Solution Manual For Beginning & Intermediate Algebra, 6th Edition by John Tobey Jeffrey Slater Jamie Blair Jennifer Crawford Anne Fischer

INSTRUCTOR
SOLUTIONS MANUAL
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BEGINNING
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AND I NTERMEDIATE A LGEBRA
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SIXTH EDITION
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John Tobey
North Shore Community College
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Jeffrey Slater
North Shore Community College
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Jamie Blair
Orange Coast College
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Jennifer Crawford
Normandale Community College
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Anne Fischer
Tulsa Community College
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Contents


Chapter 0 1
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Chapter 1 18

Chapter 2 37
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Chapter 3 74

Chapter 4 109
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Chapter 5 140
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Chapter 6 162

Chapter 7 185
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Chapter 8 214

Chapter 9 240
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Chapter 10 288

Chapter 11 316
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Chapter 12 338
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Practice Final Examination 364
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, Chapter 0
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0.1 Exercises 8
25 1
26.  3 25  8
13 3 24 3
2. : 17 is the denominator
17 1
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3 17 11
4. 1: or 79 2
3 17 28.  7 79  11
7 7 7
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3 09
6. 3
4 7
2

20 5  4 5
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8.   15
24  4 6 124 4 1
30.  8 124  15  15
8 8 8 2
8 1 8 1 44
10.  
48 6  8 6 40
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4
72 18  4
12.  4
18 18  1
2 (4  5)  2 20  2 22
32. 4   
30 5   6 5 5 5 5
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14.  
85 5  17 17
1 (5  12)  1 60  1 61
34. 5   
33 11  3 3 12 12 12 12
16.  
55 11  5 5 5 (1  6)  5 6  5 11
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36. 1   
6 6 6 6
63 9  7 7
18.   7
9  1 2 (6  3)  2 18  2 20
38. 6   
3 3 3 3 3
19 4
20.  5 19  3
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5 5 1 (10  9)  1 90  1 91
15 40. 10   
4 9 9 9 9

78
7 42.  78  6
54 5 6
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22.  7 54  7
7 49 7 13
5 6 78
6
6 18
41 5
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24.  6 41  6 18
6 36 6 0
5 78
 13
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, Chapter 0: Prealgebra Review ISM: Beginning and Intermediate Algebra


5 ? 5  6 30 0.2 Exercises
44.   
9 54 9  6 54
2. You must change the fractions to equivalent
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fractions with a common denominator.
5 ? 5  5 25
46.   
9 45 9  5 45 21 17
4. and
30 20
13 ? 13  3 39 30 = 2  3  5
48.   
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17 51 17  3 51 20 = 2  2  5
LCD = 2  2  3  5 = 60
10 ? 10  4 40
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50.   
15 60 15  4 60 5 1
6. and
18 24
7 ? 7  5 35 18 = 2  3  3
52.   
8 40 8  5 40 24 = 2  2  2  3
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LCD = 2  2  2  3  3 = 72
45 ? 45  2 90
54.   
50 100 50  2 100 5 7
8. and
16 48
20 5 4 5 16 = 2  2  2  2
56.  
424 106  4 106 48 = 2  2  2  2  3
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LCD = 2  2  2  2  3 = 48
1400 70  20 70
58.  
2420 121 20 121 11 7
10. and
12 20
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3 3 12 = 2  2  3
60.  is raisins
1 3 1 5 20 = 2  2  5
LCD = 2  2  3  5 = 60
1 3 4
62.  does not contain nuts
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1 3 1 5 7 8
12. and
30 45
6 6 3 2 3 30 = 2  3  5
64.   
6  8 14 7  2 7 45 = 3  3  5
3 LCD = 2  3  3  5 = 90
of the students enrolled in 1994 were male.
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7 1 3 9
14. , , and
Classroom Quiz 0.1 7 14 35
7=7
56 7  8 7 14 = 2  7
1.  
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88 11 8 11 35 = 5  7
LCD = 2  5  7 = 70
11 (5  14)  11 70  11 81
2. 5    3 5 11
14 14 14 14 16. , , and
8 12 42
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109 1 8=222
3. 9 12 = 2  2  3
12 12
9 42 = 2  3  7
12 109 LCD = 2  2  2  3  7 = 168
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108
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