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Chapter 3 Systems of Linear Equations

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Chapter 3
Systems of Linear Equations
Homework 3.1
1.
5. Write y = 3 ( x в€’ 1) in slope-intercept form.
y
y = 3 ( x в€’ 1)
y = 2x +2
y = 3x в€’ 3
4
в€’2
(1,4)
y
в€’2
8
x
4
y = 3 x в€’3
(2, 3)
y = в€’3 x + 7
в€’8
Verify that (1, 4 ) satisfies both equations.
y = 2x + 2
y = в€’3 x + 7
4 = 2 (1) + 2
4 = в€’3(1) + 7
4 = 2+2
4 = в€’3 + 7
4 = 4 true
4 = 4 true
3.
в€’8
y = в€’2 x + 7
Verify that ( 2,3) satisfies both equations.
(в€’6, 6)
y = 3 ( x в€’ 1)
y = в€’2 x + 7
3 = 3 (1)
3 = в€’4 + 7
3 = в€’2 ( 2 ) + 7
3 = 3 ( 2 в€’ 1)
y
8
1
y = x +8
3
8
в€’8
в€’8
3 = 3 true
3 = 3 true
7. Write 4 x + 2 y = 6 in slope-intercept form.
4x + 2 y = 6
x
1
y = в€’ x +3
2
2 y = в€’4 x + 6
y = в€’2 + 3
Verify that ( в€’6, 6 ) satisfies both equations.
1
y = в€’ x+3
2
1
6 = в€’ ( в€’6 ) + 3
2
6 = 3+3
6 = 6 true
x
8
y
1
y = x+8
3
1
6 = ( в€’6 ) + 8
3
6 = в€’2 + 8
y = 3 x в€’7
2
в€’4
(2, в€’1)
x
в€’6
6 = 6 true
y = в€’2 x + 3
y = 2x + 2
y = в€’3 x + 7
Verify that ( 2, в€’1) satisfies both equations.
4x + 2 y = 6
2x + y = 3
4 ( 2 ) + 2 ( в€’1) = 6
2 ( 2 ) + ( в€’1) = 3
8в€’2 = 6
4 в€’1 = 3
6 = 6 true
51
3 = 3 true
Homework 3.1
SSM: Intermediate Algebra
13.
9. Write both equations in slope-intercept form.
5 ( y в€’ 2 ) = 21 в€’ 2 ( x + 3) y = 3 ( x в€’ 1) + 8
5 y в€’ 10 = 21 в€’ 2 x в€’ 6
The solution is (1, 4 ) .
y = 3x в€’ 3 + 8
5 y = в€’2 x + 25
y = 3x + 5
15.
2
y = в€’ x+5
5
y
y = 3 x +5
The solution is ( 480, 460 ) .
(0, 5)
4
2
y = в€’ x +5
5
в€’4
17. Write 4 x в€’ 2 y = 2 in slope-intercept form.
4x в€’ 2 y = 2
x
4
в€’2 y = в€’4 x + 2
y = 2x в€’1
The equations are identical.
Verify that (0,5 ) satisfies both equations.
5 ( y в€’ 2 ) = 21 в€’ 2 ( x + 3) y = 3 ( x в€’ 1) + 8
5 (5 в€’ 2 ) = 21 в€’ 2 (0 + 3 ) 5 = 3 (0 в€’ 1) + 8
5 (3 ) = 21 в€’ 2 (3 )
5 = 3 ( в€’1) + 8
15 = 15 true
5 = 5 true
The system is dependent. The solution set
contains all ordered pairs ( x, y ) such that
11. Write both equations in slope-intercept form.
1
1
1
1
xв€’ y =1
x+ y =2
2
2
4
2
xв€’ y = 2
x + 2y = 8
в€’ y = в€’x + 2
y = xв€’2
1
y = в€’ x +4
2
y = 2x в€’1 .
19. Write 0.2 y в€’ x = 3 in slope-intercept form.
0.2 y в€’ x = 3
0.2 y = x + 3
2 y = в€’x + 8
1
y = в€’ x+4
2
y = 5 x + 15
y
4
в€’4
(4, 2)
4
x
The system is inconsistent. The solution set is
the empty set, в€… .
в€’4
y = x в€’2
Verify that ( 4, 2 ) satisfies both equations.
1
1
xв€’ y =1
2
2
1
1
(4 ) в€’ (2 ) = 1
2
2
2 в€’1 = 1
1 в€’ 1 true
1
1
x+ y = 2
4
2
1
1
(4 ) + (2 ) = 2
4
2
1+1 = 2
2 = 2 true
52
Homework 3.1
SSM: Intermediate Algebra
21. Write both equations in slope-intercept form.
1
1
1
2
xв€’ y =1
x+ y =2
2
2
3
3
xв€’ y = 2
x + 2y = 6
в€’ y = в€’x + 2
2 y = в€’x + 6
y = xв€’2
25. a.
1
y = в€’ x+3
2
Start by plotting the data. Then find the
regression lines for the data.
Women:
W (t ) = 0.14t + 4.52
Men:
The solution is roughly (3.33,1.33) .
23. a.
In 2002, t = 32 .
W (32 ) = в€’0.19 (32 ) + 43.73
M (t ) = 0.083t + 4.60
= 37.65
M (32 ) = в€’0.16 (32 ) + 39.90
b.
= 34.78
The women’s time is 37.65 seconds and the
men’s time is 34.78 seconds. The error for
the women’s time estimate is 0.275 seconds
and the error for the men’s time estimate is
0.165 seconds.
b.
c.
The enrollment for men and women were
equal in 1971 with roughly 4.72 million
each.
The absolute value of the slope of W is more
than the absolute value of the slope of M.
This shows that the winning times of women
decrease at a greater rate than the winning
times of men.
c.
In 2009, t = 39.
W (39 ) = 0.14 (39 ) + 4.52
= 9.98
M (39 ) = 0.083 (39 ) + 4.60
Since the winning time for women is
decreasing at a faster rate than for men, the
winning time for men may equal the
winning time for women in an upcoming
year since these times are only a few
seconds apart in the year 2002 and they are
getting closer as each year passes.
= 7.837
The total enrollment will be roughly
9.98 + 7.837 = 17.817 million students.
27. a.
Start by plotting the data. Then find the
regression lines for the data.
Punch Cards:
d.
f (t ) = в€’2.45t + 72.82
Men and women will have a winning time of
19.47 seconds in the year 2098.
53
Homework 3.1
SSM: Intermediate Algebra
37.
Optical Scan:
f ( x ) = g ( x ) when x = в€’1 .
39. Answers may vary. Possible answers:
a.
y
1
y =в€’ x +2
2
f (t ) = 3.08t + 12.58
y = 2 x в€’3
4
(2, 1)
b.
в€’4
x
4
в€’4
b.
y
The percentage of voters using punch cards
will equal the percentage using electronic
devices in 2001. This is not a national
election year. The percentages will be
closest in the year 2000.
In 2000, t = 10.
f (10 ) = в€’2.45 (10 ) + 72.82
в€’4
c.
3
x
2
1
y = (3 x + 2 ) в€’ 1
2
y
4
3
x
2
x
4
в€’4
41. Graph the equations.
y = x+3
y = в€’2 x + 9
31. The slope of f ( x ) is в€’3 and the f-intercept is
(0,30 ) . Therefore, f ( x ) = в€’3x + 30 . The slope
of g ( x ) is 5 and the g-intercept is (0, 2 ) .
Therefore, g ( x ) = 5 x + 2 . The solution for this
system is (3.5,19.5 ) . You may verify your
y = 3x в€’ 1
y
4
answer using the slope or substituting the values
for x in the functions.
f ( x ) = 3 when x = 5 .
y=
в€’4
29. The solution of this system is estimated to be
( в€’1.9, в€’2.8) .
35.
x
4
y=
100 в€’ ( 48.32 + 43.38 ) = 8.3
In 2000, 8.3% of voters used the other two
methods of voting.
f ( в€’4 ) = 0
x
в€’4
= 43.38
In 2000, the percentage of voters using
punch cards or lever machines was 48.32%
and the percentage using optical scans or
other electronic devices was 43.38%.
33.
y
4
= 48.32
g (10 ) = 3.08 (10 ) + 12.58
c.
y = x +1
в€’2
в€’2
y = 3 x в€’1
y = x +3
(2, 5)
4
x
y = в€’2 x + 9
The solution of the system is ( 2,5 ) .
54
Homework 3.2
SSM: Intermediate Algebra
5. Substitute 2 ( x в€’ 5 ) for y in 3 x в€’ 5 y = 29 .
43. Answers may vary. One possible answer:
3
y=в€’ x
4
y = x+7
3 x в€’ 5 ( 2 ( x в€’ 5 )) = 29
3 x в€’ 5 ( 2 x в€’ 10 ) = 29
3 x в€’ 10 x + 50 = 29
y = в€’2 x в€’ 5
y
в€’7 x = в€’21
y = x +7
x=3
Let y = 3 in y = 2 ( x в€’ 5 ) .
4
(в€’4, 3)
в€’4
в€’2
y = 2 (3 в€’ 5 )
y = в€’2 x в€’ 5
= 2 ( в€’2 )
x
2
= в€’4
The solution is (3, в€’4 ) .
3
y =в€’ x
4
45. The solution of a system of two equations is an
ordered pair that satisfies both equations. This
solution is the intersection point of the graphs
since this is the point that satisfies both
equations.
7. Substitute в€’4 y в€’ 8 for x in 3 x в€’ 2 y = 18 .
3 ( в€’4 y в€’ 8 ) в€’ 2 y = 18
в€’12 y в€’ 24 в€’ 2 y = 18
в€’14 y = 42
y = в€’3
Let y = в€’3 in x = в€’4 y в€’ 8 .
Homework 3.2
1. Substitute x в€’ 5 for y in x + y = 9 .
x = в€’4 ( в€’3 ) в€’ 8
x + ( x в€’ 5) = 9
= 12 в€’ 8
2x в€’ 5 = 9
=4
The solution is ( 4, в€’3) .
2 x = 14
x=7
Let x = 7 in y = x в€’ 5 .
y = 7в€’5
9. Substitute 99x for y in y = 100 x .
99 x = 100 x
=2
The solution is (7, 2 ) .
в€’x = 0
x=0
Let x = 0 in y = 99 x .
3. Substitute 4 y + 7 for x in 2 x в€’ 3 y = 1 .
y = 99 (0 )
2 ( 4 y + 7 ) в€’ 3 y = в€’1
=0
The solution is (0, 0 ) .
8 y + 14 в€’ 3 y = в€’1
5 y + 14 = в€’1
11. Substitute 0.2 x + 0.6 for y in 2 y в€’ 3 x = в€’4 .
5 y = в€’15
2 (0.2 x + 0.6 ) в€’ 3 x = в€’4
y = в€’3
Let y = в€’3 in x = 4 y + 7 .
0.4 x + 1.2 в€’ 3 x = в€’4
x = 4 ( в€’3 ) + 7
в€’2.6 x = в€’5.2
x=2
Let x = 2 in y = 0.2 x + 0.6 .
= в€’12 + 7
= в€’5
The solution is ( в€’5, в€’3) .
y = 0.2 ( 2 ) + 0.6
= 0.4 + 0.6
=1
The solution is ( 2,1) .
55
Homework 3.2
SSM: Intermediate Algebra
19. Rewrite the second equation so the variables are
on the left side of the equation and the constant
is on the right side.
3x + 5 y = 3
1
x в€’ 5 for y in 2 x + 3 y = в€’1 .
2
пЈ«1
пЈ¶
2 x + 3 пЈ¬ x в€’ 5 пЈ· = в€’1
пЈ­2
пЈё
3
2 x + x в€’ 15 = в€’1
2
7
x = 14
2
x=4
1
Let x = 4 in y = x в€’ 5 .
2
1
y = (4 ) в€’ 5
2
= 2в€’5
13. Substitute
x + 4y = 8
Multiply the second equation by в€’3 and add the
equations.
3x + 5 y = 3
в€’3 x в€’ 12 y = в€’24
в€’ 7 y = в€’21
y=3
Substitute y = 3 into x = 8 в€’ 4 y and solve for x.
x = 8 в€’ 4 (3)
= 8 в€’ 12
= в€’3
The solution is ( 4, в€’3) .
= в€’4
The solution is ( в€’4,3) .
15. Multiply the first equation by в€’1 and then add
the equations.
4x + y = 9
21. Multiply the first equation by в€’3 and the second
equation by 2, then add the equations.
в€’24 x + 27 y = 129
в€’3 x в€’ y = в€’5
24 x + 30 y = 42
x=4
Substitute x = 4 into 4 x + y = 9 and solve for y.
57 y = 171
y=3
Substitute y = 3 into 8 x в€’ 9 y = в€’43 and solve
for x.
8 x в€’ 9 (3 ) = в€’43
4 (4 ) + y = 9
16 + y = 9
y = в€’7
The solution is ( 4, в€’7 ) .
8 x в€’ 27 = в€’43
8 x = в€’6
17. Multiply the first equation by 2 and add the
equations.
6 x в€’ 4 y = 14
x = в€’2
The solution is ( в€’2,3) .
в€’6 x в€’ 5 y = 4
23. Multiply the second equation by 2 and add the
equations.
0.9 x + 0.4 y = 1.9
в€’ 9 y = 18
y = в€’2
Substitute y = в€’2 into 3 x в€’ 2 y = 7 and solve
for x.
3 x в€’ 2 ( в€’2 ) = 7
0.6 x в€’ 0.4 y = 2.6
1.5 x = 4.5
x=3
Substitute x = 3 into 0.9 x + 0.4 y = 1.9 and
solve for y.
0.9 (3) + 0.4 y = 1.9
3x + 4 = 7
3x = 3
x =1
The solution is (1, в€’2 ) .
2.7 + 0.4 y = 1.9
0.4 y = в€’0.8
y = в€’2
The solution is (3, в€’2 ) .
56
Homework 3.2
SSM: Intermediate Algebra
25. Use the distributive property to simplify both
equations.
3 ( 2 x в€’ 1) + 4 ( y в€’ 3) = 1
29. Multiply the first equation by 3 and the second
equation by в€’2 , then add the equations.
3
1
2x + y =
2
2
в€’2 x в€’ 5 y = в€’11
6 x в€’ 3 + 4 y в€’ 12 = 1
6 x + 4 y = 16
4 ( x + 5 ) в€’ 2 ( 4 y + 1) = 18
в€’
4 x + 20 в€’ 8 y в€’ 2 = 18
4x в€’ 8 y = 0
The system can be rewritten as
6 x + 4 y = 16
7
21
y=в€’
2
2
y=3
Substitute y = 3 into
2
1
1
x + y = and solve
3
2
6
for x.
2
1
1
x + (3) =
3
2
6
2
3 1
x+ =
3
2 6
2
4
x=в€’
3
3
x = в€’2
The solution is ( в€’2,3) .
4x в€’ 8 y = 0
Multiply the first equation by 2 and add the
equations.
12 x + 8 y = 32
4x в€’ 8 y = 0
16 x = 32
x=2
Substitute x = 2 into 4 x в€’ 8 y = 0 and solve
for y.
4 (2 ) в€’ 8 y = 0
31. Substitute 2 x + 5 for y in 6 x в€’ 3 y = в€’3 .
6 x в€’ 3 ( 2 x + 5 ) = в€’3
8 в€’8y = 0
в€’8 y = в€’8
6 x в€’ 6 x в€’ 15 = в€’3
y =1
в€’15 = в€’3 false
This is a contradiction. The system is
inconsistent. The lines are parallel so the solution
set is the empty set, в€… .
The solution is ( 2,1) .
27. Multiply the first equation by 3 and add the
equations.
3
9
x + y = 21
5
2
2
9
x в€’ y = в€’16
5
2
x=5
1
3
Substitute x = 5 into x + y = 7 and solve
5
2
for y.
1
3
(5 ) + y = 7
5
2
3
1+ y = 7
2
3
y=6
2
y=4
33. Multiply the first equation by 3 and the second
equation by 2, then add the equations.
39 x + 30 y = в€’21
34 x в€’ 30 y = 94
73 x = 73
x =1
Substitute x = 1 into 13 x + 10 y = в€’7 and solve
for y.
13 (1) + 10 y = в€’7
13 + 10 y = в€’7
10 y = в€’20
y = в€’2
The solution is (1, в€’2 ) .
The solution is (5, 4 ) .
57
Homework 3.2
SSM: Intermediate Algebra
35. Multiply the first equation by 3 and add the
equations.
12 x в€’ 15 y = 9
43. Substitute
1
x + 3 for y in 2 y в€’ x = 6 .
2
пЈ«1
пЈ¶
2 пЈ¬ x + 3пЈ· в€’ x = 6
пЈ­2
пЈё
x+6в€’x = 6
в€’12 x + 15 y = в€’9
0 = 0 true
This is an identity. The system is dependent. The
solution set is the set of ordered pairs ( x, y )
6 = 6 true
This is an identity. The system is dependent. The
solution set is the set of ordered pairs ( x, y ) such
such that 4 x в€’ 5 y = 3 .
37. Substitute 2.3 x в€’ 7 for y in y = в€’0.6 x + 4 .
2.3 x в€’ 7 = в€’0.6 x + 4
that y =
2.9 x = 11
1
x +3.
2
45. Multiply the first equation by 2 and the second
equation by 5, then add the equations.
5
1
x+ y = 6
3
2
5
25
в€’ x+
y = 20
3
2
13 y = 26
x ≈ 3.79
Substitute x = 3.79 into y = 2.3 x в€’ 7 and solve
for y.
y = 2.3 (3.79 ) в€’ 7
≈ 1.72
The solution is roughly (3.79,1.72 ) .
y=2
39. Multiply the first equation by 3 and the second
equation by 4, then add the equation.
в€’24 x + 27 y = в€’21
Substitute y = 2 into
5
1
x + y = 3 and solve
6
4
for x.
5
1
x + (2 ) = 3
6
4
5
1
x+ =3
6
2
5
5
x=
6
2
x=3
The solution is (3, 2 ) .
24 x в€’ 60 y = в€’12
в€’ 33 y = в€’33
y =1
Substitute y = 1 into 6 x в€’ 15 y = в€’3 and solve
for x.
6 x в€’ 15 (1) = в€’3
6 x в€’ 15 = в€’3
6 x = 12
x=2
The solution is ( 2,1) .
47. Elimination:
Rewrite the second equation so the variables are
on the left side of the equation.
3 x + y = 11
41. Multiply the first equation by 2 and add the
equations.
8x в€’ 6 y = 2
2x + y = 9
Multiply the second equation by в€’1 and add the
equations.
3 x + y = 11
в€’8 x + 6 y = в€’5
0 = в€’3 false
This is a contradiction. The system is
inconsistent so the solution set is the empty set.
в€’2 x в€’ y = в€’9
x=2
Substitute x = 2 into y = в€’2 x + 9 and solve
for y.
y = в€’2 ( 2 ) + 9
=5
The solution is ( 2,5 ) .
58
Homework 3.2
SSM: Intermediate Algebra
Substitute 4 x + 3 for y in y = в€’6 x + 50 and
solve for x.
4 x + 3 = в€’6 x + 50
Substitution:
Substitute в€’2 x + 9 for y in the first equation.
3 x + ( в€’2 x + 9 ) = 11
10 x = 47
3 x в€’ 2 x + 9 = 11
x = 4.7
Substitute x = 4.7 into y = 4 x + 3 and solve
for y.
y = 4 ( 4.7 ) + 3
x=2
Substitute x = 2 into y = в€’2 x + 9 and solve
for y.
y = в€’2 ( 2 ) + 9
= 18.8 + 3
= в€’4 + 9
= 21.8
The solution is ( 4.7, 21.8 ) .
=5
The solution is ( 2,5 ) .
53. The coordinate for A is (0, 0 ) since it lies at the
origin. The coordinate for B is the y-intercept of
l 1 , which is (0,3) . The coordinate for C is the
Graphing:
y
8
y = в€’3 x +11
point of intersection of l 1 and l 2 . Solve the
following system:
l 1 : y = 2x + 3
(2, 5)
4
y = в€’2 x + 9
4
8
l 2 : 3 y + x = 30
Substitute 2 x + 3 for y in the second equation.
3 ( 2 x + 3) + x = 30
x
49. The student is not correct. A system is
inconsistent (i.e. the solution set is the empty set)
when the lines are parallel. Parallel lines must
have the same slope. The slopes are different yet
they are close enough to where the lines look
parallel around the origin. To find the correct
solution, substitute 2 x + 3 for y in y = 2.01x + 1
and solve for x.
2 x + 3 = 2.01x + 1
6 x + 9 + x = 30
7 x = 21
x=3
Substitute x = 3 into the first equation and solve
for y.
y = 2 (3 ) + 3
= 6+3
=9
The solution is (3,9 ) so the coordinate of C is
в€’0.01x = в€’2
x = 200
Substitute x = 200 into y = 2 x + 3 and solve
for y.
y = 2 ( 200 ) + 3
(3,9 ) . The coordinate for D is the point of
intersection of l 2 and l 3 . Solve the following
system:
l 2 : 3 y + x = 30
= 403
The solution is ( 200, 403) .
l 3 : y + 3x = 26
Multiply the second equation by в€’3 and add the
equations.
3 y + x = 30
51. The function f has a slope of 4 and an f-intercept
of (0,3) . Therefore, f ( x ) = 4 x + 3 . The
в€’3 y в€’ 9 x = в€’78
function g has a slope of в€’6 and a g-intercept of
(0,50 ) . Therefore, g ( x ) = в€’6 x + 50 . The
в€’ 8 x = в€’48
x=6
Substitute x = 6 into 3 y + x = 30 and solve
for y.
system that describes the functions f and g is:
y = 4x + 3
y = в€’6 x + 50
Solve using substitution.
59
Homework 3.2
SSM: Intermediate Algebra
3 y + (6 ) = 30
пЈ« в€’ax + c пЈ¶
dx + e пЈ¬
пЈ·= f
пЈ­ b пЈё
aex ce
dx в€’
+ = f
b
b
bdx в€’ aex + ce = bf
3 y = 24
y =8
The solution is (6,8) so the coordinate of D is
(6,8) . The coordinate for E is the point of
(bd в€’ ae ) x + ce = bf
(bd в€’ ae ) x = bf
intersection of l 3 and l 4 . Solve the following
system.
l 3 : y + 3x = 26
bf в€’ ce
ce в€’ bf
or
bd в€’ ae
ae в€’ bd
Substitute this result for x in ax + by = c and
solve for y.
пЈ« ce в€’ bf пЈ¶
aпЈ¬
пЈ· + by = c
пЈ­ ae в€’ bd пЈё
ace в€’ abf
+ by = c
ae в€’ bd
ace в€’ abf
by = c в€’
ae в€’ bd
ace в€’ bcd ace в€’ abf
by =
в€’
ae в€’ bd
ae в€’ bd
в€’bcd + abf
by =
ae в€’ bd
abf в€’ bcd
by =
ae в€’ bd
af в€’ cd
y=
ae в€’ bd
Therefore, the solution is
пЈ« ce в€’ bf af в€’ cd пЈ¶
,
пЈ¬
пЈ· , assuming ae в€’ bd в‰ 0 .
пЈ­ ae в€’ bd ae в€’ bd пЈё
x=
l 4 : y = 2 x в€’ 10
Substitute 2 x в€’ 10 in for y in y + 3 x = 26 .
2 x в€’ 10 + 3 x = 26
5 x = 36
x = 7.2
Substitute x = 7.2 into y = 2 x в€’ 10 and solve
for x.
y = 2 ( 7.2 ) в€’ 10
= 4.4
The solution is (7.2, 4.4 ) so the coordinate of E
is (7.2, 4.4 ) . The coordinate of F is the x-
intercept of l 4 . Let y = 0 in l 4 and solve for x.
y = 2 x в€’ 10
0 = 2 x в€’ 10
2 x = 10
x=5
The coordinate of F is (5, 0 ) .
y
C (3 , 9 )
D (6 , 8 )
8
b.
E (7 .2, 4 .4 )
4
B (0 , 3 )
F (5 , 0 )
A (0 , 0 )
55. a.
4
8
в€’ ce
x
Substitute 3 for a, 5 for b, 2 for c, 4 for d, 3
for e, and 4 for f in the solution from part a.
2 (3 ) в€’ 5 ( 4 ) в€’14 14
x=
=
=
3 (3) в€’ 5 ( 4 ) в€’11 11
y=
You may use more than one method to solve
the system. For example, you may use
substitution as follows:
Solve the first equation for y.
ax + by = c
3 (4 ) в€’ 2 (4 )
3 ( 3) в€’ 5 ( 4 )
=
4
4
=в€’
11
в€’11
пЈ« 14 4 пЈ¶
The solution is пЈ¬ , в€’ пЈ· .
пЈ­ 11 11 пЈё
57. Written response. Answers may vary.
by = в€’ ax + c
в€’ax + c
b
Substitute this result for y in the second
equation and solve for x.
y=
60
Homework 3.3
SSM: Intermediate Algebra
7. a.
Homework 3.3
1. Solve the system
y = в€’0.19t + 43.73
Solve the system
y = 13.5t + 229
y = в€’5t + 365
Substitute 13.5t + 229 for y in the second
equation and solve for t.
13.5t + 229 = в€’5t + 365
y = в€’0.16t + 39.90
Substitute в€’0.19t + 43.73 for y in the second
equation and solve for t.
в€’0.19t + 43.73 = в€’0.16t + 39.90
18.5t = 136
t ≈ 7.35
Substitute this result into the first equation
and solve for y.
y = 13.5 (7.35 ) + 229
в€’0.03t = в€’3.83
t ≈ 127.67
Substitute this result into the first equation and
solve for y.
y = в€’0.19 (127.67 ) + 43.73
≈ 328.23
According to the models, the two
newspapers had equal circulations of
roughly 328 thousand in 1997.
≈ 19.47
According to the models, the winning times for
women and men will both be 19.47 seconds in
the year 2098.
b.
Since the circulations were roughly equal in
1997, competition heated up as each
newspaper tried to overtake the other.
c.
D (10 ) = 13.5 (10 ) + 229
3. Solve the system
E = 0.14t + 4.52
E = 0.083t + 4.60
Substitute 0.14t + 4.52 for E in the second
equation and solve for t.
0.14t + 4.52 = 0.083t + 4.60
= 135 + 229
= 364
R (10 ) = в€’5 (10 ) + 365
0.057t = 0.08
= в€’50 + 365
t ≈ 1.40
Substitute this result into the first equation and
solve for E.
E = 0.14 (1.4 ) + 4.52 ≈ 4.71
According to the models, the enrollments for
men and women were approximately equal in
1971 (roughly 4.71 million).
= 315
826 в€’ (364 + 315 ) = 826 в€’ 679 = 147
According to the models, the combined
increase due to bonus copies was roughly
147 thousand bonus copies.
d.
5. Solve the system
p = в€’2.45t + 72.82
D (11) = 13.5 (11) + 229
= 377.5
R (11) = в€’5 (11) + 365
p = 3.08t + 12.58
Substitute в€’2.45t + 72.82 for p in the second
equation and solve for t.
в€’2.45t + 72.82 = 3.08t + 12.58
= 310
377.5 + 310 = 687.5
According to the models, the combined
circulation from the two newspapers was
roughly 688 thousand copies in 2001.
в€’5.53t = в€’60.24
t ≈ 10.89
Substitute this result into the first equation and
solve for p.
p = −2.45 (10.89 ) + 72.82 ≈ 46.14
According to the models, the percentage of
voters using optical scan or other electronic
devices equaled the percentage of voters using
punch cards or lever machines in 2001. This was
not a national election year. The percentages
were closest in 2000.
e.
61
The estimate in part d. was an overestimate.
After joining revenue streams, the
competition for subscribers ceased (or at
least reduced if there were other
competitors). The end of bonus copies, or
just the merger in general, may have caused
some subscribers to cancel subscriptions.
Homework 3.3
9. a.
SSM: Intermediate Algebra
c.
Start by plotting the data, then compute the
regression line for each data set.
Women:
Women:
W (t ) = в€’1.01t + 323.57
Men:
W (t ) = в€’1.22t + 338.47
M (t ) = в€’0.36t + 240.73
Men:
Solve the system
y = в€’1.01t + 323.57
y = в€’0.36t + 240.73
Substitute в€’1.01t + 323.57 for y in the
second equation and solve for t.
в€’1.01t + 323.57 = в€’0.36t + 240.73
в€’0.65t = в€’82.84
t ≈ 127.45
Substitute this result into the first equation
and solve for y.
y = в€’1.01(127.45 ) + 323.57
M (t ) = в€’0.36t + 240.44
b.
≈ 194.85
Now the models predict that the record
times for men and women will be the same
in 2027.
Solve the system
y = в€’1.22t + 338.47
y = в€’0.36t + 240.44
Substitute в€’1.22t + 338.47 for y in the
second equation and solve for t.
в€’1.22t + 338.47 = в€’0.36t + 240.44
в€’0.86t = в€’98.03
t ≈ 113.99
Substitute this result into the first equation
and solve for y.
y = в€’1.22 (113.99 ) + 338.47
≈ 199.40
According to the models, the record times
for men and women will both be
approximately 199.40 seconds in 2014.
d.
For the most part the data appear to be
linear. However, the data value for women
in 1926 seems to deviate somewhat from the
linear pattern. Removing this value makes
the linear model a better fit.
11. a.
Since a 2001 Taurus’ value decreases by a
constant $1582 each year, the function T (t )
is linear and its slope is –1582. The Tintercept is (0,12939 ) , since the car is
worth $12,939 at year t = 0 (2002). Doing
similar work for E (t ) , we get the following
equations:
T (t ) = в€’1582t + 12,939
E (t ) = в€’1024t + 9330
62
Homework 3.3
SSM: Intermediate Algebra
b.
Solve the system
V = в€’1582t + 12,939
c.
Find the intersection point using a graphing
utility.
15. a.
Since Jenny Craig’s program fees increase
by a constant $72 each week, the function J
is linear and its slope is 72. The J-intercept
is (0,19 ) , since the start-up fee is $19 at
V = в€’1024t + 9330
Substitute в€’1582t + 12,939 in for V in the
second equation and solve for t.
в€’1582t + 12,939 = в€’1024t + 9330
в€’558t = в€’3609
t ≈ 6.468
Substitute this result into the first equation
and solve for V.
V = в€’1582 (6.468 ) + 12,939
c.
≈ 2707
The cars will both be worth roughly $2707
in 2008.
t = 0 . So, an equation for J (t ) is:
Find the intersection point using a graphing
utility.
Since Weight Watchers’ program fees
increase by a constant $77 each week ($17
fee + $60 food), the function W is linear and
its slope is 77. The W-intercept is (0, 0 )
J (t ) = 72t + 19 .
since there is no start-up fee at t = 0 . An
equation for W (t ) is: W (t ) = 77t .
13. a.
b.
Since college A’s tuition increases by a
constant $670 each year, the function A (t )
y = 77t
Substitute 72t + 19 for y into the second
equation.
72t + 19 = 77t
is linear and its slope is 670. The A-intercept
is (0, 6100 ) , since tuition is $6100 in year
t = 0 (2000). Similar work for college B
gives the following equations:
A (t ) = 670t + 6100
5t = 19
t = 3.8
Substitute this result into the first equation
and solve for y.
y = 72 (3.8 ) + 19
B (t ) = 440t + 8500
b.
Solve the system
y = 72t + 19
Solve the system
y = 670t + 6100
= 292.6
The total cost at both Jenny Craig and
Weight Watchers is approximately $293 in 4
weeks.
y = 440t + 8500
Substitute 670t + 6100 for y in the second
equation and solve for t.
670t + 6100 = 440t + 8500
c. Find the intersection point using a graphing
utility.
230t = 2400
t ≈ 10.435
Substitute this result in the first equation and
solve for y.
y = 670 (10.435 ) + 6100
= 13091.45
The tuition at both colleges will be $13,091
in approximately 10 years (in 2010).
63
Homework 3.4
SSM: Intermediate Algebra
17. Let P (t ) represent the average price (in dollars)
order to predict when the family will be able to
pay a 10% down payment on an average-priced
house, solve the following system for t when
y = 0.1( P (t )) = 0.1(8000t + 214000 )
of a home in a community and S (t ) represent
the amount of money (in dollars) a family has
saved at t years since 2000. Since the average
price of a home increases at a constant $8000 per
year, the function P is linear and its slope is
8000. the P-intercept is (0, 214000 ) since the
y = S (t ) = 3600t + 10000
Substitute 0.1(8000t + 214000 ) for y in the
second equation and solve for t.
0.1(8000t + 214000 ) = 3600t + 10000
price of a home is $214,000 in year t = 0 . So, an
equation for P is : P (t ) = 8000t + 214000 .
800t + 21400 = 3600t + 10000
Similar work in finding the equation for the
function S gives: S (t ) = 3600t + 10000 . (The
slope of S is 3600 since the family plans to save
$300 each month which is $3600 each year.) In
в€’2800t = в€’11400
t ≈ 4.07
The family will be able to pay a 10% down
payment in 4 years (2004).
Homework 3.4
1.
3.
Words
Inequality
Notation
Numbers greater than 3
x>3
Numbers less than or
equal to 4
x≤4
Numbers less than 5
x<5
Numbers greater than or
equal to в€’1
x ≥ −1
0
( в€’в€ћ,5)
5
в€’1
7.
( в€’в€ћ, в€’4]
0
в€’4
x + 2− 2 ≥ 5− 2
[в€’1, в€ћ )
0
2 x + 7 < 11
2 x + 7 в€’ 7 < 11 в€’ 7
x≥3
Interval: [3, в€ћ )
5.
(3,в€ћ )
3
0
x+2≥5
0
Interval
Notation
Graph
2x < 4
2x 4
<
2 2
x<2
Interval: ( в€’в€ћ, 2 )
3
−x + 2 ≥ 5
−x + 2 − 2 ≥ 5 − 2
0
−x ≥ 3
в€’x 3
≤
в€’1 в€’1
x ≤ −3
Interval: ( в€’в€ћ, в€’3]
в€’3
0
64
2
Homework 3.4
SSM: Intermediate Algebra
9x < 4 + 5x
9.
17.
9x в€’ 5x < 4 + 5x в€’ 5x
3 в€’ 2x + 8 > 4x +1
4x < 4
в€’2 x + 11 > 4 x + 1
в€’2 x + 11 в€’ 11 > 4 x + 1 в€’ 11
4x 4
<
4 4
x <1
Interval: ( в€’в€ћ,1)
в€’2 x > 4 x в€’ 10
в€’2 x в€’ 4 x > 4 x в€’ 10 в€’ 4 x
в€’6 x > в€’10
0
в€’6 x в€’10
<
в€’6
в€’6
5
x<
3
5пЈ¶
пЈ«
Interval: пЈ¬ в€’в€ћ, пЈ·
3пЈё
пЈ­
1
2.1x − 7.4 ≤ 10.4
11.
2.1x − 7.4 + 7.4 ≤ 10.4 + 7.4
2.1x ≤ 17.8
2.1x 17.8
≤
2.1
2.1
x ≤ 8.48
Interval: ( в€’в€ћ,8.48]
0
13.
в€’3
19.
0
3
в€’6.23 x + 2.35 < 1.76 (3 в€’ 2.73 x )
в€’6.23 x + 2.35 < 5.28 в€’ 4.8048 x
8
4
в€’6.23 x + 2.35 в€’ 2.35 < 5.28 в€’ 4.8048 x в€’ 2.35
2 x в€’ 3 > 7 x + 22
в€’6.23 x < 2.93 в€’ 4.8048 x
2 x в€’ 3 + 3 > 7 x + 22 + 3
в€’6.23 x + 4.8048 x < 2.93 в€’ 4.8048 x + 4.8048 x
2 x > 7 x + 25
в€’1.4252 x < 2.93
2 x в€’ 7 x > 7 x + 25 в€’ 7 x
в€’1.4252 x
2.93
>
в€’1.4252 в€’1.4252
x > в€’2.06
Interval: ( в€’2.06, в€ћ )
в€’5 x > 25
в€’5 x 25
<
в€’5 в€’5
x < в€’5
Interval: ( в€’в€ћ, в€’5 )
в€’3
в€’5
15.
3 в€’ 2 ( x в€’ 4) > 4x + 1
0
3
21. 7 ( x + 1) − 8 ( x − 2 ) ≤ 0
0
7 x + 7 − 8 x + 16 ≤ 0
7 x − 3 ≤ −2 x − 21
− x + 23 ≤ 0
7 x − 3 + 3 ≤ −2 x − 21 + 3
− x ≤ −23
7 x ≤ −2 x − 18
в€’ x в€’23
≥
в€’1 в€’1
x ≥ 23
7 x + 2 x ≤ −2 x − 18 + 2 x
9 x ≤ −18
Interval: [23, в€ћ )
9 x в€’18
≤
9
9
x ≤ −2
Intervals: ( в€’в€ћ, в€’2]
23
в€’2
0
65
26
Homework 3.4
23.
SSM: Intermediate Algebra
2
в€’ x>4
3
3 2
3
в€’ в‹…в€’ x < в€’ в‹…4
2 3
2
x < в€’6
Interval: ( в€’в€ћ, в€’6 )
29.
0
в€’6
25.
1
xв€’2 <3
4
1
x в€’2+ 2 < 3+ 2
4
1
x<5
4
1
4в‹… x < 4в‹…5
4
x < 20
Interval: ( в€’в€ћ, 20 )
в€’1
31.
T >E
в€’1582t + 12, 939 в€’ 12, 939 > в€’1024t + 9330 в€’ 12, 939
в€’1582t > в€’1024t в€’ 3609
в€’1582t + 1024t > в€’1024t в€’ 3609 + 1024t
в€’558t > в€’3609
в€’558t
в€’3609
<
в€’558
в€’558
t < 6.47
0
20
2 3
5
− x≤
3 4
2
2 3
2 5 2
− x− ≤ −
3 4
3 2 3
3
11
− x≤
4
6
4 3
4 11
− ⋅− x ≥ − ⋅
3 4
3 6
22
x≥−
9
пЈ® 22 пЈ¶
Interval: пЈЇ в€’ , в€ћ пЈ·
пЈ° 9
пЈё
в€’
0
в€’1582t + 12, 939 > в€’1024t + 9330
15
27.
1
1 1
− x− ≥
2
6 3
1
1 1 1 1
− x− + ≥ +
2
6 6 3 6
1
1
− x≥
2
2
1
1
−2 ⋅ − x ≤ −2 ⋅
2
2
x ≤ −1
Interval: ( в€’в€ћ, в€’1]
22
9
The value of the Taurus is more than the value of
the Escort for years up until 2008 ( t < 6 ).
33. a.
Since U-Haul’s charge increases at a
constant rate of $0.69 per mile, the equation
is linear with slope 0.69 . The U-intercept is
19.95 since U-Haul charges a flat fee of
$19.95. An equation for U-Haul’s charge is:
U ( x ) = 0.69 x + 19.95 .
After similar work for Penske, an equation
for Penske’s charge is:
P ( x ) = 0.39 x + 29.95 .
0
b.
U<P
0.69 x + 19.95 < 0.39 x + 29.95
0.69 x + 19.95 в€’ 19.95 < 0.39 x + 29.95 в€’ 19.95
0.69 x < 0.39 x + 10
0.69 x в€’ 0.39 x < 0.39 + 10 в€’ 0.39 x
0.3 x < 10
0.3x 10
<
0.3 0.3
100
x<
3
U-Haul will be cheaper for miles driven less
than 33.3 miles.
66
Homework 3.4
SSM: Intermediate Algebra
35. a.
W ( 28 ) = 0.098 ( 28 ) + 77.58
37. The student made a mistake. It is not necessary
to switch the direction of the inequality when
dividing by a positive number.
≈ 80.3
M ( 28 ) = 0.192 ( 28 ) + 69.98
39. a.
≈ 75.4
80.3 в€’ 75.4 = 4.9
Women born in 2008 will live roughly 5
years longer, on average, than men born in
2008.
b.
Solve the inequality for x.
3( x − 2) + 1 ≥ 7 − 4x
3x − 6 + 1 ≥ 7 − 4 x
3x − 5 ≥ 7 − 4 x
3x − 5 + 5 ≥ 7 − 4 x + 5
W <M
3 x ≥ 12 − 4 x
0.098t + 77.58 < 0.192t + 69.98
3 x + 4 x ≥ 12 − 4 x + 4 x
0.098t + 77.58 в€’ 77.58 < 0.192t + 69.98 в€’ 77.58
7 x ≥ 12
0.098t < 0.192t в€’ 7.6
7 x 12
≥
7
7
12
x≥
7
Any three numbers that are greater than or
12
equal to
are possible solutions.
7
0.098t в€’ 0.192t < 0.192t в€’ 7.6 в€’ 0.192t
в€’0.094t < в€’7.6
в€’0.094t
в€’7.6
>
в€’0.094 в€’0.094
t > 80.9
Men will have a longer average life
expectancy than women for birth years
starting with 2061.
c.
i.
ii.
b.
The y-intercept represents the average
life expectancy for 1980. Since this
value is larger for women than men, she
should marry a younger man.
From part a., any number less than
12
is
7
not a solution.
41. We have mx < c and x > 2 . This implies that m
must be negative so that dividing both sides by m
produces:
mx < c
The average life expectancy for a
woman born in 1980 is 77.58 years. She
wants to marry a younger man, but each
year later decreases her average life
expectancy. So, t years after 1980, the
average life expectancy of a born in
1980 is 77.58 в€’ t . This value needs to
be less than or equal to the life
expectancy of the man she will marry.
Thus, we have
77.58 − t ≤ 0.192t + 69.98
mx c
>
m m
c
x>
m
c
= 2 , this means c must also be
m
negative. Thus, c and m can be any negative
c
numbers such that
= 2.
m
Since we want
77.58 − t − 77.58 ≤ 0.192t + 69.98 − 77.58
в€’t < 0.192t в€’ 7.6
в€’t в€’ 0.192t < 0.192t в€’ 7.6 в€’ 0.192t
43. A solution of the form 2 < x < 5 requires a
compound inequality and is not possible for a
single linear inequality.
в€’1.192t < в€’7.6
в€’1.192t
в€’7.6
>
в€’1.192 в€’1.192
t > 6.4
45. True. When x = в€’4 , the graph of f is above the
graph of g.
47. False. When x = в€’1 , the graph of f is above the
graph of g.
She must marry a man who is born in
1986 or later.
67
Chapter 3 Review Exercises
SSM: Intermediate Algebra
y = 2 (1) + 5
49. The graph of f is above the graph of g for all
values of x such that x < 2.8 . So, f ( x ) > g ( x )
= 2+5
when x < 2.8 .
=7
The solution is (1, 7 ) .
51. Written response. Answers may vary.
4. Solve using elimination.
Multiply the first equation by 3 and the second
equation by в€’4 , then add the equations.
12 x в€’ 15 y = в€’66
в€’12 x в€’ 8 y = 20
Chapter 3 Review Exercises
3
1. y = в€’ x + 1
2
1
y = xв€’6
4
y
в€’4
4
в€’ 23 y = в€’46
y=2
Substitute y = 2 into 3 x + 2 y = в€’5 and solve
for x.
3 x + 2 ( 2 ) = в€’5
x
в€’4
3 x + 4 = в€’5
(4, в€’5)
3 x = в€’9
The solution is ( 4, в€’5 ) .
y = в€’2 ( x в€’ 4 )
2. 3 x в€’ 5 y = в€’1
в€’5 y = в€’3 x в€’ 1
y
x = в€’3
The solution is ( в€’3, 2 ) .
= в€’2 x + 8
5. Solve using elimination.
Rewrite the second equation so all the variables
are on the left side of the equation and the
constant is on the right side.
в€’2 x + 3 y = 7
3
1
y = x+
5
5
8
4 x в€’ 6 y = в€’14
Multiply the first equation by 2 and add the
equations.
в€’4 x + 6 y = 14
4
(3, 2)
4
8
4 x в€’ 6 y = в€’14
x
0 = 0 true
This is an identity. The system is dependent. The
solution set contains all ordered pairs ( x, y )
The solution is (3, 2 ) .
3. Solve using substitution.
Substitute 2 x + 5 for y in the second equation.
y = в€’3 x + 10
such that y =
2 x + 5 = в€’3 x + 10
2
7
x+ .
3
3
6. Solve using elimination.
Multiply the first equation by –2 and add the
equations.
в€’6 x + 14 y = в€’10
2 x + 5 в€’ 5 = в€’3 x + 10 в€’ 5
2 x = в€’3 x + 5
2 x + 3 x = в€’3 x + 5 + 3 x
6 x в€’ 14 y = в€’1
5x = 5
x =1
Substitute this result into the first equation and
solve for y.
0 = в€’11 false
This is a contradiction. The system is
inconsistent. The solution set is the empty set.
68
Chapter 3 Review Exercises
SSM: Intermediate Algebra
7. Solve using elimination.
Rewrite the second equation so all the variables
are on the left side of the equation and the
constant is on the right side.
в€’4 x в€’ 5 y = 3
10. Solve using elimination.
Multiply the first equation by 2 and add the
equations.
0.8 x + 0.6 y = 0.8
в€’0.8 x + 1.2 y = 6.4
8 x + 10 y = в€’6
Multiply the first equation by 2 and add the
equations.
в€’8 x в€’ 10 y = 6
1.8 y = 7.2
y=4
Substitute y = 4 into 0.4 x + 0.3 y = 0.4 and
solve for x.
0.4 x + 0.3 ( 4 ) = 0.4
8 x + 10 y = в€’6
0 = 0 true
This is an identity. The system is dependent. The
solution set contains all ordered pairs ( x, y )
0.4 x + 1.2 = 0.4
0.4 x = в€’0.8
x = в€’2
The solution is ( в€’2, 4 ) .
4
3
such that y = в€’ x в€’ .
5
5
11. Solve using substitution.
1
Substitute x в€’ 4 for y in the first equation.
2
пЈ«1
пЈ¶
3 x в€’ 5 пЈ¬ x в€’ 4 пЈ· = 21
2
пЈ­
пЈё
5
3 x в€’ x + 20 = 21
2
1
x + 20 = 21
2
1
x =1
2
x=2
Substitute x = 2 into the second equation and
solve for y.
1
y = (2 ) в€’ 4
2
= 1в€’ 4
= в€’3
The solution is ( 2, в€’3) .
8. Solve using substitution.
Substitute 4.2 x в€’ 7.9 for y in the second
equation.
y = в€’2.8 x + 1.1
4.2 x в€’ 7.9 = в€’2.8 x + 1.1
4.2 x в€’ 7.9 + 7.9 = в€’2.8 x + 1.1 + 7.9
4.2 x = в€’2.8 x + 9
4.2 x + 2.8 x = в€’2.8 x + 9 + 2.8 x
7x = 9
x ≈ 1.29
Substitute this result into the first equation and
solve for y.
y = 4.2 (1.29 ) в€’ 7.9
≈ −2.5
The solution is approximately (1.29, в€’2.5) .
9. Solve using substitution.
Substitute 4.9x for y in the second equation and
solve for x.
в€’3.2 y = x
в€’3.2 ( 4.9 x ) = x
12. Solve using elimination.
6
4
xв€’ y =8
5
3
6
8
в€’ x + y = в€’4
5
3
4
y=4
3
y=3
в€’15.68 x = x
в€’16.68 x = 0
x=0
Substitute this result into the first equation and
solve for y.
y = 4.9 (0 )
=0
The solution is (0, 0 ) .
Substitute y = 3 into
for x.
69
3
2
x в€’ y = 4 and solve
5
3
Chapter 3 Review Exercises
SSM: Intermediate Algebra
2 x в€’ 5 y = 15
3
2
x в€’ (3) = 4
5
3
3
xв€’2= 4
5
3
x=6
5
x = 10
The solution is (10,3) .
в€’2 x в€’ y = в€’9
в€’ 6y = 6
y = в€’1
Substitute y = в€’1 into 2 x в€’ 5 y = 15 and solve
for x.
2 x в€’ 5 ( в€’1) = 15
2 x + 5 = 15
2 x = 10
13. First use the distributive property to simplify
each equation.
2 (3 x в€’ 4 ) + 3 ( 2 y в€’ 1) = в€’5
x=5
The solution is (5, в€’1) .
6 x в€’ 8 + 6 y в€’ 3 = в€’5
Substitution:
Substitute в€’2 x + 9 for y in the first equation and
solve for x.
2 x в€’ 5 y = 15
6 x + 6 y в€’ 11 = в€’5
6x + 6 y = 6
в€’3 ( 2 x + 1) + 4 ( y + 3 ) = в€’7
2 x в€’ 5 ( в€’2 x + 9 ) = 15
в€’6 x в€’ 3 + 4 y + 12 = в€’7
в€’6 x + 4 y + 9 = в€’7
2 x + 10 x в€’ 45 = 15
в€’6 x + 4 y = в€’16
The system is:
6x + 6 y = 6
12 x = 60
x=5
Substitute this result into the second equation
and solve for y.
y = в€’2 x + 9
в€’6 x + 4 y = в€’16
Solve using elimination.
Add the two equations together.
6x + 6 y = 6
= в€’2 (5 ) + 9
= в€’10 + 9
= в€’1
The solution is (5, в€’1) .
в€’6 x + 4 y = в€’16
10 y = в€’10
y = в€’1
Substitute y = в€’1 into 6 x + 6 y = 6 and solve
for x.
6x + 6 y = 6
Graphically:
2 x в€’ 5 y = 15
y = в€’2 x + 9
в€’5 y = в€’2 x + 15
2
y = xв€’3
5
6 x + 6 ( в€’1) = 6
6x в€’ 6 = 6
y
6 x = 12
x=2
The solution is ( 2, в€’1) .
4
8
14. Elimination:
Rewrite the second equation so that all the
variables are on the left side of the equation and
the constant is on the right side.
2 x в€’ 5 y = 15
в€’4
2
y = x в€’3
5
x
(5, в€’1)
y = в€’2 x + 9
The solution is (5, в€’1) .
2x + y = 9
Multiply the second equation by в€’1 and add the
equations.
70
Chapter 3 Review Exercises
SSM: Intermediate Algebra
2 x + 3 y = 19
15. Answers may vary. Possible answers:
a.
3 y = в€’2 x + 19
x + 2y = 4
2
19
y = в€’ x+
3
3
3 x + 6 y = 12
Solving this system, you will encounter an
identity such as 0 = 0 . The solution set is
the set of ordered pairs ( x, y ) such that
6 x в€’ 4 y = 18
в€’4 y = в€’6 x + 18
y=
3
9
xв€’
2
2
x + 2y = 4 .
b.
x + 2y = 4
18. The coordinate for A is (0, 0 ) since it lies at the
2 x + 6 y = 10
Solving this system, you will encounter a
contradiction such as 0 = в€’2 . The solution
set is the empty set, в€… .
c.
origin. The coordinate for B is the y-intercept of
l 1 , which is (0, 4 ) . The coordinate for C is the
point of intersection of l 1 and l 2 . Solve the
following system:
l 1 : y = 3x + 4
x + y = 10
3 x в€’ 3 y = в€’6
l 2 : 3 y + 2 x = 34
Substitute 3 x + 4 for y in the second equation.
3 (3 x + 4 ) + 2 x = 34
The point ( 4, 6 ) satisfies both equations.
16. The approximate solutions for this system are
( 2.6, 21.9 ) and (0.4,15.1) . Although answers
9 x + 12 + 2 x = 34
11x = 22
may vary, your answers should be close to these
points.
x=2
Substitute x = 2 into the first equation and solve
for y.
y = 3 (2 ) + 4
17. Begin by substituting 5 for x and 3 for y in the
equations and solve for a and b.
2 (5 ) + 3 ( 3 ) = a
6 (5 ) в€’ 4 (3 ) = b
10 + 9 = a
= 6+ 4
30 в€’ 12 = b
= 10
The solution is ( 2,10 ) so the coordinate of C is
19 = a
18 = b
Substitute 19 for a and 18 for b in the original
system. This gives:
2 x + 3 y = 19
6 x в€’ 4 y = 18
Verify:
2 (5 ) + 3 (3 ) = 19
6 (5 ) в€’ 4 (3 ) = 18
10 + 9 = 19
30 в€’ 12 = 18
( 2,10 ) . The coordinate for D is the point of
intersection of l 2 and l 3 . Solve the following
system:
l 2 : 3 y + 2 x = 34
l 3 : y + 4 x = 28
Multiply the second equation by в€’3 and add the
equations.
3 y + 2 x = 34
19 = 19 true
18 = 18
To verify graphically, put each equation in slopeintercept form and then find the intersection
point using a graphing utility.
в€’3 y в€’ 12 x = в€’84
в€’ 10 x = в€’50
x=5
Substitute x = 5 into 3 y + 2 x = 34 and solve
for y.
3 y + 2 (5 ) = 34
3 y = 24
y =8
71
Chapter 3 Review Exercises
SSM: Intermediate Algebra
y
The solution is (5,8 ) so the coordinate of D is
y = 3 x в€’6
(5,8) . The coordinate for E is the point of
4
intersection of l 3 and l 4 . Solve the following
system.
l 3 : y + 4 x = 28
в€’4
l 4 : y = 3 x в€’ 14
Substitute 3 x в€’ 14 in for y in y + 4 x = 28 .
y = в€’2 x + 7
(3x в€’ 14 ) + 4 x = 28
20. −2 x ≤ 18
в€’2 x 18
≥
в€’2 в€’2
x ≥ −9
Interval: [в€’9, в€ћ )
x=6
Substitute x = 6 into y = 3 x в€’ 14 and solve
for x.
y = 3 ( 6 ) в€’ 14
= 28 в€’ 14
=4
The solution is (6, 4 ) so the coordinate of E is
21.
3 x − 8 ≤ 13
3 x − 8 + 8 ≤ 13 + 8
l 4 . Let y = 0 in l 4 and solve for x.
y = 3 x в€’ 14
3 x ≤ 21
3 x 21
≤
3
3
x≤7
Interval: ( в€’в€ћ, 7 ]
0 = 3 x в€’ 14
3 x = 14
14
3
пЈ« 14 пЈ¶
The coordinate of F is пЈ¬ , 0 пЈ· .
пЈ­ 3 пЈё
4
C (2 , 1 0)
D (5 , 8 )
в€’8 x + 5 x > в€’5 x + 15 + 5 x
в€’3 x > 15
E (6 , 4 )
F
пЈ« 14 пЈ¶
пЈ¬ ,0пЈ·
пЈ­3 пЈё
8
10
в€’8 x > в€’3 x + 6 в€’ 2 x + 9
в€’8 x > в€’5 x + 15
8
B (0 , 4 )
7
в€’8 x > в€’3 ( x в€’ 2 ) в€’ 2 x + 9
22.
16
A (0 , 0 )
в€’4
в€’9
(6, 4) . The coordinate of F is the x-intercept of
y
y = в€’x + 2
The three graphs do not share a common point.
Thus, the solution set is the empty set.
7 x = 42
x=
x
8
в€’3 x 15
<
в€’3 в€’3
x < в€’5
Interval: ( в€’в€ћ, в€’5 )
x
19. Graph each of the equations.
y = в€’x + 2
в€’10
y = в€’2 x + 7
y = 3x в€’ 6
72
в€’5
0
Chapter 3 Review Exercises
SSM: Intermediate Algebra
23.
4.2 − 3.6 x ≥ 3.9 ( x + 2.1)
26.
4.2 − 3.6 x ≥ 3.9 x + 8.19
4.2 − 3.6 x − 4.2 ≥ 3.9 x + 8.19 − 4.2
−3.6 x ≥ 3.9 x + 4.01
−3.6 x − 3.9 x ≥ 3.9 x + 4.01 − 3.9 x
−7.5 x ≥ 3.99
в€’7.5 x 3.99
≤
в€’7.5 в€’7.5
x ≤ −0.532
Interval: ( в€’в€ћ, в€’0.532]
в€’4
в€’
27.
в€’0.532 0
24. −5 ( 2 x + 3) ≥ 2 (3 x − 4 )
−10 x − 15 ≥ 6 x − 8
−10 x − 15 + 15 ≥ 6 x − 8 + 15
−10 x ≥ 6 x + 7
−10 x − 6 x ≥ 6 x + 7 − 6 x
−16 x ≥ 7
7
в€’16 x
≤
в€’16 в€’16
7
x≤−
16
7пЈ№
пЈ«
Interval: пЈ¬ в€’в€ћ, в€’ пЈє
16 пЈ»
пЈ­
в€’4
3
5
− x≤
4
2
4 3
4 5
− ⋅− x ≥ − ⋅
3 4
3 2
10
x≥−
3
пЈ® 10 пЈ¶
Interval: пЈЇ в€’ , в€ћ пЈ·
пЈ° 3
пЈё
в€’
7
16
0
10
3
в€’2
2
3
x > xв€’
5
2
2
3
3в€’ x в€’3 > x в€’ в€’3
5
2
2
9
в€’ x > xв€’
5
2
2
9
в€’ xв€’x > xв€’ в€’x
5
2
7
9
в€’ x>в€’
5
2
5 7
5 9
в€’ в‹…в€’ x < в€’ в‹…в€’
7 5
7 2
45
x<
14
45 пЈ¶
пЈ«
Interval: пЈ¬ в€’в€ћ, пЈ·
14
пЈ­
пЈё
3в€’
0
2
5 7
25.
в€’ x+ <
3
2 3
2
5 5 7 5
в€’ x+ в€’ < в€’
3
2 2 3 2
2
14 15
в€’ x< в€’
3
6 6
2
1
в€’ x<в€’
3
6
3 2
3 1
в€’ в‹…в€’ x > в€’ в‹…в€’
2 3
2 6
1
x>
4
пЈ«1 пЈ¶
Interval: пЈ¬ , в€ћ пЈ·
пЈ­4 пЈё
0
1
4
2
0
28. a.
45
14
Solve the inequality for x.
7 в€’ 2 (3 x + 5 ) < 4 x + 1
7 в€’ 6 x в€’ 10 < 4 x + 1
в€’6 x в€’ 3 < 4 x + 1
в€’6 x в€’ 3 + 3 < 4 x + 1 + 3
в€’6 x < 4 x + 4
в€’6 x в€’ 4 x < 4 x + 4 в€’ 4 x
в€’10 x < 4
4
в€’10 x
>
в€’10 в€’10
2
x>в€’
5
Any three numbers that are greater than в€’
1
are possible solutions.
73
2
5
Chapter 3 Review Exercises
b.
SSM: Intermediate Algebra
From part a., any number less than or equal
2
to в€’ is not a solution.
5
c.
0.22 x + 75 < 0.69 x + 29.95
0.22 x + 75 в€’ 75 < 0.69 x + 29.95 в€’ 75
0.22 x < 0.69 x в€’ 45.05
29. The student’s work is incorrect. When dividing
both sides of an inequality by a negative number,
you must switch the direction of the inequality.
30.
0.22 x в€’ 0.69 x < 0.69 в€’ 45.05 в€’ 0.69 x
в€’0.47 x < в€’45.05
в€’0.47 x в€’45.05
>
в€’0.47
в€’0.47
x > 95.9
Rent A Wreck will be cheaper for miles
driven more than 95.9 miles.
f (4) = 1
31. g ( 4 ) = в€’3
32.
f ( x ) = 0 when x = 1 .
38. Let P (t ) represent the average price (in dollars)
33. g ( x ) = 0 when x = в€’5 .
34.
of a home in a community and S (t ) represent
the amount of money (in dollars) a family has
saved at t years since 2000.
f ( x ) = g ( x ) when x = в€’2 .
35. The graph of f ( x ) is above the graph of g ( x )
Since the average price of a home increases at a
constant $9000 per year, the function P is linear
and its slope is 9000. the P-intercept is
(0, 250000) since the price of a home is
for values of x that are greater than в€’2 . Thus,
f ( x ) > g ( x ) when x > в€’2 .
$250,000 in year t = 0 . So, an equation for P is :
P (t ) = 9000t + 250000 .
cв€’b
=5,
a
a < 0 , and b > c . For example, a = в€’1 , b = 6 ,
and c = 1 will work.
36. Any values for a, b, and c so that
37. a.
b.
R <U
Similar work in finding the equation for the
function S gives: S (t ) = 4800t + 12000 . (The
Since U-Haul’s charg e increases at a
constant rate of $0.69 per mile, the equation
is linear with slope 0.69 . The U-intercept is
29.95 since U-Haul charges a flat fee of
$29.95. An equation for U-Haul’s charge is:
U ( x ) = 0.69 x + 29.95 .
After similar work for Rent A Wreck, an
equation for Rent A Wreck’s charge is:
R ( x ) = 0.22 x + 75.00 .
slope of S is 4800 since the family plans to save
$400 each month which is $4800 each year.)
In order to predict when the family will be able
to pay a 10% down payment on an averagepriced house, solve the following system for t
when
y = 0.1( P (t )) = 0.1(9000t + 250000 )
y = S (t ) = 4800t + 12000
Substitute 0.1(9000t + 250000 ) for y in the
U (x) = R (x)
second equation and solve for t.
0.1(9000t + 250000 ) = 4800t + 12000
0.69 x + 29.95 = 0.22 x + 75
0.69 x + 29.95 в€’ 29.95 = 0.22 x + 75 в€’ 29.95
900t + 25000 = 4800t + 12000
0.69 x = 0.22 x + 45.05
в€’3900t = в€’13000
0.69 x в€’ 0.22 x = 0.22 x + 45.05 в€’ 0.22 x
t ≈ 3.33
The family will be able to pay a 10% down
payment in 3 years (2003).
0.47 x = 45.05
x ≈ 95.9
The two charges will be the same when the
number of miles driven is roughly 95.9
miles.
74
Chapter 3 Test
SSM: Intermediate Algebra
39. a.
40. a.
Start by plotting the data sets, then find the
regression line for each region.
Solve the system
y = H (t ) = 21.14t + 100.00
y = A (t ) = в€’15.63t + 220.68
Solve using substitution.
Substitute 21.14t + 100.00 for y in the
second equation and solve for t.
y = в€’15.63t + 220.68
North America:
21.14t + 100.00 = в€’15.63t + 220.68
n (t ) = 0.52t + 35.04
21.14t + 100 в€’ 100 = в€’15.63t + 220.68 в€’ 100
21.14t = в€’15.63t + 120.68
Far East:
21.14t + 15.63t = в€’15.63t + 120.68 + 15.63t
36.77t = 120.68
t ≈ 3.3
Substitute t = 3.3 into y = 21.14t + 100.00
and solve for y.
y = 21.14t + 100.00
f (t ) = 1.50t + 13.89
b.
Both slopes are positive indicating an
increase in consumption each year. The
slope for the Far East is larger than the slope
for North America. The consumption of
petroleum is increasing at a faster rate in the
Far East than in North America.
c.
n (t ) = f (t )
= 21.14 (3.3) + 100.00
≈ 169.8
According to the models, there were the
same number of jobs in Hollywood as in the
aerospace industry in the year 1993 ( t = 3 ).
b.
inequality is true when t > 3.3 . Therefore,
there are more jobs in Hollywood than in the
aerospace industry in years after 1993
( t > 3 ).
0.52t + 35.04 = 1.50t + 13.89
0.52t + 35.04 в€’ 35.04 = 1.50t + 13.89 в€’ 35.04
0.52t = 1.50t в€’ 21.15
0.52t в€’ 1.50t = 1.50t в€’ 21.15 в€’ 1.50t
в€’0.98t = в€’21.15
c.
t ≈ 21.6
The consumption of petroleum was the same
for the Far East and North America in 2002.
d.
Solve for t when H (t ) > A (t ) . This
n ( t ) < f (t )
Hollywood has experienced the most change
in terms of employment. This is evident
from the greater difference in the number of
jobs in Hollywood compared to the
difference in the number of jobs in the
aerospace industry between the years 1992
and 1996.
Chapter 3 Test
0.52t + 35.04 < 1.50t + 13.89
0.52t + 35.04 в€’ 35.04 < 1.50t + 13.89 в€’ 35.04
1. Solve using substitution.
Substitute 3 x в€’ 1 for y in the second equation
and solve for x.
3 x в€’ 2 y = в€’1
0.52t < 1.50t в€’ 21.15
0.52t в€’ 1.50t < 1.50t в€’ 21.15 в€’ 1.50t
в€’0.98t < в€’21.15
3 x в€’ 2 (3 x в€’ 1) = в€’1
в€’0.98t в€’21.15
>
в€’0.98
в€’0.98
t > 21.6
The consumption of petroleum in North
America will be less than the consumption
of petroleum in the Far East for years after
2002.
3 x в€’ 6 x + 2 = в€’1
в€’3 x + 2 = в€’1
в€’3 x = в€’3
(cont.)
75
x =1
Chapter 3 Test
SSM: Intermediate Algebra
Substitute x = 1 into the first equation and solve
for y.
y = 3 (1) в€’ 1
3
1
(5 ) +
5
4
1
3+
4
1
4
1
4
= 3 в€’1
=2
The solution is (1, 2 ) .
2. Solve using elimination.
First write the second equation so that all the
variables are on the left side of the equation and
the constant is on the right side.
2x в€’ 5 y = 3
y =1
y =1
y = 1в€’ 3
y = в€’2
y = в€’8
The solution is (5, в€’8) .
5. First use the distributive property to simplify
each equation.
в€’4 ( x + 2 ) + 3 ( 2 y в€’ 1) = 21
6 x в€’ 15 y = 9
Multiply the first equation by в€’3 and add the
equations.
в€’6 x + 15 y = в€’9
в€’4 x в€’ 8 + 6 y в€’ 3 = 21
в€’4 x + 6 y в€’ 11 = 21
6 x в€’ 15 y = 9
в€’4 x + 6 y = 32
2 x в€’ 3 y = в€’16
0 = 0 true
This is an identity. The system is dependent. The
solution set is the set of ordered pairs ( x, y )
5 (3 x в€’ 2 ) в€’ ( 4 y + 3) = в€’59
such that 2 x в€’ 5 y = 3 .
15 x в€’ 10 в€’ 4 y в€’ 3 = в€’59
15 x в€’ 4 y в€’ 13 = в€’59
3. Solve using elimination.
Multiply the first equation by 3 and the second
equation by в€’2 , then add the equations.
12 x в€’ 18 y = 15
15 x в€’ 4 y = в€’46
The system can be rewritten as:
2 x в€’ 3 y = в€’16
15 x в€’ 4 y = в€’46
Solve using elimination.
Multiply the first equation by 4 and the second
equation by в€’3 , then add the equations.
8 x в€’ 12 y = в€’64
в€’12 x + 18 y = 4
0 = 19 false
This is a contradiction. The system is
inconsistent. The solution set is the empty set.
4. Solve using elimination.
Multiply the second equation by 3 and add the
equations.
2
3
xв€’ y =8
5
4
9
3
x+ y =3
5
4
11
x = 11
5
x=5
9
3
Substitute x = 5 into x + y = 3 and solve
5
4
for y.
в€’45 x + 12 y = 138
в€’ 37 x = 74
x = в€’2
Substitute x = в€’2 into 2 x в€’ 3 y = в€’16 and solve
for y.
2 ( в€’2 ) в€’ 3 y = в€’16
в€’4 в€’ 3 y = в€’16
в€’3 y = в€’12
y=4
The solution is ( в€’2, 4 ) .
6. Answers may vary. One possible answer:
x в€’ 2y =1
3 x + y = 17
76
Chapter 3 Test
SSM: Intermediate Algebra
7. Elimination:
First rewrite the second equation so that all the
variables are on the left side and the constant is
on the right side.
4x в€’ 3y = 9
8. If the solution set is the empty set, the system is
two parallel lines. Therefore, m in y = mx + b is
5 since this is the slope in y = 5 x в€’ 13 and
parallel lines have the same slope.
The y-intercept in y = mx + b is any number
other than в€’13 since y = 5 x в€’ 13 and
в€’2 x + y = в€’5
Multiply the second equation by 2 and add the
equations.
4x в€’ 3 y = 9
в€’4 x + 2 y = в€’10
y = mx + b intersect the y-axis at different points
( b в‰ в€’13 ).
9.
в€’ y = в€’1
2 − 10 x − 2 ≥ 3 x + 14 − 2
y =1
Substitute y = 1 into 4 x в€’ 3 y = 9 and solve
for x.
4 x в€’ 3 (1) = 9
−10 x ≥ 3 x + 12
−10 x − 3 x ≥ 3 x + 12 − 3 x
−13 x ≥ 12
в€’13 x 12
≤
в€’13 в€’13
12
x≤−
13
12 пЈ№
пЈ«
Interval: пЈ¬ в€’в€ћ, в€’ пЈє
13 пЈ»
пЈ­
4x в€’ 3 = 9
4 x = 12
x=3
The solution is (3,1) .
Substitution:
Substitute 2 x в€’ 5 for y in the first equation.
4x в€’ 3 y = 9
в€’2
3 x + 13 < 5 x в€’ 10
3 x + 13 в€’ 13 < 5 x в€’ 10 в€’ 13
в€’2 x + 15 = 9
в€’2 x = в€’6
3 x < 5 x в€’ 23
x=3
Substitute x = 3 into y = 2 x в€’ 5 and solve for y.
3 x в€’ 5 x < 5 x в€’ 23 в€’ 5 x
в€’2 x < в€’23
y = 2 (3 ) в€’ 5
в€’2 x в€’23
>
в€’2
в€’2
23
x>
2
пЈ« 23 пЈ¶
Interval: пЈ¬ , в€ћ пЈ·
пЈ­ 2
пЈё
=1
The solution is (3,1) .
Graphically:
4x в€’ 3 y = 9
y = 2x в€’ 5
4
y = xв€’3
3
0
y = 2 x в€’5
4
(3, 1)
y=
в€’2
4
2
3 x + 12 + 1 < 5 x в€’ 10
4 x в€’ 6 x + 15 = 9
в€’4
0
10. 3 ( x + 4 ) + 1 < 5 ( x в€’ 2 )
4 x в€’ 3 (2 x в€’ 5) = 9
y
2 − 10 x ≥ 3 x + 14
x
4
x в€’3
3
The solution is (3,1) .
77
6
12
Chapter 3 Test
SSM: Intermediate Algebra
2.6 ( x в€’ 3.1) > 4.7 x в€’ 5.9
11.
18. a.
2.6 x в€’ 8.06 > 4.7 x в€’ 5.9
2.6 x в€’ 8.06 + 8.06 > 4.7 x в€’ 5.9 + 8.06
3 x в€’ 11 + 11 < 7 в€’ 6 x + 11
2.6 x > 4.7 x + 2.16
3 x < 18 в€’ 6 x
3 x + 6 x < 18 в€’ 6 x + 6 x
2.6 x в€’ 4.7 x > 4.7 x + 2.16 в€’ 4.7 x
в€’2.1x > 2.16
в€’2.1x 2.16
<
в€’2.1 в€’2.1
x < в€’1.03
Interval: ( в€’в€ћ, в€’1.03)
в€’2
в€’1
9 x < 18
9 x 18
<
9
9
x<2
Any number less than 2 will satisfy the
inequality.
0
b.
5
1 7
12.
− x+ ≤ x
3
6 4
5
1 5
7
5
− x+ + x ≤ x+ x
3
6 3
4
3
1 41
x
≤
6 12
12 1 12 41
⋅ ≤ ⋅ x
41 6 41 12
2
≤x
41
2
x≥
41
пЈ®2
пЈ¶
Interval: пЈЇ , в€ћ пЈ·
пЈ° 41 пЈё
0
13.
1
19. a.
Answers may vary. From part a., any
number greater than or equal to 2 does not
satisfy the inequality.
In the year 2000, t = 100 . Compare A (100 )
and B (100 ) .
A (100 ) = в€’0.12 (100 ) + 31.75
= 19.75
B (100 ) = 0.08 (100 ) + 9.43
= 17.43
Since 19.75 в€’ 17.43 = 2.32 , College A had
an enrollment of 2320 more than College B.
b.
The slopes of A and B are в€’0.12 and 0.08 ,
respectively. The enrollment at college A is
decreasing by 120 students per year while
the enrollment at college B is increasing by
80 students per year.
c.
Solve for t when A (t ) = B (t ) .
в€’0.12t + 31.75 = 0.08t + 9.43
2
f (5 ) = 3
14. g ( x ) = 3 when x = в€’4 .
15.
Answers may vary.
Solve the inequality for x.
3 x в€’ 11 < 7 в€’ 6 x
в€’0.12t + 31.75 в€’ 31.75 = 0.08t + 9.43 в€’ 31.75
в€’0.12t = 0.08t в€’ 22.32
f ( x ) = g ( x ) when x = 2 .
в€’0.12t в€’ 0.08t = 0.08t в€’ 22.32 в€’ 0.08t
в€’0.2t = в€’22.32
t = 111.6
16. The graph of f is below the graph of g for x < 2
and the two graphs are the same for x = 2 .
Therefore, f ( x ) ≤ g ( x ) when x ≤ 2 .
According to the models, the colleges will
have the same enrollment in the year 2012
( t = 112 ).
17. The graph of f is lower than the graph of g for all
values of x that are greater than 13. Thus,
f ( x ) < g ( x ) for x > 13 .
78
Cumulative Review Chapters 1-3
SSM: Intermediate Algebra
d.
Solve for t when A (t ) > B (t ) .
d.
в€’0.12t + 31.75 > 0.08t + 9.43
в€’0.12t + 31.75 в€’ 31.75 > 0.08t + 9.43 в€’ 31.75
в€’0.12t > 0.08t в€’ 22.32
в€’0.12t в€’ 0.08t > 0.08t в€’ 22.32 в€’ 0.08t
в€’0.2t > в€’22.32
в€’0.2t в€’22.32
<
в€’0.2
в€’0.2
t < 111.6
According to the models, college A will
have a larger enrollment than college B for
years before 2012 ( t < 112 ).
20. a.
The slope for the two-year colleges is larger
than for the four-year colleges. This means
that the number of two-year colleges is
growing at a faster rate. However, the slopes
are not very different so it will take many
years for the number of two-year colleges to
overcome the difference in numbers.
Cumulative Review Chapters 1-3
1.
T
O rig in a l
te m p .
Start by plotting each set of data, then find
the regression lines for the data.
R e frig e ra to r
te m p .
2-Year Colleges:
t
2
2. y = в€’ x + 4
3
y
f ( x ) = 25.59t + 929.18
4
4-Year Colleges:
в€’4
4
x
в€’4
3. 5 x в€’ 3 y = 15
g (t ) = 23.14t + 1685.30
b.
в€’3 y = в€’5 x + 15
5
y = xв€’5
3
In 2008, t = 38 .
f (38 ) = 25.59 (38 ) + 929.18
y
= 1901.6
g (38 ) = 23.14 (38 ) + 1685.30
4
= 2564.62
The models predict that there will be a total
of 4466 two-year and four-year colleges in
2008.
c.
в€’4
4
в€’4
Solve for t when f (t ) = g (t ) .
25.59t + 929.18 = 23.14t + 1685.30
25.59t = 23.14t + 756.12
2.45t = 756.12
t ≈ 308.62
The number of two-year and four-year
colleges will be the same in 2279 ( t = 309 ).
79
x
Cumulative Review Chapters 1-3
SSM: Intermediate Algebra
4. 3 ( x в€’ 4 ) = в€’2 ( y + 5 ) + 4
8. First find the slope:
3 в€’ ( в€’2 )
3+ 2 5
m=
=
=
в€’2 в€’ ( в€’5 ) в€’2 + 5 3
3 x в€’ 12 = в€’2 y в€’ 10 + 4
3 x в€’ 12 = в€’2 y в€’ 6
Next use the slope and either point (we will use
the first point) to find b.
5
m = ; ( в€’5, в€’2 )
3
y = mx + b
2 y = в€’3 x + 6
y
3
y = в€’ x+3
2
3
5
( в€’5 ) + b
3
25
в€’2 = в€’ + b
3
19
b=
3
The equation of the line is:
5
19
y = x+
3
3
в€’2 =
2
1
в€’1
1
в€’1
2
3
x
5. x + 2 = 0
x = в€’2
y
9. 2 x в€’ 5 y = 20
4
в€’4
4
в€’5 y = в€’2 x + 20
2
y = xв€’4
5
x
в€’4
6. m =
=
Since the slope of the given line is
2
, our line
5
5
since perpendicular lines
2
have opposite-reciprocal slopes.
5
m = в€’ ; ( в€’5,3)
2
y = mx + b
y2 в€’ y1
x2 в€’ x1
must have slope в€’
в€’1 в€’ 2
3 в€’ ( в€’4 )
в€’3
7
3
=в€’
7
=
5
( в€’5 ) + b
2
25
3=
+b
2
19
b=в€’
2
The equation of the line is:
5
19
y = в€’ xв€’
or y = в€’2.5 x в€’ 9.5
2
2
3=в€’
3
7. m = в€’ ; ( 2, в€’3)
5
y = mx + b
3
(2 ) + b
5
6
в€’3 = в€’ + b
5
9
b=в€’
5
The equation of the line is:
3
9
or y = в€’0.6 x в€’ 1.8
y = в€’ xв€’
5
5
в€’3 = в€’
10. Answers may vary.
The lines y = 2 x + 3 and y = 2 x + 3.1 are
parallel. A line between these lines and parallel
to both is y = 2 x + 3.05 . We can select any three
points on this line. Some possible points:
(0,3.05) , (1,5.05) , and ( 2, 7.05)
80
Cumulative Review Chapters 1-3
SSM: Intermediate Algebra
11. Determine the slopes from the given data and use them to obtain the remaining values.
k (x) :
f (x) :
g (x) :
h (x) :
m=
58 в€’ 97 в€’39
=
= в€’13
3в€’0
3
Equation 1
x
12.
f (x)
Equation 2
m=
43 в€’ 4 39
=
= 13
9в€’6
3
m=
Equation 3
x
g (x)
x
в€’22 в€’ 23 в€’45
=
= в€’9
6 в€’1
5
15.
в€’16 в€’ ( в€’28 )
14 в€’ 10
=
12
=3
4
Equation 4
h (x)
x
k (x)
0
97
4
-22
1
23
10 -28
1
84
5
-9
2
14
11 -25
2
71
6
4
3
5
12 -22
3
58
7
17
4
-4
13 -19
4
45
8
30
5
-13
14 -16
5
32
9
43
6
-22
15 -13
f (5) = 32
17. The x-intercept is found by solving f ( x ) = 0 .
3
0= в€’ x+7
2
13. g ( x ) = 30 when x = 8 .
14.
m=
3
x=7
2
2 3
2
в‹… x = в‹…7
3 2
3
14
x=
3
3
f (x) = в€’ x + 7
2
3
f ( в€’4 ) = в€’ ( в€’4 ) + 7
2
= 6+7
= 13
пЈ« 14 пЈ¶
The x-intercept is пЈ¬ , 0 пЈ· .
пЈ­ 3 пЈё
3
f (x) = в€’ x + 7
2
3
в€’4 = в€’ x + 7
2
3
в€’11 = в€’ x
2
2
2 3
в€’ в‹… в€’11 = в€’ в‹… в€’ x
3
3 2
22
x=
3
18. Since the equation is in the form y = mx + b , the
y-intercept is (0,b ) or (0, 7 ) .
19.
3
f (x) = в€’ x + 7
2
y
8
4
16. Since the equation is in the form y = mx + b , the
slope is the coefficient on x.
3
m=в€’
2
4
20.
81
f ( в€’6 ) = в€’3 .
8
x
Cumulative Review Chapters 1-3
SSM: Intermediate Algebra
21. g ( x ) = 0 when x = в€’6 .
28.
22. The y-intercept of f is (0,1) .
23. To find the slope of g, pick two points on the
graph. Two possible points are the intercepts
( в€’6, 0 ) and (0, в€’2 ) .
m=
в€’2 в€’ 0
в€’2
1
=
=в€’
0 в€’ ( в€’6 ) 6
3
1
The slope of g is в€’ .
3
24. To find an equation for f, first determine the
slope by finding two points on the graph. Two
possible points are (0,1) and (3,3) . Use the
29. Yes, the graph is that of a function. The graph
passes the vertical line test.
points to find the slope of the line.
3 в€’1 2
m=
=
3в€’ 0 3
Since the y-intercept is (0,1) , the equation of the
line is y =
25.
7 1
1 3
в€’ x= + x
8 4
2 8
7 1
7 1 3
7
в€’ xв€’ = + xв€’
8 4
8 2 8
8
1
3
3
в€’ x = xв€’
4
8
8
1
3
3
3 3
в€’ xв€’ x = xв€’ в€’ x
4
8
8
8 8
5
3
в€’ x=в€’
8
8
3
x=
5
30. x =
y
y
8
2
x +1.
3
4
f ( x ) = g ( x ) when x = в€’3 .
2
x
The relation is a function. It passes the vertical
line test. For each input value, there is only one
output value.
26. The graph of f is at or below the graph of g for
all values of x that are less than or equal to в€’3 .
Therefore, f ( x ) ≤ g ( x ) when x ≤ −3 .
31. Answers may vary. One possible answer:
Equation:
y = 3x + 4
27. в€’5 x в€’ 3 ( 2 x + 4 ) = 8 в€’ 2 x
в€’5 x в€’ 6 x в€’ 12 = 8 в€’ 2 x
в€’11x в€’ 12 = 8 в€’ 2 x
Table:
x
y
в€’2 в€’2
в€’1 1
0 4
1 7
2 10
в€’11x в€’ 12 + 12 = 8 в€’ 2 x + 12
в€’11x = 20 в€’ 2 x
в€’11x + 2 x = 20 в€’ 2 x + 2 x
в€’9 x = 20
x=в€’
4
20
9
Graph:
y
4
в€’4
4
в€’4
82
x
Cumulative Review Chapters 1-3
SSM: Intermediate Algebra
32. Solve using elimination.
Multiply the first equation by 3 and the second
equation by 4, then add the equations.
6 x + 12 y = в€’24
Substitution:
x + 3y = 9
x = в€’3 y + 9
Substitute в€’3 y + 9 for x in the first equation and
solve for y.
2 ( в€’3 y + 9 ) в€’ 5 y = в€’4
20 x в€’ 12 y = 76
26 x = 52
x=2
Substitute x = 2 into 2 x + 4 y = в€’8 and solve
for y.
2 ( 2 ) + 4 y = в€’8
в€’6 y + 18 в€’ 5 y = в€’4
в€’11 y + 18 = в€’4
в€’11 y = в€’22
y=2
Substitute y = 2 in x = в€’3 y + 9 and solve for x.
4 + 4 y = в€’8
4 y = в€’12
x = в€’3 ( 2 ) + 9
y = в€’3
= в€’6 + 9
The solution is ( 2, в€’3) .
=3
The solution is (3, 2 ) .
33. Solve using substitution.
3
Substitute x в€’ 2 for y in the first equation.
7
3 x в€’ 7 y = 14
Graphically:
2 x в€’ 5 y = в€’4
в€’5 y = в€’2 x в€’ 4
пЈ«3
пЈ¶
3 x в€’ 7 пЈ¬ x в€’ 2 пЈ· = 14
7
пЈ­
пЈё
3 x в€’ 3 x + 14 = 14
3y = в€’x + 9
2
4
y = x+
5
5
1
y = в€’ x+3
3
y
14 = 14 true
This is an identity. The system is dependent. The
solution set consists of all ordered pairs ( x, y )
such that y =
x + 3y = 9
4
3
xв€’2.
7
(3, 2)
в€’2
4
x
в€’4
34. Elimination:
Multiply the second equation by в€’2 and add the
equations.
2 x в€’ 5 y = в€’4
в€’2 x в€’ 6 y = в€’18
35. −2 ( 4 x + 5 ) ≥ 3 ( x − 7 ) + 1
−8 x − 10 ≥ 3 x − 21 + 1
−8 x − 10 ≥ 3 x − 20
−8 x − 10 + 10 ≥ 3x − 20 + 10
в€’ 11 y = в€’22
−8 x ≥ 3 x − 10
y=2
Substitute y = 2 into 2 x в€’ 5 y = в€’4 and solve
for x.
2 x в€’ 5 ( 2 ) = в€’4
−8 x − 3 x ≥ 3x − 10 − 3x
−11x ≥ −10
в€’11x в€’10
≤
в€’11 в€’11
10
x≤
11
10 пЈ№
пЈ«
Interval: пЈ¬ в€’в€ћ, пЈє
11 пЈ»
пЈ­
2 x в€’ 10 = в€’4
2x = 6
x=3
The solution is (3, 2 ) .
0
83
10
11
2
Cumulative Review Chapters 1-3
36.
SSM: Intermediate Algebra
d.
4
1
5
xв€’ < 2в€’ x
3
6
6
4
1 1
5
1
xв€’ + < 2в€’ x+
3
6 6
6
6
4
13 5
x< в€’ x
3
6 6
4
5
13 5
5
x+ x < в€’ x+ x
3
6
6 6
6
13
13
x<
6
6
6 13
6 13
в‹… x< в‹…
13 6
13 6
x <1
Interval: ( в€’в€ћ,1)
в€’3 x + 6 = 4 x в€’ 1
в€’3 x = 4 x в€’ 7
в€’7 x = в€’7
x =1
Substitute this result into the first equation
and solve for y.
y = в€’3 x + 6
y = в€’3 (1) + 6
=3
The solution is (1,3) .
39. Answers may vary. Possible answers:
1
0
Solve using substitution.
Substitute в€’3 x + 6 for y in the second
equation and solve for x.
y = 4x в€’ 1
a.
37. Decrease m to make the slope more negative.
Increase b to raise the y-intercept. For example:
3x + 2 = 8
2x = 6
y
x=3
3
b.
2
y = в€’ x+2
3
y
x
38. a.
4
в€’3 x + 6 = 0
в€’3 x = в€’6
x=2
b.
в€’4
4
в€’3 x + 6 < 0
x
в€’4
в€’3 x < в€’6
в€’3 x в€’6
>
в€’3 в€’3
x>2
Interval: ( 2, в€ћ )
c.
y = 2x в€’1
y = в€’x + 2
y
y = 2 x в€’1
4
2
c.
(1, 1)
в€’4
y
4
в€’4
4
в€’2
в€’2
d.
4
y = в€’x + 2
4 x + 2 ≤ 10
4x ≤ 8
x
x≤2
0
84
x
2
Cumulative Review Chapters 1-3
SSM: Intermediate Algebra
40. a.
The grass height is declining at a constant
rate. Thus, h (t ) is linear and has the form
e.
h (t ) = m в‹… t + b .
Selecting two ordered pairs allows us to find
1пЈ¶
пЈ«
пЈ« 1пЈ¶
the slope. We will use пЈ¬ 55, пЈ· and пЈ¬ 95, пЈ· .
4
пЈ­
пЈё
пЈ­ 8пЈё
1 1
в€’
1
в€’1/ 8
m= 8 4 =
=в€’
95 в€’ 55
40
320
Using the slope and the first point, we can
find the y-intercept.
y = mx + b
1
1
=в€’
(55) + b
4
320
1
11
= в€’ +b
4
64
27
b=
64
1
27
Therefore, h (t ) = в€’
t+
.
320
64
b.
c.
d.
1
27
t+
320
64
1
27
t=
320
64
t = 135
The t-intercept is (135, 0 ) .
0=в€’
The putting greens will have no grass in
2035.
41. a.
Since the number of bicyclists younger than
13 hit by motorists declines at a constant
rate, f (t ) is linear.
The slope is в€’12 since the number hit
decreases by 12 each year. The n-intercept is
(0, 446 ) since 446 bicyclists younger than
13 were hit by motorists in 1975. Thus, an
equation for f (t ) is:
f (t ) = в€’12t + 446
1
27
(105 ) + ≈ 0.094
320
64
The grass on the green will have a height of
about 0.094 inches in 2005.
h (105 ) = в€’
b.
(see a.) The slope of f is в€’12 . The number
of bicyclists younger than 13 hit by
motorists decreases by 12 each year.
c.
(see a.) The n-intercept is (0, 446 ) . There
were 466 bicyclists younger than 13 hit by
motorists in 1975.
1
1
27
t+
=в€’
16
320
64
23
1
t
в€’
=в€’
64
320
t = 115
The grass on the green will have a height
1
that is
of an inch in 2015.
16
d.
Find the t-intercept by solving f (t ) = 0 .
0 = в€’12t + 446
12t = 446
t ≈ 37.17
The t-intercept is (37.17,0 ) .
The number of bicyclists younger than 13
hit by motorists will decrease to 0 in 2012.
42. a.
1 1
в€’
1
в€’1/ 8
m= 8 4 =
=в€’
95 в€’ 55
40
320
1
The slope of h is в€’
. The height of the
320
grass on the green gets shorter each year by
1
1
of an inch. Note:
≈ 0.0031
320
320
Since Company A’s sales increase by a
constant 1.3 million dollars each year, the
slope for A (t ) is 1.3. The A-intercept is
(0,9.5)
since the total sales are 9.5 million
dollars in year t = 0 . Therefore, the
equation for A (t ) is A (t ) = 1.3t + 9.5 .
Similar work for B (t ) gives the equation
B (t ) = 1.8t + 5.2 .
85
Cumulative Review Chapters 1-3
b.
M illio n s o f d olla rs
y
30
О’
SSM: Intermediate Algebra
43. a.
О‘
24
Women:
18
12
6
t
3
6
9
Ye ars
W (t ) = в€’0.064t + 27
12 15
The intersection of A and B is (8.6, 20.68 ) .
Men:
According to the sketch, sales at the
companies will both equal $20.68 million
dollars in 2006 ( t = 9 ).
c.
Solve for t when A (t ) = B (t ) .
M (t ) = в€’0.029t + 22.10
A (t ) = B (t )
1.3t + 9.5 = 1.8t + 5.2
b.
1.3t + 9.5 в€’ 9.5 = 1.8t + 5.2 в€’ 9.5
W (107 ) = в€’0.064 (107 ) + 27
= 20.15
M (107 ) = в€’0.029 (107 ) + 22.10
1.3t = 1.8t в€’ 4.3
1.3t в€’ 1.8t = 1.8t в€’ 4.3 в€’ 1.8t
= 19
The models predict that 200-Meter Run
record time in 2007 will be 20.15 seconds
for women and 19 seconds for men.
в€’0.5t = в€’4.3
t = 8.6
Substitute 8.6 for t in A (t ) = 1.3t + 9.5 .
A (8.6 ) = 1.3 (8.6 ) + 9.5 = 20.68
In 2009 ( t = 9 ), the sales at both companies
will equal $20.68 million dollars. This is the
same result as in part b.
d.
Start by plotting each set of data and then
find the regression lines.
Solve the inequality A (t ) < B (t ) .
c.
The absolute value of the slope for the
women is larger than for the men. This
means that the record time for women is
decreasing at a faster rate than for men.
d.
Since the slopes are not equal, the graphs of
the two equations will cross eventually. The
record times are getting closer which
indicates that the times will eventually be
equal sometime in the future.
e.
Solve for t in W (t ) = M (t ) .
A (t ) < B (t )
1.3t + 9.5 < 1.8t + 5.2
1.3t + 9.5 в€’ 9.5 < 1.8t + 5.2 в€’ 9.5
1.3t < 1.8t в€’ 4.3
1.3t в€’ 1.8t < 1.8t в€’ 4.3 в€’ 1.8t
W ( t ) = M (t )
в€’0.5t < в€’4.3
в€’0.064t + 27 = в€’0.029t + 22.1
в€’0.5t в€’4.3
>
в€’0.5 в€’0.5
t > 8.6
According to the models, Company A’s
sales will be less than Company B’s sales
after 2009 ( t = 9 ).
в€’0.064t = в€’0.029t в€’ 4.9
в€’0.035t = в€’4.9
t = 140
The models predict that the record times will
be equal in 2040.
86
Cumulative Review Chapters 1-3
SSM: Intermediate Algebra
W ( t ) > M (t )
f.
в€’0.064t + 27 > в€’0.029t + 22.1
в€’0.064t > в€’0.029t в€’ 4.9
в€’0.035t > в€’4.9
в€’0.035t
в€’4.9
<
в€’0.035 в€’0.035
t < 140
Women’s record times are greater than
men’s record times for years before 2040.
g.
Women:
0 = в€’0.064t + 27
0.064t = 27
t ≈ 421.88
Men:
0 = в€’0.029t + 22.1
0.029t = 22.1
t ≈ 762.07
The t-intercept for women is ( 421.88, 0 )
and the t-intercept for men is (762.07, 0 ) .
According to the models, the women’s
record time will be 0 seconds in 2322 and
the men’s record time will be 0 seconds in
2662. This is not feasible so model
breakdown has occurred.
h.
Answers may vary. One possible answer:
r
W
M
t
87
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