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Exam (elaborations)

Beginning Algebra – 9th Edition | John Tobey | Instructor’s Solutions Manual | Complete Worked-Out Answers

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This Instructor’s Solutions Manual is designed to accompany Beginning Algebra (9th Edition) by John Tobey. It contains comprehensive, step-by-step solutions to all exercises across every chapter, including chapter reviews and tests. This manual is a valuable resource for instructors and students seeking detailed explanations to reinforce foundational algebra skills.

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Institution
Beginning Algebra, 9th Edition
Course
Beginning Algebra, 9th edition











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Institution
Beginning Algebra, 9th edition
Course
Beginning Algebra, 9th edition

Document information

Uploaded on
May 15, 2025
Number of pages
296
Written in
2024/2025
Type
Exam (elaborations)
Contains
Questions & answers

Subjects

Content preview

INSTRUCTOR’S
YTREW



SOLUTIONS MANUAL

BEGINNING ALGEBRA
NINTH EDITION
ST

John Tobey
U
D
ALL CHAPTERS INCLUDED

ALL ANSWERS INCLUDED
YL
AB

, Table of Contents


Chapter 0 1

Chapter 1 18

Chapter 2 37
ST
Chapter 3 71

Chapter 4 92

Chapter 5 114
U
Chapter 6 135
D
Chapter 7 165
YL
Chapter 8 198

Chapter 9 232
AB
Chapter 10 256

Practice Final Examination 287

, YTREW




Chapter 0
0.1 Exercises 8
25 1
26. = 3 25 = 8
13 3 3
2. : 17 is the denominator 24
17 1
3 17 11
4. 1: or 79 2
3 17 28. = 7 79 = 11
7 7 7
3 09
6. 3
4 7
2
ST
20 5 4 5 15
8. = = 124 4 1
24  4 6 30. = 8 124 = 15 = 15
8 8 8 2
8 1 8 1
10. = = 44
48 68 6 40
U
4
72 18 4
12. = =4
18 181
2 = (4  5) + 2 = 20 + 2 22
32. 4 =
5 5 5 5
D
32 32 1 1
14. = =
64 32  2 2
1 (512) + 1 60 + 1 61
34. 5 = = =
33 11 3 3 12 12 12 12
= =
YL
16.
55 11 5 5 5 (1 6) + 5 6 + 5 11
36. 1 = = =
6 6 6 6
63 9  7 7
18. = =
81 9  9 9
38. 6 2 = (6  3) + 2 = 18 + 2 = 20
3 3 3 3 3
19 4
= 5 19 = 3
AB
20.
5 5 1 (10  9) + 1 90 + 1 91
15 40. 10 = = =
4 9 9 9 9

78
54
7
5 42. = 78  6
22. = 7 54 = 7 6
7 49 7 13
5 6 78
6
6
41 5 18
24. = 6 41 = 6 18
6 36 6 0
5 78
= 13
6




Copyright © 2017 Pearson Education, Inc. 1

, Chapter 0: A Brief Review of Arithmetic Skills ISM: Beginning Algebra


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


2 Copyright © 2017 Pearson Education, Inc.

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