Section 3.2 Graphing Linear Equations Using Intercepts
39. 23 11xy
x-intercept:
11
2
40. 32 7xy
x-intercept:
7
3
y-intercept:
7
41. The equation for this horizontal line is 3.y
42. The equation for this horizontal line is 5y .
43. The equation for this vertical line is
3.x
47. 4y
All ordered pairs that are solutions will have a value
of y that is 4. Any value can be used for x. Three
ordered pairs that are solutions are (−2,4), (0,4), and
All ordered pairs that are solutions will have a value
of y that is 2. Any value can be used for x. Three
ordered pairs that are solutions are
4, 2 , 0, 2 ,
and
3, 2 .
Plot these points and draw the line
through them. The graph is a horizontal line.
Chapter 3 Linear Equations in Two Variables
51.
2x
All ordered pairs that are solutions will have a value
of x that is 2. Any value can be used for y. Three
52.
4x
All ordered pairs that are solutions will have a value
of x that is 4. Any value can be used for y. Three
53.
10
1
x
x
Three ordered pairs are (−1,−3), (−1,0), and (−1,3).
The graph is a vertical line.
54.
50
5
x
x
Three ordered pairs are
5, 2
,
5, 0
,
and
5, 3 .
55.
3.5 0
3.5
y
y
56.
2.5 0
2.5
y
y
Three ordered pairs are
2, 2.5 , 0, 2.5 ,
and
The graph is a vertical line, the y-axis.
Section 3.2 Graphing Linear Equations Using Intercepts
59.
39
3
y
y
60.
520
4
y
y
Three ordered pairs are
3, 4 , 0, 4 ,
and
2, 4 .
61.
12 3 0
312
4
x
x
x
62.
12 4 0
412
3
x
x
x
Three ordered pairs are
3, 2 , 3, 0 ,
and
3, 3 .
63. Using intercepts, we see that 32 6xy
corresponds to Exercise 4.
3
y
64. Using intercepts, we see that 24xy
corresponds to Exercise 3.
x-intercept:
20 4
x
66. Since
3x
is a vertical line at
3
, it corresponds
to Exercise 8.
67. Using intercepts, we see that 4312xy
4
y
68. Using intercepts, we see that 25 10xy
corresponds to Exercise 2.
x-intercept:
2510
xy
Chapter 3 Linear Equations in Two Variables
69. a. Let x + 5 + 5 = x + 10 = the length.
Let y + 8 = width.
Using the formula for the perimeter of a
b. x and y must be non-negative because they are
dimensions.
70. a. The base of the trapezoid has length x and the
top has length 2xy. The two sides each have
b. The total perimeter is 84 feet and two sides are
each 25 feet. This leaves 34 feet for the
remaining two sides. Thus, the largest value
71. The eagle’s height is decreasing from 3 seconds to
12 seconds.
74. The equation is 45.y
This means that the eagle’s height remains constant
3000 24, 000
8
x
x
After 8 years, the car is worth nothing.
c. x and y must be non-negative because they
represent time and the car’s value.
After 9 years, the car is worth nothing.
b.
5000 45,000
yx
Section 3.2 Graphing Linear Equations Using Intercepts
c. x and y must be non-negative because they
represent time and the car’s value.
78. – 86. Answers will vary.
87. makes sense
88. does not make sense; Explanations will vary.
Sample explanation: The checkpoint must be a point
91. Since the x-intercept is 5, y = 0 when x = 5.
? 10
?5 0 10
?5 10
xy
92. Since the x-intercept is −2, y = 0 when x = −2.
? 12
?2 012
?212
xy
So, the coefficient of y is 3.
The equation of the line is 6312xy .
93. Answers will vary.
94.
24
xy
The y–intercept is 4.
The x–intercept is 2.
95.
39
39
1139
xy
yx
yx
Chapter 3 Linear Equations in Two Variables
96.
23 30
3230
xy
yx
The y–intercept is 20.
The x–intercept is
10.
98.
13.4 13.4
99.
73573545xx xx x
100.
8( 2) 2( 3) 8
816268
xxx
xxx
101.
21
21
13 3 10 2
61 5
yy
xx
102.
21
21
4(2) 2 1
64 2
yy
xx
3.3 Check Points
1. a. Let
11 2 2
,3,4 and , 4,2.xy x y
21
The slope is
7
5
. Since the slope is negative,
the line falls from left to right.
2. a. Let
11 2 2
, 6,5 and , 2,5 .xy xy
21
21
Change in 5 5 0 0
Change in 2 6 4
yyy
mxxx
Since the slope is 0, the line is horizontal.
b. Let
11 2 2
,1,6 and , 1,4.xy x y
21
21
Change in 4 6 2
Change in 1 1 0
yyy
mxxx
Because division by 0 is undefined the slope is
Since their slopes are equal, the lines are parallel.
Section 3.3 Slope
4. Line through (–1,4) and (3,2):
Change in 2 4 2 1
Change in 3 ( 1) 4 2
y
mx
5. Let
11 2 2
, 2000,11.2 and , 2013,15.0 .xy x y
3.3 Concept and Vocabulary Check
1.
21
21
yy
xx
2. y; x
3.3 Exercise Set
1. Let
11 2 2
,4,7 and ,8,10.xy x y
21
21
Change in 10 7 3
Change in 8 4 4
yyy
mxxx
3. Let
11 2 2
, 2,1 and , 2,2 .xy x y
21
Change in 2 1 1
yyy
mxxx
21
Change in 2 1 3
Since the slope is positive, the line rises from left to
right.
6. Let
11 2 2
, 4, 1 and , 3, 1 .xy xy
21
21
11
Change in 0 0
Change in 3 4 1
yyy
mxxx
Since the slope is zero, the line is horizontal.
7. Let
11 2 2
,2,4 and , 1,1.xy x y
Change in 1 4 5 5
yyy
9. Let
11 2 2
, 5,3 and , 5, 2 .xy xy
21
21
Change in 2 3 5
Change in 5 5 0
yyy
mxxx
Since the slope is undefined, the line is vertical.
Chapter 3 Linear Equations in Two Variables
12. Line through
2, 3 and 3,2 :
23
51
325
m
15. Line through (−2,1), (0,0), and (2,−1)
Use any two of these points to find the slope.
01 1 1
022 2
m
16. Line through
3,4 , 0, 2 and 3,0 :
Use any two of these points to find the slope.
19. Line through (−2,1) and (4,1):
11 0 0
426
m
(Since the line is horizontal, it is not necessary to do
this computation. The slope of every horizontal line
is 0.)
20. Line through
2, 3 and 3, 3 :
21. Line through (−3,4) and (−3,−2):
24 6
;
330
m
undefined
(Since the line is vertical, it is not necessary to do
undefined.)
23. Line through (−2,0) and (0,6):
60 3
02
m
Line through (1,8) and (0,5):
58 3 3
01 1
m
25. Line through (0,3) and (1,5):
53 2 2
10 1
m
Line through (−1,7) and
1,10
:
10 7 3
112
m
Since their slopes are not equal, the lines are not
Section 3.3 Slope
27. Line through
1,5 and 0,3:
35 2
01
m
28. Line through
3, 2 and 2, 2 :
22 4
m
29. Line through
1, 6 and 2, 9 :
9(6) 5
2(1)
m
30. Line through
1, 6 and 2, 6 :
6(6) 4
2(1)
m
Line through
8, 1 and 4, 2 :
31. Line through
2, 5 and 3,10 :
10 ( 5) 3
3(2)
m
13 ( 7) 4
3(2)
m
Line through
1, 9 and 5,15 :
Line through
0, 5 and 2, 4 :
4(5) 1
20 2
m
6(8) 1
12 0 6
m
Since their slopes are not equal, nor are the slopes
negative reciprocals, the lines are neither parallel
nor perpendicular.
Chapter 3 Linear Equations in Two Variables
36. Line through
2, 15 and 0, 3 :
3(15) 6
0(2)
m
37.
38.
39.
21
21
31 4 4
30 3 3
yy
mxx
21
11 2 2
yy
mxx
40.
21
1
21
36 9
23 5
23
5
yy
mxx
yy
3, 6
and
2, 3
is parallel to the line connecting
11,2
and
6,11
. Since
2
m
and
4
m
are the same,
the line connecting
3, 6
and
6,11
is parallel to
the line connecting
2, 3
and
11,2
.
x
y
10-10
10
and (−2, 1).
21
21
31 2 1
2242
yy
mxx
Now, use the slope formula, the slope and the points
(5, y) and (1, 0) to find y.
10
y
Section 3.3 Slope
42. First find the slope of the line passing through
3, 4
and
5, 2
.
43. Find the slope of the line passing through
1, y
and
1, 0 .
21
21
0
1(1) 2
yy y y
mxx
4
y
44. Find the slope of the line passing through
2, y
and
4, 4 .
21
21
44
2(4) 2
yy y y
mxx
Find the slope of the line passing through
1, 2
Since the lines are perpendicular, the product of
their slopes is
1.
.
by 0.5%. The rate of change is –0.5% per year
of aging.
46. a.
Change in 42 26 16 0.16
Change in 180 80 100
y
mx
b. For each minute of brisk walking, the percentage
49. The grade of an access ramp is
1 foot 1 0.083 8.3%.
12 feet 12
50. The grade of this ramp is
1 foot 1 0.071 7.1%
14 feet 14
Chapter 3 Linear Equations in Two Variables
60. does not make sense; Explanations will vary.
Sample explanation: The slopes of the lines are both
negative so they can not be perpendicular to each
other.
64. false; Changes to make the statement true will vary.
A sample change is: The slope of the line is
undefined.
65. The positive slopes are
1
m
and
2
m
, and of these,
the line with slope
m
is the steeper one so has the
66. Use the graph to observe where each line crosses
the y-axis. In order of decreasing size, the y–
intercepts are
2143
,, ,.bbbb
67. 24yx
68. 36yx
Two points on the graph are (0, 6) and (1, 3).
63 3 3
01 1
m
70.
32
4
yx
Two points on the graph are (−4,−5) and (8,4).
4593
336
12
x
x
The pieces are 12 inches and 24 inches.
73.
10 16 2 4 10 8 4
10 32
,4
75.
1 unit 4 units
right up
(0 1 , 3 4 ) (1,1)
225 02
52
2
5
xxy x
yx
x
y
Section 3.4 The Slope-Intercept Form of the Equation of a Line
3.4 Check Points
1. a. 53yx
The slope is the x-coefficient, which is 5.
m
The y-intercept is the constant term, which is –3.
b. 24
3
yx
2. 32yx
The y-intercept is –2, so plot the point (0, 2).
The slope is 3
3or .
mm
Find another point by
3. 31
5
yx
The y-intercept is 1, so plot the point (0,1).
The slope is 3.
5
m
Find another point by going up
3 units and to the right 5 units.
4.
34 0
43
3
4
xy
yx
yx
The y-intercept is 0, so plot the point (0,0).
The slope is
3.
4
m
Find another point by going
The model projects that 3% of Americans will
improve their mood by shopping in 2023.
3.4 Concept and Vocabulary Check
Chapter 3 Linear Equations in Two Variables
3.
35
35
yx
yx
6.
36
4
3; –intercept = 6
4
yx
my
7.
7
yx
9.
10
010
0; -intercept 10
y
yx
my
12.
5
515
1; –intercept =5
yx
yx x
my
14.
95
9959
xy
xy x x
17.
60
660
xy
yxx
6; -intercept 0my
18.
80
xy
20.
39
3
3; -intercept = 0
yx
yx
my
92
2
9
yx
yx
2; -intercept = 0
Section 3.4 The Slope-Intercept Form of the Equation of a Line
23.
32 3
233
xy
yx
24.
43 4
344
344
xy
yx
yx
25.
34 12
4312
xy
yx
26.
52 10
2510
xy
yx
27. 24yx
Step 1. Plot (0,4) on the y-axis.
Step 2.
2rise
1run
m
Start at (0,4). Using the slope, move 2 units up (the
28. 31yx
Step 1 Plot
0,1
on the y-axis.
Step 3 Draw a line through
0,1 and 1,4.
29. 35yx
Slope =
3
31
; y-intercept = 5.
1
Plot
0, 4
on the y-axis. From this point, move 2
units down (because −2 is negative) and 1 unit to
the right to reach the point
1, 2 .
Draw a line
through
0,4 and 1, 2 .
Chapter 3 Linear Equations in Two Variables
31.
11
2
yx
Slope =
1
; y-intercept = 1
32.
12
3
yx
Slope =
1
3
;y-intercept = 2
Plot
0, 2 .
From this point, move 1 unit up and 3
33.
25
3
yx
Slope =
2;
3
y-intercept = −5
Plot (0,−5). From this point move 2 units up and 3
34.
34
4
yx
Slope =
3;
y-intercept = −4
35.
32
4
yx
Slope =
33
;
44
y-intercept = 2
36.
24
3
yx
22
;
33
m
y-intercept = 4
Section 3.4 The Slope-Intercept Form of the Equation of a Line
37.
5
3
yx
Slope =
55
;
33
y-intercept = 0
Plot (0,0). From this point, move 5 units down and 3
38.
4
3
yx
44
33
m
;y-intercept = 0
Plot
0,0 .
From this point (the origin), move 4
39. a.
30
3
xy
yx
b.
3; m
y-intercept = 0
c. Plot (0,0). Since
3
31
m
, move 3 units
40. a.
20
2
xy
yx
b.
2;m
y-intercept = 0
2
41. a.
34
4
3
yx
yx
42. a.
45
5
4
yx
yx
b.
5;
m
y-intercept = 0
Chapter 3 Linear Equations in Two Variables
43. a.
23
23
xy
yx
b.
2;m
y-intercept = 3
44. a.
34
34
xy
yx
b.
3;m
y-intercept = 4
45. a.
72 14
2714
2714
xy
yx
yx
c. Plot (0,7). Since
77
22
m
, move 7 units
down and 2 units to the right to reach the point
(2,0).
Draw a line through (0,7) and (2,0).
55
3
yx
b.
5
3
m
; y-intercept = 5
3; -intercept 1
33:
3; -intercept 3
my
yx
my
The lines are parallel because their slopes are equal.
Section 3.4 The Slope-Intercept Form of the Equation of a Line
48.
24:
2; -intercept = 4
23
yx
my
yx
49.
32:
3; -intercept 2
32:
3; -intercept 2
yx
my
yx
my
50.
21
2; -intercept = 1
21
yx
my
yx
51.
3
1; -intercept = 3
1
yx
my
yx
52.
2
1; -intercept = 2
1
yx
my
yx
53.
1
22 1
2
13
xy y x
Chapter 3 Linear Equations in Two Variables
54.
1
39 3
3
xy y x
55.
21 21
1
26 3
2
xy y x
xy y x
The lines are perpendicular because the product of
56.
32 32
1
39 3
3
xy y x
xy y x
The lines are perpendicular because the product of
57. Find the slope of the parallel line.
36
36
xy
yx
58. Find the slope of the parallel line.
28
28
xy
yx
60. The slope of the line 83yx
is 5. The negative
reciprocal of 8 is
1.
8
We are given that the y-intercept is 7, so using the
Find the slope of the parallel line.
33 9
339
3
xy
yx
yx
44 20
4420
xy
yx