Solution Manual – College Algebra: Graphs and Models, 7th Edition – Marvin L. Bittinger, Judith A. Beecher, Judith A. Penna, Barbara L. Johnson – ISBN 9780138240691 (All Chapters Covered 1–8)

College Algebra:
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Graphs and Models
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7th Edition
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SOLUTIONS
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MANUAL
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Marvin L. Bittinger
Judith A. Beecher
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Judith A. Penna
Barbara L. Johnson

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Comprehensive Solutions Manual for Instructors and Students

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9780138240691
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© Marvin L. Bittinger, Judith A. Beecher, Judith A. Penna, & Barbara L. Johnson. All rights
reserved. Reproduction or distribution without permission is prohibited.




© MEDCONNOISSEUR

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College Algebra: Graphs and Models — Solutions Manual
Marvin L. Bittinger, Judith A. Beecher, Judith A. Penna and Barbara L. Johnson
ISBN: 9780138240691
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Chapter 1: Graphs, Functions, and Models
Chapter 2: More on Functions
Chapter 3: Quadratic Functions and Equations; Inequalities
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Chapter 4: Polynomial Functions and Rational Functions
Chapter 5: Exponential Functions and Logarithmic Functions
Chapter 6: Systems of Equations and Matrices
Chapter 7: Conic Sections
Chapter 8: Sequences, Series, and Combinatorics
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© MEDCONNOISSEUR

,Chapter 1
Graphs, Functions, and Models
To graph (−1, 4) we move from the origin 1 unit to the
Check Your Understanding Section 1.1 left of the y-axis. Then we move 4 units up from the
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x-axis.
1. The point (−5, 0) is on an axis, so it is not in any quadrant. To graph (0, 2) we do not move to the right or the left of
The statement is false. the y-axis since the first coordinate is 0. From the origin
we move 2 units up.
2. The ordered pair (1, −6) is located 1 unit right of the origin
To graph (2, −2) we move from the origin 2 units to the
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and 6 units below it. The ordered pair (−6, 1) is located 6
units left of the origin and 1 unit above it. Thus, (1, −6) right of the y-axis. Then we move 2 units down from the
and (−6, 1) do not name the same point. The statement x-axis.
is false. y

3. True; the first coordinate of a point is also called the
( 1, 4) 4
abscissa.
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2 (0, 2)
(4, 0)
4. True; the point (−2, 7) is 2 units left of the origin and
4 2 2 4 x
7 units above it. 2 (2, 2)

5. True; the second coordinate of a point is also called the ( 3, 5) 4
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ordinate.

6. False; the point (0, −3) is on the y-axis. 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.
To graph (5, 1) we move from the origin 5 units to the right
Exercise Set 1.1 of the y-axis. Then we move 1 unit up from the x-axis.
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To graph (2, 3) we move from the origin 2 units to the right
1. Point A is located 5 units to the left of the y-axis and of the y-axis. Then we move 3 units up from the x-axis.
4 units up from the x-axis, so its coordinates are (−5, 4).
To graph (2, −1) we move from the origin 2 units to the
Point B is located 2 units to the right of the y-axis and right of the y-axis. Then we move 1 unit down from the
2 units down from the x-axis, so its coordinates are (2, −2). x-axis.
Point C is located 0 units to the right or left of the y-axis
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To graph (0, 1) we do not move to the right or the left of
and 5 units down from the x-axis, so its coordinates are the y-axis since the first coordinate is 0. From the origin
(0, −5). we move 1 unit up.
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5). y

Point E is located 5 units to the left of the y-axis and 4
(2, 3)
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4 units down from the x-axis, so its coordinates are
2
(−5, −4). ( 5, 1) (0, 1) (5, 1)
4 2 4 x
Point F is located 3 units to the right of the y-axis and 2 (2, 1)
0 units up or down from the x-axis, so its coordinates are
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(3, 0).
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3. To graph (4, 0) we move from the origin 4 units to the right
7. The first coordinate represents the year and the corre-
of the y-axis. Since the second coordinate is 0, we do not
sponding second coordinate represents the number of cities
move up or down from the x-axis.
served by Southwest Airlines. The ordered pairs are
To graph (−3, −5) we move from the origin 3 units to the (1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72),
left of the y-axis. Then we move 5 units down from the and (2021, 121).
x-axis.




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c 2025 Pearson Education, Inc.

, 14 Chapter 1: Graphs, Functions, and Models


9. To determine whether (−1, −9) is a solution, substitute 2a + 5b = 3
−1 for x and −9 for y. 3
2·0+5·
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y = 7x − 2 5 

−9 ? 7(−1) − 2 0+3 
 
 −7 − 2 3  3 TRUE
  3

−9 −9 TRUE
The equation 3 = 3 is true, so 0, is a solution.
The equation −9 = −9 is true, so (−1, −9) is a solution. 5
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To determine whether (0, 2) is a solution, substitute 0 for 15. To determine whether (−0.75, 2.75) is a solution, substi-
x and 2 for y. tute −0.75 for x and 2.75 for y.
y = 7x − 2 x2 − y 2 = 3

2 ? 7 · 0 − 2 (−0.75)2 − (2.75)2 ? 3
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 0−2 0.5625 − 7.5625 
 

2 −2 FALSE −7  3 FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution. The equation −7 = 3 is false, so (−0.75, 2.75) is not a
2 3 solution.
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11. To determine whether , is a solution, substitute To determine whether (2, −1) is a solution, substitute 2
3 4 3
for x and −1 for y.
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3
for x and for y.
4 x2 − y 2 = 3
6x − 4y = 1
22 − (−1)2 ? 3
2 3 
6· −4· ? 1 4−1 

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3 4  3  3 TRUE

4−3  The equation 3 = 3 is true, so (2, −1) is a solution.

1  1 TRUE
2 3 17. Graph 5x − 3y = −15.
The equation 1 = 1 is true, so , is a solution. To find the x-intercept we replace y with 0 and solve for
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x.
To determine whether 1, is a solution, substitute 1 for 5x − 3 · 0 = −15
2
3
x and for y. 5x = −15
2
x = −3
6x − 4y = 1
The x-intercept is (−3, 0).
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3
6·1−4· ? 1 To find the y-intercept we replace x with 0 and solve for
2 
 y.
6−6 
 5 · 0 − 3y = −15
0  1 FALSE
 3 −3y = −15
The equation 0 = 1 is false, so 1, is not a solution. y=5
2
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 1 4 The y-intercept is (0, 5).
13. To determine whether − , − is a solution, substitute We plot the intercepts and draw the line that contains
2 5
1 4 them. We could find a third point as a check that the
− for a and − for b.
2 5 intercepts were found correctly.
2a + 5b = 3
y
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 1  4 (0, 5)
2 − +5 − ? 3 4
2 5 
5x 3y 15
 2
−1 − 4  ( 3, 0)
 4 2 2 4
−5  3 FALSE 2
x
 1 4
4
The equation −5 = 3 is false, so − , − is not a solu-
2 5
tion.
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To determine whether 0, is a solution, substitute 0 for
5
3
a and for b.
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c 2025 Pearson Education, Inc.