Solution Manual For College Algebra 13th Edition by R. David Gustafson, Jeff Hughes by digitaldownload87

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?

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

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

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

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

,

f , e , c , a =

, B 

d n a



g , f , e , d =

e , d , c , b , a =

t e L



 . 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}

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

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

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 

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]

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]

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]

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, )

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.

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 .

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

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.

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

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.)

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 



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

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

Solution

 x    x  x  x   x 2

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.

2x 5 x 10 x 25

Solution a. 2x 5  x 5  x 5 b.

x 10  x 5  x 5

x 25  x 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.

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

Solution

x   x m

15.

 xy   _________ n

Solution

 xy   x y n

16.

xm  __________ xn

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

Solution x0  1

18. x  n  __________

Solution 1 x n  n x

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 

Solution

 5   5 5  25 2

23. 4 x 3

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

Solution

 4 x    4 x  4 x  4 x  3

25.  5x 

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

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 

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

   

32. 2a 2a 2a

Solution

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

  

33. 3t 3t 3t



Solution

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

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

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

Solution

y 3 y 4  y 3 4  y 7

 

43. z 2

Solution

z   z 44.  t  2

23

 z6

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

Solution

t   t 6

45.

y y  5

67

 t 42

Solution

y y   y   y 5





46. a3a6 a 4

Solution

a a  a  a a  a 3

  z 

47. z 2

Solution

z  z   z z  z 2

  t 

48. t 3

Solution

t  t   t t  t 3

  a 

49. a2

12 10

Solution

a  a   a a  a 2

  a 

50. a2

Solution

a  a   a a  a 2

51.

3 x 

Solution

 3 x   3 x  27 x 3

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

52.  2 y 

Solution

 2 y    2 y  16 y 4



53. x 2 y



Solution

x y  x  y  x y 3



54. x 3 z 4



Solution

x z   x  z   x z 3

 a2  55.    b

Solution

  a

3 a2  a2     b3  b 

 x  56.  3  y 

Solution 4

 x  x4 x4   3  4 y 12 y  y3

 

57.   x 

Solution

x   1 0

58. 4 x 0

Solution 4x0  4  1  4

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

59.  4 x 

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

y 2 y 3  y 5 

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

4

Solution

y y   y   y 2

4



1 y4

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

x2 x

Solution

x   x  x 2

x x

72.

4 3

 x1  x

s9 s 3

s  2

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

 

 m3  73.  2  n 

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

Solution

   

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

 t4  74.  3  t 

Solution 3

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

a  75. 3



  

2

aa2

Solution

a   a 3

76.

2

6

 a 6  3  a 9 

1 a9

r 9r 3

r  2

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

2

Solution

 t 4   3  t 

2

t   t  t  t  t 4 3

2

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 

Solution 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

5

 

3

Solution

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

5

1

3

 x 5 y 2  81.  3 2  x y 

3

2

3

11



1 x 11

Solution 4

 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 

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

 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   

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

 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

 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 

Solution 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

3

1

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

3

1

2

1

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

2

4

2

6

1

4

1

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

2

4

2

1

6

8

14

4

12

1

2

1

5

13

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

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



5[36  4] 4  1



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





, 0

, 2

Let x  

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 

Solution

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

96.  xz

Solution

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

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

97.



 x2 z3 z y 2



Solution



 x2z3

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

z2  y 2

98.

90

32  02



z2 x2  y 2



x z

Solution



z2 x2  y 2

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

 8 3

 2  3

x3z

24

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

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

 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

5 xz 2

Solution

 5x z   25x z 2

3

5 xz

2

5 xz

6

2



5x 4 2

xz z



5x 3 z



5  2  4



5  8  81



40 40  81 81

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

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

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 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

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

8  4  5

600,000

6  10

6  0.288  109

 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

4

10  5   4 

1.8  5.3  1019

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

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

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:

 3x y  3

2

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

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

  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.

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

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

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

  x 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

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

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



x  y

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.

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

 64

2/3



 64  or  64 3

2. Write the expressions in exponential form. 9 64



 16 

Solution

  9 64

1/2



 16   16 . 4

5/4

3. Simplify

a. b. c. d.

Solution

x2  x

a. b.

x3  x

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

x4  x

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

16 9 = 144

144  12

and

5. Simplify. a.

5 2  10 2  15 2

5 x  10 x  15 x

Solution a.

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

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

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

11.

1/3

a  _________.

Solution a 1/n

a2  _________.

12.

Solution a

13.

a n b  ________.

Solution n

14.

ab a  _________. b

Solution n

x  y ________

15.

x

Solution



16.

m n

x or

n m

x can be written as _______.

Solution mn

Practice Simplify each expression. 17. 9 1/ 2

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



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

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

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

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

 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



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 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

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

42. 82/3

Solution

  2 4

82/3  81/3

43. 163/2

Solution



163/2   161/2 44.  8

   4  64 3

2/3

Solution

 8 

2/3

   8  

1/3

  2 2  4   

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

 100 46. 100 3/2

Solution



1003/2  1001/2

  10  1,000 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



64  1/2



1 3



1 512



1 343

50. 493/2

Solution 493/2 

1 493/2



 49  1/2



1 73

51. 93/2

Solution 93/2  

1 3/2



9  1/2



1 3



1 27

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



 27 



2/3

1  27 1/3     



 3 



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

4/3

Solution

 125     8 

4/3

4/3

 8     125 

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

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





59. 32 y 10 z 5

  v  1/3

 641/3 u6

1/3

 4u2v

1/5

Solution

32 y z 

1/5



1/4

32 y z  10



60. 625a4 b8



1/5



  z 

321/5 y 10

1/5



1/5

1 2 y 2z

Solution

625a b  4

61.

x y  10

1/4



625a b  4

1/4

 625

1/4

a   b  1/4

1/4



1 5ab2

3/5

Solution

x y  10

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



5/6



 645/6 a30/6 b60/6  641/6

 a b  2 a b  32a b 5

3/4

Solution

r s  8

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

  8 

2/3

x 18/3 y 24/3 

 8 

2/3

x 6 y 8 

 2  x y 2



1 6

4x y 8

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

52 b6

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

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

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

Simplify each radical expression.

71.

Solution

49  72  7 81

72.

Solution

81  92  9 73.

125

Solution

74.

125  3 53  5

64

Solution 3

75.

64  3  4   4 3

125

Solution

76.

125  3  5  5

243

Solution 5

243  5  3   3 5

77. 5 

32 100, 000

Solution 5

78. 4

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

256 625

Solution 4

256 4  4  4     625 5 5

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 

a4b8

82.

Solution a4 b8 

a b   a b  a b 2

83. 3 8 y 3

Solution 3

84.

8 y 3  3 2 y   2 y 3

27z9

Solution 3

85. 4



27 z 9  3 3z 3

  3 z 3

x4 y 8 z 12

Solution 4

x4 y 8 z 12

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

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

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

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

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

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.

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

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

100.

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



6 5 5

6 5

Solution

6 5



6 5



103.

Solution 2 x







x x

104.

y Solution 8 y

105.







8 y y

2 3

Solution

2 3



2 3





23 4 3



23 4 3  4 2

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

106.

4d 3

Solution

4d 3

107.



4d 3



4d 3 3



4d 3 3 3



5a 3

25a

Solution

5a 3

108.

25a



5a 3

25a



5a2

6c2

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.

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

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

112.



3s 5 4r 2

Solution



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

9 3

115.

Solution

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

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

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







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







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



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

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



2y 3





3 3

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 

Simplify each radical expression. 123.

Solution

124.

 

9  91/4  32

1/4

 32/4  31/2  3

Solution

125.

 

27  27 1/6  33

16x6

1/6

 33/6  31/2  3

Solution

126.









16 x 6  16 x 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

Step 2:  10 4 Step 3: 10,000

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

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

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.

3 2

Solution 4

3 2

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

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 x x  n y y

Solution x x n   y y

1/ n



x 1/ n

 1/ n

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 y m/ n

y m/ n

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

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

2

x 87 x

143. 0111/19

c. 2

144.  1

d. 0

12/19

x 86 x

145.

1

146.

x88

undefined

147.

148.

1 87

3 3

512

h. –1

Solution 141.

 16 

142.

 1024

1/4

is undefined. g

1/10

 

 210

1/10

 2. c

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.

1  1. h

146.

x88 

147.

148.

3 3



x87 87 x  x 87 x. b 

x 86



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

 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

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

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

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

Solution yes, monomial, 5th degree 23.

15 Solution yes, monomial, 0th degree

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

Solution

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



 

28. 2 x 4  5 x 3  7 x 3  x 4  2 x



Solution

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

Solution

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

 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

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

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

 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

 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

 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

 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

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

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

 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 

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





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 

Solution

a  6  a  6a  6 2

 a2  6a  6a  36  a2  12a  36

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

50.  t  9 

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

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 

Solution

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





60. 4a  5b 4a  5b



Solution

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

61.

 3m  4n 3m  4n Solution

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

62.  4r  3s 

Solution

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





63. 2 y  4 x 3 y  2 x



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



Solution

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

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





66. 8a2  b  a  2b 

Solution

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



67.  5z  2t  z 2  t



Solution

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



68. y  2 x 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

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

 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 

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

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

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

 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 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











 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



74.  4r  3s  2r 2  4rs  2s2



Solution

 4r  3s   2r  4rs  2s  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

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

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



77. 3  x



Solution

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



78. 5 y  2 x



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

 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  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 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

n    n

 15 x

 10 x y  15 x y 4n

2 n   2 n 

 10 x

9 a



nn

 15 y 2 n

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

2n 1

 2a

3n 1

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





86. a n  5 an  3



Solution

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





87. 2r n  7 3r n  2



Solution

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

 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

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



92. x 3/2  y 1/2



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





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



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  

14 y 2 3 Solution 14 y 2 3

97.

Solution

96.

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

x x 3

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

Solution x x 3

98.





x 3

x 3 x 3





x x 3 x2 

 3

  x x  3

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

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.



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

 3  2  6  3  3 2

13

  

2  6  3 3 2  2  6  3 3





2 3 3 2  6 2

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  3  6  2  2 2

12

3 6 2 2 1   3 6 22 





 6  2  3 2 103.

x

x

Solution x x

104.

y y



x x



x x

y y

x 2  xy  xy 



 x  y 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



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



  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

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

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



a  ab  ab  b 2

ab



a  2 ab  b

Solution

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

x3 x  3

x3 x  3

x 3 x



 x 3  x 3



110.

x3 x



 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h2

 2  h  2 h

 2  h  2



1 2h  2

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.



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

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

 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 

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

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 

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

2x 2  0 x 

2x  2x 2

2x 

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

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

Solution x

x  x  1 x  2x 2  4 x  3 2

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

 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

 2x 3  x 2  0x  9  2x 3

4  x

128.

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

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 

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

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 

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

134. Rationalize the denominator.

4 2 3

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 =





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

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

∙

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

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.

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

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.

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

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

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

3. Fill in the boxes to complete each factorization.



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

8 x  11  1  8 x 11

Solution





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

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   

12 x 2  5 x  2   3 x  2   1

Solution a.

x 2  10 x  9   x  9 x  1

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      

125 x 3  8z 3      

27 y 6  64 z 3      

Solution a.



49 x 2  100z 4   7 x   10z 2 2



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

125 x 3  8z 3   5 x    2z 

27 y 6  64z 3  3 y 2



6. Fill in the box. x  7

   4z  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 12. x  2xy  y  _________

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

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

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



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

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.

 x  z   25 2

Solution

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

32.  x  y   9 2

Solution

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

34. a 2  12a  36

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

35. b2  10b  25

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

2 36. y  14 y  49

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

37. m2  4mn  4n2

Solution

m2  4mn  4n2   m  2n m  2n

  m  2n

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 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 

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  



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



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 





Solution

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

64. z 2   y  3 

Solution

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

101

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

65.  x  y    x  y  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

66.  2a  3    2a  3  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

4 4 67. x  y

Solution

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

x4  y 4  x2

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

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 

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

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 

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

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

 9 y 2n

Solution

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

4 x 2n  9 y 2n  2 x 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

 11 y n  6

Solution





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

96. 16 y



 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







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



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

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

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

102. a 6  b6

Solution

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

a6  b6  a2

2 2

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

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 2

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

  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

  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

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

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

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

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

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

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

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

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

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

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



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

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

 5x  3

  x  1



5 x 10 x



10 x



5 x 10  10 x





5 x 10  2 x



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

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

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

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

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

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.

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.





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

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

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

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

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

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

115

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

Solution a.

 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

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

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

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

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





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



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

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

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

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

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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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

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

x 3  23

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



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

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.



 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

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 

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

x1

61.

3 x2  x3 x3

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

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  4 a  4  a  4 a  4  a4





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

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 2





 x  2 x  2  x  2 x  2



 x  2 x  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 

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 

 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





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

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

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



 x  2 x  1 x  3

0

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

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



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



 x  5 x  4  x  5 x  4 



 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

 x  5 x  4  x  5 x  4 

 x  3 x  43 x 2  58 x  697 4

 x  5 x  4  x  5 x  4 

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

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    y  x 1 x

1 y

1 x

xy  xy 

1 y

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 

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  

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.



abx

    abx    abx    3a x  4a b  a 3 x  4ab  3a b

4 a2 x

3a b

abx  b1  ax1 

abx  b1   abx  ax1 

ax  b

x y

x2 1 y2 Solution

1  xy 2

x y2

1





y 2 1  xy y

  y  1  y    y  xy  2

x y

  1 y    y  1 2

x y2

x y 2

y  y  x

 x  y  x  y 



y xy

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







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

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

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

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

 

 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

Solution 2 x  x  3  3  x  2 1

 x  3  x  2 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

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

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

113.

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

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

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

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



k1k2



k1k2

 kk  kk   k k 1 2

1 k1

1 k2

1 2

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 R1

1 R2

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.

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

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

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

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

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.

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 

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

Simplify each expression. 34. n2 n4

Solution n 2 n 4  n 2  4  n6

148

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

 

35. p 3

Solution

p   p 3



36. x 3 y 2



32

 p6

Solution

x y   x   y   x y 3

 a4  37.  2  b 

Solution

   

3 a4  a4  a 12   2   3 b6 b  b2



38. m 3 n0



Solution

 m n    m  1  m  m1 3

3

6

 p2q2  39.    2 

Solution

   

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

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

 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

 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 

 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 

 xy 3   y2  y4        3   3x 2  9x 4  3 x y    3

Solution

 2m2 n0    2 1    4m n 

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

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

1.05  104

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

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 

1/2

  y 

 x 12

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

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

63. 32 3/5

Solution 1

323/5 

 16  64.    81 

3/5



32  1/5



1 8

3/4

Solution

 16     81 

3/4

 32  65.    243 

 

 

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

2/3

Solution

 8     27 

2/3

 27     8 

2/3

   

27 1/3 272/3 32 9  2/3    2 8 22 4 81/3

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

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.

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

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 

 x y2

75.

Solution 3

76. 4

x3  x m8 n4 p16

Solution 4

77. 5

m8 n4 p16



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.

24 x 4  3 3x 4

Solution 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

155

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

Rationalize each denominator.

81.

5 Solution

7 5 82.





35 5



8 8 Solution

8 8 83.







8 2 16

8 2 2 2 4



1 3

Solution

1 3

84.



1 3



3 3

4 4



4 2

2 3

Solution

2 3



2 3





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

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

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



Solution

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





97. 4a  2b 2a  3b



Solution

 4a  2b 2a  3b  8a  12ab  4ab  6b  8a  8ab  6b 2



98.  z  3  3z 2  z  1



Solution

 z  3  3z  z  1  3z  z  z  9z  3z  3  3z  10z  2z  3 2







99. an  2 an  1

Solution

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

100.

 2  x

Solution

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

101.

 2  1 3  1 Solution

 2  1 3  1  6  2  3  1

102.

 3  2 9  2 3  4 3

Solution

 3  2 9  2 3  4  27  2 9  4 3  2 9  4 3  8  3  8  5 3

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



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

2x x 2

Solution

2x x 2

106.

2

Solution

105.



x

x

x 2



x 2 x 2



 x  2  2 x  x  2 x 4  x 2

Solution x x

y y



x x

y y



x x

y y



x 2  xy  xy  y

 x  y 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

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 

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 

2x  3 2x  7 x  8x  3 3

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

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

2x 3  3x 2  2x  2x

2x 3  3x  3x

3

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

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

123. 121z 2  4  44 z

Solution 121z 2  4  44 z  121z 2  44 z  4   11z  2  11z  2    11z  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

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

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

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

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



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

x 2  x  1

x1



144.





x 2  x  1

5 x 3  5 x 2  15 x  15 x  x  1 2



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

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 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

167

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

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

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

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

a 3

Solution

a a   a   a  a 1

2

a3

2

a3

2

a3

 x0 x 2  14.  2   x 

Solution 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

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

2/3

Solution

 8t 6   27 s9    22.

2/3

 27 s9     6   8t 

2/3

   8 t  272/3 s9 2/3

2/3

 27  s  9s  8  t 4t 1/3

1/3

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

170

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

Solution

x x 2



x 2

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

x 2





26. Rationalize the numerator:

x

x

Solution

 x  y 2

x

x



x

x



x



x  xy  xy  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

 a2  7



28. 3a 3 b2

 2a b  3

Solution

 3a b  2a b   6a b 3





29. 3 x  4 2 x  7



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

 a2n  an  6 31.

 x  4  x  4  2

Solution

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

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

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 

2x  1 2x  3x  0x  1 3

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

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

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

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.

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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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

1.688

1.929

1.205

0.655

0.335

0.224

0.168

0.135

0.112

0.096

0.084

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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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.

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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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 



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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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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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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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

 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

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

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

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 

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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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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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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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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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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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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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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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

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

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

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)

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

 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

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

 5x  6



3x  2 x

 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

 x



2x x

 5x



4 2   10 5

x  2 x

 6x  5

197

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 3x





 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



 8

3 x

 4



2(3 x  2) ( x  2)( x 2  2 x  4)

Solution 3x  5



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

 8

( 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

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

 13a  48

Solution 4 2



2 a





 18a  32



 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

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

 6  5a



2  3a a

 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

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

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

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

f (Pn  L)  f 

si f

f (Pn  L)  si f (Pn  L) si  Pn  L Pn  L f 

si Pn  L

203

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

mMg

101. F 

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

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

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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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

111. R 

1 1 r1



1 r2

 r1

; r1

Solution R  R 

R 

1 1 r1



 r1

1 r2

r1r2 r3 (1)



r1r2 r3 r1  r1  r1 1



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

Solution

R  R 

R 

1 1 r1



1 r2



 r1

r1r2 r3 (1)

r1r2 r3 r1  r1  r1 1



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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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 x 4

  1 x 3



  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

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

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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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

Assessment

Program development and instruction

Professional knowledge and legal issues

AVERAGE SCORE

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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

.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 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

 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



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.

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 

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.

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

239

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

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

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

 7i 



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

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

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 

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

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 

 3  4i  4  6i  5  12i  4  2i



58. 1 

Solution

1 

 

81  8 

 

121

 

81  8 

 

121

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

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



Solution

2  3i 



66. 3  4i

 2  3i 2  3i   4  12i  9i 2  4  12i  9 1  4  12i  9  5  12i



Solution

3  4i 



67. 11 



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 

 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



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

 i



23  i  9   1



23  i  10



 3  i  5

 

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

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 



6i  15 4  25i



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.

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 

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

 

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



3  i  4  i 2 

4  i 24  i 2 



12  3i

2  4i  i 2 16  2i 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

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 

6 2

89. i 38

Solution

i  i  1 i  i  1

i 38  i 36 i 2 

9 2

90. i 99

Solution

i 99  i 96 i 3 

 i  i  1 i  i  i 4

24 3

91. i 87

Solution

i 87  i 84 i 3 

 i  i  1 i  i  i 4

21 3

92. i 44

Solution

 

i 44  i 4

 111  1

93. i 100

Solution

 

i 100  i 4

 125  1

94. i 201

Solution

  i1 ii

i 201  i 200 i  i 4

95. i 6

Solution

i 6 

1 i6



1  i2 i6  i2



i2 i8



i2  i 2  1 1

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

i 10 

i 10

1  i2



i 10  i 2



i 12

i2  i 2  1 1

98. i 31

Solution 1

i 31 

99.

1  i



 i



i i



i  i 1

1 i3

Solution 1 i 100.



1  i i

 i



i i

i  i  i 1



3 i5

Solution

3 i5 101.



3  i3 i5  i3

3i 3



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

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 

106. 5  i

Solution

5  i  107. 7 



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 



1 1  i 2 2 Solution

1 1  i  2 2

 1    2

 1     2



1 1   4 4

1  2

2 2

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.

 1    2

 1     4

1 1   4 16



5  16

5 4

6i

Solution

02   6

6i  0  6i 



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 1  i 



 1  i 

12   1



 3     10 

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 



81 9  100 100



90  100

9 3 10  10 10

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

116.

 6     5

45 3 5  5 5

5i i  2 Solution

5i 5i   i  2 2  i

5i  2  i 

2  i 2  i  4  i



5i  2  i  5

10i  5i 2 5



 1  2i

117.



1   2



1  4 

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

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

4    5



9 16  25 25



25  25

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

 4a  3i 4a  3i 

121. 25p2  36q2

Solution 25 p2  36q2  5 p



 36q2

  5p  6qi   5p  6qi 5p  6qi  2

122. 100r 2  49s2

Solution 100r 2  49s2   10r 



 49s2

  10r   7si   10r  7si 10r  7si  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

124. 12b2  75c2

Solution



12b2  75c2  3 4b2  25c2

  32b  25c   32b  5ci   32b  5ci2b  5ci 2

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

126. 64a4  4b2

Solution

64a4  4b2  4 16a4  b2

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

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

1  i   i  1  i 1  i   i  1  i  i  i2  i  i

i   i  i 2  i  1  i

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.

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.

1 100 3  10i 3

125  5i

Solution False. 141.

 i

125  5

  i

Solution True.

  i i     i i i i i2





142. 2  3i

 8  27i

Solution



False. 2  3i



 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

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 



 10 y 2

  9x   10 yi   9x  10 yi 8x  10 yi  2

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

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    5



4 25

5. Simplify. a.

4 

( 4)2  4(1)(5) 2(1)

4 

( 4)2  4(1)( 5) 2(1)

Solution a.

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

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

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

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

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 

7 2

19. x2  x  6  0 Solution

x2  x  6  0

 x  2 x  3  0 x  2  0

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

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

x  2  0

2 x  5

x  2

 52

x  2

x 

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

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

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

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

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

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

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

x  2i

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

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

7 9

x    x  i x  

7 9

7 9 7 i 3

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 

x  i x 

11 25



11 25

x   

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 

26 2

or x  

13 2

x  

2 2



13 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

270

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 



15 2

x  i



x  

15 2

x  i

30 i 2

x 

x   



30 i 2

48. 5x2  11 Solution

5 x 2  11 x2   x 



11 5

x  i

x   

or 5



x  i

x 

55 i 5



 8  0

11 5

x  

11 2

11 5



5 5

55 i 5

49. x  1

Solution

 x  1  8  0 2  x  1  8 2

x  1 

x  1   8 x  1   4

x  1  2 2



x  1  2 2

  98  0

50. y  2

Solution

 y  2  98  0 2  y  2  98 2

y  2 

49 2

y  2  7 2 51.

y  2   98 y  2   49 2 y  2  7 2

 x  1  12  0 2

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

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

Solution

 y  2  120  0 2  y  2  120 2

y  2 



200

y  2   200

y  2  i 4 30

y  2  i 4 30

y  2  2i 30

y  2  2i 30



53. 2x  1

 27

Solution

2 x  1 2x  1 

 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

Solution

5 y  2  48  0 2 5 y  2  48 2

5y  2 

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

272

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities





55. 5x  1

 8

Solution

5x  1 5x  1 

8



 8

or 5 x  1   8

5 x  1  i x  

5 x  1  i

1 2i 2  5 5

x  

1 2i 2  5 5

  48  0

56. 7 y  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  





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

10 x  1 

10 x  1 

x 

1  5 10



 14



58. 7 3x  4

x 

1  5 10

Solution

7 3 x  4

 14

3x  4

 2

2 2

3x  4 

or 3 x  4   2

3 x  4 

4  3

x 

3 x  4 

4  3

x 

273

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities





59. 3 8x  11

 9

Solution

8 x  11 

38 x  11

 9

8x  11

 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

Solution

46 x  5

 3  19

46 x  5

 16

6x  5

 4

2 2

6x  5 

4

6 x  5  2i

6 x  5   4 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

 x 2  6 x  32  x 2  6x  9

62. x2  8x

Solution

x 2  8x   21 8

 x 2  8x  42  x 2  8x  16

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

 x2  4x  4 64. x2  12x

Solution

x 2  12 x   21  12

 x 2  12 x   6

 x 2  12 x  36 65. a2  5a

Solution 2

5  a  5a     2 25  a2  5a  4

9  t  9t    2 81  t 2  9t  4

 5a   21 5

66. t 2  9t

Solution

 9t   21 9

67. r 2  11r

Solution r 2  11r   21  11

 r 2  11r 

   r  11r 

121 4

 s2  7s 

   s  7s 

49 4

11 2

68. s2  7s

Solution s2  7 s   21  7

69. y 2 

7 2

3 y 4

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 

 y2 

3 3 y    4 8

 y2 

3 9 y  4 64

3 p 2

Solution 2

 1  3  3  p     2  2  4 

3 3  p    2 4

 p

 p2 

3 9 p  2 16

1 q 5

71. q2 

Solution

 1  1  1  q      5  2  5 

72. m2 

 q

 1  1  q    5  10 

 q2 

1 1 q  5 100

2 m 3

Solution 2

 1  2  2  m      3  2  3 

2 2

 m

 1  2  m    3 3

 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

x  6 

x  6  2 7 x  6  2 7

 28 x  6   28 x  6  2 7 x  6  2 7

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 

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

 12

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

 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

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

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 

 3

or x  2   3

x  2 

x  2 

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

y  

11 5 3  i 2 2

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

x 

5 3 7  i 2 2

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

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 

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

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



7 9

7 3 2  x  3

x  x 

2   3

7 9

2 7   3 3 2  x  3

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



19 9

or x  x  19

2   3

19 9

2 19   3 3 2  x  3

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



x 

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



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

281

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x 



b 

 18  b2  4ac  2a

18 

180



63  i 5 18  6i 5  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



14  4230 14   22 2



27  i 4

196  240 4

  7  i 11  7  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

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

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

36  12 6  24  6 6

  3  6 3

96. 2 x  x  3  1

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

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

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

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

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 



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 

x2 a2  x2

a2 b2 a2  x 2





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





 x2

 1; x

Solution

x2 a2



 1

b2 x2

 1 

a2 x2



b2  y2

b2 a2 b2  y 2



x2 

b 2

x  

x2 a2





a b2  y 2

x  

113.



b  y  2

 1; a

Solution



 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

b x 2

 y2

bx b2  y 2 b2  y 2



 a2  a  a

288

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

114.

x2 a2



 1; b

Solution



 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

a2 y 2

b2 

x 2  a2 a2 y 2

b   b  

x 2  a2 x 2  a2

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



y  y 2

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

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  

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

 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

 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

 4949

 1764  1764  0 one repeated rational number

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

 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

 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

 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

 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:

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 

x  3  0 x  0

15 x

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

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

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



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

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

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

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

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

y  2  0 y  2

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

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

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

 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

 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

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 

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

4a2

b   b2  4ac  2



4a2





b2  b2  4ac 4a2

  b  b  4ac  4ac  c 2

4a2

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.

300

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

16t1t2  16 

v0 

and t2 

v02  64h



v0 

v02  64h

32 v0 

v02  64h

. 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

301

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

 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

 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.

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

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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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

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 

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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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

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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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

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

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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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 

 64

144  48 x  4 x

 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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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

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 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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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 

 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 

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

 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 



10 26 

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

 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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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

 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

r  10  0

r  20

First trip Return trip

r  10 Rate

Time

Dist.

40 r

r  10

40 r  10

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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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

r  40  0

r  30

First trip Return trip

r  40 Rate

Time

Dist.

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

r  70  0

r  60

Slower trip Faster trip

r  70 Rate

Time

Dist.

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

r  75  0

r  50

r  75 Rate

Time

Dist.

25 r

r  25

25 r  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 

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

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

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

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

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

 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 

Chloe

240

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

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

Laura

400

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

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

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

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

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

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

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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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

x  20  0

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 

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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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 

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



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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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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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

 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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Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

4. Simplify:  32 5

Solution

32 5   5 32  3

  2

 8

5. If u  x 3 what is u2?

Solution 1

If u  x 3 , then u2 

x   x 13

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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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

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

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

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

y 2  25  0

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

y2  4  0

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

y 2  18  0

y 2  18 y  

y2  5  0 y2  5

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

x2  2  0

x 2  49

x2  2

x  7

x   2

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

x2  9  0

x2  1  0

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

x 2  16  0

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

4y2  9  0 y2 

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

333

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

9 y 2  25  0 y2 

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

x2  4

1 or 3

x  2

3x  1 x 

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

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

x 2  49  0

x2  1  0

x 2  49

x2  1

x  7

x  1

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



0

2m1 3  1  0 m1 3 

 1 m1 3  2

or m1 3  2  0 m 1 3  2

1 2

m     3

1 2

m   2

m  8

1 8

m 

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    

t   243 32

1 t   243

3 2

1 3

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

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

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

p 1 2  5

p1 2  4  0 p1 2  4

 p   5

 p   4 

p  25

p  16

p  25 does not check and is extraneous. 12

33. 6p  p

 1

336

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

2 p

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

1 2

1 4

p 

p    

1 4

p 

1 3

1 9

p 

does not check and is extraneous.

34. 3r  r 1 2  2

Solution

3r

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 3

4 9

r   1 2

r  1

4 9

r  r 

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



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

t 

1 2

t   2 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

z   8 32

z3  1

z 3  64

z  1

z  4

z  1 does not check and is extraneous.

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

x 1  8

 8 x 1  2

x  1

1

x 1  2

 8

x 

x 1  2  0

x 

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



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

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

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

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

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

a  3  25 a  28 The solution checks. 43. 3 x  1 

Solution

3 x  1 

3 x  1   6  2

9 x  1  6 9x  9  6

9 x  3 x   31 The solution checks.

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  3   2 x  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

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

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





4 x 2  3  16 x  3 4x

 12  16 x  3

4 x 2  16 x  15  0

2x  32x  5  0

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.

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

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

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

5  x 2   x  1

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

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

x  37  x 2  10 x  25 0  x 2  11x  12

0   x  12 x  1 x  12  0

x  1  0

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

10  x  x 2  8 x  16 0  x2  7 x  6

0   x  6 x  1 x  6  0

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

3z  1  z 2  2z  1

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

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

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

x2  4x  4

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

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 

  x  1

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

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 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



Solution

x2  1 3x  5    



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

The solutions check.

345

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

61.

2p  1  1 

Solution

2p  1  1 

 2p  1  1   p  2

2p  1  2 2p  1  1  p p  2  2 2p  1



 p  2

 2 2p  1

p2  4 p  4  42p  1



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



r  4  4 r  2  r  2 4 r  2  6

4 r  2  6 2

16r  2  36 16r  32  36 16r  4 r 

4 16



1 4

The solutions check. 63.

x  3 

2x  8  1

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

x  3  2x  8  2 2x  8  1 2 2x  8  x  6

2 2x  8   x  6 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

x  2  0

x  2

The solution checks. 64.

x  2  1 

2x  5

Solution

x  2  1 

2x  5

 x  2  1   2x  5 2

x  2  2 x  2  1  2x  5 2 x  2  x  2

2 x  2   x  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

x  2  0

x  2

x  2

Both solutions check. 65.

y  8 

y  4  2

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

y  4  4

y  4  8

4 y  4   8 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

z  5  22 z  5  4  z  3

12  4 z  5



122  4 z  5 144  16 z  5



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

2b  3b  1  b  1  b  2

3b  4  2 2b2  5b  3  b  2 2b  6  2 2b2  5b  3

348

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

2b  6



 2 2b2  5b  3





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

5a  1

   5a  1 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



4 3a2  3a

a

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  

 22

b  8  4 b  8  4 

349

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

 b  8   4  b  2

b  8  16  8 b  b 8 b  8 b  1

 b  1 2

b  1  The solution checks.

x  19 

70.

x  2 

Solution

x  19    

x  2  x  2  

x  19  x  19 



 3

x  2  3 x  19  3 

x  2

 x  19  3 

x  2



x  19  9  6 x  2  x  2 12  6 x  2 2 

x  2

22 

 x  2

4  x  2 6  x  The solutions checks. 71.

7x  1  4

Solution 3

7x  1  4

 7 x  1  4 3

7 x  1  64 7 x  63 x  9 The solution checks.

350

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

72.

11a  40  5

Solution 3

11a  40  5

 11a  40   5 3

11a  40  125 11a  165 a  15 The solution checks. 73.

x3  7  x  1

Solution 3

x3  7  x  1

 x  7    x  1 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.

x3  7  1  x

Solution 3

x3  7  1  x 3

x3  7  x  1

 x  7    x  1 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.

351

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

75.

8x 3  61  2x  1

Solution 3

8x 3  61  2 x  1

 8x  61  2x  1 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.

8x 3  37  2x  1

Solution 3



8x 3  37  2 x  1

8x 3  37

  2x  1 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

30t  25  626 30t  600 t  20 The solution checks.

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

3z  1  16 3z  15 z  5 The solution checks. 79.

2 x  11 

2x  11 

Solution 5

 2x  11   14  5

2 x  11  14 2 x  25 x 

25 2

The solution checks. 80.

x 2  24  1

Solution 5



x 2  24  1

x 2  24

  1 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

353

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Step 2:

 2 x  1    x  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

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

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?

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



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).

 

P 0.02

To the nearest 10 watts, find the power generated when the velocity of the wind is 31 mph.

Solution

v 

P 0.02

31 

P 0.02

 P   31   3  0.02    P 29791  0.02 297910.02  p

600  P The power generated is about 600 watts.

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 

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 

A  1 P

0.065 

4000  1 P

1.065 

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.

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



 w  56.25 2 1

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.

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

97. To solve the equation 5 y 3  4 y 3  1  0, we can make the substitution u  y 3 .

Solution True. 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 .

359

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

True.



999

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

x  3  0

3x  2

x  3

x 

2 3

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.

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, 

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,  

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

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

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

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

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 

 18

12 



9 

6, 9

40. 3  2 x  2  6

Solution 3  2x  2  6 1 

 4

1 2

 2   21 , 2



41. 9 



x  4  2 2

Solution x  4  2 2 18  x  4  2 9 

22 

8 

 8

 22  8, 22

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 

 38  32, 38

4  x  5 3

43. 0 

Solution 4  x 0   5 3 0  4  x  15 4 

x

 11

4 

 11

11 



4  11, 4

5  x  10 2

44. 0 

Solution 5  x  10 2 0  5  x  20 0 

5 

x

 25



 25  5, 25

45. 2 

1  x  10 2

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



 21  5, 21

46. 2 

1  x  10 2

Solution 1  x 2   10 2 4  1  x  20 5 

x

 19



 19

19 



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   , 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  0, 

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  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   , 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



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

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

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

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,  

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

 7 x  12

–5

–3.5

–0.25

+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

+12

–10

+12

Solution set: 1, 12

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

2.5

–0.25

, 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

 5x  6

3 2

–2

3 2 2 3

–6

2 3



  , 

Solution: ,  32 

2 3

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

 5x  6

–4

–2.5

–0.25

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

 9 x  20

–6

–4.5

–0.25

+20

Solution:  , 5   4,  

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

 1    , 3  2 

–34

3, 

 1 Solution:  ,    3,  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

 1  2,  3 

–2

1   ,  3 

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

 5x  1

1 2

−1

1 3

–0.4

–0.04

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

–4.5

–0.25

+20

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

value of

test number

interval

1 3

1 ,x 2

 5x  1

1 3

1 2

0.4

–0.04

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 2

–3

1, 

interval

,    , 1 3 2

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  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

 24 x  16

–2

+16

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

test number

,   ,  4 3

4 3

Solution: x 



  , 

intervals: , 43 ,

interval

4 , x 3

value of 9x

 24 x  16

+16

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

 2    ,   5 

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

, 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

 6x  9

Solution:  , 

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, 

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, 

value of x2  7 +2 –7 +2

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  





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





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, 

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

–

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

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

–

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

sign of x 2  4 x 9

+ – + – +

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

 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

–

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

 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

–

x  x  12

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0.

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

–

interval

, 2

2,    ,   ,  1 3

1 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

–

x  2x  8

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

–

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

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

–

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

–

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

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

–

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

–

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

–

, 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

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

–

11 5

–

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

–

2,    ,  5 4

5 4

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

–

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

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 

–

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

 1

 1

Solution 6 x

 1

 1

 1  0  1 x2  1 6   0 x2  1 x2  1 7  x2  0 x2  1 7  x2  0  x  1 x  1 x

factors  0: x   7, x  1





 

intervals: ,  7 ,  7, 1 , 1, 1 , 1,

  7, 

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, 

x2  1 –

−2

–





Solution: ,  7  1, 1 

100.

7  x2

−3





sign of

 7, 

 1

x2  1

Solution 6 x

 1

 1

 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

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  x2 x2  1

−3

–

−2

–

397

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

 



Solution:  7, 1  1,



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?

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?

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

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.

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  2w P  80 , or . 2 2

200

180  80  P  80  200  80 100

 P  80 

120

100 2



P  80 2



120 2



length



The length is between 50 and 60 inches.

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



 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



25 4



P 4



60 4



152

 



625 16

 Area  225

25 4

 P 4

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

 10t  9  0

t  1t  9  0

factors  0: t  1, t  9

intervals:  , 1,  1, 9, 9, 

403

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

interval

value of

test number

t  10t  9

–7

, 1 1, 9 9, 

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

 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

+36

–8

+10

, 3 3, 12 12, 

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.

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

–100

+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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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

–

, 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

3x  1  8 4

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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

3x  1  8 4 3 x  1  32 3 x  31 3x 

31 3

4. Solve each equation. a.

6x  2  2x  5

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

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

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

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

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

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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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

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

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.

x  3  5 4 Solution

x  3  2 4 An absolute value can never equal a negative number. no solution

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

x  7  0 9 x  7  0

x  7 39.

x  7

4x  2  3 x Solution 4x  2 x

 3

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 

Solution x  3  x

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

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

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

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 

3, 9

 9

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, 

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 

 9

1 



1, 

9 5

59. 3 x  5  1  9

Solution 3x  5  1  9 3x  5  8

8  3x  5  8 13 

 3

 13 3

 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, 

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

x  3

x  3

, 3  3, 

62. x  3  0

Solution

x  3  0 0

 x  3 0



 3

x 3, 3

63.

5x  2  1 3 Solution 5x  2 3

1 

 1

5x  2 3

 1

3  5x  2  3 5 

 1

1 



1, 

1 5

64.

3x  2  2 4

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

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

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



9  16



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.

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, 

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 

 2

x   21

x   21

2 

 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 

 4

3 2

1 

 2

x 

(1)

(2) (1)

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 

 9

 73

 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 

 9

3 2

x  1

 47 



x 

(1)

9 4

(2)

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

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 

 17

(2) (1) (2)

(1) and (2)

74. 3 



7,  1   11, 17

x  3  5 2

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

(2) (1) (2)

(1) and (2) 

75. 10 

7, 3   9, 13

x  2  4 2

Solution

10 

x  2  4 2

and

1 x 2 2  4 x  2  4 2 x  2  8 x  10

(1)

x  2  4 2 x  2  8 x  6

10 

x  2 2

 2

 10

x  2  10 2 20  x  2  20

10  18 

 22

(2)

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

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 

 13

(2) (1) (2)

(1) and (2)

77. 2 



17, 5   1, 13

x  1  3 3

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

(2) (1)

(2) (1) and (2)

78. 8 



10, 7   5, 8

3x  1  2 2

Solution

8 

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 

 15

x  1

x   53

 17  3

 5

(1)

(2)

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 

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

x 2  2x  1  x 2  4 x  4 2x  3 x   32 Solution:  32 , 



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

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

9x 2  12x  4  9x 2  6x  1 18x  3 1 6

x 



Solution: , 61  83. x  1 

Solution

x  1  x

 x  1  x2 2  x  1  x2 2

x 2  2x  1  x 2 2x  1



x   21

Solution: ,  21



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

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

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

9x 2  12x  4  9x 2  6x  1 18x  3 x  Solution:

 , 

1 6

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.

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 

 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 

 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 

 1.1

0.6  0.5 

 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 

 0.265

0.25  0.015 

 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.

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 

 72

55  17 

 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 

 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 

 26.46

a. 24.76% and 26.45% are within the range.

b. The error is less than 1%.

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.

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, 

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 

 x  1

 8

Solution

1  4 x  1 restrictions: x  1

437

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

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

438

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

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

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

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 Pn  l

f 

si Pn  l

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.

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.

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.

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%.

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

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



Solution

8  3i 

 8  3i 8  3i   64  24i  24i  9i 2  64  48i  9 1  55  48i





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



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

 i



2  4i 2 4    i 5 5 5

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

44. Simplify: i 103 .

Solution

i 103  i 100 i 3 

i  i  1 i  0  i 4

25 3

45.  3

Solution



2 i

 

2 i 3

i  i

 

2i i

 

2i  0  2i 1

46. 3  i

Solution

3  i 

47.

32   1



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

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

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

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 

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

 32

Solution

4z  5  32 2 4z  5   32 2

4z  5  4 2 4z  5  4 2 z 

5  4 2 4

450

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities





56. 5x  7

 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 

 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 

 1

1 or x  3   1

x  3  1 x  2

x  3  1 x  4

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 

 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  



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.

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 

 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

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

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

 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

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

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

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

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

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

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

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

x2  1  0

or x 2  1  0

x2  0

x2  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

x 2  36  0

or x 2  1  0

x 2  36

x2  1

x  6i

x  1

81. a  a

6  0

Solution

a

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

a   3

a  4

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

Both answers check.

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

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

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

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  9   3  a  2

a  9  9  6 a  a 0  6 a



02  6 a



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

5  x  16  8 5  x  5  x 8 5  x  16  2 x

8 5  x   16  2x 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  5   1 



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 y   4 2

4 y  16 y  4 The solution does not check.  No solution. 89.

4x  9  3  2

Solution 3

4x  9  3  2 3

4 x  9  1

 4x  9  1 3

4 x  9  1 4x  8 x  2 The solution checks. 90.

x  2  3  5

Solution 4

x  2  3  5 4

x  2  2

 x  2  2 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

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 

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

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

–8

Solution:  ,  2  4,  

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

–4

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

–3

Solution: 1, 3

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 2

–3

1, 

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

–

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,  

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

–

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

–

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0.

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

 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

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.

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

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

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 

5, 1

 1

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

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 

 1

x  1

x  2

 72 



1 2

(2)

(1) (1)

(2) (1) and (2)



 , 2  1,  7 2

1 2

115. 0  3 x  4  7

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 

 11

4 3

1 



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  1 Solution x

x  x  1

 2

restrictions: x  0, x  1 2.

4  3  7 3x  2 Solution

4  3  7 3x  2 restrictions: x  23

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 

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

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%.

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

475

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution

 

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

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

x  7 

 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

 5x  14



5  439 5   23 2

25  108 5  133  6 6

4 x

 5x  6

Solution

3 x

 5x  14 3

 x  7 x  2



4 x



 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:

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



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 

 1     10 



9 1  100 100



10  100

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

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

1 2

 p   1 15

1 p   32

p  1

Both answers check.

x  5  12

29.

Solution

x  5  12



x  5

  12 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



2z  3  1  2 z  1  z  1 2 z  1  z  1

2 z  1  z  1 2

4 z  1  z 2  2z  1 4 z  4  z 2  2z  1

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 

3 

 8

 4  3, 4

34. 1  x  3 x  3  4 x  2

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

–8

+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, 

481

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

interval

test number

sign of xx  21

–3

–

, 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

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 

 12

9 

 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 

483

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

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

Solution





 

81a4

81 a4

 

2    9a2   

 9a2

Solution

 

81 a4

2   81 a2   

a b  3 2

 81a2

2

Solution

a b  3 2

2

 4x4  10.   2   12x y 

  b 

 a3

2

 a6b4

2

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 

 12x 2 y     4 x 4   

 3y     2 x 

9y2



2

Solution

 4x0 y 2   2   x y 

2

 x2 y     4 x0 y 2   

 x2     4 y   



x4 16 y 2

 4 x 5 y 2  12.  2 3   6x y  Solution

 4 x 5 y 2   2 3   6x y  13.

 2y5     3x 3   



4 y 10 9x 6

a b ab  12

Solution

a b ab   ab a b  a b 12

14.

a b c 12 12

3 3

Solution

a b c  abc 12 12

Rationalize each denominator and simplify. 15.

3 3 Solution

3 3 16.



3 3 3 3



3 3  3

2 3

485

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 1: Equations and Inequalities

Solution 2 3

17.



2 2x 2 3

2x 2



2 2x 2 3

8x 3

2 2x 2  2x



2x 2 x

3 y 

Solution

3 y 



3 y 





 y  3 y  3



  3 y  3 y  3 y   3 

3 y 

18.

x  1 Solution

3x x  1

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



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

Solution

 2  3   2  3 2  3  2



22. 3 



5 3 

4  2 6 

9  5  2 6



Solution

3  53  5  9  25  9  5  4

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











  10x  5x  x  x  9x  4x

24. 5x 2 2x2  x  x x2  x3

Solution

5x 2 2 x 2  x  x x 2  x 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

 2x 3x  2 3x  2 0

28. x 2  2 3 x 4  7 x 2  x  2

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

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

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

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 

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

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%?

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 



1  3i 2i   2i  6i 2  6  2i  3  1 i 4

1  2i 2  i 4

47. 3  4i

Solution

32  42 

3  4i  48.

5 i7

9  16 

25  5

 5i

Solution

5 i

 5i 

5i 7

i i

 5i 

5i i

 5i 

i  4

 5i 

5i 12

 5i  5i  5i  10i  0  10i

Solve each equation. 49. 15x2  16x  7  0

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

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

x2  4  0

or x 2  9  0

x2  4

x2  9

x  2

x  3

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

y  2  25  11  y 10

y  2  2 y  16

10 y  2  2 y  16 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





 4 z1 3  9  0

 4  0

or z 1 3  9  0

z1 3  4

z1 3  9

z   4

z   9

z  64

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 

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

+15

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

–

Include endpoints which make the numerator equal to 0. Do not include endpoints which make the denominator equal to 0. Solution:

3,  1  2, 

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

–

, 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

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



 9

1 3



 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

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

17.9

118

68 inches

30.8

240

74 inches

130

66 inches

166

67 inches

28.8

224

74 inches

22.1

150

69 inches

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:

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.

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 

y x

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.

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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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.

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

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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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

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 

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

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

y 2  16   x  1 y 

16   x  1

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

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

f  5 

Solution

f  x   5 x  2 f  4   5  4   2  20  2  18

f  5   5  5   2  25  2  27

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

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

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

 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

 27  9  15  1 2

49. f  x   2  x  1  2 3

a. b.

f (3)

f  2 

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

 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 

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

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

 1 f  5

 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

f 6

f  4 

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 

2x x 9 f 4

f  2 

 

59. f x 

f  4  

Solution

f  x 

2x x 9 2

f 4 

2 4

4 9 8  7

 

60. f x  

f  2   

2  2 

 2  9 2

4 5

x x  25 2

f 6

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 

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 

f  3x 

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 

f  4 x 

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 

f x2

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

 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

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 

f x2

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

f  x  5   x  5  2  x  5  1 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 

f 2x 3

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

 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 

f 3x 2

f  x  2

 

Solution

f  x   3 x 2  5 x  12

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

 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 

f x2

f  x  1

 

Solution

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

  x 3  3x 2  x  4

     3 x   x   4

f x2  x2

 x6  3x 4  x 2  4

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

 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 

f x3

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

 2x  2x  5x  1 3

 

   2x   5x   1

f x 3  2 x 3

 2 x 9  2 x 6  5 x 3  1

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 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 x 3  4 x 2  7 x  6

69. f  x   x 4  3 x 2  4 a.

f x 

f 2x 

f x2

 

Solution

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

 x4  3x2  4



 

f x2  x2

f 2x    2x   3 2x   4 4

 16 x 4  12 x 2  4

  3 x   4 4

 x 8  3x 4  4 70. f  x   2 x 4  2 x 2  5 a.

f x 

f  3 x 

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

 

   2 x   5

f x 4  2 x 4

 2 x 16  2 x 8  5

517

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

71. f  x  

1 2

x 2

f x 

f 4x 

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 

f 9x 2  3

 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

f x 

f  8 x 

f x3

 

Solution

f  x   23 x  5

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

 

f x3  2 x3  5  2x  5

 

3 74. f x  3 x  2

f x 

f  27 x 

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 x 2

3x x2  9 f x 

f x2

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



3 x x2  9

 

f x2  

3x 2

x   9 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

f x 

f x3

f  x  4

 

519

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

Solution

5x x 9

f  x  

f x    

5 x 

x   9 2

 

f x3  

5x x2  9



5x 3

x   9 3

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

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   ,  

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,  

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    

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



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    ,  

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

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

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 

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

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





  x  h   x 3

   x  3x h  3xh  h   x 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





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



x  x  4

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.



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

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.

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

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

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

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

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).

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

 , 0  0,  

146. f  x   x 2  1

 , 0

147. f  x    x 2

  1,  

148. f  x   x 3

 1,  

149. f  x   x

 ,  

0,  

 

150. f x 

x 1

 

3 151. f x  x

152. f  x  

1 x

Solution 145. e 146. d 147. b

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 

2x  7

–2

–1

2 x  5, find the y-values that correspond to the x-values of –6, –3, 0, 3, and 6. 3

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

–5

–3

–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

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

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

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

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)

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

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

–2

38. y = –4x – 3

Solution y  3  4 x

y  4 x  3 X

–3

–1

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

40. y – 3x = 6

Solution y  3x  6

y  3x  6

41. y 

–2

1 x1 3

Solution

y

1 x1 3

540

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

42. y  

1 x2 4

Solution 1 y   x2 4 x

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



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

1 8



15 8

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

–2

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 

–3

–2

3 x 3 2

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

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

544

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

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

–2

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

545

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

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

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

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

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

–1

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

–2

61. 2 x  3 y  9  0

Solution 2x  3 y  9  0

3 y  2x  9 y

2 x 3 3

550

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

–2

–3

–5

62. 4 x  7 y  12  0

Solution 4 x  7 y  12  0

7 y  4 x  12 y  x

–4

4 12 x 7 7

63. y = 3

Solution y=3

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

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

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

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

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

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



16  9  25  5

78. P(–5, 12)

Solution

x  x    y  y    5  0    12  0    5    12 

d

 25  144 

169  13

79. P(5, 0)

Solution

x  x    y  y   5  0  0  0  5  0

d

 25  0  25  5

80. P(6, –8)

Solution

x  x    y  y    6  0    8  0    6    8 

d

 36  64 

100  10

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

 94 

82. P(1, 1)

Solution

x  x    y  y    1  0   1  0   1   1 2

d



1 1  2

83. P  3,  6

Solution

x  x    y  y    3  0    6  0    3    6  2

d

 9  36  45  3 5

84. P  6,  2

Solution

x  x    y  y    6  0    2  0    6    2 

d 

 36  4  40  4  10  2 10

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

 3  0   1  0   3    1 2



 31  4  2 86. P

 7, 2 

Solution

d

x  x    y  y  2

 7  0   2  0   7    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

 9  16  25  5

88. P(4, 9); Q(9, 21)

Solution

x  x    y  y    4  9    9  21   5    12 

d

 25  144 

169  13

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

 25  144  169  13 90. P(0, 5); Q(6, –3)

Solution

x  x    y  y   0  6  5   3     6    8 

d

 36  64 

100  10

91. P(–2, –15); Q(–6, –21)

Solution

x  x    y  y    2   6     15   21 

d



 4   6



16  36  52  2 13

92. P(–7, 11); Q(–11, 7)

Solution

x  x    y  y    7   11    11  7   4  4

d



16  16  32  4 2

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

 64  64 

128  8 2

94. P(6, –3); Q(–3, 2)

Solution

x  x    y  y   6   3     3  2   9    5 

d

 81  25 

106

3  95. P  1,  3  ; Q  ,  6  2  Solution d

x  x    y  y  2

2  3   1    3   6  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    3       1   2   2  





2  5       1  2



25 1 4

29  4

29 2

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 

 0  49  49  7

98. P

 5, 0 ; Q 0, 2

Solution

d

x  x    y  y 



 5  0  0  2 2    



 5    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      

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     

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 

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

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 

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.

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

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 

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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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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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:

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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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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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

 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 

 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?

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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

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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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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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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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)

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

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   

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

m

y2  y1 x2  x 1

52 10 3  3 1



34. f(x) = 5x – 8

Solution y  5x  8 x

–8

–3

m

y2  y 1 x2  x 1

 

 3   8  10 5 5 1

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

–6

–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

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



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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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

5 8

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

–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

10 3



–2

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

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

5 2

m

y2  y1

x2  x 1 5 5   2 2 10 0  0 1

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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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.

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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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

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

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

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

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)

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

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

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



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

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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



 1  0   1  1  5 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



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



4  4; d  H, E   1

 1   2    1  3  17 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

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.

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 



c0

a  b  0



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

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:

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.)

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 .

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

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.)

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.

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.

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

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

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 

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

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

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

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

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  

2  y  3

5 15 x  20  2  y  3 

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

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

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

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

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

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

The lines are parallel.

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

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

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

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

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

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 

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

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

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 

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.

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

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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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  

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.

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.

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?

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.

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.

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)

62.5 cm

200 cm

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.

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  

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

115

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

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 

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)

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

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 

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

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.

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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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 

5 x  10 y



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

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.

 y  x2  5

y   x 2  5, so no. c.

 y  x   5 2

y   x 2  5, so no.

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 

 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

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

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

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

Practice Find the x- and y-intercepts of each graph. Do not graph the equation. 17. y = x2 – 4

Solution

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 

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

y 0

1  x-int:  0, 0  ,  , 0  y -int:  0, 0  2 

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

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

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

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

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 

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

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 

y 0

y -int:  0, 0 

Graph each equation. 29. y = x2 Solution

y  x2

x-int:  0, 0 

y -int:  0, 0 

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

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 

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

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

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

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

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 

 y   x 

y 2  x2

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

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

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

equivalent: symmetry

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

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

y  3x 3  7 x equivalent: symmetry

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 

 y   3 x  2

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

equivalent: symmetry

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

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

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

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

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

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

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

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

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

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

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

74. x 2  y 2  121

Solution x 2  y 2  121

 x  0    y  0   11 C:  0, 0  ; r  11 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

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

 x  0    y   3     2 2  C:  0,  3  ; r  2 2 2

77.  x  6   y 2  2

1 4

Solution

 x  6   y  41 2

 x   6     y  0    21  2

C:  6, 0  ; r 

78.  x  5   y 2  2

1 2

16 25

Solution 16  x  5   y  25 2









 x  5    y  0    45  2

4 C:  5, 0  ; r  5

79.  x  4    y  1  25 2

Solution

 x  4    y  1  25  x  4    y  1  5 C:  4, 1 ; r  5 2

80.  x  11   x  7   169 2

Solution

 x  11   y  7   169  x   11    y   7    13 C:  11,  7  ; r  13 2

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  1  x    y   2   3 5 4  1  C:  ,  2  ; r  3 5 4 



82. x  5



   y  3  225 2

Solution

 x  5    y  3  225  x    5    y  3   15 C:   5, 3  ; r  15 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

x 2  y 2  25

84. Center at the origin; r 

Solution

 x  0   y  0   3  2

x2  y 2  3

85. Center at (0, –6); r = 9

Solution

 x  0    y   6    9 x   y  6   81 2

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

87. Center at (8, 0); r 

1 5

Solution  

 x  8    y  0    51  2

 x  8

  1  y2  25

88. Center at (–10, 0); r 

Solution

 x   10    y  0    11  2

 x  10   y  11 2

89. Center at (–2, 12), r = 20

Solution

 x   2    y  12  20 2

 x  2    y  12   400 2

2  90. Center at  , 5  ; r  6 7 

Solution 2

2  2 2  x    y   5   6 7 





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

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

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

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 

Solution

 x  3   y  4   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

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

 x  3   y  3  5 2

x 2  6 x  9  y 2  6 y  9  25 x2  y 2  6x  6 y  7  0

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

 x  0    y  0    106  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

 x  3   y  4  5 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

 x  2    y  6    40  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

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

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

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

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

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

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

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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

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

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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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

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

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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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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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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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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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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

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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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

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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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

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

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

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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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

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

 x  7   y  4  5 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

 x  36   y  12  12 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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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

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

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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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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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 

302 k 100  900 90,000  k

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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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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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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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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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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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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 

m  3n2 q

24  k  4  2  24  8k

m  3  5

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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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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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

Coca-Cola Classic

Pepsi

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.

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

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

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

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

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 

 

 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

P  5  3

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 

k  9m1m2

4d 2 9 km1m2   4 d2 The force is multiplied by

9 . 4

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



k  2m1  3m2 

  d 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

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:

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.

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.

B

kst 2 r3

B

kt 2 r3

B  kr 3t 2

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

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

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   ,  

f x 

3x x 5

Solution

3x x 5 domain   , 5    5,   f x  y 

f x 

3x 4 x  16 2

Solution

f x 

3x 4 x  16 2



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   ,  

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

f  5 

Solution

f  x   5 x  4

f  4   5  4   4  20  4  16

f  5   5  5   4  25  4  29

14. f  x   x 2  2 x  9 a.

f  3

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  1

3x x 9 f 4

f  2 

 

15. f x 

Solution

f  x  a.

3x x 9 2

f  4 

3  4

42  9 12  7

695

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

f  2  

3  2

 2  9 2

6 6  5 5

16. f  x   x 2  6 x  1 a.

f x 

f x2

f  x  5

 

Solution

f  x   x 2  6x  1

f x   x   6 x   1 2

 x 2  6x  1

     6x   1

f x2  x2

 x4  6x2  1

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

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

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

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

–6

–2

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

–2

–5

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

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 

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

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

 36  64 

100  10

40. P  8, 6  ; Q  12, 10 

Solution

x  x    y  y    12   8     10  6    4   4

d 



41. P

16  16  32  4 2

 3, 9 ; Q  3, 7 

Solution d 

x  x    y  y 



 3  3   9  7 

 02   2   04 

4 2

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 

 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

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

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

–6

–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.

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.

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

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

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

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.

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

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

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

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 

y 0

y -int:  0, 0 

92. y  x 2  10 x  24

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

y 2  8x equivalent: 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

not equivalent: no symmetry

equivalent: symmetry

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

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

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

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

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.

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

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.

 x  7   y  41 2

Solution

 x  7   y  41 2

 x   7     y  0    21  2

C:  7, 0  ; r 

112.

1 2

 x  5   y  1  9 2

Solution

 x  5    y  1  9  x  5    y    1   3 C:  5,  1 ; r  3 2

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

x 2  y 2  49

114. Center at (3, 0); r 

1 5

Solution  

 x  3    y  0    51  2

 x  3

  1  y  25 2

115. Center at (  2, 12); r  5

Solution

 x   2     y  12   5 2

 x  2    y  12   25 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

or x 2  y 2  6 x  8 y  119  0

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 



 1  5     5    8   2  2  2

121 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

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

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

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

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

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

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

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

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 

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.

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

 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 

1000k 0.0025 0.0005  k 200 

R V 

0.0005L D2 0.0005  1500 

0.08 

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 

727

Solution and Answer Guide: Gustafson/Hughes, College Algebra 2023, 9780357723654; Chapter 2: Functions and Graphs

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 

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

0,  8 Solution

0,  8  negative y -axis

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 

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

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.

–2

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

1 2

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

1 4



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 



16  25 

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 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

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

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

 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

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

equivalent: symmetry

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

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

37. Center at (2, 4); passing through (6, 8)

Solution r  

2  6  4  8 2

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

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

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

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

y  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

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

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