Instructor’s Solutions Manual for College Algebra 5th Edition by Beecher, Penna & Bittinger

College Algebra – 5th Edition

INSTRUCTOR’S
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SOLUTIONS
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MANUAL
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Judith A. Beecher, Judith A. Penna, Marvin L. Bittinger
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Complete Solutions Manual for Instructors and
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Students
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© Judith A. Beecher, Judith A. Penna & Marvin L. Bittinger

All rights reserved. Reproduction or distribution without permission is prohibited.
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©Medexcellence ✅��

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Contents
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
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Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . 57
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Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 107

Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 163

Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . 249
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Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 305

Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 357
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Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 399

Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 451

Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . 551
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Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . 635

Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . 677

Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 687
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, Chapter 1
Graphs, Functions, and Models
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4. y
Exercise Set 1.1
4 (1, 4)
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1. Point A is located 5 units to the left of the y-axis and 2
(
5, 0) (4, 0)
4 units up from the x-axis, so its coordinates are (−5, 4).
4
2 2 4 x
Point B is located 2 units to the right of the y-axis and
2
(
4,
2)
2 units down from the x-axis, so its coordinates are (2, −2).
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4 (2,
4)
Point C is located 0 units to the right or left of the y-axis
and 5 units down from the x-axis, so its coordinates are
(0, −5). 5. To graph (−5, 1) we move from the origin 5 units to the
left of the y-axis. Then we move 1 unit up from the x-axis.
Point D is located 3 units to the right of the y-axis and
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5 units up from the x-axis, so its coordinates are (3, 5). To graph (5, 1) we move from the origin 5 units to the right
of the y-axis. Then we move 1 unit up from the x-axis.
Point E is located 5 units to the left of the y-axis and
4 units down from the x-axis, so its coordinates are To graph (2, 3) we move from the origin 2 units to the right
(−5, −4). of the y-axis. Then we move 3 units up from the x-axis.
Point F is located 3 units to the right of the y-axis and To graph (2, −1) we move from the origin 2 units to the
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0 units up or down from the x-axis, so its coordinates are right of the y-axis. Then we move 1 unit down from the
(3, 0). x-axis.
To graph (0, 1) we do not move to the right or the left of
2. G: (2, 1); H: (0, 0); I: (4, −3); J: (−4, 0); K: (−2, 3); the y-axis since the first coordinate is 0. From the origin
L: (0, 5) we move 1 unit up.
3. To graph (4, 0) we move from the origin 4 units to the right
y
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of the y-axis. Since the second coordinate is 0, we do not
move up or down from the x-axis.
4
To graph (−3, −5) we move from the origin 3 units to the (2, 3)
2
left of the y-axis. Then we move 5 units down from the (
5, 1) (0, 1) (5, 1)
x-axis.
4
2 4 x
To graph (−1, 4) we move from the origin 1 unit to the left
2 (2,
1)
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of the y-axis. Then we move 4 units up from the x-axis.
4

To graph (0, 2) we do not move to the right or the left of
the y-axis since the first coordinate is 0. From the origin y
6.
we move 2 units up.
To graph (2, −2) we move from the origin 2 units to the 4
right of the y-axis. Then we move 2 units down from the (
5, 2)
2
x-axis. (
5, 0) (4, 0)
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4
2 2 4 x
y
2


4 (4,
3)
(
1, 4) 4 (
1,
5)
2 (0, 2)
(4, 0) 7. The first coordinate represents the year and the second co-

4
2 2 4 x
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ordinate represents the number of Sprint Cup Series races

2 (2,
2)
in which Tony Stewart finished in the top five. The or-
(
3,
5)
4 dered pairs are (2008, 10), (2009, 15), (2010, 9), (2011, 9),
(2012, 12), and (2013, 5).

8. The first coordinate represents the year and the second
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coordinate represents the percent of Marines who are
women. The ordered pairs are (1960, 1%), (1970, 0.9%),
(1980, 3.6%), (1990, 4.9%), (2000, 6.1%), and (2011, 6.8%).


Copyright 
c 2016 Pearson Education, Inc.

, 2 Chapter 1: Graphs, Functions, and Models


9. To determine whether (−1, −9) is a solution, substitute 12. For (1.5, 2.6): x2 + y 2 = 9
−1 for x and −9 for y.
(1.5)2 + (2.6)2 ?
9
y = 7x − 2

2.25 + 6.76

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−9 ?
7(−1) − 2 9.01
9 FALSE



−7 − 2 (1.5, 2.6) is not a solution.


−9
−9 TRUE For (−3, 0): x2 + y 2 = 9
The equation −9 = −9 is true, so (−1, −9) is a solution. (−3)2 + 02 ?
9
To determine whether (0, 2) is a solution, substitute 0 for

9+0



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x and 2 for y.
9
9 TRUE
y = 7x − 2
(−3, 0) is a solution.
2 ?
7 · 0 − 2  1 4
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13. To determine whether − , −

0−2 is a solution, substitute

2 5
2
−2 FALSE 1 4
− for a and − for b.
The equation 2 = −2 is false, so (0, 2) is not a solution. 2 5
  2a + 5b = 3
1  1  4
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10. For , 8 : y = −4x + 10
2 2 − +5 − ? 3
2 5


1

8 ? −4 · + 10 −1 − 4



2


−5
3 FALSE

−2 + 10

 1 4
8
8
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TRUE The equation −5 = 3 is false, so − , − is not a solu-
  2 5
1 tion.
, 8 is a solution.  3
2 To determine whether 0, is a solution, substitute 0 for
5
For (−1, 6): y = −4x + 10 3
a and for b.
5
6 ?
−4(−1) + 10

2a + 5b = 3
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4 + 10

3
6
14 FALSE 2·0+5· ? 3
5


(−1, 6) is not a solution.

0+3

2 3

11. To determine whether , is a solution, substitute
2 3
3 TRUE
3 4 3  3
3
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for x and for y. The equation 3 = 3 is true, so 0, is a solution.
4 5
6x − 4y = 1  3
14. For 0, : 3m + 4n = 6
2 3 2
6· −4· ? 1 3
3 4

3·0+4· ? 6

2


4−3



0+6

1
1 TRUE

6
6 TRUE
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2 3
The equation 1 = 1 is true, so , is a solution.  3
3 4 0, is a solution.
 3 2
To determine whether 1, is a solution, substitute 1 for 2 
2
3 For ,1 : 3m + 4n = 6
x and for y. 3
2 2
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6x − 4y = 1 3·
+4·1 ? 6
3


3

6·1−4· ? 1 2+4

2



6
6 TRUE


6−6
2 


0
1 FALSE The equation 6 = 6 is true, so , 1 is a solution.
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3
 3
The equation 0 = 1 is false, so 1, is not a solution.
2


Copyright 
c 2016 Pearson Education, Inc.