Applications of Linear Systems

4.4. APPLICATIONS OF LINEAR SYSTEMS
259
Take the answer y = 8 and substitute 8 for y in the first equation.
x + 9y = 73
First equation.
x + 9(8) = 73
x + 72 = 73
Substitute 8 for y.
Multiply.
x=1
Subtract 72 from both sides.
Hence, the solution is (x, y) = (1, 8).
4.4
Applications of Linear Systems
1. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Our variable dictionary will take the form
of a diagram, naming the two complementary angles α and β.
β
α
2. Set up a System of Equations. The “second angle is 42 degrees larger
than 3 times the first angle” becomes
β = 42 + 3α
Secondly, the angles are complementary, meaning that the sum of the
angles is 90◦ .
α + β = 90
Thus, we have a system of two equations in two unknowns α and β.
3. Solve the System. Substitute 42 + 3α for β in α + β = 90.
α + β = 90
α + (42 + 3α) = 90
4α + 42 = 90
4α = 48
α = 12
Substitute 42 + 3α for β.
Combine like terms.
Subtract 42 from both sides.
Divide both sides by 4.
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4. Answer the Question. The first angle is α = 12 degrees. The second
angle is:
β = 42 + 3α
β = 42 + 3(12)
Substitute 12 for α.
β = 78
Simplify.
5. Look Back. Certainly 78◦ is 42◦ larger than 3 times 12◦ . Also, note that
12◦ + 78◦ = 90◦ , so the angles are complementary. We have the correct
solution.
3. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Our variable dictionary will take the form
of a diagram, naming the width and length W and L, respectively.
L
W
W
L
2. Set up a System of Equations. The perimeter is found by summing the
four sides of the rectangle.
P =L+W +L+W
P = 2L + 2W
We’re told the perimeter is 116 inches, so we can substitute 116 for P in
the last equation.
116 = 2L + 2W
We can simplify this equation by dividing both sides by 2, giving the
following result:
L + W = 58
Secondly, we’re told that the “length is 28 inches more than twice the
width.” This translates to:
L = 28 + 2W
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3. Solve the System. As the last equation is already solved for L, let use
the substitution method and substitute 28 + 2W for L in the equation
L + W = 58.
L + W = 58
(28 + 2W ) + W = 58
3W + 28 = 58
3W = 30
W = 10
Perimeter equation.
Substitute 28 + 2W for L.
Combine like terms.
Subtract 28 from both sides.
Divide both sides by 3.
4. Answer the Question. The width is W = 10 inches. To find the length,
substitute 10 for W in the equation L = 28 + 2W .
L = 28 + 2W
L = 28 + 2(10)
Length equation.
Substitute 10 for W .
L = 28 + 20
L = 48
Multiply.
Add.
Thus, the length is L = 48 inches.
5. Look Back. Perhaps a picture, labeled with our answers might best
demonstrate that we have the correct solution. Remember, we found
that the width was 10 inches and the length was 48 inches.
48
10
10
48
Note that the perimeter is P = 48 + 10 + 48 + 10 = 116 inches. Secondly,
note that the length (48 inches) is 28 inches more than twice the width.
So we have the correct solution.
5. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Let N represent the number of nickels and
let Q represent the number of quarters.
2. Set up a System of Equations. Using a table to summarize information is
a good strategy. In the first column, we list the type of coin. The second
column gives the number of each type of coin, and the third column
contains the value (in cents) of the number of coins in her pocket.
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262
Number of Coins
Value (in cents)
Nickels
N
5N
Quarters
Q
25Q
Totals
59
635
Note that N nickels, valued at 5 cents apiece, are worth 5N cents. Similarly, Q quarters, valued at 25 cents apiece, are worth 25Q cents. Note
also how we’ve change $6.35 to 635 cents.
The second column of the table gives us our first equation.
N + Q = 59
The third column of the table gives us our second equation.
5N + 25Q = 635
3. Solve the System. Because both equations are in standard form Ax+By =
C, we’ll use the elimination method to find a solution. Because the
question asks us to find the number of quarters in her pocket, we’ll focus
on eliminating the N -terms and keeping the Q-terms.
−5N
5N
−
+
5Q = −295
25Q =
635
20Q
=
Multiply first equation by −5.
Second equation.
340
Add the equations.
Dividing both sides of the last equation by 20 gives us Q = 17.
4. Answer the Question. The previous solution tells us that Maria has 17
quarters in her pocket.
5. Look Back. Again, summarizing results in a table might help us see if
we have the correct solution. First, because we’re told that Maria has 59
coins in all, and we found that she had 17 quarters, this means that she
must have 42 nickels.
Number of Coins
Value (in cents)
Nickels
42
210
Quarters
17
425
Totals
59
635
42 nickels are worth 210 cents, and 17 quarters are worth 425 cents.
That’s a total of 59 coins and 635 cents, or $6.35. Thus we have the
correct solution.
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7. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Let x be the number of pounds of cashews
used and let y be the number of pounds of raisins used.
2. Set up a System of Equations. The following table summarizes the information given in the problem:
Cost per pound
Amount (pounds)
Cost
cashews
$6.00
x
6.00x
raisins
$7.00
y
7.00y
Totals
$6.42
50
6.42(50) = 321.00
At $6.00 per pound, x pounds of cashews cost 6.00x. At $7.00 per pound,
y pounds of raisins cost 7.00y. Finally, at $6.42 per pound, 50 pounds of
a mixture of cashews and raisins will cost $6.42(50), or $321.00.
The third column of the table gives us our first equation. The total
number of pounds of mixture is given by the following equation:
x + y = 50
The fourth column of the table gives us our second equation. The total
cost is the sum of the costs for purchasing the cashews and raisins.
6.00x + 7.00y = 321.00
Therefore, we have the following system of equations:
x + y = 50
6.00x + 7.00y = 321.00
3. Solve the System. We can solve this system by substitution. Solve the
first equation for x.
x + y = 50
First Equation.
x = 50 − y
Subtract y from both sides.
Next, substitute 50 − y for x in the second equation and solve for y.
6.00x + 7.00y = 321.00
6.00(50 − y) + 7.00y = 321.00
300.00 − 6.00y + 7.00y = 321.00
300.00 + 1.00y = 321.00
1.00y = 21.00
y = 21
Second Equation.
Substitute 50 − y for x.
Distribute the 6.00.
Combine like terms.
Subtract 300.00 from both sides.
Divide both sides by 1.00.
Thus, there are 21 pounds of raisins in the mix.
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4. Answer the Question. The question asks for both amounts, cashews and
raisins. Substitute 21 for y in the first equation and solve for x.
x + y = 50
First Equation.
x + 21 = 50
x = 29
Substitute 21 for y.
Subtract 21 from both sides.
Thus, there are 29 pounds of cashews in the mix.
5. Look Back. First, note that the amount of cashews and raisins in the
mix is 29 and 21 pounds respectively, so that the total mixture weighs 50
pounds as required. Let’s calculate the costs: for the cashews, 6.00(29),
or $174.00, for the raisins, 7.00(21), or $147.00.
Cost per pound
Amount (pounds)
Cost
cashews
$6.00
29
$174.00
raisins
$7.00
21
$147.00
Totals
$6.42
50
$321.00
Note that the total cost is $321.00, as required in the problem statement.
Thus, our solution is correct.
9. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Let D represent the number of dimes and
let Q represent the number of quarters.
2. Set up a System of Equations. Using a table to summarize information is
a good strategy. In the first column, we list the type of coin. The second
column gives the number of each type of coin, and the third column
contains the value (in cents) of the number of coins in his pocket.
Number of Coins
Value (in cents)
Dimes
D
10D
Quarters
Q
25Q
Totals
38
545
Note that D times, valued at 10 cents apiece, are worth 10D cents. Similarly, Q quarters, valued at 25 cents apiece, are worth 25Q cents. Note
also how we’ve change $5.45 to 545 cents.
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265
The second column of the table gives us our first equation.
D + Q = 38
The third column of the table gives us our second equation.
10D + 25Q = 545
3. Solve the System. Because both equations are in standard form Ax+By =
C, we’ll use the elimination method to find a solution. Because the
question asks us to find the number of dimes in his pocket, we’ll focus on
eliminating the Q-terms and keeping the D-terms.
−25D
10D
−
+
25Q =
25Q =
−950
545
Multiply first equation by −25.
Second equation.
=
−405
Add the equations.
−15D
Dividing both sides of the last equation by −15 gives us D = 27.
4. Answer the Question. The previous solution tells us that Roberto has 27
dimes in his pocket.
5. Look Back. Again, summarizing results in a table might help us see if we
have the correct solution. First, because we’re told that Roberto has 38
coins in all, and we found that he had 27 dimes, this means that he must
have 11 quarters.
Number of Coins
Value (in cents)
Dimes
27
270
Quarters
11
275
Totals
38
545
27 dimes are worth 270 cents, and 11 quarters are worth 275 cents. That’s
a total of 38 coins and 545 cents, or $5.45. Thus we have the correct
solution.
11. In geometry, two angles that sum to 180◦ are called supplementary angles.
If the second of two supplementary angles is 40 degrees larger than 3 times the
first angle, find the degree measure of both angles.
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13. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Let C represent the amount invested in the
certificate of deposit and M represent the amount invested in the mutual
fund.
2. Set up a System of Equations. The following table summarizes the information given in the problem:
Rate
Amount invested
Interest
Certificate of Deposit
3%
C
0.03C
Mutual Fund
5%
M
0.05M
20,000
780
Totals
At 3%, the interest earned on a C dollars investment is found by taking
3% of C (i.e., 0.03C). Similarly, the interest earned on the mutual fund
is 0.05M .
The third column of the table gives us our first equation. The total
investment is $20,000.
C + M = 20000
The fourth column of the table gives us our second equation. The total
interest earned is the sum of the interest earned in each account.
0.03C + 0.05M = 780
Therefore, we have the following system of equations:
C + M = 20000
0.03C + 0.05M = 780
3. Solve the System. We can solve this system by substitution. Solve the
first equation for C.
C + M = 20000
First Equation.
C = 20000 − M
Subtract M from both sides.
Next, substitute 20000 − M for C in the second equation and solve for
M.
0.03C + 0.05M = 780
0.03(20000 − M ) + 0.05M = 780
600 − 0.03M + 0.05M = 780
600 + 0.02M = 780
0.02M = 180
M = 9000
Second Edition: 2012-2013
Second Equation.
Substitute 20000 − M for C.
Distribute the 0.03.
Combine like terms.
Subtract 600 from both sides.
Divide both sides by 0.02.
4.4. APPLICATIONS OF LINEAR SYSTEMS
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Thus, the amount invested in the mutual fund is M = $9, 000.
4. Answer the Question. The question asks us to find the amount invested
in each account. So, substitute 9000 for M in the first equation and solve
for C.
C + M = 20000
First Equation.
C + 9000 = 20000
C = 11000
Substitute 9000 for M .
Subtract 9000 from both sides.
Thus, the amount invested in the certificate of deposit is $11, 000.
5. Look Back. First, note that the investments in the certificate of deposit
and the mutual fund, $11000 and $9000 respectively, total $20,000. Let’s
calculate the interest on each investment: 3% of $11000 is $330 and 5%
of $9000 is $450.
Rate
Amount invested
Interest
Certificate of Deposit
3%
11,000
330
Mutual Fund
5%
9,000
450
20,000
780
Totals
Note that the total interest is $780, as required in the problem statement.
Thus, our solution is correct.
15. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Our variable dictionary will take the form
of a diagram, naming the width and length W and L, respectively.
L
W
W
L
2. Set up a System of Equations. The perimeter is found by summing the
four sides of the rectangle.
P =L+W +L+W
P = 2L + 2W
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We’re told the perimeter is 376 centimeters, so we can substitute 376 for
P in the last equation.
376 = 2L + 2W
We can simplify this equation by dividing both sides by 2, giving the
following result:
L + W = 188
Secondly, we’re told that the “length is 12 centimeters less than three
times the width.” This translates to:
L = 3W − 12
3. Solve the System. As the last equation is already solved for L, let use
the substitution method and substitute 3W − 12 for L in the equation
L + W = 188.
L + W = 188
Perimeter equation.
(3W − 12) + W = 188
4W − 12 = 188
Substitute 3W − 12 for L.
Combine like terms.
4W = 200
W = 50
Add 12 to both sides.
Divide both sides by 4.
4. Answer the Question. The width is W = 50 centimeters. To find the
length, substitute 50 for W in the equation L = 3W − 12.
L = 3W − 12
L = 3(50) − 12
Length equation.
Substitute 50 for W .
L = 150 − 12
L = 138
Multiply.
Subtract.
Thus, the length is L = 138 centimeters.
5. Look Back. Perhaps a picture, labeled with our answers might best
demonstrate that we have the correct solution. Remember, we found
that the width was 50 centimeters and the length was 138 centimeters.
138
50
50
138
Note that the perimeter is P = 138 + 50 + 138 + 50 = 376 centimeters.
Secondly, note that the length (138 centimeters) is 12 centimeters less
than three times the width. So we have the correct solution.
Second Edition: 2012-2013