Mid-Chapter Check Point
6. 1
3
621
y
x
xy
Substitute 1
3
y for x in the second equation and
solve for y.
621
xy
7.
34 6
56 8
xy
xy
Multiply the first equation by −5, the second
8.
32 32
31
5
xy
xy
Multiply the second equation by 5 to clear the
fraction.
Multiply the second equation by −3 and solve by
addition.
3 2 32
34515
47 47
1
xy
xy
y
y
Back-substitute −1 for y and solve for x.
3
xy
Substitute 3y for x in the second equation and
solve for y.
242
xy
440 7
42 5 40 7
42 10 40 7
84040 7
87
xy
yy
yy
yy
yy
Chapter 4 Systems of Linear Equations
11. 32
29
yx
yx
Substitute 32x for y in the second equation and
12. 23 4
34 0
xy
xy
Multiply the first equation by 4, the second equation
by 3 and add to eliminate y.
812 16
912 0
17 16
xy
xy
x
13. 27
4214
yx
xy
Solve the first equation for y.
Substitute
27x for y and solve for x.
4214
422714
xy
xx
14.
4337
25 5
xy
yx
First, rewrite both equations in the form
.
A
xBy C
4337
41237
xy
xy
215
211 15
22 15
xy
x
x
15.
1
25
8
3
xy
x
y
−=
−=
Multiply the first equation by 10 and the second
equation by 3 to clear fractions.
()
10 10 10 1
25
5210
xy
xy
−=
−=
Back-substitute and solve for x.
()
324
310 24
xy
x
−+ =
−+ =
4.4 Check Points
1. Let x = average time per day women spend
socializing.
Let y = average time per day men spend socializing.
138
8
xy
xy
+=
−=
2. Let x = the number of calories in a Quarter Pounder.
Let y = the number of calories in a Whopper with
cheese.
2 3 2607
1000 9
xy
xy
+=
+= +
Solve the second equation for x.
1000 9
1009
xy
xy
+= +
=− +
Substitute 1009y−+ for x to find y.
write the first equation.
22
360 2 2
Plw
xy
=+
=+
Use the other information in the problem to write
20 16 3280
xy
+=
Multiply the first equation by −8 and add the result
to the second equation.
16 16 2880
20 16 3280
4400
100
xy
xy
x
x
−− =−
+=
=
=
Back-substitute to find y.
Chapter 4 Systems of Linear Equations
4. Let x=the number of years the heating system is used.
Let
y=
the total cost of the heating system.
5000 1100
12,000 700
yx
yx
=+
=+
5. Let x = the amount invested at 9%.
Let y = the amount invested at 11%.
5000
0.09 0.11 487
xy
xy
This system can be solved by substitution.
Solve for y in terms of x.
5000
5000
xy
yx
6. Let x = the number of ounces of 12% acid solution.
Let y = the number of ounces of 20% acid solution.
160
0.12 0.20 0.15(160)
xy
xy
Section 4.4 Problem Solving Using Systems of Equations
0.12 0.20 0.15(160)
0.12 0.20 24
xy
xy
7. Let x = the rate of the motorboat in still water.
Let y = the rate of the current.
Rate Time = Distance
Trip with the Current 2 2( )
Trip against the Current 3 3( )
xy xy
xy xy
This gives,
2( ) 84
3( ) 84
xy
xy
This system simplifies to:
42
xy
4.4 Concept and Vocabulary Check
1.
56xy+
2.
10 2 15 2 or 20 30xyxy⋅+⋅ +
Chapter 4 Systems of Linear Equations
4.4 Exercise Set
1. Let x = one number.
Let y = the other number.
17
xy
+=
2.
5
13
2 18
xy
xy
x
+=
−=
=
3. Let x = one number.
Let y = the other number.
31
223
xy
xy
−=−
+=
Solve the second equation for x.
223
223
xy
xy
+=
=− +
4.
32 43
24
xy
xy
+=
−=−
Multiply the second equation by 2.
32 43
xy
+=
The numbers are 5 and 14.
5. Let x = the average number of minutes 20- to
24-year-old women spend grooming.
Let y = the average number of minutes 20- to
24-year-old men spend grooming.
298
49
x
x
=
=
Back-substitute to find y.
86
49 86
37
xy
y
y
+=
+=
=
20- to 24-year-old women averaged 49 minutes
Section 4.4 Problem Solving Using Systems of Equations
7. Let x = the number of calories in a Mr. Goodbar.
Let y = the number of calories in a Mounds bar.
2780
2786
xy
xy
+=
+=
Multiply the bottom equation by –2 and then add
calories in a Mounds bar.
8. Let x = the number of calories in a Snickers bar.
Let y = the number of calories in a Reese’s Peanut
Butter Cup.
2737
2778
xy
xy
+=
+=
Multiply the bottom equation by –2 and then add
9. Let x = the number of Mr. Goodbars.
Let y = the number of Mounds bars.
5
16.3 14.1 70 7.1
xy
xy
+=
+−=
Solve the first equation for y in terms of x.
Back-substitute to find y.
5
35
2
xy
y
y
+=
+=
=
There are 3 Mr. Goodbars and 2 Mounds bars.
10. Let x = the number of Snickers bars.
Let y = the number of Reese’s Peanut Butter Cups.
7
x
=
Back-substitute to find y.
12
712
5
xy
y
y
+=
+=
=
There are 7 Snickers bars and 5 Reese’s Peanut
Butter Cups.
Chapter 4 Systems of Linear Equations
11. Let x = the price of one sweater.
Let y = the price of one shirt.
342
32 56
xy
xy
+=
+=
Multiply the first equation by −3 and add the result
12. Let x = the cost of one tablecloth.
Let y = the cost of one napkin.
85106
624
xy
xy
+=
+=
Multiply the second equation by −8 and add the
13. Let x = the length of a badminton court.
Let y = the width of a badminton court.
Use the formula for the perimeter of a rectangle to
write the first equation.
22
Plw
=+
Multiply the first equation by −3 and add the result
to the second equation.
6 6 384xy−− =−
69 444xy+=
360
y
=
Let y = the width of a tennis court.
22
228 2 2
Plw
xy
=+
=+
Use the other information in the problem to write
the second equation.
7 4 690xy+=
156 2 228
272
36
y
y
y
+=
=
=
The length is 78 ft and the width is 36 ft, so the
dimensions of a standard tennis court are 78 ft by
Section 4.4 Problem Solving Using Systems of Equations
15. Let x = the length of the lot.
Let y = the width of the lot.
Use the formula for the perimeter of a rectangle to
write the first equation.
22 320xy+=
Use the other information in the problem to write
the second equation.
()
16 5 2 2140xy+=
16. Let x = the length of the lot.
Let y = the width of the lot.
y
y
x
Use the formula for the perimeter of a rectangle to
write the first equation.
2 2 1600xy+=
Back-substitute to find y.
2 2 1600
2(500) 2 1600
1000 2 1600
2600
300
xy
y
y
y
y
+=
+=
+=
=
=
The length is 500 feet and the width is 300 feet, so
b. Plan A: 15(6) 40 130y=+=
Plan B: 20(6) 30 150y=+=
The monthly cost would be $130 for Plan A and
$150 for Plan B, so Plan A is the better deal by
$20.
18. a. Let x = the number of gigabytes used.
Let y = the total monthly cost.
Plan A: 18 52yx=+
Plan B: 22 32yx=+
Chapter 4 Systems of Linear Equations
19. Let x = the number of dollars of merchandise
purchased in a year.
Let y = the total cost for a year.
Plan A: 100 0.80yx=+
Plan B: 40 0.90yx=+
20. Let x = the number of dollars of merchandise
purchased in a year.
Let y = the total cost for a year.
Plan A: 300 0.70yx=+
Plan B: 40 0.90yx=+
21. Let x = the number of adult tickets sold.
Let y = the number of student tickets sold.
301
3487
xy
xy
+=
+=
Multiply the first equation by –1 and add it to the
second equation.
301
xy
−− =−
22. Let x = the number of adult tickets sold.
Let y = the number of student tickets sold.
1281
53425
xy
xy
+=
+=
Multiply the first equation by –1 and add it to the
tickets sold.
23. Let x = the cost of an item from column A.
Let y = the cost of an item from column B.
5.49
xy
+=
3.99
x
=
The cost of an item from column A is $3.99 and the
cost of an item from column B is $1.50.
24. a.
Rectangle Triangle
22 34and (1)30xy xyy+= +++=
Solve the second equation for x.
12
y
=
Section 4.4 Problem Solving Using Systems of Equations
Back-substitute to find x.
229
2(12) 29
5
xy
x
x
=− +
=− +
=
The value of x is 5 and the value of y is 12.
25. Let x = the number of servings of macaroni.
Let y = the number of servings of broccoli.
32 14
16 4 48
xy
xy
+=
+=
Multiply the first equation by −2 and add to second
26. Let x = the number of apples.
Let y = the number of avocados.
100 350 1000
24 14 100
xx
xy
+=
+=
Multiply the first equation by 6.
27. The sum of the measures of the three angles of any
triangle is 180°, so
()()()
8 1 3 4 7 5 180.xy y x+−++++=
Simplify this equation.
8118180
xy
++=
811 172
73 1
xy
xy
+=
−=−
Multiply the first equation by 3 and the second
equation by 11; then add the results.
24 33 516xy+=
77 33 11xy−=−
28. The sum of the measures of the three angles of any
triangle is 180°, so
()()( )
4 2 9 2 10 10 180.xy y x++−+ +=
Simplify this equation.
Chapter 4 Systems of Linear Equations
Multiply the first equation by 5.
Multiply the second equation by −7.
70 55 860
70 63 84
xy
xy
+=
−+ =
29. Let x = the amount invested at 6%.
Let y = the amount invested at 8%.
7000
0.06 0.08 520
xy
xy
Solve the first equation for x.
7000xy
Substitute this result for x in the second equation.
30. Let x = the amount invested in stocks.
Let y = the amount invested in bonds.
11000
0.05 0.08 730
xy
xy
Back-substitute to solve for x.
11000
6000 11000
xy
x
Back-substitute to solve for y.
0.10 0.01 860
0.10 8000 0.01 860
800 0.01 860
0.01 60
60 6000
0.01
xy
y
y
y
y
9000
y
Back-substitute to solve for y.
0.12 0.02 1500
0.12 0.02 1500
xy
xy
Section 4.4 Problem Solving Using Systems of Equations
33. Let x = amount invested with 12% return.
Let y = amount invested with the 5% loss.
20,000
0.12 0.05 1890
xy
xy
Multiply the first equation by
0.05
and add the two
34. Let x = amount invested with 14% return.
Let y = amount invested with the 6% loss.
30,000
0.14 0.06 200
xy
xy
Multiply the first equation by
0.06
and add the two
35. Let x = gallons of 5% wine.
Let y = gallons of 9% wine.
200
0.05 0.09 0.07 200
xy
xy
Back-substitute and solve for x.
200
200 100
100
xy
The wine company should mix 100 gallons of the 5%
Substitute this result for x into the second equation
and solve for y.
0.16 32 0.28 8
5.12 0.16 0.28 8
0.12 2.88
yy
yy
y
0.75 0.5 174
xy
Solve the first equation for x.
300xy
Substitute this result for x into the second equation
and solve for y.
Chapter 4 Systems of Linear Equations
38. Let x = gallons of cream.
Let y = gallons of milk.
50
0.25 0.35 0.125(50)
xy
xy
or
50
xy
39. Let x = pounds of cheaper candy.
Let y = pounds of more expensive candy.
75
1.6 2.1 1.9 75
xy
xy
or
75
xy
40. Let x = pounds of raisins.
Let y = pounds of granola.
10
2 3.25 2.50(10)
xy
xy
Substitute this result for x into the second equation
and solve for y.
2(10 ) 3.25 25
20 2 3.25 25
1.25 5
4
yy
yy
y
y
0.75 0.05 0.1 1.10
0.05 0.35
7
dd
d
d
Back-substitute to solve for n.
715
8
n
n
12 15
3
d
d
The purse has 3 dimes and 12 quarters.
Section 4.4 Problem Solving Using Systems of Equations
43. Let x = the speed of the plane in still air.
Let y = the speed of the wind.
Rate Time = Distance
Trip with the Wind 5 5( )
Trip against the Wind 8 8( )
xy xy
xy xy
44. Let x = the speed of the plane in still air.
Let y = the speed of the wind.
Rate Time = Distance
Trip with the Wind 6 6( )
Trip against the Wind 7 7( )
xy xy
xy xy
64200
xy
Chapter 4 Systems of Linear Equations
45. Let x = the crew’s rowing rate.
Let y = the rate of the current.
Rate Time = Distance
Trip with the Current 2 2( )
Trip against the Current 4 4( )
xy xy
xy xy
46. Let x = the boat’s rate in still water.
Let y = the rate of the current.
Rate Time = Distance
Trip with the Current 1.5 1.5( )
Trip against the Current 2 2( )
xy xy
xy xy
1.5 36
236
xy
xy
Divide the first equation by 1.5 and the second equation by 2 and solve by addition.
Section 4.4 Problem Solving Using Systems of Equations
47. Let x = the speed in still water.
Let y = the speed of the current.
Rate Time = Distance
4
Rewrite the system in
A
xByC
form.
44 24
6618
xy
xy
Multiply the first equation by –3 and the second equation by 2.
48. Let x = the speed in still water.
Let y = the speed of the current.
Rate Time = Distance
Trip with the Current 3 3( )
Trip against the Current 4 4( )
xy xy
xy xy
324
xy
49. – 53. Answers will vary.
54. makes sense
Chapter 4 Systems of Linear Equations
55. does not make sense; Explanations will vary.
Sample explanation: The model is 40 19 .yx=+
(4 2 4) (12 6 12) 180.xy xy
−++ ++ =
58. Let x = the first lucky number.
Then y = the second lucky number.
36 12
25
xy
xy
+=
+=
Solve this system. Multiply the second equation by
59. Let x = the number of birds.
Let y = the number of lions.
Since each bird has one head and each lion has one
head, 30.xy+=
Since each bird has two feet and each lion has four
feet, 2 4 100.xy+=
60.
()
90
320180.
xy
xy
+=
++=
To solve this system by the addition method,
multiply the first equation by −1 and add to the
second equation.
90
3160
2 70
xy
xy
y
−− =−
+=
=
61. Let x = the number of people in the downstairs
apartment.
Let y = the number of people in the upstairs
apartment.
If one of the people in the upstairs apartment goes
downstairs, there will be the same number of people
in both apartments, so 11yx−= +.
Back-substitute to find y.
52 7y=+=
There are 5 people downstairs and 7 people upstairs.
Section 4.4 Problem Solving Using Systems of Equations
62. Let x = the weight of Tweedledum.
Let y = the weight of Tweedledee.
2361
2362
xx
xy
+=
+=
63.
Principal Rate = Interest
8% Investment 0.08 0.08
12% Investment 0.12 0.12
9% Investment 70,000 0.09 0.09(70,000)
xx
yy
×
Since the total investment is $70,000 the first equation is 70,000.xy+=
Since the total interest is
0.09(70, 000)
the second equation is
0.08 0.12 0.09(70,000).xy+=
Back-substitute to find y.
70,000
52,500 70,000
17,500
xy
y
y
+=
+=
=
$52,500 should be invested at 8% and $17,500 should be invested at 12% to obtain an overall return of 9%.
64. Answers will vary.
Chapter 4 Systems of Linear Equations
65.
2( 3) 24 2( 4)
262428
xx
xx
+= − +
+= − −
66.
56( 1) 56 6
611
xx
x
++=++
=+
67. Find slope:
21
21
10 6 16 2
3(5) 8
yy
mxx
−−− −
== ==−
−−−
68.
248
2(3) (2) 4( 3) 8
88,true
xy z
Yes, the ordered triple satisfies the equation.
69.
5243
332 3
xyz
xyz