Introductory and Intermediate Algebra 5th Edition Bittinger Solutions Manual

SOLUTION MANUAL




SOLUTION MANUAL

,Chapter 2
Solving Equations and Inequalities
26. 4
Exercise Set 2.1 28. 25

RC2. The correct answer is (c). 30. −16
7
RC4. The correct answer is (a). 32. 24
10
2. t + 17 = 53 34. 8.2
35 + 17 ?
53 1
36.
52
FALSE 4
35 is not a solution. 2 5
38. x+ =−
3 6
4. a − 19 = 17 5 4 9
x=− − =−
36 − 19 ?
17 6 6 6
17
TRUE x=−
3
2
36 is a solution.
3 5
6. 8y = −72 40. y− =
4 6
8(−9) ?
−72 10 9
y= +
−72
TRUE 12 12
19
−9 is a solution. y=
12
8. y
=6 1 3
8 42. − +y = −
8 4
49 6 1
? 6 y=− +
8

8 8


1
5
6

FALSE y=−
8 8
49 is not a solution. 44. 2.7
10. 9x + 5 = 86 46. 16
9 · 9 + 5 ?
86 48. −10.6
86
TRUE
1 2
9 is a solution. 50. 5 = 4 +x
4 3
12. 6(y − 2) = 18 5
3
−4
8
=x
12 12
6(−5 − 2) ?
18


15
−4
8
6(−7)

4
12 12
=x



−42
FALSE 7
=x
−5 is not a solution. 12
3
14. 7 52. 136
8
16. 34 54. −5.2
18. −23 56. 172.72
20. −31 58. 65t miles
22. 23 60. x+x = x
2x = x
24. −11 x=0

,18 Chapter 2: Solving Equations and Inequalities

2 4
62. x+4 = 5+x 30. y=−
5 15
4=5  
5 2 5 4
No solution · y= · −
2 5 2 15
|x| + 6 = 19 20
64. y=−
|x| + 6 − 6 = 19 − 6 30
|x| = 13 y=−
2
x represents a number whose distance from 0 is 13. Thus 3
x = −13 or x = 13. 3 15
32. − x=−
The solutions are −13 and 13. 8 16
8  3  8  15 
− · − x =− · −
3 8 3 16
Exercise Set 2.2 120
x=
48
RC2. The correct answer is (d). 5
x=
2
RC4. The correct answer is (b).
34. 20
2. 17
36. −2
4. 9
38. 8
6. 7
9
40. − y = 12.06
8. −53 7

7 9 7
10. 47 − · − y = − · (12.06)
9 7 9
12. −7 84.42
y=−
9
14. −7 y = −9.38
16. 8 −x
42. = −16
18. −30 8
 
−x
20. −88 8 = 8 · (−16)
8
−x = −128
4
22. x = 16 −1 · (−x) = −1 · (−128)
5
5 4 5 x = 128
· x = · 16
4 5 4 m
x = 20 44. = 10
−3
 
3 m
24. − x = 12 −3 · = −3 · 10
8 −3
  m = −30
8 3 8
− − x = − · 12
3 8 3 46. −x + 5
x = −32
48. −32y
−x
26. =9 50. 2 − 5(x + 5) = 2 − 5x − 25 = −5x − 23
6
−x = 54
x = −54 52. −2a − 4(5a − 1) = −2a − 20a + 4 = −22a + 4
1
28.
1
=−
y 54. · b · 10 m2 , or 5b m2
8 5 2
5 56. All real numbers
= −y
8
58. 4|x| = 48
5
− =y |x| = 12
8
The distance of x from 0 is 12. Thus, x = 12 or x = −12.

60. 5

Copyright 
c 2015 Pearson Education, Inc.

, Exercise Set 2.3 19


a2 + 1 24. −5y − 7y = 144
62.
c −12y = 144
64. To “undo” the last step, divide 22.5 by 0.3. y = −12

22.5 ÷ 0.3 = 75 26. −10y − 3y = −39
−13y = −39
Now divide 75 by 0.3.
y=3
75 ÷ 0.3 = 250 1
The answer should be 250 not 22.5. 28. x + x = 10
4
5
x = 10
Exercise Set 2.3 4
4
x = · 10
5
RC2. The correct answer is (a). x=8
RC4. The correct answers are (a) and (e). We would usually 30. 6.8y − 2.4y = −88
multiply by 100. 4.4y = −88
y = −20
2. 7x + 6 = 13
7x = 7 32. 4x − 6 = 6x
x=1 −6 = 2x
4. 4y + 10 = 46 −3 = x
4y = 36 34. 5y − 2 = 28 − y
y=9 6y = 30
y=5
6. 5y − 2 = 53
5y = 55 36. 4 − 3x = 6 − 7x
y = 11 4x = 2
1
8. 4x − 19 = 5 x=
2
4x = 24 38. 14 − 6a = −2a + 3
x=6 11 = 4a
10. 5x + 4 = −41 11
=a
5x = −45 4
x = −9 40. −7z + 2z − 3z − 7 = 17
−8z − 7 = 17
12. −91 = 9t + 8 −8z = 24
−99 = 9t z= −3
−11 = t
42. 5 + 4x − 7 = 4x − 2 − x
14. −5x − 7 = 108 4x − 2 = 3x − 2
−5x = 115 x=0
x = −23
44. 5y − 7 + y = 7y + 21 − 5y
3 6y − 7 = 2y + 21
16. x − 24 = −36 4y = 28
2
3 y = 7
x = −12
2 7 1 3 1
46. x− + x = + x, LCM is 16
2 3 2 8 4 4 16
· x = (−12)
3 2 3 14x − 4 + 12x = 1 + 16x
x = −8 26x − 4 = 1 + 16x
10x =5
18. 8x + 3x = 55
1
11x = 55 x=
x=5 2
3 5 4
20. 8x + 5x = 104
48. − +x = − − , LCM is 6
2 6 3
13x = 104 −9 + 6x = −5 − 8
x=8 −9 + 6x = −13
22. 6x = −4
7x + 18x = 125
25x = 125 2
x= −
x=5 3

Copyright 
c 2015 Pearson Education, Inc.