours for the two bacteria to fill the dish, for a total of 2 hours and 5 minutes. 71. Pest control To reduce the population of a destructive moth, biologists release 1000 sterilized male moths each day into the environment. If 80% of these moths alive one day survive until the next, then after a long time the population of sterile males is the sum of the infinite geometric series 1000  1000  0.8   1000  0.8   1000  0.8    2

3

Find the long-term population.

Solution a  1000, r  0.8 1000 a   5000 S  1  r 1  0.8 72. Pest control If mild weather increases the day-to-day survival rate of the sterile male moths in Exercise 71 to 90%, find the long-term population.

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1882


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution a  1000, r  0.9 1000 a   10, 000 S  1  r 1  0.9 73. Mathematical myth A legend tells of a king who offered to grant the inventor of the game of chess any request. The inventor said, “Simply place one grain of wheat on the first square of a chessboard, two grains on the second, four on the third, and so on, until the board is full. Then give me the wheat.” The king agreed. How many grains did the king need to fill the chessboard?

Solution a  1, r  2, n  64 a  ar n 1  1  2   S 1 r 12  1.8447  1019 grains 64

74. Mathematical myth Estimate the size of the wheat pile in Exercise 73. (Hint: There are about one-half million grains of wheat in a bushel.)

Solution 1.8447  10 19  3.689  10 13 bushels 500, 000 Discovery and Writing 75. What is a geometric sequence? Give an example to support your description.

Solution Answers may vary. 76. What is an infinite geometric series? Give an example to support your description.

Solution Answers may vary. 77. What is the common ratio r in a geometric sequence or series?

Solution Answers may vary. 78. How do you determine the sum of the first n terms of a geometric series?

Solution Answers may vary. 79. How do you know whether or not an infinite geometric series has a sum?

Solution Answers may vary. 80. If an infinite geometric series has a sum, how do you determine the sum?

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1883


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution Answers may vary. 81. Does 0.999999 = 1? Explain.

Solution no 82. Does 0.999 . . . = 1? Explain.

Solution yes Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 83. The sequence 5, 10, 20, 40, 80, ⋯ is a geometric sequence.

Solution True. 84. The sequence 5, 10, 15, 20, 25, ⋯ is a geometric sequence.

Solution False. There is not a common ratio. 85. The nth term of a geometric sequence is arn.

Solution False. The nth term is arn – 1. 86. The common ratio of a geometric series is always positive.

Solution False. The common ratio can be any real number except 1. 87. The sum of an infinite geometric series can always be found.

Solution False. The common ratio must be between –1 and 1. 88. The nth term an of the geometric sequence

1,

 1 1 1 1 , , ,  is an   1   5 25 125 5

n 1

1 5n 1

.

Solution

 1 . False. a  n

n

5n 1

89. The sum of the first n terms of the geometric series 2 + 4 + 8 + 16 + ⋯ is 2(2n – 1).

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1884


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution True. 

90.  2  5 

n 1



n 1

1 2

Solution False. The sum does not exist, because the common ratio of 5 is greater than 1.

EXERCISES 8.5 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Given 1  2  3    n 

n  n  1 2

. Show that the formula is true for n = 10.

Solution

1  2  3  4  5  6  7  8  9  10 

10  10  1

55  55

2

2. Given 1  3  5    2n  1  n2 . Show that the formula is true for n = 9.

Solution 1  2  3  4  5   17  92 81  81

3. Given 2  4  6    2n  n n  1 . Show that the formula is true for n = 8.

Solution

2  4  6  8   16  8  8  1 72  72

4. Given 1  3  32    3n 1 

3n  1 . Show that the formula is true for n = 6. 2

Solution 36  1 2 364  364

1  3  32  33  34  35 

5. Given

1 1 1 1      n  1. Show that the formula is true for n = 4. 2 4 8 2

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1885


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution 1 1 1 1    1 2 4 8 16 15 1 16 6. Add and simplify:

k 1  k  1  k  1 k  2 

Solution

 k  1  k  1 k  1 k  k  2  1 k k 2  2k  1 1      k  1  k  1 k  2  k  1 k  2  k  1 k  2  k  1  k  2 k  2 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. Any proof by induction requires __________ parts.

Solution two 8. Part 1 is to show that the statement is true for __________.

Solution n1 9. Part 2 is to show that the statement is true for __________ whenever it is true for n = k.

Solution nk1 10. When we assume that a formula is true for n = k, we call the assumption the induction __________.

Solution hypothesis Practice Verify each formula for n = 1, 2, 3, and 4. 11. 5  10  15    5n 

5n  n  1 2

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1886


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution n1

n2

? 5  1 1  1 5  1  2 55

? 5  2  2  1 5  5  2  2 10 3   ? 15  2 15  15 n4

n3

? 5  3  3  1 5  10  5  3  2 15 4   ? 30  2 30  30 12. 12  22  32    n2 

? 5  4  4  1 5  10  15  5  4   2 20 5   ? 50  2 50  50

n  n  1 2n  1 6

Solution

n1

n2

? 1  1  1 2  1  1 12  6 ? 1  2 3  1 6 11

? 2  2  1 2  2  1 12  22  6 ? 2  3 5  5 6 55

n3

n4

? 3  3  1 2  3  1 12  22  32  6 3 4 7    ? 14  6 14  14

13. 7  10  13    3n  4 

? 4  4  1 2  4   1 12  22  32  42  6 4 5 9    ? 30  6 30  30

n  3n  11 2

Solution

n1

n2

 7  3  2  4 ? 2  3 2  11

? 1 3  1  11 3  1  4  2 ? 1  14  7 2 77

2 2 17   ? 17  2 17  17

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1887


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

n3

n4

 7  10  13  3 4  4 ? 4 3  4  11

? 3 3  3   11 7  10  3  3  4  2 ? 3  20  30  2 30  30

2

? 4  23 46  2 46  46 n  n  1 2n  7  6

14. 1  3   2  4   3  5     n  n  2  

Solution n1

n2

? 1 1  1  2   1  1 2  1  7 6 ? 1 3   2 9  6 33 n3

?2 3  2  2  2    2  1 2  2   7 6 ?2 11   3  11 6 11  11 n4

?3 3  8  3  3  2   3  1 2  3  7 6 ?3 26   4  13  6 26  26

 3  8  15  4  4  2 ? 46  4  1 2  4  7  ?4 50   5  15  6 50  50

Prove each formula by mathematical induction, if possible.

15. 2  4  6    2n  n n  1

Solution Check n  1:

? 2  1  1  1

True for n  1

22

Assume for n  k Show for n  k  1

2  4  6    2k  k  k  1

2  4  6    2k  2  k  1  k  k  1  2  k  1 2  4  6    2  k  1  k 2  k  2k  2 2  4  6    2  k  1  k 2  3k  2

2  4  6    2  k  1   k  1 k  2 Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

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1888


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

16. 1  3  5    2n  1  n2

Solution ? 2  1  1  12

Check n  1:

True for n  1

1 1

1  3  5     2k  1  k 2

Assume for n  k

1  3  5     2k  1  2  k  1  1  k 2  2  k  1  1

Show for n  k  1

1  3  5    2  k  1  1  k 2  2k  2  1 1  3  5    2  k  1  1  k 2  2k  1 1  3  5    2  k  1  1   k  1

2

Since this is what results when n  k  1 in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

17. 3  7  11    4n  1  n 2n  1

Solution

Check n  1:

? 4  1  1  1 2  1  1

True for n  1

33 Assume for n  k and show for n  k  1:

3  7  11     4k  1  k  2k  1

3  7  11     4k  1  4  k  1  1  k  2k  1  4  k  1  1 3  7  11    4  k  1  1  2k 2  k  4k  4  1 3  7  11    4  k  1  1  2k 2  5k  3

3  7  11    4  k  1  1   k  1 2k  3 

3  7  11    4  k  1  1   k  1 2  k  1  1

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

18. 4  8  12    4n  2n n  1

Solution Check n  1:

? 4  1  2  1 1  1

True for n  1

44

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1889


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Assume for n  k and show for n  k  1: 4  8  12    4k  2k  k  1

4  8  12    4k  4  k  1  2k  k  1  4  k  1 4  8  12    4  k  1  2k 2  2k  4k  4 4  8  12    4  k  1  2k 2  6k  4

4  8  12    4  k  1   2k  2 k  2  4  8  12    4  k  1  2  k  1 k  2

Since this is what results when n  k  1 in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

19. 10  6  2    14  4n  12n  2n2

Solution 2 ? 14  4  1  12  1  2  1

Check n  1:

True for n  1

10  10

Assume for n  k and show for n  k  1:

10  6  2     14  4k   12k  2k 2

10  6  2     14  4k   14  4  k  1  12k  2k 2  14  4  k  1

  10  6  2     14  4  k  1   12k  12  2k  4k  2 10  6  2     14  4  k  1   12  k  1  2  k  2k  1 10  6  2     14  4  k  1   12  k  1  2  k  1 10  6  2    14  4  k  1  12k  2k 2  14  4k  4 2

2

2

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

20. 8  6  4    10  2n  9n  n2

Solution Check n  1:

2 ? 10  2  1  9  1   1

True for n  1

88

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1890


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Assume for n  k and show for n  k  1:

8  6  4     10  2k   9k  2k 2

8  6  4     10  2k   10  2  k  1  9k  k 2  10  2  k  1

  8  6  4     10  2  k  1   9k  9  k  2k  1 8  6  4     10  2  k  1   9  k  1   k  2k  1 8  6  4     10  2  k  1   9  k  1   k  1 8  6  4    10  2  k  1  9k  k 2  10  2k  2 2

2

2

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

21. 2  5  8    3n  1 

Solution

Check n  1:

n  3n  1 2

? 1 3  1  1 3  1  1  2 22

True for n  1

Assume for n  k and show for n  k  1: 2  5  8     3k  1  2  5  8     3k  1  3  k  1  1 

2  5  8    3  k  1  1 

k  3k  1 2 k  3k  1 2

k  3k  1

 3  k  1  1

2  3  k  1  1

2 2 3k 2  k  6k  6  2 2  5  8    3  k  1  1  2 3k 2  7k  4 2  5  8    3  k  1  1  2 1 k    3k  4  2  5  8    3  k  1  1  2  k  1 3  k  1  1 2  5  8    3  k  1  1  2

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k . 22. 3  6  9    3n 

3n  n  1 2

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1891


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

Check n  1:

? 3  1 1  1 3  1  2 33

True for n  1

Assume for n  k and show for n  k  1: 3  6  9    3k  3  6  9    3k  3  k  1  3  6  9    3  k  1  3  6  9    3  k  1  3  6  9    3  k  1 

3k  k  1

2 3k  k  1 2 3k  k  1

 3  k  1 2  3  k  1

2 2 3  k  1 k  3  k  1  2 2 3  k  1 k  2  2

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k . 23. 12  22  32    n2 

Solution Check n  1:

n  n  1 2n  1 6

? 1  1  1 2  1  1 12  6 11

True for n  1

Assume for n  k and show for n  k  1: 12  22  32    k 2 

k  k  1 2k  1 6 k  k  1 2k  1

12  22  32    k 2   k  1  2

12  22  32     k  1  2

12  22  32     k  1  2

12  22  32     k  1  2

12  22  32     k  1  2

12  22  32     k  1  2

6

  k  1

2k 2  k  k  1

6 6 2k 2  k  6  k  1  k  1

6 2 2k  7k  6  k  1

2

6  k  1 k  1

6 2 3 k k     2  k  1 6  k  1 k  2 2k  3 6

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1892


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

24. 1  2  3    n  1  n  n  1    3  2  1  n2

Solution

Check n  1:

? 1  12

True for n  1

Assume for n  k and show for n  k  1:

1  2  3     k  1  k   k  1    1  k 2

1  2  3    k   k  1  k 

 k  1    1  k  k  k  1 1  2  3    k   k  1  k    1  k  2k  1 1  2  3    k   k  1  k    1   k  1 2

2

2

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k . 25.

5 5 1 11 4 1  2      n    n n   3 3 3 2 3 6 Solution Check n  1:

5 4 ? 5 1 1   1   1    3 3 2 6 1 1  3 3

True for n  1

Assume for n  k and show for n  k  1: 5 5 1 11 4 1  2    k    k k   3 3 3 2 3 6 5 5 5 5 1 11 4 4 1 4  2      k      k  1    k  k      k  1   3 3 3 3 2 3 3 4 6 3 5 1 11 4 5 1 5  4  2       k  1    k 2  k  k   3 3 3 6 2 3 3 3 4 5 7 1 1 11 4 5  2       k  1    k 2  k  6 6 3 3 3 4 3   5 5 1 11 4 1  2       k  1     k  1  k   3 3 3 3 4 6 5 5 1 11 4 5 1  2       k  1     k  1  k    3 3 4 3 6 6 2    5 5 1 11 4 1  2       k  1     k  1   k  1   3 3 3 2 4 6

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1893


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

26.

1 1 1 1 n     1 2 2  3 3  4 n 1 n  n  1

Solution Check n  1:

1 ? 1  1 2 1  1 1 1  2 2

True for n  1

Assume for n  k and show for n  k  1: k 1 1 1 1     1 2 2  3 3  4 k  k  1 k  1 k 1 1 1 1 1 1       1 2 2  3 3  4 k  k  1  k  1 k  2 k  1  k  1 k  2 k  k  2  1 1 1 1 1     1 2 2  3 3  4  k  1 k  2  k  1 k  2

k 2  2k  1 1 1 1 1     1 2 2  3 3  4  k  1 k  2  k  1 k  2

 k  1 1 1 1 1     1 2 2  3 3  4  k  1 k  2  k  1 k  2 2

k1 1 1 1 1     k 1 2 2  3 3  4   k k 1 2     2 Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

27.

n

n

1

1

 1  1 1 1 1       1   2 4 8 2 2 Solution

Check n  1:

 1 ?  1    1   2 2 1 1  2 2

True for n  1

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1894


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Assume for n  k and show for n  k  1: k

 1  1 1 1 1       1   2 4 8 2 2

k

k 1

 1  1  2

 1 1 1 1      2 4 8 2

k 1

 1  1   1  1  2      2 2    2

 1 1 1 1      2 4 8 2

k 1

 1  1  2  2

 1 1 1 1      2 4 8 2

k 1

 1  1  2

 1 1 1 1      2 4 8 2

k

 1   2

k

 1   2

k 1

k

k 1

 1   2

k 1

k 1

k 1

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

28.

n 1

2  1  3

1 1

1

1 2 4 1 2      3 9 27 33

n

Solution

Check n  1:

1 2   33

2 ?  1  3 1 1  3 3

True for n  1

Assume for n  k and show for n  k  1: 1 2 4 1 2      3 9 27 33 1 2 4 1 2      3 9 27 33

k 1

k 1

2  1  3

k

k

2 1 2     1  33 3

k

k

1 2    33

k

k

1 2 4 1 2 3 22 3 2 1 2        1         3 9 27 33 2 33 2 333 k

1 2  2 1 2 4       1   3 9 27 33 23 k

2 1 2 4 1 2       1   3 9 27 33 3

k 1

1 2    23

k

k 1

k 1

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

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1895


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

29. 20  2 1  2 2  2 3    2 n  1  2 n  1

Solution Check n  1:

? 2 1 1  2 1  1 1 1

True for n  1

Assume for n  k and show for n  k  1: 20  21  22    2k  1  2k  1 20  21  22    2k  1  2k  2k  1  2k 20  21  22    2k  2  2k  1 20  21  22    2k  2k  1  1

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .  n  n  1   30. 1  2  3    n   2   3

3

3

2

3

Solution

Check n  1:

?  1  1  1   1   2  11

2

True for n  1

3

Assume for n  k and show for n  k  1:  k  k  1   1  2  3  k   2   3

3

1  2  3  k 3

3

3

3

3

  k  1

3

 k  k  1     2  

1  2  3     k  1  3

3

3

2

3

3

2

  k  1

k 2  k  1  4  k  1 2

3

3

4   k  1 k 2  4  k  1 2 12  22  32     k  1  4 2

 k  1  k  2 1  2  3     k  1  4 2

2

2

2

2

2

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k . 31. Prove by induction that x – y is a factor of x n  y n . (Hint: Consider subtracting and adding xy k to the binomial x k  1  y k  1 . )

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1896


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution Check n  1:

x  y is a factor of x 1  y 1 .

True for n  1

Assume for n  k and show for n  k  1:

Thus, we assume that x k  y k   x  y  SOMETHING x k  1  y k  1  x k  1  xy k  xy k  y k  1

 x xk  y k  y k  x  y 

 x  x  y  SOMETHING  y k  x  y    x  y   x  SOMETHING  y k 

We have shown that x – y is a factor of x k  1  y k  1 if it is a factor of x k  y k . 32. Prove by induction that n < 2n.

Solution Check n  1:

1  21

True for n  1

Assume for n  k and show for n  k  1: Certainly, 2k  1  2k  2  2k  2k . k  2k k  1  2k  1  2k  2k  2  2k  2k  1 k  1  2k  1

We have shown that k  1  2 k  1 if k  2k . 33. There are 180° in the sum of the angles of any triangle. Prove by induction that (n – 2)180° is the sum of the angles of any simple polygon when n is its number of sides. (Hint: If a polygon has k + 1 sides, it has k – 2 sides plus three more sides.)

Solution

The formula is true for n  3, since a triangle has 180  3  2  180. Next, assume

that a polygon with k sides has an angle sum of k  2  180. Take a polygon with k  1 sides. Consider two adjacent sides with a common endpoint. Connect the endpoints which are NOT common to both sides. The figure is now a polygon with k sides with a triangle adjacent to it.

Sum of angles of

 k  1 -sided polygon

Sum of angles of Sum of angles of  k-sided polygon triangle

  k  2   180  180   k  1  180   k  1  2  180 Thus, the formula works for n = k + 1 if it works for n = k. 34. Consider the equation 1 + 3 + 5 + ⋯ + 2n – 1 = 3n – 2 a. Is the equation true for n = 1?

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1897


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

b. Is the equation true for n = 2? c. Is the equation true for all natural numbers n?

Solution The formula works for n  1 and n  2, but it does not work for n  3. Therefore, it does not work for all natural numbers.

n  n  1  1 were true for n = k, show that it would be true for 2 n = k + 1. Is it true for n = 1?

35. If 1  2  3    n 

Solution Assume for n  k and show for n  k  1: k 1  2  3    k   k  1  1 2 k 1  2  3    k   k  1   k  1  1   k  1 2 1 1  2  3     k  1  k  k  1   k  1  1 2 1  1  2  3     k  1   k  1  k  1  2   1 1  2  3     k  1   k  1 k  2  2  k  1 k  2 1  2  3     k  1    2 The formula works for n = k + 1 if it works for n = k. However, the formula does not work for n = 1. Thus, the formula does not work for all natural numbers. 36. Prove by induction that n + 1 = 1 + n for each natural number n.

Solution

Check n  1:

1  1  1  1 True for n  1

Assume for n  k and show for n  k  1: k  1  1 k k  1 1  1 k  1

k  1 1  1 1 k

 k  1  1  1   1  k  We have shown that the formula works for n  k  1 if it works for n  k. 37. If n is any natural number, prove by induction that 7n – 1 is divisible by 6.

Solution

Check n  1:

71  1  7  1  6

True for n  1

Thus, 7  1 is divisible by 6. 1

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1898


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Assume for n  k and show for n  k  1: 7k  1 is divisible by 6, so 7k  1  6  x, where x is some natural number. Then 7k  1  1  7k  7  1   6 x  1  7  1  42 x  6  6  7 x  1 Thus, 7k  1  1 is divisible by 6. We have shown that 7 k  1  1 is divisible by 6 if 7 k  1 is divisible by 6. 38. Prove by induction that 1 + 2n < 3n for n > 1.

Solution

Check n  1:

1  2  2  32

True for n  1

59 Assume for n  k and show for n  k  1: Certainly, 3k  2  3k  1  1  3k  3k  3k . 1  2k  3k 1  2k  2  3k  2  3k  3k  3k  3  3k  3k  1

1  2  k  1  3k  1

We have shown that 1  2 k  1  3k  1 if 1  2k  3k . 39. Prove by induction that, if r is a real number where r

1  r  r2    rn 

1, then

1  r n 1 . 1 r

Solution Check n  1:

? 1 r 1  r1  True for n  1 1 r ?  1  r  1  r  1 r  1r 1 r  1 r 2

Assume for n  k and show for n  k  1: 1  r k 1 1 r 1  r k 1 1  r  r 2  r 3    r k  r k 1   r k 1 1 r 1  r k 1  r k 1  1  r  1  r  r 2  r 3    r k 1  1 r k 1 1  r  r k  1  r k 2 1  r  r 2  r 3    r k 1  1 r  k  1  1 1 r 1  r  r 2  r 3    r k 1  1 r 1  r  r2  r3    rk 

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1899


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Since this is what results when n = k + 1 is in the formula, we have shown that the formula works for n = k + 1 if it works for n = k. 40. Prove the formula for the sum of the first n terms of an arithmetic series:

a  a  d   a  2d   a   n  1 d  

n  a  an  2

where an  a   n  1 d. Solution Check n  1:

 

? 1 a  a   1  1 d a   1  1 d  2 aa

 True for n  1

Assume for n  k and show for n  k  1: a   a  d    a  2d     a   k  1 d 

 

k a  a   k  1 d



2 2ak  k d  kd  2 2ak  k 2d  kd a   a  d    a  2d     a   k  1 d  a  kd   a  kd 2 2ak  k 2d  kd  2a  2kd a   a  d    a  2d      a  kd   2 2a  k  1  k 2d  kd a   a  d    a  2d      a  kd   2 2a  k  1  kd  k  1 a   a  d    a  2d      a  kd   2 k  1 2a  kd   a   a  d    a  2d      a  kd   2  k 1   a  a  kd  a   a  d    a  2d      a  kd   2 2

Since this is what results when n = k + 1 is in the formula, we have shown that the formula works for n = k + 1 if it works for n = k.

Fix It In exercises 41 and 42, identify the step where the first error is made and fix it. 41. Given

1 1 1 1      n  1. Show that the formula is holds for n = 5. 2 4 8 2

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1900


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution Step 3 was incorrect. Step 3:

16  8  4  2  1 1 32

Step 4:

15 1 16

42. Prove by induction that if x  1 then x n  1 for all natural numbers n.

Solution Step 4 was incorrect. Step 4: Note the statement is true for k  1: x k  1  1 because x  1

Discovery and Writing 43. Describe how to apply the Principle of Mathematical Induction to prove that a statement is true for every natural number n.

Solution Answers may vary. 44. Explain why proofs in mathematics are very important.

Solution Answers may vary. 45. The expression am, where m is a natural number, was defined in Section R.2. An alternative definition of am is (part 1) a 1  a and (part 2) a m 1  a m  a. Use induction on n to prove the Product Rule for Exponents, a man  a m n .

Solution Check n  1:

a m a 1  a m  a  a m  1 , by definition

True for n  1

Assume for n  k and show for n  k  1: amak  am k a ma k a  a m  k a amak  1  am k  1 amak  1  a

m   k  1

We have shown that the formula works for n  k  1 if it works for n  k.

   a . (See

46. Use induction on n to prove the Power Rule for Exponents, am

n

mn

Exercise 37.)

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1901


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

a   a  a

Check n  1:

m

1

m

m1

True for n  1

Assume for n  k and show for n  k  1:

a   a a  a  a a a   a a   a   m

m

k

m

m

k

m

k 1 k 1

mk

mk

m

mk  m m k1

We have shown that the formula works for n  k  1 if it works for n  k. 47. Tower of Hanoi A well-known problem in mathematics is “The Tower of Hanoi,” first attributed to Edouard Lucas in 1883. In this problem, several disks, each of a different size and with a hole in the center, are placed on a board, with progressively smaller disks going up the stack. The object is to transfer the stack of disks to another peg by moving only one disk at a time and never placing a disk over a smaller one. a. Find the minimum number of moves required if there is only one disk. b. Find the minimum number of moves required if there are two disks. c. Find the minimum number of moves required if there are three disks. d. Find the minimum number of moves required if there are four disks.

Solution a. 1 b. 3 c. 7 d. 15 48. Tower of Hanoi The results in Exercise 47 suggest that the minimum number of moves required to transfer n disks from one peg to another is given by the formula 2n –1. Use the following outline to prove that this result is correct using mathematical induction. a. Verify the formula for n = 1. b. Write the induction hypothesis. c. How many moves are needed to transfer all but the largest of k + 1 disks to another peg? d. How many moves are needed to transfer the largest disk to an empty peg? e. How many moves are needed to transfer the first k disks back onto the largest one? f. How many moves are needed to accomplish steps c, d, and e?  k  1

g. Show that part f can be written in the form 2

 1.

h. Write the conclusion of the proof.

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1902


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution a. Number of moves for 1 disk: 2 n  1  2 1 1  20  1 move b. It takes 2 k  1 moves to transfer k disks. c. All but the largest  k disks  2k  1 moves d. 1 e. 2 k  1 to transfer back f. 2 k  1  1  2 k  1 g. 2 k  1  1  2 k  1  2 k  2 k  1  2  2 k  1  2 k  1  1 h. Since the number of moves works for k  1 disks when it works for k disks, and since it works for 1 disks, it works for any number of disks.

Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 49. Mathematical induction is used to prove that statements are true for all real numbers.

Solution False. It is used for natural numbers. 50. To prove that if x  1, then x n  1 for all natural numbers n, we would use mathematical induction and begin by showing that the statement is true for n = 1.

Solution True. 51. When mathematical induction is used, we assume that the statement is true for n = k + 1.

Solution False. Assume it is true for n = k. 52. 3n > 3n + 1 is true for all natural numbers n greater than 2.

Solution True.

EXERCISES 8.6 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. 1.

Evaluate: 7!

Solution 7!  5040

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1903


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

2. Evaluate: 0!

Solution 0!  1 3. Evaluate

n!

n  r  !

if n  8 and r  3

Solution 8!

 8  3 !

 336

4. Evaluate

n!

n  r  !

if n  6 and r  6

Solution 6!

6  6 !

5. Evaluate

6!  720 0! n!

r ! n  r  !

if n  10 and r  6

Solution 10!

6!  10  6  !

10!  210 6! 4 !

6. How many ways can you arrange a spider plant, a peace lily, and an African Violet, on a 3-tier plant stand in your house?

Solution 3!  6 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. If E1 and E2 are two events and E1 can be done in 4 ways and E2 can be done in 6 ways, then the event E1 followed by E2 can be done in ______ ways.

Solution 6  4  24 8. An arrangement of n objects is called a __________.

Solution permutation 9. P(n, r) = __________

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1904


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution n!

n  r  ! 10. P(n, n) = __________

Solution n! 11. P(n, 0) = __________

Solution 1 12. There are __________ ways to arrange n things in a circle. Solution

 n  1 !

13. C(n, r) = __________

Solution n!

r ! n  r  ! 14. Using combination notation, C(n, r) = __________

Solution  n   r 15. C(n, n) = __________

Solution 1 16. C(n, 0) = __________

Solution 1 17. If a word with n letters has a of one letter, b of another letter, and so on, the number of different words that can be formed is ____________.

Solution n! a !  b !

18. Where the order of selection is not important, we are interested in ____________, not __________.

Solution combinations, permutations

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1905


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Practice Evaluate each expression. 19. P(7, 4)

Solution P  7, 4  

7!

7  4 !

7 ! 7  6  5  4  3!   7  6  5  4  840 3! 3!

8! 8  7  6  5!   8  7  6  336 5! 5!

5! 5!   5  4  3  2  1  120 0! 1

5! 1 5!

20. P(8, 3)

Solution P  8, 3  

8!

8  3 !

21. P(5, 5)

Solution P  5, 5  

5!

5  5 !

22. P(5, 0)

Solution P  5, 0  

5!

5  5 !

23. 9 P2

Solution 9

p2 

9!

 9  2 !

9!  9  8  72 7!

11!  11  10  9  990 8!

24. 11 P3

Solution P 

11 3

11!

 11  3 !

25. 15 P4

Solution P 

15 4

15!

 15  4  !

15!  15  14  13  12  32, 760 11!

26. 12 P4

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1906


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution P 

12 4

12!

 12  4  !

12!  12  11  10  9  11, 880 8!

27. C(7, 4)

Solution C  7, 4  

7! 7  6  5  4! 7  6  5    35 4! 3! 4 ! 3! 321

8! 8  7  6  5! 8  7  6    56 3!5! 3!5! 321

7!

4 ! 7  4 !

28. C(8, 3)

Solution C  8, 3  

8!

3!  8  3  !

29. 10 C6

Solution C  10, 6  

10!

6!  10  6  !

10! 10  9  8  7  6! 10  9  8  7    210 6! 4! 6! 4! 4321

30. 11 C4

Solution C  11, 4   31.

9

11!

4!  11  4  !

11! 11  10  9  8  7 ! 11  10  9  8    330 4! 7 ! 4! 7 ! 4321

C7

Solution C  9, 7  

9!

7 ! 9  7  !

9  8  7! 9  8   36 7 ! 2! 2

6  5  4! 6  5   15 2! 4! 21

32. 6 C2

Solution C  6, 2  

6!

2!  6  2  !

5 33.   4 Solution 5 5! 5  4! 5   5    1  4  4!  5  4  ! 4! 1!

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1907


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

8 34.   4 Solution 8 8! 8  7  6  5  4! 8  7  6  5    70    4! 4! 4321  4  4!  8  4  !

5 35.   0 Solution 5 5! 5! 1   1    0 0!5! 1   0!  5  0  !

5 36.   5 Solution 5 5! 5! 1   1    5 5!0! 1 5! 5  5 !    

9 37.   6 Solution 9 9! 9! 987    84    32  6  6!  9  6  ! 6! 3!

 13  38.   9 Solution  13  13! 13! 13  12  11  10    715    4321  9  9!  13  9 ! 9! 4!

 11 39.   2 Solution  11  11! 11! 11  10    55    21  2  2!  11  2  ! 2!9!

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1908


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

 14  40.    10  Solution  14  14! 14! 14  13  12  11    1001    4321  10  10!  14  10  ! 10! 4!

 68  41.    66  Solution  68  68! 68  67  66! 68  67    2278    66! 2! 21  66  66!  68  66  !

 100  42.    99  Solution  100  100! 100  99!   100    99! 1!  99  99!  100  99 !

 5  4  3 43.        3  3   3 Solution 5  4  3 5! 4! 4!    10  4  1  40         3   3   3  3!  5  3  ! 3!  4  3  ! 3!  3  3  !

5 6  7  8 44.         5 6  7  8 Solution 5 6  7  8          1  1  1  1  1 5 6  7  8

 

45. P 5, 4  C 5, 3

Solution P  5, 4   C  5, 3  

 

46. P 3, 2  C 4, 3

5!

5!

5  4  ! 3! 5  3 !

5! 5!   120  10  1200 1! 3! 2!

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1909


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution P  3, 2   C  4, 3  

3!

4!

 3  2 ! 3!  4  3 !

3! 4!   6  4  24 1! 3! 1!

Fix It In exercises 47 and 48, identify the step where the first error is made and fix it. 47. How many different ways can five types of nuts be selected from ten for making a homemade trail mix. To begin, state if order matters and this is a permutation problem or that order doesn’t matter and it is a combination problem. Then write the permutation or combination and solve the problem.

Solution Step 3 was incorrect. Step 3:

Step 4:

10!

5!  10  5  ! 10 ! 5!5!

Step 5: 252 48. An ice cream shop has 18 flavors of ice cream. How many cones with two different flavors can you order if it is important to you which flavor goes on the top and bottom? To begin, state if order matters and this is a permutation problem or that order doesn’t matter and it is a combination problem. Then write the permutation or combination and solve the problem.

Solution Step 5 was incorrect. Step 5: 306

Applications 49. Choosing lunch A lunchroom has a machine with eight kinds of sandwiches, a machine with four kinds of soda, a machine with both white and chocolate milk, and a machine with three kinds of ice cream. How many different lunches can be chosen? (Consider a lunch to be one sandwich, one drink, and one ice cream.)

Solution 8  6  3  144 50. Manufacturing license plates How many six-digit license plates can be manufactured if no license plate number begins with 0?

Solution 9  10  10  10  10  10  900, 000

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1910


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

51. Available phone numbers How many different seven-digit phone numbers can be used in one area code if no phone number begins with 0 or 1?

Solution 8  10  10  10  10  10  10  8,000, 000 52. Arranging letters

In how many ways can the letters of the word number be arranged?

Solution 6!  720 53. Arranging letters with restrictions In how many ways can the letters of the word number be arranged if the e and r must remain next to each other?

Solution Consider the e and the r to be a block that cannot be divided, say x. Then the problem becomes finding the number of ways to rearrange the letters in the word numbx. This can be done in 5!, or 120 ways. For each of these possibilities, the e and the r could be reversed, doubling the number of possibilities. The answer is 2 ∙ 120, or 240 ways. 54. Arranging letters with restrictions In how many ways can the letters of the word number be arranged if the e and r cannot be side by side?

Solution The total number of ways, without restrictions, of rearranging the letters is 6!, or 720. Since there are 240 ways of rearranging the letters so that the e and the r ARE next to each other (see #53) then there are 720 – 240, or 480 ways of rearranging the letters so that the e and the r are NOT next to each other. 55. Arranging letters with repetitions How many ways can five Scrabble tiles bearing the letters, F, F, F, L, and U be arranged to spell the word fluff ?

Solution The word must appear as

LU

, where one of the Fs must appear in each box.

This can be done in 3!, or 6 ways. 56. Arranging letters with repetitions How many ways can six Scrabble tiles bearing the letters B, E, E, E, F, and L be arranged to spell the word feeble?

Solution The word must appear as F

BL

, where one of the Es must appear in each box.

This can be done in 3!, or 6 ways. 57. Placing people in line

In how many arrangements can 8 women be placed in a line?

Solution 8! = 40,320 58. Placing people in line In how many arrangements can 5 women and 5 men be placed in a line if the women and men alternate?

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1911


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution The line might look like this:

5 5 W M

4 W

4 3 M W

3 M

2 W

2 M

1 W

1 M

Then there would be 5  5  4  4  3  3  2  2  1  1  14, 400 ways. However, the line could start with a man instead of a woman, so there are 2  14, 400  28,800 possible arrangements. 59. Placing people in line In how many arrangements can 5 women and 5 men be placed in a line if all the men line up first?

Solution The line will look like this:

5 M

4 M

3 M

2 M

1 M

5 W

4 W

3 W

2 W

1 W

Then there are 5  4  3  2  1  5  4  3  2  1  14, 400 arrangements. 60. Placing people in line In how many arrangements can 5 women and 5 men be placed in a line if all the women line up first?

Solution The line will look like this:

5 W

4 W

3 W

2 W

1 W

5 M

4 M

3 M

2 M

1 M

Then there are 5  4  3  2  1  5  4  3  2  1  14, 400 arrangements. 61. Combination locks How many permutations does a combination lock have if each combination has 3 numbers, no two numbers of the combination are the same, and the lock dial has 30 notches?

Solution P  30, 3  

30 !  24, 360 27 !

62. Combination locks How many permutations does a combination lock have if each combination has 3 numbers, no two numbers of the combination are the same, and the lock dial has 100 notches?

Solution 100 !  970, 200 97 ! 63. Seating at a table In how many ways can 8 people be seated at a round table? P  100, 3  

Solution

8  1 !  7 !  5040

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1912


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

64. Seating at a table

In how many ways can 7 people be seated at a round table?

Solution (7  1) !  6 !  720

65. Seating at a table In how many ways can 6 people be seated at a round table if 2 of the people insist on sitting together?

Solution Consider the two people who must sit together as a single person, so that there are 5 “people” who must be arranged in a circle. This can be done in (5  1) !  4 !  24 ways. However, the two people who have been seated next to each other could be switched, so that the number of arrangements is doubled. There are 2  24  48 possible arrangements. 66. Seating arrangements with conditions In how many ways can 6 people be seated at a round table if 2 of the people refuse to sit together?

Solution Without restrictions, 6 people can be seated at a round table in (6  1) !  5 !  120 ways. Since there are 48 ways to sit the 6 people so that 2 MUST be together (see #65), there are 120 – 48, or 72 ways of seating the people so that the 2 are NOT seated next to each other. 67. Arrangements in a circle In how many ways can 7 children be arranged in a circle if Ella and Eli want to sit together and Jayden and Jackson want to sit together?

Solution Consider Ella and Eli as a single person and Jayden and Jackson as a single person, so that there are 5 “people” who must be arranged in a circle. This can be done in (5  1) !  4 !  24 ways. However, each group of 2 people could be switched, so that the number of arrangement will equal 2  2  24, or 96. 68. Arrangements in a circle In how many ways can 8 children be arranged in a circle if Laura, Scott, and Grace want to sit together?

Solution Consider the three people as a single person, so that there are 6 “people” who must be arranged in a circle. This can be done in (6  1) !  5 !  120 ways. However, the group of three people can be rearranged in 3!, or 6 ways, so the total number of arrangements is 6  120, or 720. 69. Selecting candy bars In how many ways can 4 candy bars be selected from 10 different candy bars?

Solution  10  10!  210    4 4!6!   70. Selecting hiking trails In how many ways can a family select 2 hiking trails to walk from 10 different hiking trails?

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1913


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution  24  24!  134,596    6 6! 18!   71. Circuit wiring A wiring harness containing a red, a green, a white, and a black wire must be attached to a control panel. In how many different orders can the wires be attached?

Solution 4!  24 72. Grading homework A professor grades homework by randomly checking 7 of the 20 problems assigned. In how many different ways can this be done?

Solution  20  20!  77,520    7 7 ! 13!   73 Forming words with distinct letters How many words can be formed from the letters of the word plastic if each letter is to be used once?

Solution 7!  5040 74. Forming words with distinct letters How many words can be formed from the letters of the word computer if each letter is to be used once?

Solution 8!  40, 320 75. Forming words with repeated letters How many words can be formed from the letters of the word banana if each letter is to be used once?

Solution 6!  60 3! 2! 1! 76. Forming words with repeated letters How many words can be formed from the letters of the word laptop if each letter is to be used once?

Solution 6!  360 2! 1! 1! 1! 77. Manufacturing license plates How many license plates can be made using two different letters followed by four different digits if the first digit cannot be 0 and the letter O is not used?

Solution 25  24  9  9  8  7  2, 721, 600

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1914


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

78. Planning class schedules If there are 7 class periods in a school day, and a typical student takes 5 classes, how many different time patterns are possible for the student?

Solution 7    21 5 79. Selecting golf balls From a bucket containing 6 red and 8 white golf balls, in how many ways can we draw 6 golf balls of which 3 are red and 3 are white?

Solution 6 8      20  56  1120 3 3 80. Selecting a committee In how many ways can you select a committee of 3 Republicans and 3 Democrats from a group containing 18 Democrats and 11 Republicans?

Solution  11   18       165  816  134,640  3 3  81. Selecting a committee In how many ways can you select a committee of 4 Democrats and 3 Republicans from a group containing 12 Democrats and 10 Republicans?

Solution  12   10       495  120  59, 400  4  3  82. Drawing cards In how many ways can you select a group of 5 red cards and 2 black cards from a deck containing 10 red cards and 8 black cards?

Solution  10   8       252  28  7056  5  2 83. Planning dinner In how many ways can a husband and wife choose 2 different dinners from a menu of 17 dinners?

Solution

17 16   272 H W 84. Placing people in line In how many ways can 7 people stand in a row if 2 of the people refuse to stand together?

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1915


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution There are a total of 7!, or 5040 ways to arrange the people without restrictions. Count the number of arrangements with the two people together. To do this, consider them as one person, so that there are 6!, or 720 arrangements. This needs to be doubled, since the two people considered as a group can switch places. Then there are 1440 arrangements with the two people together. The number of arrangements in which they are NOT together is 5040  1440  3600. 85. Geometry line?

How many lines are determined by 8 points if no 3 points lie on a straight

Solution 8    28 2 86. Geometry line?

How many lines are determined by 10 points if no 3 points lie on a straight

Solution  10     45 2 87. Coaching basketball How many different teams can a basketball coach start if the entire squad consists of 10 players? (Assume that a starting team has 5 players and each player can play all positions.) Solution  10     252 5 88. Managing baseball How many different teams can a manager start if the entire squad consists of 25 players? (Assume that a starting team has 9 players and each player can play all positions.)

Solution  25   9   2,042,975   89. Selecting job applicants There are 30 qualified applicants for 5 openings in the sales department. In how many different ways can the group of 5 be selected?

Solution  30     142,506  5 90. Sales promotions If a customer purchases a new stereo system during the spring sale, he may choose any 6 CDs from 20 classical and 30 jazz selections. In how many ways can the customer choose 3 of each?

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1916


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

 20   30       1140  4060  3  3   4,628, 400 91. Guessing on matching questions Ten words are to be paired with the correct 10 out of 12 possible definitions. How many ways are there of guessing?

Solution  12     66  10  92. Guessing on true-false exams How many possible ways are there of guessing on a 10-question true-false exam, if it is known that the instructor will have 5 true and 5 false responses?

Solution

 10  Pick 5 of the 10 as true:    252 5 93. Number of Wendy’s® hamburgers Wendy’s® Old Fashioned Hamburgers offers eight toppings for their single hamburger. How many different single hamburgers can be ordered?

Solution For each topping, you have two choices, select or do not select:

2 2 2 2 2 2 T1 T2 T3 T4 T5 T6 2  2  2  2  2  2  2  2  256

2 T7

2 T8

94. Number of ice cream sundaes A restaurant offers ten toppings for their ice cream sundaes. How many different sundaes can be ordered?

Solution For each topping, you have two choices, select or do not select:

2 2 2 2 2 2 2 2 T1 T2 T3 T4 T5 T6 T7 T8 2  2  2  2  2  2  2  2  2  2  1024

2 T9

2 T10

Practice Use Pascal’s Triangle to compute each combination. 95. C(8, 3)

Solution 9th row of triangle: 1 8 28 56 70 56 28 8 1; 4th number in row:

56

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1917


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

96. C(7, 4)

Solution 8th row of triangle: 1 7 21 35 35 21 7 1; 5th number in row:

35

Discovery and Writing 97. Describe the Fundamental Counting Principle.

Solution Answers may vary. 98. What is a permutation?

Solution Answers may vary. 99. What is a combination?

Solution Answers may vary. 100. Explain the difference between a permutation and a combination.

Solution Answers may vary.

101. Prove that C n, n  1.

Solution C  n, n  

n!

n !  n  n !

n! n!  1 n !0! n !

102. Prove that C n, 0  1.

Solution C  n, 0  

n! n!  1 0! n! n!

 n  n  103. Prove that      .  r  n  r  Solution  n   n n! n! n!          n  r  n  r  ! n  n  r  ! n  r  ! r ! n  r  !  r 

104. Show that the Binomial Theorem can be expressed in the form n

 n

k 0

 

a  b    k  a n

n k

bk

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1918


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution Answers may vary. 105. Explain how to use Pascal’s Triangle to find C(8, 5).

Solution Answers may vary. 106. Explain how to use Pascal’s Triangle to find C(10, 8).

Solution Answers may vary. Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 107. Permutation problems involve situations in which the order of the items makes no difference.

Solution False. Permutations are used when order matters. 108. Combination problems involve situations in which order matters. Solution False. Combinations are used when order does not matter. 109. The number of permutations of n distinct objects is greater than the number of combinations of those n objects.

Solution True. 110. The number of permutations of n things taken r at a time can be found using the Fundamental Counting Principle.

Solution True. 111. The number of combinations of n things taken r at a time cannot be found using the Fundamental Counting Principle.

Solution True. 112. The number of ways to choose 11 people out of 25 to form a soccer team is 25P11.

Solution False. Use 25C11. 113. The number of ways to choose 3 company employees out of 25, one to work in Italy, one to work in Aruba, and one to work in Hawaii, is 25P3.

Solution True.

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1919


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

114. The digits 1–9 are used to create a four-digit ATM personal identification number (PIN) for your debit bank card. If no digits are repeated, the number of possible numbers is 10C4.

Solution False. Use 10P4.

EXERCISES 8.7 Getting Ready Complete these just-in-time review problems to prepare you to successfully work the practice exercises. For Exercises 1 and 2, use the following information. There are six equally likely outcomes if one die is rolled. The outcomes are 1, 2, 3, 4, 5, and 6. 1.

Identify the fraction of the outcomes that are less than 6.

Solution

5 6 2. Identify the fraction of the outcomes that are less than or equal to 3.

Solution

3 1  6 2 For Exercises 3 and 4, use the following information. A standard deck of 52 cards has two red suits, hearts and diamonds, and two black suits, clubs and spades. Each suit has 13 cards, including an ace, a king, a queen, a jack, and cards numbered from 2 to 10. If one card is drawn there are 52 equally likely outcomes. 3. Identify the fraction of the outcomes that are hearts?

Solution

13 1  52 4 4. Identify the fraction of the outcomes that are aces?

Solution

4 1  52 13 For Exercises 5 and 6, use the following information. An American roulette wheel has 38 slots, numbered 1 through 36, 0, and 00. 18 slots are red, 18 slots are black, and two slots are green. The dealer spins the wheel in one direction and rolls a small ball in the opposite direction until both the wheel and the ball come to a stop. The ball is equally likely to stop in any one of the 38 slots. On one spin of the roulette wheel, determine the probability of each event.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1920


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

5. Identify the fraction of outcomes that are 0 or 00?

Solution

2 1  38 19 6. Identify the fraction of outcomes that are red?

Solution

18 9  38 19 Vocabulary and Concepts You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 7. An __________ is any process for which the outcome is uncertain.

Solution experiment 8. A list of all possible outcomes for an experiment is called a __________.

Solution sample space

 

9. The probability of an event E is defined as P E  __________ 

s n

Solution

n E  n  s

10. P A  B  __________

Solution

P  A  P  B | A

Practice List the sample space of each experiment. 11. Rolling one die and tossing one coin

Solution

 1, H ,  2, H , 3, H , 4, H , 5, H , 6, H ,  1, T  , 2, T  , 3, T  , 4, T  , 5, T  , 6, T 

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1921


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

12. Tossing three coins

Solution

H, H, H , H, H, T  , H, T , H  , H, T , T  , T , H, H  , T , H, T  , T , T , H  , T , T , T 

13. Selecting a letter of the alphabet

Solution

A, B, C, D, E, F , G, H, I, J, K , L, M, N, O, P, Q, R, S, T , U, V , W , X , Y , Z

14. Picking a one-digit number

Solution

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

An ordinary die is rolled. Find the probability of each event. 15. Rolling a 2

Solution 1 6 16. Rolling a number greater than 4

Solution 2 1  6 3 17. Rolling a number greater than 1 but less than 6

Solution 4 2  6 3 18. Rolling an odd number

Solution 3 1  6 2 Balls numbered from 1 to 42 are placed in a container. If one is drawn at random, find the probability of each result. 19. The number is less than 20.

Solution

19 42

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1922


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

20. The number is less than 50.

Solution

42 1 42 21. The number is a prime number.

Solution

13 42 22. The number is less than 10 or greater than 40.

Solution

9  2 11  42 42 If the spinner shown below is spun, find the probability of each event. Assume that the spinner never stops on a line.

23. The spinner stops on red.

Solution 3 8 24. The spinner stops on green.

Solution 2 1  8 4 25. The spinner stops on orange.

Solution 0 0 8 26. The spinner stops on yellow.

Solution 1 8

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1923


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Find the probability of each event. 27. Rolling a sum of 4 on one roll of two dice

Solution Rolls of 4:

 1, 3 ,  2, 2  , 3, 1

Probability 

3 1  36 12

28. Drawing a diamond on one draw from a card deck

Solution

# diamonds # cards

13 1  52 4

29. Drawing two aces in succession from a card deck if the card is replaced and the deck is shuffled after the first draw

Solution

# aces

# aces

# cards # cards

4 4 1   52 52 169

30. Drawing two aces from a card deck without replacing the card after the first draw

Solution

# aces

# aces

# cards # cards

4 3 1   52 51 221

31. Drawing a red egg from a basket containing 5 red eggs and 7 blue eggs

Solution

# red # eggs

5 12

32. Getting 2 red eggs in a single scoop from a bucket containing 5 red eggs and 7 yellow eggs

Solution

5   2 10 5      total # ways to  12  66 33   get 2 eggs 2 # ways to get 2 red eggs

33. Drawing a bridge hand of 13 cards, all of one suit

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1924


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

 13  4     13   # ways to get 13 cards  52    from the deck of 52  13 

# ways to get 13 cards of the same suit

4 6.350136  1011  6.3  1012 

34. Drawing 6 diamonds from a card deck without replacing the cards after each draw

Solution  13    6 1716     # ways to get 6 cards  52  20, 358, 520   from the deck of 52 6 # ways to get 6 diamonds

33 391, 510

35. Drawing 5 aces from a card deck without replacing the cards after each draw

Solution impossible  0 36. Drawing 5 clubs from the black cards in a card deck

Solution

 13    5 # ways to get 5 clubs 1287 9      # ways to get 5 cards  26  65780 460   from the 26 black cards  5  37. Drawing a face card (king, queen, or jack) from a card deck

Solution

# face cards # cards in deck

12 3  52 13

38. Drawing 6 face cards in a row from a card deck without replacing the cards after each draw

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1925


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution  12    from the 12 face cards  6  924   # ways to get 6 cards  52  20, 358, 520   from the deck of 52 5 # ways to get 6 cards

33 727, 090

39. Drawing 5 orange cubes from a bowl containing 5 orange cubes and 1 beige cube

Solution

5   # ways to get 5 orange  5  1   # ways to get 5 cubes  6  6   5 40. Rolling a sum of 4 with one roll of three dice

Solution

rolls of 4:  1, 1, 2 ,  1, 2, 1 ,  2, 1, 1

Probability 

3 1  216 72

41. Rolling a sum of 11 with one roll of three dice

Solution

rolls of 11:  1, 4, 6  ,  1, 5, 5  ,  1, 6, 4  ,  2, 3, 6  ,  2, 4, 5  ,  2, 5, 4  ,  2, 6, 3  ,  3, 2, 6  ,  3, 3, 5 

 3, 4, 4  ,  3, 5, 3 ,  3, 6, 2 ,  4, 1, 6  ,  4, 2, 5 ,  4, 3, 4  ,  4, 4, 3 ,  4, 5, 2 ,  4, 6, 1 , 5, 1, 5 , 5, 2, 4  , 5, 3, 3 , 5, 4, 2 , 5, 5, 1 , 6, 1, 4  , 6, 2, 3 , 6, 3, 2 , 6, 4, 1

Probability 

27 1  216 8

42. Picking, at random, 5 Republicans from a group containing 8 Republicans and 10 Democrats

Solution 8    5   56  1  18  8568 153   5

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1926


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

43. Tossing 3 heads in 5 tosses of a fair coin

Solution 5  3  10 5    32 16 25 44. Tossing 5 heads in 5 tosses of a fair coin

Solution 5   5  1 32 25 An American roulette wheel has 38 slots, numbered 1 through 36, 0, and 00. 18 slots are red, 18 slots are black, and two slots are green. The dealer spins the wheel in one direction and rolls a small ball in the opposite direction until both the wheel and the ball come to a stop. The ball is equally likely to stop in any one of the 38 slots. On one spin of the roulette wheel, determine the probability of each event. 45. The ball stops on a red slot.

Solution

18 9  38 19 46. The ball stops on a black slot.

Solution

18 9  38 19 47. The ball stops on green slot.

Solution

2 1  38 19 48. The ball lands on an even number.

Solution

18 9  38 19 49. The ball lands on an odd number.

Solution

18 9  38 19

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1927


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

50. The ball stops on 0.

Solution

1 38 51. The ball stops on 00.

Solution

1 38 52. The ball does not stop on 00.

Solution

1 37  38 38 Assume that the probability that an airplane engine will fail during an emergency training torture test is 21 and that the aircraft in question has 4 engines. 1

53. Construct a sample space for the torture test. Use S for survive and F for fail.

Solution SSSS, SSSF , SSFS, SSFF , SFSS, SFSF , SFFS, SFFF ,

FSSS, FSSF , FSFS, FSFF , FFSS, FFSF , FFFS, FFFF 54. Find the probability that all engines will survive the test.

Solution 1 16 55. Find the probability that exactly 1 engine will survive.

Solution 4 1  16 4 56. Find the probability that exactly 2 engines will survive.

Solution 6 3  16 8 57. Find the probability that exactly 3 engines will survive.

Solution 4 1  16 4

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1928


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

58. Find the probability that no engines will survive.

Solution 1 16 59. Find the sum of the probabilities in Exercises 54 through 58.

Solution 16 1 16 Assume that a survey of 282 people is taken to determine the opinions of doctors, teachers, and lawyers on a proposed piece of legislation, with the results as shown in the table. A person is chosen at random from those surveyed. Refer to the table to find each probability. Number that Favor

Number that Oppose

Number with No Opinion

Doctors

70

32

17

119

Teachers

83

24

10

117

Lawyers

23

15

8

46

Total

176

71

35

282

Total

60. The person favors the legislation.

Solution 176 88  282 141 61. A doctor opposes the legislation.

Solution 32 119 62. A person who opposes the legislation is a lawyer.

Solution

15 71 63. Quality control In a batch of 10 tires, 2 are known to be defective. If 4 tires are chosen at random, find the probability that all 4 tires are good.

Solution 8    4   70  1  10  210 3   4

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1929


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

64. Medicine Out of a group of 9 patients treated with a new drug, 4 suffered a relapse. Find the probability that 3 patients of this group, chosen at random, will remain disease-free.

Solution 5    3   10  5  9  84 42    3 Use the Multiplication Property of Probabilities.

 

65. If P A  0.3 and P B | A  0.6, find P A  B .

Solution

P  A  B   P  A   P  B | A

 0.3  0.6  0.18

 

66. If P A  B  0.3 and P B | A  0.6, find P A .

Solution

P  A  B   P  A  P  B | A 0.3  P  A 0.6 0.5  P  A

67. Conditional probability The probability that a person owns a luxury car is 0.2, and the probability that the owner of such a car also owns a personal computer is 0.7. Find the probability that a person, chosen at random, owns both a luxury car and a computer.

Solution

P  A  B   P  A   P  B | A

 0.2  0.7   0.14

68. Conditional probability If 40% of the population have completed college, and 85% of college graduates are registered to vote, what percent of the population are both college graduates and registered voters?

Solution

P  A  B  P  A  P  B | A  0.40  0.85

 0.34  34% 69. Conditional probability About 25% of the population watches the evening television news coverage as well as the soap operas. If 75% of the population watches the news, what percent of those who watch the news also watch the soaps?

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1930


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

P  A  B  P  A  P  B | A 0.25  0.75P  B | A 0.33  P  B | A 33%  P  B | A

70. Conditional probability The probability of rain today is 0.40. If it rains, the probability that Abishola will forget her raincoat is 0.70. Find the probability that Abishola will get wet. Solution

P  A  B   P  A  P  B | A

 0.4  0.7   0.28

Fix It In exercises 71 and 72, identify the step where the first error is made and fix it. 71. Find the probability of the event of “rolling a sum of 10” on one roll of two dice. First identify n(S), the number of equally likely outcomes that form the sample space S. Next identify E, the set of favorable outcomes. Then determine the probability of favorable outcomes.

Solution Step 2 was incorrect. Step 2: E 

6, 4 , 5, 5 , 4,6

 

Step 3: P E  P rolling a sum of 10  Step 4:

n E  n  s

3 1  36 12

72. Determine the probability of drawing three face cards (jack, queen, or king) from a standard deck of playing cards. The cards are drawn without replacement.

Solution Step 4 was incorrect. Step 4: P  three f ace cards in a row  

3 11 1   13 51 5

Step 5: P  three f ace cards in a row  

11 1105

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1931


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Discovery and Writing 73. What is an experiment? Give two examples.

Solution Answers may vary. 74. What is meant by the sample space of an experiment?

Solution Answers may vary. 75. Describe how to determine the probability of an event.

Solution Answers may vary. 76. Explain the Multiplication Property of Probabilities.

Solution Answers may vary.

77. If P A  B  0.7, is it possible that P B | A  0.6 ? Explain.

Solution no

 

78. Is it possible that P A  B  P A ? Explain.

Solution yes Critical Thinking Determine if the statement is true or false. If the statement is false, then correct it and make it true. 79. The probability that an event occurs can be a negative number.

Solution False. Probabilities are between 0 and 1. 80. The probability of a certain event is 0.

Solution False. A certain event has probability 1. 81. The probability of an impossible event is 1.

Solution False. An impossible event has probability 0. 82. If the probability that you will graduate is 0.63, then the probability that you will not graduate is –0.63.

Solution False. The probability is 1 – 0.63 = 0.37.

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1932


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

83. Events that cannot occur simultaneously are called mutually exclusive events. If one card is randomly selected from a deck of cards, drawing a jack or a queen would be mutually exclusive events.

Solution True. 84. Events that can occur simultaneously are called non–mutually exclusive events. If one card is randomly selected from a deck of cards, drawing a jack or a heart would be non–mutually exclusive events.

Solution True. 85. The probability of a couple having four boys is 21 because the probability each time of having a boy is 21 .

Solution False the probability is

1 1 1 1 1     . 2 2 2 2 16

86. If the probability of a couple of five girls is 321 , then the probability of having five boys 31 is 32 .

Solution False. The probability is

1 . 32

CHAPTER REVIEW SOLUTIONS Exercises Find each value. 1.

6!

Solution

6!  6  5  4  3  2  1  720

2.

7!  0!  1!  3! Solution 7 !  0!  1!  3!  5040  1  1  6  30, 240

3.

8! 7! Solution 8! 8  7 !  8 7! 7!

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1933


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

4.

5!  7!  8! 6!  9! Solution

5!  7 !  8! 6!  9!

5!  7  6!  8! 6!  9  8!

5!  7 9

280 3

Expand each expression. 5.

x  y

3

Solution

 x  y   x  1!3!2! x y  2!3!1! xy  y  x  3x y  3xy  y 3

6.

 p  q

3

2

2

3

3

2

2

3

4

Solution

 p  q   P  1!4!3! p q  2!4!2! p q  3!4!1! pq  q  p  4p q  6p q  4pq  q 4

7.

4

3

2

2

3

4

4

3

2

2

3

4

a  b 

5

Solution

5! 5! 5! 5! a  b  a  b  a  b  a  b   b a  b  a  1!4! 2!3! 3!2! 4! 1! 5

5

4

3

2

2

3

4

5

 a5  5a4 b  10a3b2  10a2b3  5ab4  b5

8.

 2a  b 

3

Solution

 2a  b   2a   1!3!2!  2a   b  2!3!1!  2a  b   b  8a  12a b  6ab  b 3

3

2

2

3

3

2

2

3

Find the required term of each expansion. 9.

a  b  ; 4th term 8

Solution The 4th term will involve b3.

8! 5 3 a b  56a5 b3 5! 3! 10.  2 x  y  ; 3rd term 5

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1934


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution The 3rd term will involve (  y )2 . 3 2 5!  2x    y   80x 3 y 2 3! 2!

11.

 x  y  ; 7th term 9

Solution The 7th term will involve (  y )6 . 6 9! 3 x   y   84 x 3 y 6 3!6!

12.

 4 x  7  ; 4th term 6

Solution The 4th term will involve 73. 3 6! 4 x  73  439,040 x 3  3! 3!

Write the fourth term in each sequence. 3 13. 0, 7, 26, , n  1,

Solution

43  1  63 14.

n2  2 3 11 , 3, , , , 2 2 2 Solution

42  2 18  9 2 2 Find the first five terms of the sequence and then find a10. 2 15. an  2n  2

Solution

a1  2  1  2  0; a2  2  2  2  6 2

2

a3  2  3  2  16; a4  2  4  2  30 2

2

a5  2  5  2  48 2

a10  2  10  2  198 2

Find the first four terms of the sequence. © 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1935


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

16. a1  2 and an1  2an

2

Solution

a1  2

a2  2a12  2  2  8 2

a3  2a22  2  8  128 2

a4  2a32  2  128  32, 768 2

Evaluate each expression. 17.

4

 3k

2

k 1

Solution

 3k  3 k  3  1  2  3  4  4

2

k 1

18.

4

2

k 1

2

2

2

2

 3  30   90

10

6 k 1

Solution 10

 6  10 6   60 k 1

19.

  k  3k  8

3

2

k 5

Solution

  k  3k    k  3 k  5  6  7  8   3 5  6  7  8  8

3

8

2

k 5

3

k 5

8

2

3

3

3

3

2

2

2

2

k 5

 1718 30 3  3 30 20.   k  12    k k 1  2  2 k 1

Solution 30

3

3 30

3 30

30

3 30

30

k 1

k 1

k 1

k 1

k 1

k 1

  2 k  12   2  k  2  k   12  2  k   12  360

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1936


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Find the required term of each arithmetic sequence. 21. 5, 9, 13,  ; 29th term

Solution a  5, d  4

a29  a   n  1 d

 5   29  1 4  117

22. 8, 15, 22,  ; 40th term

Solution a  8, d  7

a40  a   n  1 d

 8   40  1 7  281

23. 6,  1,  8,  ; 15th term

Solution a  6, d  7

a15  a   n  1 d

 6   15  1 7   92

24.

1 3 7 ,  ,  ,; 35th term 2 2 2 Solution 1 a  , d  2 2 a35  a   n  1 d 

1 135   35  1 2    2 2

25. Find three arithmetic means between 2 and 8.

Solution a  2, a5  8

8  2  4d 6  4d 3 7 13  d  2, , 5, ,8 2 2 2

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1937


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

26. Find five arithmetic means between 10 and 100.

Solution a  10, a7  100 100  10  6d 90  6d 15  d  10, 25, 40, 55, 70, 85 , 100

Find the sum of the first 40 terms in each sequence. 27. 5, 9, 13, 

Solution a  5, d  4

a40  a   n  1 d  5  39  4   161 S40 

n  a  a40  2

40  5  161 2

 3320

28. 8, 15, 22

Solution a  8, d  7

a40  a   n  1 d  8  39  7   281 S40 

n  a  a40  2

40  8  281 2

 5780

29. 6,  1,  8, 

Solution a  6, d  7

a40  a   n  1 d  6  39  7   267; S40  30.

n  a  a40  2

40  6  267  2

 5220

1 3 7 ,  ,  , 2 2 2 Solution 1 a  , d  2 2

a40  a   n  1 d 

1 155 ; S40   39  2   2 2

n  a  a40  2

 1 155  40    2 2     1540 2

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1938


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Find the required term of each geometric sequence. 31. 81, 27, 9,  ; 11th term

Solution

1 3

a  81, r 

 1 a11  ar n 1  81   3

10

1 729

32. 2, 6, 18,  ; 9th term

Solution a  2, r  3

a9  ar n 1  2  3  13, 122 8

33. 9,

9 9 , ,  ; 15th term 2 4

Solution

1 2

a  9, r  a15  ar

34. 8, 

 1  9  2

n 1

14

9 16, 384

8 8 , ,  ; 7th term 5 25

Solution

a  8, r  

1 5 6

a7  ar

n 1

 1 8  8   15,625  5

35. Find three positive geometric means between 2 and 8.

Solution a5  ar 4

Use r   2:

8  2r

2, 2 2, 4, 4 2 , 8

4

4  r4  44 r r  2

36. Find four geometric means between –2 and 64.

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1939


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

a6  ar 5 64  2r 5 32  r 5 2  r 2, 4,  8, 16,  32 , 64 37. Find the positive geometric mean between 4 and 64.

Solution a3  ar 2 64  4r 2

16  r 2  r  4 problem states positive 

 4, 16 , 64

Find the sum of the first 8 terms in each geometric sequence. 38. 81, 27, 9, 

Solution a  81, r 

1 ,n8 3

a  ar n 81  81  3  S8   1r 1  31

8

1

81  811 2 3

3280 27

39. 2, 6, 18, 

Solution a  2, r  3, n  8

a  ar n 2  2  3 S8   1 r 13 13, 120   6560 2 8

40. 9,

9 9 , , 2 4

Solution

a  9, r 

1 ,n8 2

a  ar n 9  9  2  S8   1 r 1  21 1

2295

 256  1 2

8

2295 128

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1940


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

41. 8, 

8 8 , 5 25

Solution 1 a  8, r   , n  8 5

a  ar n 8  8   5  S8   1 r 1    51  1

3,124,992 390,625 6 5

8

520, 832 78, 125

42. Find the sum of the first eight terms of the geometric sequence

1 , 1, 3,  . 3

Solution

a

1 , r  3, n  8 3

1  31  3  a  ar n 3  S8  1 r 13  6560 3280 3   2 3 8

43. Find the seventh term of the geometric sequence 2 2, 4, 4 2,  .

Solution

a7  ar 6 2 2

 2

6

 16 2 Find the sum of each infinite geometric sequence, if possible. 44.

1 1 1 , , , 3 6 12

Solution 1 1 a ,r  3 2 1 1 a 2  3 1  31  S  1 r 1 2 2 3 45.

1 2 4 , , , 5 15 45

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1941


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution 1 2 a ,r  5 3 1 1 a 3 5 S    55  2 1  r 1    3  3 25 46. 1,

3 9 , , 2 4

Solution

a  1, r 

3  1  no sum 2

47. 0.5, 0.25, 0.125, 

Solution 1 1 ,r  2 2 1 1 a  2 1  21  1 S  1 r 1 2 2

a

Change each decimal into a common fraction. 48. 0.3

Solution 3 1 ,r  a 10 10 3 3 a 1  10 1  109  S  1  r 1  10 3 10 49. 0.9

Solution 9 1 a ,r  10 10 9 9 a S   10 1  109  1 1  r 1  10 10 50. 0.17

Solution 17 1 ,r  a 100 100 17 17 a 17  100 1  100  S  99 1  r 1  100 99 100

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1942


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

51. 0.45

Solution 45 1 a ,r  100 100 45 45 a 5 S   100 1  100  99 1  r 1  100 11 100 52. Investment problem If Landon invests $3000 in a 6-year certificate of deposit at the annual rate of 7.75%, compounded daily, how much money will be in the account when it matures?

Solution

0.0775 , n  2190 365 2190  0.0775  365  3000  1  ar  365    $4775.81 a  3000, r  1 

53. College enrollments The enrollment at Hometown College is growing at the rate of 5% over each previous year’s enrollment. If the enrollment is currently 4000 students, what will it be 10 years from now? What was it 5 years ago?

Solution a  4000, r  1.05, n  10 ar 10  4000  1.05 

10

ar 5  4000  1.05 

5

 6516 in 10 years  3134 5 years ago

54. Mobile home depreciation A mobile home that originally cost $10,000 depreciates in value at the rate of 10% per year. How much will the mobile home be worth after 10 years?

Solution a  10, 000; r  0.90; ar 10  10, 000  0.90 

10

 $3486.78 in 10 years

55. Verify the following formula for n  1, n  2, n  3, and n  4: n2  n  1

1  2  3  n  3

3

3

3

2

4

Solution

n1 3 ?

1 

1  1  1 2

11

4

n2 2 3 ?

1 2  3

22  2  1

2

4

  9 ?4 9

99

4

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1943


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

n3 3 ?

1 2 3  3

3

n4

3  3  1 2

2 3 ?

1 2 3 4  3

4 ? 9  16  36  4 36  36

3

3

42  4  1

2

4

 

? 16 25

100 

4 100  100

56. Prove the formula given in Exercise 55 by mathematical induction.

Solution Check n  1:

12  1  1

1  3

1 1

2

True for n  1

4

Assume for n  k and show for n  k  1: 1  2  3  k  3

3

3

3

1  2  3    k   k  1  3

3

3

3

3

1  2  3     k  1  3

3

3

3

12  22  32     k  1  2

k 2  k  1

2

4 k  k  1 2

2

4

  k  1

k 2  k  1  4  k  1 2

4

 k  1  k  2 1  2  3     k  1  4 2

3

4 2   k k  1  2  4  k  1 2

2

3

2

2

2

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

Evaluate each expression.

57. P 8, 5

Solution

P  8, 5  

8!

8  5 !

8!  87 654 3!  6720

58. C 7, 4

Solution

C  7, 4  

7!

4!  7  4! 

7! 7  6  5  4!  4! 3! 4!  3  2  1  35

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1944


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

59. 0!  1!

Solution 0!  1!  1  1  1

 

60. P 10, 2  C 10, 2

Solution P  10, 2   C  10, 2  

 

10 ! 10 !   90  45 8! 2! 8!  4050

61. P 8, 6  C 8, 6

Solution P  8, 6   C  8, 6  

 

62. C 8, 5  C 6, 2

8! 8!   20, 160  28 2! 6! 2!  564, 480

Solution C  8, 5   C  6, 2  

 

63. C 7, 5  P 4, 0

8! 6!   56  15 5! 3! 2! 4 !  840

Solution C  7, 5   P  4, 0  

 

64. C 12, 10  C 11, 0

7! 4!   21  1  21 5! 2! 4 !

Solution C  12, 10   C  11, 0  

65.

12! 11!   66.1 10! 2! 0! 11!  66

P  8, 5 

C  8, 5 

Solution P  8, 5 

C  8, 5 

6720  120 56

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1945


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

66.

C  8, 5

C  13, 5

Solution C  8, 5 

C  13, 5 

67.

56 1287

20 1  120 6

1287 33  2, 598, 960 66, 640

C  6, 3 

C  10, 3 

Solution C  6, 3

C  10, 3

68.

C  13, 5 

C  52, 5 

Solution C  13, 5 

C  52, 5 

69. In how many ways can 10 teenagers be seated at a round table if 2 girls wish to sit with their boyfriends?

Solution Consider each set of two people who must sit together as a single person, so that there are 8 “people” who must be arranged in a circle. This can be done in (8  1) !  7 !  5040 ways. However, each pair seated next to each other could be switched, so that the number of arrangements is multiplied by 4. There are 4  5040  20, 160 possible arrangements. 70. How many distinguishable words can be formed from the letters of the word casserole if each letter is used exactly once?

Solution 9!  90, 720 2! 2! 71. Make a tree diagram (like Figure 8-2 on page 832) to illustrate the possible results of tossing a coin four times.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1946


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Solution

72. In how many ways can you draw a 5-card poker hand of 3 aces and 2 kings?

Solution 4 4      4  6  24  3  2 73. Find the probability of drawing the hand described in Exercise 72.

Solution 4 4     24 1 3 2   2, 598,960 108, 290  52    5 74. Find the probability of not drawing the hand described in Exercise 72.

Solution 1 108, 289 1  108, 290 108, 290 75. Find the probability of having a 13-card bridge hand consisting of 4 aces, 4 kings, 4 queens, and 1 jack.

Solution 4 4 4 4         4 4 4 4  1    6.3  1012  52  6.35  1011    13  76. Find the probability of choosing a committee of 3 men and 2 women from a group of 8 men and 6 women.

Solution 8 6      3   2   840  60 2002 143  14    5

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1947


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

77. Find the probability of drawing a club or a spade on one draw from a card deck.

Solution 13  13 1  52 2 78. Find the probability of drawing a black card or a king on one draw from a card deck.

Solution 26  4  2 7  52 13 79. Find the probability of getting an ace-high royal flush in hearts (ace, king, queen, jack, and ten of hearts) in poker.

Solution 1 1   52  2, 598, 960   5 80. Find the probability of being dealt 5 cards of one suit in a poker hand.

Solution  13  4   5148 33 5   2,598,960 16,660  52    5

CHAPTER TEST SOLUTIONS Find each value. 1.

3!  0!  4 !  1!

Solution 3!  0!  4 !  1!  6  1  24  1  144 2.

2!  4!  6!  8! 3!  5!  7 ! Solution

2!  4!  6!  8! 3!  5!  7 !

 2! 

4! 6! 8!   3! 5! 7 !

 2  4  6  8  384

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1948


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

Find the required term in each binomial expansion. 3.

 x  2 y  ; 2nd term 5

Solution The 2nd term will involve (2 y )1 . 1 5! 4 x  2 y   10 x 4 y 4 ! 1!

4.

 2a  b  ; 7th term 8

Solution The 7th term will involve (b) . 6

2 6 8! 2a   b   112a2 b6  2!6!

Find each sum. 5.

3

  4k  1 k 1

Solution 3

3

3

k 1

k 1

k 1

4

4

4

k 2

k 2

k 2

  4k  1  4 k   1  4  1  2  3  3  1  24  3  27 6.

4

  3k  21 k 2

Solution

  3k  21  3 k   21  3  2  3  4   3  21  27  63  36 Find the sum of the first ten terms of each sequence. 7.

2, 5, 8, 

Solution a  2, d  3

a10  a   n  1 d  2  9  3   29 S10 

n  a  a10  2

10  2  29 2

 155

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1949


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

8.

5, 1,  3, 

Solution a  5, d  4

a10  a   n  1 d  5  9  4   31 S10 

n  a  a10  2

10  5  31 2

 130

9. Find three arithmetic means between 4 and 24.

Solution a  4, a5  24 24  4  4d 20  4d 5  d  4, 9, 14, 19 , 24

10. Find two geometric means between –2 and –54.

Solution

a4  ar 3 54  2r 3 27  r 3 3  r  2,  6,  18 ,  54 Find the sum of the first ten terms of each sequence. 11.

1 1 , , 1, 4 2 Solution 1 a  , r  2, n  10 4 10 1 1   2 a  ar n S10   4 4 1 r 12 1023  4  1 1023   255.75 4

2 12. 6, 2, ,  3

Solution

a  6, r 

1 , n  10 3

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1950


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

a  ar n 6  6  3  S10   1 r 1  31 1

10

354,288

 59,049 2 3

177, 144  9 19, 683

13. A car is purchased for $C and depreciates in value 25% of the previous year’s value each year. In terms of C, how much is the car worth after 3 years?

Solution a  C; r  0.75

ar 3  C  0.75   $0.42C in 3 years 3

14. A house is purchased for $C and appreciates in value 10% of the previous year’s value each year. In terms or C, how much will the house be worth after 4 years?

Solution a  C; r  1.10

ar 4  C  1.10   $1.46C in 4 years 4

15. Prove by mathematical induction:

3  4  5     n  2 

1 n  n  5 2

Solution

Check n  1:

1  1 1  5 2 33 3

True for n  1

Assume for n  k and show for n  k  1: 1 k  k  5 2 1 3  4  5     k  2    k  1  2  k  k  5    k  1  2 2 1 7 3  4  5     k  1  2  k 2  k  3 2 2 1 2 3  4  5     k  1  2  k  7k  6 2 1 1 3  4  5     k  1  2   k  1 k  6    k  1  k  1  5 2 2 3  4  5     k  2 

  

  

Since this is what results when n  k  1 is in the formula, we have shown that the formula works for n  k  1 if it works for n  k .

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1951


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

16. How many six-digit license plates can be made if no plate begins with 0 or 1?

Solution 8  10  10  10  10  10  800, 000 Find each value.

17. P 7, 2

Solution P  7, 2  

18. P 4, 4

7!  42 5!

Solution 4!  24 0!

P  4, 4  

19. C 8, 2

Solution C  8, 2  

20. C 12, 0

8!  28 2!6!

Solution C  12, 0  

12! 1 0! 12!

21. How many ways can 4 men and 4 women stand in line if all the women are first?

Solution 4!4!  576 22. How many different ways can 6 people be seated at a round table?

Solution

6  1 !  5!  120

23. How many different words can be formed from the letters of the word bluff if each letter is used once?

Solution 5!  60 2!

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1952


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

24. Show the sample space of the experiment: toss a fair coin three times.

Solution

H, H, H  , H, H, T  ,  H, T , H  , H, T , T  , T , H, H  , T , H, T  , T , T , H  , T , T , T 

Find each probability. 25. Rolling a 5 on one roll of a die

Solution 1 6 26. Drawing a jack or a queen from a standard card deck

Solution 44 2  52 13 27. Receiving 5 hearts for a 5-card poker hand

Solution  13     5   33  52  66,640   5 28. Rolling a sum of 9 on one roll of 2 dice.

Solution

rolls of 9:

 3, 6 , 4, 5 , 5, 4 , 6, 3

Probability 

4 1  36 9

29. A box contains 50 cubes of the same size. Of these cubes, 20 are red, and 30 are blue. If 2 cubes are drawn at random, without replacement, find the probability that 2 blue cubes are drawn.

Solution

# blue # blue 30 29 87     # all # all 50 49 245 30. In a batch of 20 tires, 2 are known to be defective. If 4 tires are chosen at random, find the probability that all 4 tires are good.

Solution  18     4   3060  12  20  4845 19   4

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1953


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

GROUP ACTIVITY SOLUTIONS Lottery Mathematics Real-World Example of Probability Powerball has been described America’s game! Since 1992 the game has inspired the country with a chance to become a millionaire, while raising $26 billion for good causes supported by lotteries. Powerball tickets are $2 per play. Tickets are sold in 45 states, the District of Columbia, Puerto Rico and the U.S. Virgin Islands. Powerball drawings are broadcast live every Monday, Wednesday, and Saturday at 10:59 p.m. ET from the Florida Lottery draw studio in Tallahassee. Drawings are also live streamed on Powerball.com. The Powerball jackpot grows until it is won.

Group Activity The Powerball lottery game is designed so that each player chooses five different numbers from 1 to 69 and one Powerball number from 1 to 26. Winning numbers are selected using two ball machines: one containing the 69 white balls and the other containing the 26 red Powerballs. Five white balls are drawn from the first machine and one red ball from the second machine. A player wins the jackpot by matching all five numbers from the white balls in any order and matching the number on the red Powerball. a. Consider the first part of the lottery game where five white balls are selected from 69 white balls in the first ball machine. Use the combination formula to determine the number of possible combinations of numbers that can result from the drawing? b. Consider the second part of the lottery game where one red ball is drawn from the second machine. How many different red Powerballs combinations are possible? c. Determine the probability of winning the Powerball jackpot if you purchase of one $2 Powerball ticket. Write your answer as a fraction. d. Powerball jackpot winners may choose to receive their prize as an annuity, paid in 30 graduated payments over 29 years, or a lump sum payment. Both advertised prize options are prior to federal and jurisdictional taxes. If you did win the jackpot how would you choose to receive your prize and why? e. Mega Millions is a similar lottery game to Powerball. Tickets cost $2.00 per play. Players pick six numbers from two separate pools of numbers—five different numbers from 1 to 70 (the white balls) and one number from 1 to 25 (the gold Mega Ball). Players win the jackpot by matching all six winning numbers in a drawing. What is the probability of winning the jackpot if one $2 ticket is purchased? Write your answer as a fraction.

Solution a.

C  69, 5  

b.

C  26, 1 

69!

5!  69  5  ! 26!

1!  26  1 !

 11, 238, 513

 26

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1954


Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 8: Sequences, Series, Induction, and Probability

c.

1 1 1   11, 238, 513 26 292, 201, 338

d. answers will vary e.

1

C  70, 5   C  25, 1

1 1  12, 103, 014  25 302, 575, 350

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1955


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