College algebra 4th edition blitzer solutions manual by testbanklive48

Chapter 1 Section 1.1

Check Point Exercises

The meaning of a [−100,100, 50] by [−100,100,10] viewing rectangle is as follows: distance between x -axis minimum maximum tick x -value x -value marks

P P P [ −100 , 100 , 50 ] by distance between y -axis minimum maximum tick y -value y -value marks

P P P [ −100 , 100 , 10 ]

x = −3, y = 7

The graph crosses the x-axis at (–3, 0). Thus, the x-intercept is –3. The graph crosses the y-axis at (0, 5). Thus, the y-intercept is 5.

The graph does not cross the x-axis. Thus, there is no x-intercept. The graph crosses the y-axis at (0, 4). Thus, the y-intercept is 4.

The graph crosses the x- and y-axes at the origin (0, 0). Thus, the x-intercept is 0 and the x =y-intercept −1, y = 0 is 0. x = 0, y = 1

x = −2, y = 6 x = −1, y = 5 x = 0, y = 4 x = 1, y = 3 x = 2, y = 2 x = 3, y = 1 3.

x = 1, y = 2 x = 2, y = 3 x = −4, y = 3 x = −3, y = 2 x = −2, y = 1

estimated by finding 60% of 160,000. N â&#x2030;&#x2C6; 60% of 160, 000 = 0.60 Ă&#x2014;160, 000 = 96, 000

The number of federal prisoners sentenced for drug offenses in 2003 is about 57% of 159,275. This can be

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.1

Exercise Set 1.1 1.

10.

11.

12.

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.1

13.

15.

x = −3, y = 7

x = −3, y = −5

x = −2, y = 2

x = −2, y = −4

x = −1, y = −1

x = −1, y = −3

x = 0, y = −2

x = 1, y = −1

x = 2, y = 2

x = 2, y = 0

x = 3, y = 7

x = 3, y = 1 16.

14.

x = –3, y = –1 x = –2, y = 0 x = –1, y = 1 x = 0, y = 2 x = 1, y = 3 x = 2, y = 4 x = 3, y = 5

x = –3, y = 11 x = –2, y = 6 x = –1, y = 3 x = 0, y = 2 x = 1, y = 3 x = 2, y = 6 x = 3, y = 11

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.1

17.

19.

x = −3, y = −5 x = −2, y = −3

x = −3, y =

2 x = −2, y = 1

x = −1, y = −1 x = 0, y = 1 x = 1, y = 3

x = −1, y =

x = 2, y = 5

1 2

x = 0, y = 0

x = 3, y = 7

x = 1, y = −

2 x = 2, y = −1

18.

x = 3, y = −

3 2

20.

x = –3, y = –10 x = –2, y = –8 x = –1, y = –6 x = 0, y = –4 x = 1, y = –2 x = 2, y = 0 x = 3, y = 2

7 2 x = –2, y = 3 5 x = –1, y = 2 x = 0, y = 2 3 x = 1, y = 2 x = 2, y = 1 1 x = 3, y = 2 x = –3, y =

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.1

21.

23.

x = −3, y = 4

x = −3, y = 6 x = −2, y = 4 x = −1, y = 2 x = 0, y = 0 x = 1, y = 2 x = 2, y = 4 x = 3, y = 6

x = −2, y = 3 x = −1, y = 2 x = 0, y = 1 x = 1, y = 2 x = 2, y = 3 x = 3, y = 4

24. 22.

x = −3, y = 2 x = −2, y = 1 x = −1, y = 0 x = 0, y = −1

x = −3, y = −6 x = −2, y = −4 x = −1, y = −2 x = 0, y = 0

x = 1, y = 0 x = 2, y = 1 x = 3, y = 2

x = 1, y = −2 x = 2, y = −4 x = 3, y = −6

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.1

25.

28.

x = −3, y = 0 x = −2, y = 5 x = −1, y = 8

x = −3, y = −28 x = −2, y = −9 x = −1, y = −2 x = 0, y = −1 x = 1, y = 0 x = 2, y = 7 x = 3, y = 26

x = 0, y = 9 x = 1, y = 8 x = 2, y = 5 x = 3, y = 0

29. (c) x-axis tick marks –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5; y-axis tick marks are the same.

26.

30. (d) x-axis tick marks –10, –8, –6, –4, –2, 0, 2, 4, 6, 8, 10; y-axis tick marks –4, –2, 0, 2, 4 31. (b); x-axis tick marks –20, –10, 0, 10, 20, 30, 40, 50, 60, 70, 80; y-axis tick marks –30, –20, –10, 0, 10, 20, 30, 40, 50, 60, 70 x = −3, y = −9 x = −2, y = −4 x = −1, y = −1 x = 0, y = 0 x = 1, y = −1 x = 2, y = −4 x = 3, y = −9

32. (a) x-axis tick marks –40, –20, 0, 20, 40; y-axis tick marks –1000, –900, –800, –700, . . . , 700, 800, 900, 1000 33. The equation that corresponds to Y in the table 2 is (c), y 2 = 2 − x . We can tell because all of the points (−3, 5) , (−2, 4) , (−1, 3) , (0, 2) , (1,1) , (2, 0) , and (3, −1) are on the line y = 2 − x , but all are not on any of the others.

27.

34. The equation that corresponds to Y1 in the table is (b), y1 = x 2 . We can tell because all of the points (−3, 9) , (−2, 4) , (−1,1) , (0, 0) , (1,1) , (2, 4) , and (3, 9) are on the graph y = x 2 , but all are not on any of the others. x = −3, y = −27

x = 3, y = 27

x = −2, y = −8 x = −1, y = 1 x = 0, y = 0 x = 1, y = 1 x = 2, y = 8

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.1

35. No. It passes through the point (0, 2) . 36. Yes. It passes through the point (0, 0) . 37.

(2, 0)

ChapterISM: 1 College Algebra

38.

ISM: College Algebra Section 1.1

49.

(0, 2)

39. The graphs of Y and Y intersect at the points 1 2 (−2, 4) and (1,1) . 40. The values of Y1 and Y2 are the same when x = −2 and x = 1 . 41. a. b.

2; The graph intersects the x-axis at (2, 0). –4; The graph intersects the y-axis at (0,–4).

42. a.

1; The graph intersects the x-axis at (1, 0).

2; The graph intersects the y-axis at (0, 2).

43. a.

1, –2; The graph intersects the x-axis at (1, 0) and (–2, 0).

b. 44. a. b. 45. a. b. 46. a. b.

50.

51.

2; The graph intersects the y-axis at (0, 2). 1, –1; The graph intersects the x-axis at (1, 0) and (–1, 0). 1; The graph intersect the y-axis at (0, 1). –1; The graph intersects the x-axis at (–1, 0). none; The graph does not intersect the yaxis. none; The graph does not intersect the x-axis. 2; The graph intersects the y-axis at (0, 2).

47.

48.

x ( x, y) −3 (−3, 5) −2 (−2, 5) −1 (−1, 5) 0 (0, 5) 1 (1, 5) 2 (2, 5) 3 (3, 5)

ChapterISM: 1 College Algebra

52.

ISM: College Algebra Section 1.1

54.

x ( x, y) −3 (−3, −1) −2 (−2, −1) −1 (−1, −1) 0 (0, −1) 1 (1, −1) 2 (2, −1) 3 (3, −1)

x −2 −1

1 ⎛ 1 ⎞ − ,2 2 ⎜⎝ 2 ⎟⎠ 1 ⎛ 1 ⎞ − ⎜ − ,3⎟ 3 ⎝ 3 ⎠ 1 3

x −2 −1

1⎞ ⎛ ⎜ −2, 2 ⎟ ⎝ ⎠ ( −1,1)

−

1 2 1

53.

( x, y)

⎛1 ⎞ ⎜ , −3 ⎟ 3 ⎝ ⎠ 1 ⎛ ⎞ ⎜ , −2 ⎟ ⎝2 ⎠ (1, −1) ⎛ 1⎞ ⎜ 2,− ⎟ ⎝ 2⎠

( x, y) 1⎞ ⎛ ⎜ −2, − ⎟ 2⎠ ⎝ ( −1, −1)

1 ⎛ 1 ⎞ ⎜ − , −2 ⎟ 2 ⎝ 2 ⎠ 1 ⎛ 1 ⎞ − ⎜ − , −3 ⎟ 3 ⎝ 3 ⎠ −

1 3

⎛1 ⎞ ⎜ , 3⎟ ⎝3 ⎠

1 2 1

⎛1 ⎞ ⎜ ,2⎟ ⎝2 ⎠ (1,1) ⎛ 1⎞ ⎜ 2, ⎟ ⎝ 2⎠

55. There were approximately 65 democracies in 1989. 56. There were 120 − 40 = 80 more democracies in 2002 than in 1973. 57. The number of democracies increased at the greatest rate between 1989 and 1993. 58. The number of democracies increased at the slowest rate between 1981 and 1985. 59. There were 49 democracies in 1977. 60. There were 110 democracies in 1997.

ChapterISM: 1 College Algebra

61.

R = 165 − 0.75A;

ISM: College Algebra Section 1.1

A = 40

64. a.

R − 165 − 0.75A = 165 − 0.75 ( 40 )

y = 4 x + 35

= 165 − 30 = 135 The desirable heart rate during exercise for a 40-year old man is 135 beats per minute. This corresponds to the point (40, 135) on the blue graph. 62.

R = 143 − 0.65A;

= 4 0 + 35 = 4 (0) + 35

A = 40

R − 143 − 0.65A = 143 − 0.65 ( 40 )

= 4 9 + 35 = 4 (3) + 35 = 47 According to the model, the head circumference at 9 months is 47 cm.

At birth we have x = 0 . y = 2.9 x + 36

= 2.9 0 + 36

At 14 months we have x = 14 . y = 4 x + 35

= 2.9 (0) + 36

= 4 14 + 35

= 36 According to the model, the head circumference at birth is 36 cm.

= 35 According to the model, the head circumference at birth is 35 cm. At 9 months we have x = 9 . y = 4 x + 35

= 143 − 26 = 117 The desirable heart rate during exercise for a 40-year old woman is 117 beats per minute. This corresponds to the point (40, 117) on the red graph. 63. a.

At birth we have x = 0 .

≈ 50 According to the model, the head circumference at 14 months is roughly 50 cm.

At 9 months we have x = 9 .

y = 2.9 x + 36 = 2.9 9 + 36

71.

The model describes severe autistic children.

y = 45.48x 2 − 334.35x + 1237.9

= 2.9 (3) + 36 = 44.7 According to the model, the head circumference at 9 months is 44.7 cm. c.

At 14 months we have x = 14 . y = 2.9 x + 36

The discharges decreased from 1990 to 1994, but started to increase after 1994. The policy was not a success.

= 2.9 14 + 36 ≈ 46.9 According to the model, the head circumference at 14 months is roughly 46.9 cm. d.

72. a.

The model describes healthy children.

False; (x, y) can be in quadrant III.

False; when x = 2 and y = 5, 3y – 2x = 3(5) – 2(2) = 11.

False; if a point is on the x-axis, y = 0.

True; all of the above are false. (d) is true.

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.2

73. (a)

74. (d) 75. (b) 76. (c)

7 x − 21 = 10 − 4x − 20 7 x − 21 = −4 x − 10 7 x + 4x = −10 + 21 11x = 11 11x 11 = 11 11 x =1 Check: x −3 5 x +5 = − 4 14 7 1 − 3 5 1+ 5 = − 4 14 7 −2 5 6 = − 4 14 7 1 1 − =− 2 2 The solution set is {1}.

77. (b) 78. (a) Section 1.2 Check Point Exercises 1.

x−3 5 x +5 = − 4 14 7 x −3 ⎛ 5 x +5 ⎞ 28 ⋅ = 28 ⎜ − 4 7 ⎟⎠ ⎝ 14 7 ( x − 3) = 2(5) − 4 ( x + 5 )

4 x + 5 = 29 4 x + 5 − 5 = 29 − 5 4 x = 24 4 x 24 = 4 4 x =6 Check: 4 x + 5 = 29 4(6) + 5 = 29 24 + 5 = 29 29 = 29 true The solution set is {6}. 4(2 x + 1) − 29 = 3(2x − 5) 8x + 4 − 29 = 6x − 15 8x − 25 = 6x − 15 8x − 25 − 6x = 6x − 15 − 6x 2 x − 25 = −15 2 x − 25 + 25 = −15 + 25 2 x = 10 2 x 10 = 2 2 x=5 Check: 4(2 x + 1) − 29 = 3(2x − 5) 4[2(5) + 1] − 29 = 3[2(5) − 5] 4[10 + 1] − 29 = 3[10 − 5] 4[11] − 29 = 3[5]

44 − 29 = 15 15 = 15 true The solution set is {5}.

5 17 1 = − , x≠0 2 x 18 3x 5 ⎛ 17 1 ⎞ 18x ⋅ = 18x ⎜ − ⎟ 2x ⎝ 18 3x ⎠ 5 17 1 18 ⋅ = 18x ⋅ − 18x ⋅ 2x 18 3x 45 = 17x − 6 45 + 6 = 17x − 6 + 6 51 = 17x 51 17x = 17 17 3= x The solution set is {3}.

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.2

Exercise Set 1.2

x 2 2 = − , x≠2 x−2 x−2 3 x 2⎤ ⎡ 2 3( x − 2) ⋅ = 3(x − 2) ⎢ − ⎥ x−2 ⎣ x − 2 3⎦ x 2 2 3( x − 2) ⋅ = (3x − 2) ⋅ − 3(x − 2) ⋅ x−2 x−2 3 3x = 6 − (x − 2) ⋅ 2 3x = 6 − 2(x − 2) 3x = 6 − 2x + 4 3x = 10 − 2x 3x + 2x = 10 − 2x + 2x 5x = 10 5x 10 = 5 5 x=2 The solution set is the empty set, ∅.

7x – 5 = 72 7x = 77 x = 11 Check: 7 x − 5 = 72 7(11) − 5 = 72 77 − 5 = 72 72 = 72 The solution set is {11}.

6x – 3 = 63 6x = 66 x = 11 The solution set is {11}. Check: 6 x − 3 = 63 6(11) − 3 = 63 66 − 3 = 63

Set y1 = y2 . 1 x+4 1 x+4 ( x +4)( x −4)

x−4 1

63 = 63 2

x − 16 22

x + 4 x − 4 ( x + 4)(x − 4) (x +4)( x −4) 22(x +4)( x −4) +( x − 4) + (x + 4)= = 22 x−4 ( x + 4)(x − 4) x − 4 + x + 4 = 22 2 x = 22

1 1 + x+4 x−4 1 1 + 11 + 4 11 − 4 1 1 + 15 7 22 105

11x − (6x − 5) = 40 11x − 6x + 5 = 40 5x = 35 5x + 5 = 40 x=7 The solution set is {7}.

x = 11

Check:

Check: 11x − (6x − 5) = 40

22 x 2 − 16 22 = 2 11 − 16 22 = 105 22 = true 105 =

11(7) − [6(7) − 5] = 40 77 − (42 − 5) = 40 77 − (37) = 40 40 = 40

4 x − 7 = 4(x − 1) + 3 4 x − 7 = 4(x − 1) + 3 4 x − 7 = 4x − 4 + 3 4 x − 7 = 4x − 1 −7 = −1 The original equation is equivalent to the statement –7 = –1, which is false for every value of x. The solution set is the empty set, ∅. The equation is an inconsistent equation.

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.2

5x – (2x – 10) = 35 5x – 2x + 10 = 35 3x + 10 = 35 3x = 25 25 x= 3

Check: 7(2) + 4 = 2 + 16 14 + 4 = 18 18 = 18

⎧ 25 ⎫ The solution set is ⎨ ⎬ . ⎩3⎭ Check: 5x − (2 x − 10) = 35

⎤ ⎛ 25 ⎞ ⎡ ⎛ 25 ⎞ 5 ⎜ ⎟ − ⎢ 2 ⎜ ⎟ − 10 ⎥ = 35 ⎝ 3 ⎠ ⎣ ⎝ 3 ⎠ ⎦ 125 ⎡ 50 ⎤ − − 10 ⎥ = 35 3 ⎢⎣ 3 ⎦ 125 20 − = 35 3 3 105 = 35 3 35 = 35 5.

13x + 14 = 12x – 5 x + 14 = –5 x = –19 The solution set is {–19}. Check: 13x + 14 = 12x − 5 13(−19) + 14 = 12(−19) − 5 − 247 + 14 = −228 − 5 − 233 = −233

2x – 7 = 6 + x x–7=6 x = 13 The solution set is {13}.

3(x – 2) + 7 = 2(x + 5) 3x – 6 + 7 = 2x + 10 3x + 1 = 2x + 10 x + 1 = 10 x=9 The solution set is {9}.

Check: 2(13) − 7 = 6 + 13 26 − 7 = 19 19 = 19 6.

7x + 4 = x + 16 6x + 4 = 16 6x = 12 x=2 The solution set is {2}.

Check: 3(9 − 2) + 7 = 2(9 + 5) 3(7) + 7 = 2(14) 21 + 7 = 28 28 = 28

3x + 5 = 2x + 13 x + 5 = 13 x=8 The solution set is {8}. Check: 3x + 5 = 2x + 13 3(8) + 5 = 2(8) + 13 24 + 5 = 16 + 13 29 = 29

10. 2(x – 1) + 3 = x – 3(x +1) 2x – 2 + 3 = x – 3x – 3 2x +1 = –2x – 3 4x + 1 = –3 4x = –4 x = –1 The solution set is {–1}. Check: 2( x − 1) + 3 = x − 3(x + 1) 2(−1 − 1) + 3 = −1 − 3(−1 + 1) 2(−2) + 3 = −1 − 3(0) − 4 + 3 = −1 + 0

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.2

−

− 1

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.2

11. 3(x – 4) – 4(x – 3) = x + 3 – (x – 2) 3x – 12 – 4x + 12 = x + 3 – x + 2 –x = 5 x = –5 The solution set is {–5}.

13. 16 = 3(x – 1) – (x – 7) 16 = 3x – 3 – x + 7 16 = 2x + 4 12 = 2x 6=x The solution set is {6}.

Check: 3(−5 − 4) − 4(−5 − 3) = −5 + 3 − (−5 − 2) 3(−9) − 4(−8) = −2 − (−7) −27 + 32 = −2 + 7 5=5

Check: 16 = 3(6 − 1) − (6 − 7) 16 = 3(5) − (−1) 16 = 15 + 1 16 = 16

12. 2 – (7x + 5) = 13 – 3x 2 – 7x – 5 = 13 – 3x –7x – 3 = 13 – 3x –4x = 16 x = –4 The solution set is {–4}.

14. 5x – (2x + 2) = x + (3x – 5) 5x – 2x – 2 = x + 3x – 5 3x – 2 = 4x – 5 –x = –3 x=3 The solution set is {3}.

Check: 2 − (7x + 5) = 13 − 3x 2 − [7(−4) + 5] = 13 − 3(−4) 2 − [−28 + 5] = 13 + 12 2 − [−23] = 15

Check: 5x − ( 2x + 2 ) = x + ( 3x − 5 ) 5(3) − [2(3) + 2] = 3 + [3(3) − 5] 15 − [6 + 2] = 3 + [9 − 5] 15 − 8 = 3 + 4 7=7

2 + 23 = 25 25 = 25

15. 25 – [2 + 5y – 3(y + 2)] 25 – [2 + 5y – 3y – 6] 25 – [2y – 4] 25 – 2y + 4 –2y + 29 6y y The solution set is {–2}.

= –3(2y – 5) – [5(y – 1) – 3y + 3] = –6y + 15 – [5y – 5– 3y + 3] = –6y + 15 – [2y – 2] = –6y + 15 – 2y + 2 = –8y + 17 = –12 = –2

Check: 25 − [2 + 5y − 3( y + 2) = −3(2 y − 5) − [5( y − 1) − 3y + 3] 25 − [2 + 5(−2) − 3(−2 + 2) = −3[2(−2) − 5] − [5(−2 − 1) − 3(−2) + 3] 25 − [2 − 10 − 3(0)] = −3[−4 − 5] − [5(−3) + 6 + 3] 25 − [−8] = −3(−9) − [−15 + 9] 25 + 8 = 27 − (−6) 33 = 27 + 6 33 = 33

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.2

16. 45 – [4 – 2y – 4(y + 7)] = –4(1 + 3y) – [4 – 3(y + 2) – 2(2y – 5)] 45 – [4 – 2y – 4y – 28] = –4 – 12y – [4 – 3y – 6 – 4y + 10] 45 – [–6y – 24] = –4 – 12y – [–7y + 8] 45 + 6y + 24 = –4 – 12y + 7y – 8 6y + 69 = –5y – 12 11y = –81 81 y=− 11 ⎧ 81 ⎫ The solution set is ⎨ − ⎬ . ⎩ 11 ⎭ 17.

20.

x x = −2 3 2 ⎡x x ⎤ 6 ⎢ = − 2⎥ ⎣3 2 ⎦ 2 x = 3x − 12 12 = 3x − 2x x = 12 The solution set is {12}.

18.

x = 15 The solution set is {15}.

x x = +1 5 6 ⎡x x 30 ⎢ = + 1⎤⎥ ⎣5 6 ⎦ 6x = 5x + 30 6 x − 5x = 30

21.

x = 30 The solution set is {30}. 19.

x 1 x − = 5 2 6 ⎡ x 1 x⎤ 30 ⎢ − = ⎥ ⎣5 2 6⎦ 6 x − 15 = 5x 6 x − 5x = 15

3x 2 x = +1 5 3 ⎡ 3x 2x ⎤ 15 ⎢ = + 1⎥ 3 ⎣5 ⎦ 9 x = 10x + 15 9x − 10 x = 15 − x = 15 x = −15 The solution set is {–15}.

x x = 3 2 x x⎤ ⎡ 6 ⎢ 20 − = ⎥ 3 2⎦ ⎣ 120 − 2x = 3x 120 = 3x + 2x 120 = 5x 120 x= 5 x = 24 20 −

22.

The solution set is {24}.

x 3x = +5 2 4 ⎡ x 3x ⎤ 4⎢ = + 5⎥ ⎣2 4 ⎦ 2 x = 3x + 20 2 x − 3x = 20 − x = 20 x = −20 The solution set is {−20}.

ChapterISM: 1 College Algebra

23.

ISM: College Algebra Section 1.2

3x x 5 −x= − 5 10 2 x 5⎤ ⎡ 3x 10 ⎢ − x = − ⎥ 5 10 2⎦ ⎣ 6 x − 10 x = x − 25 − 4x − x = −25

27.

− 5x = −25 x =5 The solution set is {5}. 28.

2x x 17 2x − = + 24. 7 2 2 2x x 17 ⎤ ⎡ 14 ⎢ 2 x − = + ⎥ 7 2 2⎦ ⎣ 28x − 4x = 7x + 119 24 x − 7x = 119

26.

The solution set is {−19} .

x +3 3 x −5 = + 6 8 4 ⎡ x +3 3 x −5 ⎤ 24⎢ = + 8 4 ⎦⎥ ⎣ 6 4 x + 12 = 9 + 6x − 30 4 x − 6x = −21 − 12 − 2x = −33 33 x= 2 ⎧ 33 ⎫ The solution set is ⎨ ⎬ . ⎩2⎭

29.

x +1 1 2 − x = + 4 6 3 ⎡ x + 1 1 2 −x ⎤ 12 = + ⎢ 4 ⎣

x −2 x +3 = 3 8 x − 2 x + 3⎤ ⎡ 24 ⎢ 5 + = 3 8 ⎥⎦ ⎣ 120 + 8x − 16 = 3x + 9 8x − 3x = 9 − 104 5+

5x = −95 x = −19

17 x = 119 x =7 The solution set is {7}. 25.

x x −3 = 2+ 4 3 x −3 ⎤ ⎡x 12 ⎢ = 2 + 4 3 ⎦⎥ ⎣ 3x = 24 + 4x – 12 3x – 4x = 12 –x = 12 x = –12 The solution set is {–12}.

30.

x +1 x +2 = 5− 3 7 x + 1 x +2 ⎤ ⎡ 21 ⎢ = 5− 7 ⎥⎦ ⎣ 3 7x + 7 = 105 – 3x – 6 7x + 3x = 99 – 7 10x = 92 92 x= 10 46 x= 5 ⎧ 46 ⎫ The solution set is ⎨ ⎬ . ⎩5⎭ 3x 5

3 ⎥ ⎦

−

x−3 2

x+2 3

⎡ 3x x −3 x +2 ⎤ 30 ⎢ − = 5 2 3 ⎥ ⎣ ⎦ 18x − 15x + 45 = 10x + 20 3x − 10 x = 20 − 45 − 7x = −25 25 x= 7

3x + 4x = 10 − 3 3x + 3 = 2 + 8 − 4x 7x = 7 x =1 The solution set is {1}.

ChapterISM: 1 College Algebra

ISM: College Algebra Section 1.2

⎧ 25 ⎫ The solution set is ⎨ ⎬ . ⎩7⎭

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