Students can use the Spectrum Math Grade 8 Answer Key Chapter 3 Lesson 3.9 Problem-Solving with Linear Equation System as a quick guide to resolve any of their doubts.
Spectrum Math Grade 8 Chapter 3 Lesson 3.9 Problem-Solving with Linear Equation System Answers Key
Linear equation systems can be used to find solutions to word problems that have a constant relationship between two variables.
The admission fee at a fair is $2.00 for children and $5.00 for adults. On a certain day, 2,400 people enter the fair and $6,801 is collected. How many children and how many adults went to the fair that day? Use a to represent the number of adults and c to represent the number of children.
a + c Step 1: Use the word problem to set up the system of equations.
5a + 2c = 6801
a = 2400 – x Step 2: Use the simplest equation to isolate one variable.
5(2400 – c) +2c = 6801
12,000 – 5c + 2c = 6801 Step 3: Use substitution to replace one of variables.
12,000 – 3c = 6801 Step 4: Use combination of like terms and inverse operations to isolate the variable in the equation.
12,000 – 3c – 12,000 = 6801 – 12,000
-3c ÷ (-3) = -5199 ÷ (-3)
c = 1,733
a + 1733 = 2400 Step 5: Find the value of one variable.
a + 1733 – 1733 = 2900 – 1733 Step 6: Use the value of the first variable in the simplest equation to find the value of the second variable.
a = 667
667 adults and 1,733 children went to the fair that day.
Set up a system of equations to solve each word problem.
Question 1.
At a convenience store, bottled water costs $1.10 and sodas cost $2.35. One day, the receipts for a total of 172 waters and sodas were $294.20. How many of each kind were sold? Use w to represent bottled water and s to represent soda.
Equation 1: ____
Equation 2: _____
w = ____; s = ____
Answer: Equation 1: 1.1w + 2.35s = 294.2
Equation 2: w + s = 172
w = 88, s = 84
At a convenience store, bottled water costs $1.10 and sodas cost $2.35. One day, the receipts for a total of 172 waters and sodas were $294.20.
Let w to represent bottled water and s to represent soda.
Equation 1: 1.1w + 2.35s = 294.2
Equation 2: w + s = 172
Use inverse operations to isolate one variable on one side of the equation.
w + s = 172
w + s -w = 172 -w
s = 172 -w
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
1.1w + 2.35s = 294.2
1.1w + 2.35(172 -w)= 294.2
1.1w + 404.2 – 2.35w= 294.2
1.1w – 2.35w= 294.2 – 404.2
1.25w = 110
w = 88
Substitute the value of the variable in one of the equations and solve.
s = 172 -w
s = 172 -88
s = 84
Therefore, w = 88, s = 84
Question 2.
Your teacher is giving you a test worth 100 points that contains 40 questions. There are 2-point questions and 4-point questions on the test. How many of each type of question are on the test? Use t to represent 2-point questions and f to represent 4-point questions.
Equation 1: ____
Equation 2: ____
t = ___; f = ____
Answer: Equation 1: t + f = 40
Equation 2: 2t + 4f = 100
t = 30, f = 10
Your teacher is giving you a test worth 100 points that contains 40 questions. There are 2-point questions and 4-point questions on the test.
Let t to represent 2-point questions and f to represent 4-point questions.
Equation 1: t + f = 40
Equation 2: 2t + 4f = 100
Use inverse operations to isolate one variable on one side of the equation.
t + f = 40
t + f -t= 40 -t
f = 40 -t
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
2t + 4f = 100
2t + 4(40 -t) = 100
2t + 160 – 4t = 100
2t – 4t = 100 – 160
-2t = -60
2t= 60
t = 30
Substitute the value of the variable in one of the equations and solve.
f = 40 -t
f = 40 – 30
f = 10
Therefore, t = 30, f = 10
Set up a system of equations to solve each word problem.
Question 1.
Jonathan has saved 57 coins in his bank. The coins are a mixture of quarters and dimes. He has saved $ 12.00 so far. How many quarters and how many dimes are in Jonathan’s bank? Use q to represent the number of quarters and d to represent the number of dimes.
Equation 1: _____
Equation 2: _____
q = ____; d = _____
Answer: Equation 1: q + d = 57
Equation 2: 0.25q + 0.1d = 12
q = 42, d = 15
Jonathan has saved 57 coins in his bank. The coins are a mixture of quarters and dimes. He has saved $ 12.00 so far.
Let q to represent the number of quarters and d to represent the number of dimes.
Equation 1: q + d = 57
Equation 2: 0.25q + 0.1d = 12
Use inverse operations to isolate one variable on one side of the equation.
q + d = 57
q + d -q= 57 -q
d = 57 -q
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
0.25q + 0.1d = 12
0.25q + 0.1(57 -q) = 12
0.25q + 5.7 – 0.1q = 12
0.25q – 0.1q = 12 – 5.7
0.15q = 6.3
q = 42
Substitute the value of the variable in one of the equations and solve.
d = 57 -q
d = 57 – 42
d = 15
Therefore, q = 42, d = 15
Question 2.
Members of the drama club held a car wash to raise funds for their spring musical. They charged $3 to wash a car and $5 to wash a pick-up truck or a sport utility vehicle. If they earned a total of $425 by washing a total of 125 vehicles, how many of each did they wash? Use c to represent the number of cars and t to represent the number of trucks.
Equation 1: _____
Equation 2: _____
c = ____; t = ____
Answer: Equation 1: 3c + 5t = 425
Equation 2: c + t = 125
c = 100, t = 25
They charged $3 to wash a car and $5 to wash a pick-up truck or a sport utility vehicle. They earned a total of $425 by washing a total of 125 vehicles
Let c to represent the number of cars and t to represent the number of trucks.
Equation 1: 3c + 5t = 425
Equation 2: c + t = 125
Use inverse operations to isolate one variable on one side of the equation.
c + t = 125
c + t -c= 125 -c
t = 125 – c
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
3c + 5t = 425
3c + 5(125 – c)= 425
3c + 625 – 5c= 425
3c – 5c= 425 – 625
-2c = -200
2c = 200
c = 100
Substitute the value of the variable in one of the equations and solve.
t = 125 – c
t = 125 – 100
t = 25
Therefore, c = 100, t = 25
Question 3.
Last Saturday 2,200 people attended an event at Fairway Gardens. The admission fee was $ 1.50 for children and $4.00 for adults. If the total amount of money collected at the event was $5,050, how many children and how many adults attended the event? Use c to represent the number of children and a to represent the number of adults.
Equation 1: ____
Equation 2: ____
c = ___; a = ____
Answer: Equation 1: c + a =2200
Equation 2: 1.5c + 4a = 5050
c = 1500, a = 700
Last Saturday 2,200 people attended an event at Fairway Gardens. The admission fee was $ 1.50 for children and $4.00 for adults. The total amount of money collected at the event was $5,050
Let c to represent the number of children and a to represent the number of adults.
Equation 1: c + a =2200
Equation 2: 1.5c + 4a = 5050
Use inverse operations to isolate one variable on one side of the equation.
c + a =2200
c + a -c=2200 -c
a = 2200 -c
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
1.5c + 4a = 5050
1.5c + 4(2200 -c)= 5050
1.5c + 8800 – 4c= 5050
1.5c – 4c= 5050 – 8800
-2.5c = -3750
2.5c = 3750
c = 1500
Substitute the value of the variable in one of the equations and solve.
a = 2200 -c
a = 2200 – 1500
a = 700
Therefore, c = 1500, a = 700