Chapter 3 Linear Equations in Two Variables
63. First we need to find the equation of the line with
x−intercept of 2 and
y−
intercept of
4.−
This
line will pass through
()
2, 0
and
()
0, 4 .−
We use
these points to find the slope.
Use the point
()
6, 4−
and the slope
1
2
−
to find the
equation of the line.
()
()
()
11
1
46
2
yy mxx
yx
−= −
−=− −−
64. First we need to find the equation of the line with
x−intercept of 3 and
y−
intercept of
9.−
This
line will pass through
()
3, 0
and
()
0, 9 .−
We use
these points to find the slope.
A
Use the point
()
5, 6−
and the slope
1
3
−
to find the
equation of the line.
A
A
()
()
()
11
1
65
3
yy mxx
yx
−= −
−=− −−
65. First put the equation 32 4xy−=
in slope-intercept
form.
32 4
234
xy
yx
−=
−=−+
y−
intercept,
2.−
So the equation is
22.
3
yx=− −
66. First put the equation 46xy−=in slope-intercept
form.
y−
intercept,
6.−
So the equation is
16.
4
yx=− −
67. To find the slope of the line whose equation is
,
A
xByC+=
put this equation in slope-intercept
The slope of this line is
A
mB
=−
so the slope of the
line that is parallel to it is the same,
.
A
B
−
68. To find the slope of the line whose equation is
,
A
xByC+=
put this equation in slope-intercept
Section 3.5 The Point-Slope Form of the Equation of a Line
69. a. Find slope:
23,300 20,000 3300 1100
63 3
m−
===
−
()
yy mxx
−= −
70. a. Find slope:
25.8 24.6 1.2 0.12
20 10 10
m−
===
−
()
11
yy mxx
−= −
71. – 72. Answers will vary.
73. does not make sense; Explanations will vary.
Sample explanation: In this situation, a better choice
would be to use the point-slope form.
74. does not make sense; Explanations will vary.
77. false; Changes to make the statement true will vary.
A sample change is: If a line has undefined slope,
then its equation is of the form
,
xc
=
where c is a
constant.
80. false; Changes to make the statement true will vary.
A sample change is: Solving this line for y shows
that its slope is
3.−
82. Answers will vary.
83. a.
010203032
x
c.
0.1358315098
22.94070022
0.9897376983
a
b
r
=
=
=
d. Graph of regression equation:
84. Let x = the number of sheets of paper.
42 29
225
25 1
x
x
+≤
≤
Chapter 3 Linear Equations in Two Variables
85. The only natural numbers in the given set are 1 and
()
42=
.
86. 35 15xy−=
x-intercept:
()
35015
x
−=
87.
22
42(1) 2
42 2
22,true
xy+=
+−=
−=
=
88.
22
42(3) 2
46 2
22,true
xy+=
−+ =
−+ =
=
89. Graph 23 6xy+=
by finding intercepts.
Find the x-intercept.
23 6
23(0)6
26
3
xy
x
x
x
+=
+=
=
=
Find the y-intercept.
22
2(0) 2
2
xy
y
y
+=−
+=−
=−
Chapter 3 Review Exercises
1. Quadrant IV
Chapter 3 Review Exercises
3. Quadrant I
4. Quadrant II
5.
(5, 6)A
(3,0)B−
(5,2)C−
(4, 2)D−−
(0, 5)E−
(3, 1)F−
6.
()
36
333 6
369
33, false
yx=+
=−+
=− +
=−
()
3,3 is not a solution.−
7.
()
312
30 4 12
412, false
xy−=
−=
−=
()
312
3 1 15 12
xy−=
−− =
8. a.
() ( )
() ( )
() ( )
23 (,)
222372,7
121351,5
020330,3
xyx xy
y
y
y
=−
−=−−=−−−
−=−−=−−−
=−=− −
9. a.
() ( )
1
2
1
1(,)
22102,0
xyx xy
y
=+
−=−+=−
Chapter 3 Linear Equations in Two Variables
11. a. The graph does not cross the x-axis, so there is
no x-intercept.
b. The graph crosses the y-axis at (0,2), so the y–
intercept is 2.
13. Find the x-intercept. Let y = 0 and solve for x.
24
204
24
2
xy
x
x
x
+=
+=
=
=
Find a checkpoint. For example, let x = 1 and solve
for y.
24
2(1) 4
24
xy
y
y
+=
+=
+=
14. Find the x-intercept. Let y = 0 and solve for x.
32 12
32(0)12
xy
x
−=
−=
The y-intercept is –6.
Find a checkpoint. For example, let x = 2 and solve
for y.
32 12
3(2) 2 12
xy
y
−=
−=
Draw the line through these three points.
2
x
=
The x-intercept is 2.
Find the y– intercept. Let x = 0 and solve for y.
362
3(0) 6 2
xy
y
=−
=−
3
y
−=
The checkpoint is (4,–3).
Draw the line through these three points.
Chapter 3 Review Exercises
16. Because the constant on the right is 0, the graph
passes through the origin. The x– and y-intercepts
are both 0.
Thus we will need to find two more points.
Let x = 1 and solve for y.
30
xy
−=
This gives the point (–1,–3).
Draw the line through these three points.
17. 3x=
Three ordered pairs are (3, −2), (3,0), and (3,2). The
graph is a vertical line.
18. 5y=−
Three ordered pairs are (−2, −5), (0, −5), and
(2, −5). The graph is a horizontal line.
19. 35
2
y
y
+=
=
Three ordered pairs are (−2,2), (0,2), and (2,2). The
graph is a horizontal line.
21. a. The minimum temperature occurred at 5 P.M.
and was −4°F.
b. The maximum temperature occurred at 8 P.M.
and was at 16°F.
c. The x-intercepts are 4 and 6. This indicates that 4
P.M. and 6 P.M., the temperature was 0°F.
22.
21
21
12 1
53 2
yy
mxx
−−
===−
−−
The slope is 1
2
−. Since the slope is negative, the
line falls from left to right.
Chapter 3 Linear Equations in Two Variables
25. 33 6
;
55 0
m
−− −
==
− undefined
Since the slope is undefined, the line is vertical.
30. Line through (−1, −3) and (2, −8):
()
()
83 55
21 3 3
m
−−− −
===−
−−
31. Line through (0,−4) and (5, −1):
1(4) 3
m
−−−
==
32. Line through (5,4) and (9,7):
74 3
95 4
m
−
==
−
33. a. 52 64 12 0.48
2010 1985 25
m
−−
===−
−
b. For each year from 1985 through 2010, the
35. 64
46
yx
yx
=−
=− +
4;
m
=−
y-intercept = 6
3
22
3
yx
=− +
2;
right to reach the point (1, −2). Draw a line through
(0, −4) and (1, −2).
39. 11
2
yx
=−
slope = 1;
40. 25
3
yx
=− +
slope = 22
;
33
−
−=
y-intercept = 5
Plot (0,5). Move 2 units down (because −2 is
41. 20
2
yx
yx
−=
=
slope = 2 = 2
1; y-intercept = 0
42. 12
3
12
xy
yx
+=
=− +
43. 14
2
yx
=− +
slope = 11
−
−=
Chapter 3 Linear Equations in Two Variables
44. a. 19.0 16.6 2.4 0.8
30 3
m
−
===
−
The y-intercept is 16.6 as shown in the graph.
ymxb
=+
45. Slope = 6, passing through (−4,7)
point-slope form:
()
11
()
76 4
yy mxx
yx
−= −
−= −−
46. Passing through (3,4) and (2,1)
First, find the slope.
14 3 3
m
−−
===
47. Rewrite 390xy+−= in slope-intercept form.
390
39
xy
yx
+−=
=− +
Since the line we are concerned with is parallel to
this line, we know it will have slope 3.
m
=−
We are
48. The line is perpendicular to 14,
3
yx
=+
so the
slope is –3. We are given that it passes through
(–2, 6). We use the slope and point to write the
636
3
yx
yx
−=− −
=−
49. a. First, find the slope.
5.3 3.7 1.6 0.08
m
−
===
3.7 0.08 1.6
0.08 2.1
yx
yx
−= −
=+
b.
0.08 2.1
yx
=+
0 10 10
10 10, true
+=
=
()
0, 5 is a solution.−
() ()
4210
42 21 10
xy−=
−− =
Chapter 3 Test
2.
() ( )
() ( )
31 (,)
232152,5
131121,2
xyx xy
y
y
=+
−=−+=−−−
−=−+=−−−
3. a. The graph crosses the x–axis at (2,0), so the
x-intercept is 2.
4. Find the x-intercept. Let y = 0 and solve for x.
42 8
42(0) 8
xy
x
−=−
−=−
The y-intercept is 4.
Find a checkpoint. For example, let x = –1 and solve
for y.
42 8
xy
−=−
5. 4y=
The graph is a horizontal line.
7.
()
31
4;
m−−
==
undefined
8. Use the points (−1, −2) and (1,1).
()
()
12
3
112
m−−
==
−−
4
lines are perpendicular.
10. Line through (2,4) and (6,1):
11.
10
110
yx
yx
=− +
=− +
The slope is the coefficient of x, which is −1. The y–
intercept is the constant term, which is 10.
Chapter 3 Linear Equations in Two Variables
13.
21
3
yx=−
slope =
2
3
; y-intercept = −1
14. 23yx=− +
slope = −2 =
2
1
−
; y-intercept = 3
Plot (0,3). Move 2 units down and 1 unit to the right
15. point-slope form:
()
yy mxx
−= −
16. Passing through (2,1) and (−1, −8)
First, find the slope.
81 9 3
m−− −
===
Solve for y to obtain slope-intercept form.
()
13 2
13 6
35
yx
yx
yx
−= −
−= −
=−
Solve for y to obtain slope-intercept form.
()
32 2
32 4
27
yx
yx
yx
−= +
−= +
=+
Therefore the slopes are the same;
1.
2
m=−
Solve for y to obtain slope-intercept form.
()
1
46
2
1
43
2
yx
yx
+=− −
+=− +
Cumulative Review
19. a. 27 21 6 0.2
2010 1980 30
m−
===
−
Cumulative Review Exercises (Chapters 1-3)
1.
()
()
2
10 6 10 6 16 2
91 8
343
−− +
===
−
−−
2.
()
623 1 4
x
−−+
The solution set is {19}.
5. 31
42
33 13
x
x
−=
−+=+
6.
ymxb
yb mxbb
=+
−= +−
7. 120; 15% 0.15AP===
APB
=
8.
4.5 46.7
133.3 4.5 46.7
133.3 46.7 4.5 46.7 46.7
180 4.5
180 4.5
4.5 4.5
40
yx
x
x
x
x
x
=−
=−
+=−+
=
=
=
2
x
≤−
(
]
,2−∞ −
10.
()
62 12
0
x
x
−>
<
()
,0−∞
Chapter 3 Linear Equations in Two Variables
12. Let x= the width of a football field.
Let
214x+=
the length of a football field.
22
346 2(2 14) 2( )
Plw
xx
=+
=++
13. Let x= the weight before the 10% loss.
0.10 180
0.9 180
xx
x
−=
=
14. Let x = the measure of the first angle.
Let x + 20 = the measure of the second angle.
Let 2x = the measure of third angle.
()
20 2 180
xx x
++ + =
15.
22
10 ( 3) 10( 3)
xx−=−−−
16.
2000 3−<−
17. x-intercept:
24
204
xy
x
−=
−=
checkpoint:
24
2(1) 4
xy
y
−=
−=
20. 1y=−