Chapter 4 Review Exercises
49.
Principal Rate = Interest
8% Investment 0.08 0.08
10% Investment 0.10 0.10
xx
yy
×
Since the total investment is $10,000 the first equation is
10,000.xy+=
Since the total interest is $940, the second equation is
0.08 0.10 940.xy+=
Back-substitute to find y.
10,000
3000 10,000
7000
xy
y
y
+=
+=
=
$3000 was invested at 8% and $7000 was invested at 10%.
50.
Gallons Percent Salt = Amount of Salt
75% Saltwater Solution 0.75 0.75
50% Saltwater Solution 0.50 0.50
60% Saltwater Solution 10 0.60 0.60(10)
xx
yy
×
Since there are a total of 10 gallons, the first equation is
10.xy+=
Chapter 4 Systems of Linear Equations
51. Let r = the speed of the plane in still air.
Let w = the speed of the wind.
Rate Time = Distance
Trip with the Wind 3 3( )
Trip against the Wind 4 4( )
rw rw
rw rw
3 2160
rw
Back-substitute 630 for r to find w.
3 3 2160
3 630 3 2160
1890 3 2160
3270
90
rw
w
w
w
w
The speed of the plane in still air is 630 miles per hour and the speed of the wind is 90 miles per hour.
52.
()
0
3250
xyz++=
−+− + =
Chapter 4 Review Exercises
53.
2 1
33 4 5
4234
xyz
xyz
xyz
−+=
−+=
−+=
Multiply the first equation by –2 and add to the
third.
The system of two equations in two variables
becomes:
2 1
33 3
xy
xy
−=−
−=−
Multiply the first equation by –3 and solve by
addition.
{
}
54.
2 5
2 30
2 1
xyz
xyz
yz
+−=
−+=
+=
Multiply the first equation by –2 and add to the
second equation.
We now have two equations in two variables.
2 1
55 10
yz
yz
+=
−+ =−
Multiply the first equation by –5 and solve by
addition.
() ( )
21 1 5
21 5
35
2
x
x
x
x
+−−=
++=
+=
=
The solution set is
()
{
}
2,1, 1 .−
448 8
12 1
xyz
xz
−+ + =−
−+=−
The system of two equations in two variables
becomes:
12 1
xz
−+ =−
Chapter 4 Systems of Linear Equations
56. Use each ordered pair to write an equation as follows:
() ()
2
2
(, ) (1,4)
41 1
4
xy
yax bxc
abc
abc
=
=++
=++
=++
The system of three equations in three variables is:
4
93 20
42 25
abc
abc
abc
++=
++=
−+=
Multiply the first equation by –1 and add to the
second equation.
The system of two equations in two variables
becomes:
8216
33 21
ab
ab
+=
−=
Multiply the first equation by 3, the second equation
by 2 and solve by addition.
()
33 3 21
93 21
312
4
b
b
b
b
−=
−=
−=
=−
Back-substitute 3 for a and –4 for b to find c.
13
53
xy
xz
−=
−=
Solve the second equation for y.
13
13
13
xy
yx
yx
−=
−=−+
=−
Solve the third equation for z.
13 53 306
3 66 306
3 372
124
xx x
x
x
x
+− +− =
−=
=
=
Back-substitute to solve for y and z.
13
yx
=−
Chapter 4 Test
Chapter 4 Test
1.
() ( )
{
}
25
25 5 5
xy+=
+− =
2.
()
57
352 7
310 7
77,true
xy+=
−+ =
−+ =
=
3.
6
44
xy
xy
+=
−=
Graph both lines on the same axes.
6:xy+=
x-intercept = 6; y-intercept = 6
{
}
4.
28
32
xy
yx
+=
=−
Graph both lines on the same axes.
5.
4
37 18
xy
xy
=+
+=−
Substitute
4y+
for x in the second equation.
4
341
xy
x
=+
=− + =
The solution set is
()
{
}
1, 3 .−
Chapter 4 Systems of Linear Equations
Back-substitute 2 for x in the equation
32 0xy+=
.
32 0
xy
+=
7.
24 3
24
xy
xy
−=
=+
Substitute
24y+
for x in the first equation.
24 3
xy
−=
8.
22xy+=
48xy−=−
6x = −6
9.
231
32 6
xy
xy
+=
+=−
Multiply the first equation by 3.
Multiply the second equation by −2.
69 3xy+=
10.
32 2
96 6
xy
xy
−=
−+ =−
dependent and the equation has infinitely many
solutions.
The solution set is
()
{
}
,32 2xy x y−=
or
()
{
}
,96 6.xy x y−+ =−
11. Let x = the number of scent receptors in dogs.
2 440
220
x
x
=
=
Back substitute to find y.
Since there are two lengths and two widths to be
fenced, the information about the cost of fencing
leads to the equation
() ()
22 12 58xy+=
.
Simplify the second equation.
42 58xy+=
Multiply this equation by −1 and add the result to
Chapter 4 Test
13. Let x = the number of gigabytes used.
Let y = the total monthly cost.
Plan A:
17 46yx=+
14. Let x = the amount invested at 6%
Let y = the amount invested at 7%
9000
0.06 0.07 610
xy
xy
+=
+=
15. Let x = the number of ounces of 6% solution
Let y = the number of ounces of 9% solution
()
36
0.06 0.09 0.08 36
xy
xy
+=
+=
Rewrite the system in standard form.
36
0.06 0.09 2.88
xy
xy
+=
+=
Chapter 4 Systems of Linear Equations
16. Let r = the speed of the paddleboat in still water.
Let c = the speed of the current.
Rate Time = Distance
×
Simplified, the system becomes
33 48
44 48
rc
rc
+=
−=
Multiply the first equation by 4, the second equation by 3, and solve by addition.
2
c
=
The speed of the paddleboat in still water is 14 miles per hour and the speed of the current is 2 miles per hour.
17. 6
3471
2 35
xyz
xyz
xyz
++=
+−=
−+=
Multiply the first equation by 7 and add to the second equation.
777 42
xyz
++=
Cumulative Review
Back-substitute 3 for y to find x.
413
xy
−− =−
Back-substitute 1 for x and 3 for y to find z.
The solution set is
()
{
}
1, 3, 2 .
Cumulative Review Exercises (Chapters 1-4)
1.
()
14 18 6 10
−− −−
3.
() ( )
17 3 13 4 10
17 51 13 4 40
17 51 4 27
13 78
6
xx
xx
xx
x
x
+=+ −
+=+−
+= −
=−
=−
The solution set is {−6}.
5.
A P Prt
=+
6. 25511
2
xx
x
−< −
>
The solution set is {2}xx>.
7.
36xy−=
x-intercept:
33 6
33
1
y
y
y
−=
−=
=−
A check point is (3, −1).
Chapter 4 Systems of Linear Equations
8.
() ( )
() ( )
04 (,)
606446,4
303443,4
xyx xy
y
y
=+
−=−+=−
−=−+=−
9. 32
5
yx=− +
slope = 33
;
55
−
−= y-intercept = 2
Plot the point (0,2). From this point, move 3 units
down (because −3 is negative) and 5 units to the
10. 34 8
45 10
xy
xy
−=
+=−
To solve this system by the addition method,
multiply the first equation by 4 and the second
equation by −3. Then add the results.
12 16 32xy−=
11. 23 9
48
xy
yx
−=
=−
To solve this system by the substitution method,
3
484 8 2
2
yx
=−= −=−
The solution set is 3,2 .
2
−
14. Use the formula for the area of a triangle.
1
2
1
80 16
2
80 8
Abh
h
h
=
=⋅⋅
=
Cumulative Review
15. Let x = the cost of one pen.
Let y = the cost of one pad.
10 15 26
510 16
xy
xy
+=
+=
Multiply the second equation by −2, and add the
result to the first equation.
16. The integers in the given set are −93, 0, 7
1 (=7) and
100 (=10).
17. The yes line has a positive slope. This means that
the percentage who answered “yes” increased