# Spectrum Math Grade 8 Chapter 3 Lesson 7 Answer Key Solving 2-Variable Linear Equation System

Students can use the Spectrum Math Grade 8 Answer Key Chapter 3 Lesson 3.7 Solving 2-Variable Linear Equation System as a quick guide to resolve any of their doubts.

## Spectrum Math Grade 8 Chapter 3 Lesson 3.7 Solving 2-Variable Linear Equation System Answers Key

Systems of equations can be solved by using the method of substitution following the steps below.
y = 7x + 10
y = 9x + 38
7x + 10 = 9x + 38 Step 1: Substitute one value of y so that there is only one variable in the new equation.
7x + 10 – 7x = 9x + 38 – 7x Step 2: Use the inverse operation and combine like terms with the x variable.
10 = 2x + 38
10 – 38 = 2x + 38 – 38 Step 3: Use the inverse operation to narrow the equation to 2 terms.
-28 = 2x
-28 ÷ 2 = 2x ÷ 2 Step 4: Use the inverse operation to isolate the x variable.
x = -14 Step 5: Find the value of the x variable.
y = 7(-14) + 10 Step 6: Substitute the value of the x variable in one of the equations.
y = -98 + 10 Step 7: Solve to find the value of the y variable.
y = -88

Use substitution to solve each equation system.

Question 1.
a. y = –$$\frac{4}{3}$$x + 6
y = 2
x = ____, y = ____
Answer: x = 3, y = 2
y = –$$\frac{4}{3}$$x + 6
y = 2
Substitute one value of y so that there is only one variable in the new equation.
2 = –$$\frac{4}{3}$$x + 6
Use the inverse operation and combine like terms with the x variable.
–$$\frac{4}{3}$$x = 2 – 6
Use the inverse operation to narrow the equation to 2 terms.
–$$\frac{4}{3}$$x = -4
$$\frac{4}{3}$$x = 4
Use the inverse operation to isolate the x variable.
4x = 12
Find the value of the x variable.
x = $$\frac{12}{4}$$
x  = 3
Therefore, x = 3, y = 2

b. y = $$\frac{1}{2}$$x + 3
y = 5
x = ____, y = ____
Answer: x = 4, y = 5
y = $$\frac{1}{2}$$x + 3
y = 5
Substitute one value of y so that there is only one variable in the new equation.
5 = $$\frac{1}{2}$$x + 3
Use the inverse operation and combine like terms with the x variable.
$$\frac{1}{2}$$x = 5-3
Use the inverse operation to narrow the equation to 2 terms.
$$\frac{1}{2}$$x = 2
Use the inverse operation to isolate the x variable.
x = 4
Therefore, x = 4, y = 5

Question 2.
a. y = 4x + 3
y = –$$\frac{1}{3}$$x – 8
x = ____, y = ____
Answer: x = $$\frac{-33}{13}$$, y = $$\frac{-93}{13}$$
y = 4x + 3
y = –$$\frac{1}{3}$$x – 8
Substitute one value of y so that there is only one variable in the new equation.
–$$\frac{1}{3}$$x – 8 = 4x + 3
Use the inverse operation and combine like terms with the x variable.
–$$\frac{1}{3}$$x – 4x = 3 + 8
Use the inverse operation to narrow the equation to 2 terms.
{-$$\frac{1}{3}$$ – 4}x = 11
$$\frac{-1-12}{3}$$x = 11
$$\frac{-13}{3}$$x = 11
Use the inverse operation to isolate the x variable.
-13x = 33
x = $$\frac{-33}{13}$$
Substitute the value of the x variable in one of the equations.
y = 4 $$\frac{-33}{13}$$ + 3
y = $$\frac{-132}{13}$$ + 3
y = $$\frac{-132 + 39}{13}$$
y = $$\frac{-93}{13}$$
Therefore, x = $$\frac{-33}{13}$$, y = $$\frac{-93}{13}$$

b. y = $$\frac{7}{2}$$x – 5
y = -5
x = ____, y = ____
Answer: x = 0, y = -5
y = $$\frac{7}{2}$$x – 5
y = -5
Substitute one value of y so that there is only one variable in the new equation.
-5 = $$\frac{7}{2}$$x – 5
Use the inverse operation and combine like terms with the x variable.
$$\frac{7}{2}$$x = -5+5
Use the inverse operation to narrow the equation to 2 terms.
$$\frac{7}{2}$$x = 0
Use the inverse operation to isolate the x variable.
7x = 0
Find the value of the x variable.
x = 0
Therefore, x = 0, y = -5

Question 3.
a. y = $$\frac{1}{3}$$x – 4
y = –$$\frac{7}{3}$$x + 4
x = ____, y = ____
Answer: x = 3, y = -3
y = $$\frac{1}{3}$$x – 4
y = –$$\frac{7}{3}$$x + 4
Substitute one value of y so that there is only one variable in the new equation.
–$$\frac{7}{3}$$x + 4 = $$\frac{1}{3}$$x – 4
Use the inverse operation and combine like terms with the x variable.
–$$\frac{7}{3}$$x –  $$\frac{1}{3}$$x= -4 -4
Use the inverse operation to narrow the equation to 2 terms.
$$\frac{-7-1}{3}$$x = -8
$$\frac{-8}{3}$$x = -8
Use the inverse operation to isolate the x variable.
-8x = -24
8x = 24
Find the value of x
x = $$\frac{24}{8}$$
x = 3
Substitute the value of the x variable in one of the equations.
y = $$\frac{1}{3}$$x – 4
y = $$\frac{1}{3}$$3 – 4
Find the value of y
y = 1-4
y = -3
Therefore, x = 3, y = -3

b. y = –$$\frac{5}{2}$$x + 10
y = $$\frac{1}{2}$$x + 4
x = ____, y = ____
Answer: Therefore, x = 2, y =5
y = –$$\frac{5}{2}$$x + 10
y = $$\frac{1}{2}$$x + 4
Substitute one value of y so that there is only one variable in the new equation.
$$\frac{1}{2}$$x + 4 = –$$\frac{5}{2}$$x + 10
Use the inverse operation and combine like terms with the x variable.
$$\frac{1}{2}$$x + $$\frac{5}{2}$$x= 10-4
Use the inverse operation to narrow the equation to 2 terms.
$$\frac{1 +5}{2}$$x = 6
$$\frac{6}{2}$$x = 6
Use the inverse operation to isolate the x variable.
6x =12
Find the value of x
x = $$\frac{12}{6}$$
x = 2
Substitute the value of the x variable in one of the equations.
y = –$$\frac{5}{2}$$x + 10
y = –$$\frac{5}{2}$$2 + 10
Find the value of y
y = -5 + 10
y = 5
Therefore, x = 2, y =5

Systems of equations can be solved by using the method of elimination following the steps below.
3x + 4y = 31
2x – y = 6
2x – y = 6 Step 1: Use inverse operations to isolate one variable on one side of the equation.
2x – y – 2x = 6 – 2x
-y = 6 – 2x
y = -6 + 2x
3x + (-6 + 2x) = 31 Step 2: Substitute the new equation ¡n place of the appropriate variable so there is only one variable in the new equation.
3x – 24 + 8x = 31
11x – 24 = 31
11x – 24 + 24 = 31 + 24 Step 3: Use inverse operations ana me distributive property to find a solution for the variable.
11x = 55
11x ÷ 11 = 55 ÷ 11
x = 5
y = -6 + 2(5) Step 4: Substitute the value of the variable ¡n one of the equations and solve.
y = 4

Use elimination to solve each system of equations.

Question 1.
a.
-4x – 2y = -12
4x + 8y = -24
x = ___, y = ___
Answer: x = 6, y = -6
-4x – 2y = -12
4x + 8y = -24
Use inverse operations to isolate one variable on one side of the equation.
-4x – 2y = -12
-4x – 2y + 4x= -12 +4x
-2y = -12 +4x
2y = 12 – 4x
y = 6 – 2x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
4x + 8y = -24
4x + 8 (6 – 2x) = -24
4x + 48 -16x = -24
-16x + 4x = -24 – 48
-12x = -72
12x = 72
x = 6
Substitute the value of the variable in one of the equations and solve.
y = 6 – 2x = 6 – 2(6) = 6 – 12
y = -6
Therefore, x = 6, y = -6

b.
4x + 8y = 20
-4x + 2y = -30
x = ___, y = ____
Answer: x = 7, y =-1
4x + 8y = 20
-4x + 2y = -30
Use inverse operations to isolate one variable on one side of the equation.
4x + 8y = 20
4x + 8y – 4x= 20 -4x
8y = 20 -4x
y = $$\frac{20}{8}$$ – $$\frac{4}{8}$$x
y = $$\frac{5}{2}$$ – $$\frac{1}{2}$$x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-4x + 2y = -30
-4x + 2{$$\frac{5}{2}$$ – $$\frac{1}{2}$$x}= -30
-4x + 5 – x = -30
-4x – x = -30 -5
-5x = -35
5x = 35
x = 7
Substitute the value of the variable in one of the equations and solve.
y = $$\frac{5}{2}$$ – $$\frac{1}{2}$$x
y = $$\frac{5}{2}$$ – $$\frac{1}{2}$$7
y = $$\frac{5}{2}$$ – $$\frac{7}{2}$$
y = $$\frac{5-7}{2}$$
y = $$\frac{-2}{2}$$
y = -1
Therefore, x = 7, y =-1

Question 2.
a.
x – y = 11
2x + y = 19
x = ___, y = ____
Answer: x = 10, y =-1
x – y = 11
2x + y = 19
Use inverse operations to isolate one variable on one side of the equation.
x – y = 11
x – y -x = 11 – x
-y = 11 – x
y = x – 11
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
2x + y = 19
2x + x – 11= 19
2x + x = 19 + 11
3x = 30
x = 10
Substitute the value of the variable in one of the equations and solve.
y = x – 11
y = 10 – 11
y = -1
Therefore, x = 10, y =-1

b.
-6x + 5y = 1
6x + 4y =-10
x = ___, y = ___
Answer: x = -9, y = $$\frac{59}{5}$$
-6x + 5y = 1
6x + 4y =-10
Use inverse operations to isolate one variable on one side of the equation.
-6x + 5y = 1
-6x + 5y + 6x= 1 + 6x
5y = 1 + 6x
y = $$\frac{1}{5}$$ – $$\frac{6}{5}$$x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
6x + 4y =-10
6x + 4{$$\frac{1}{5}$$ – $$\frac{6}{5}$$x} =-10
6x + $$\frac{4}{5}$$ – $$\frac{24}{5}$$x =-10
6x  – $$\frac{24}{5}$$x =-10 – $$\frac{4}{5}$$
$$\frac{30 – 24}{5}$$x = $$\frac{-50-4}{5}$$
$$\frac{6}{5}$$x = $$\frac{-54}{5}$$
6x = -54
x = -9
Substitute the value of the variable in one of the equations and solve.
y = $$\frac{1}{5}$$ – $$\frac{6}{5}$$x
y = $$\frac{1}{5}$$ – $$\frac{6}{5}$$9
y = $$\frac{1}{5}$$ – $$\frac{54}{5}$$
y = $$\frac{5 + 54}{5}$$
y = $$\frac{59}{5}$$
Therefore, x = -9, y = $$\frac{59}{5}$$

Question 3.
a.
-2x – 9y = -25
-4x – 9y = -23
x = ____, y = ____
Answer: x = -1, y =3
-2x – 9y = -25
-4x – 9y = -23
Use inverse operations to isolate one variable on one side of the equation.
-2x – 9y = -25
-2x – 9y + 2x= -25 + 2x
-9y = -25 + 2x
9y = 25 -2x
y = $$\frac{25}{9}$$ – $$\frac{2}{9}$$x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-4x – 9y = -23
-4x – 9{$$\frac{25}{9}$$ – $$\frac{2}{9}$$x} = -23
-4x – $$\frac{225}{9}$$ + $$\frac{18}{9}$$x = -23
-4x + $$\frac{18}{9}$$x = -23 + $$\frac{225}{9}$$
$$\frac{-36 +18}{9}$$x = $$\frac{-207 + 225}{9}$$
$$\frac{-18}{9}$$x = $$\frac{18}{9}$$
-2x = 2
x = -1
Substitute the value of the variable in one of the equations and solve.
y = $$\frac{25}{9}$$ – $$\frac{2}{9}$$x
y = $$\frac{25}{9}$$ – $$\frac{2}{9}$$(-1)
y = $$\frac{25}{9}$$ + $$\frac{2}{9}$$
y = $$\frac{25 + 2}{9}$$
y = $$\frac{27}{9}$$
y = 3
Therefore, x = -1, y =3

b.
8x + y = -16
-3x + y = -5
x = ____, y = ____
Answer:  x = -1, y =-8
8x + y = -16
-3x + y = -5
Use inverse operations to isolate one variable on one side of the equation.
8x + y = -16
8x + y -8x= -16 -8x
y = -16 -8x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-3x + y = -5
-3x -16 -8x= -5
-3x-8x = -5+16
-11x = 11
x = -1
Substitute the value of the variable in one of the equations and solve.
y = -16 -8x
y = -16 -8(-1)
y = -16 +8
y = -8
Therefore, x = -1, y =-8

Use substitution or elimination to solve each system of equations.

Question 1.
a.
y = $$\frac{2}{3}$$x – 5
y = -x + 10
x = ____, y = ___
Answer: x =9, y = 1
y = $$\frac{2}{3}$$x – 5
y = -x + 10
Use inverse operations to isolate one variable on one side of the equation.
y = -x + 10
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
y = $$\frac{2}{3}$$x – 5
-x + 10 = $$\frac{2}{3}$$x – 5
-x – $$\frac{2}{3}$$x = – 5 -10
$$\frac{-3- 2}{3}$$x = -15
$$\frac{-5}{3}$$x = -15
-5x = -45
5x = 45
x = 9
Substitute the value of the variable in one of the equations and solve.
y = -x + 10
y = -9 + 10
y = 1
Therefore, x =9, y = 1

b. x + y = -3
x – y = 1
x = ___, y = ____
Answer: x = -1, y =-2
x + y = -3
x – y = 1
Use inverse operations to isolate one variable on one side of the equation.
x + y = -3
x + y -x = -3 -x
y = -3-x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
x – y = 1
x – (-3-x) = 1
x + 3 + x = 1
2x + 3 = 1
2x = 1-3
2x = -2
x = -1
Substitute the value of the variable in one of the equations and solve.
y = -3-x
y = -3-(-1)
y = -3 +1
y = -2
Therefore, x = -1, y =-2

Question 2.
a.
3x – y = 0
$$\frac{1}{4}$$x + $$\frac{3}{4}$$y = $$\frac{5}{2}$$
Answer: x = 1, y =3
3x – y = 0
$$\frac{1}{4}$$x + $$\frac{3}{4}$$y = $$\frac{5}{2}$$
Use inverse operations to isolate one variable on one side of the equation.
3x – y = 0
3x – y -3x= 0 -3x
-y = -3x
y = 3x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
$$\frac{1}{4}$$x + $$\frac{3}{4}$$y = $$\frac{5}{2}$$
$$\frac{1}{4}$$x + $$\frac{3}{4}$$3x= $$\frac{5}{2}$$
$$\frac{1}{4}$$x + $$\frac{9}{4}$$x= $$\frac{5}{2}$$
$$\frac{1 + 9}{4}$$x = $$\frac{5}{2}$$
$$\frac{10}{4}$$x = $$\frac{5}{2}$$
x = $$\frac{5}{2}$$ x $$\frac{4}{10}$$
x = 1
Substitute the value of the variable in one of the equations and solve.
y = 3x
y = 3(1)
y =3
Therefore, x = 1, y =3

b. -6x + 6y = 6
-6x + 3y = -12
x = ___, y = ____
Answer: x = 5, y =6
-6x + 6y = 6
-6x + 3y = -12
Use inverse operations to isolate one variable on one side of the equation.
-6x + 6y = 6
-6x + 6y + 6x= 6 +6x
6y = 6 +6x
y = 1 + x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-6x + 3y = -12
-6x + 3(1 + x)= -12
-6x + 3 +3x= -12
-6x + 3x = -12 -3
-3x = -15
3x = 15
x = 5
Substitute the value of the variable in one of the equations and solve.
y = 1 + x
y = 1 + 5
y = 6
Therefore, x = 5, y =6

Question 3.
a.
y = 3 – x
y – 3x = 5
x = ___, y = ____
Answer: x = $$\frac{-1}{2}$$, y = $$\frac{7}{2}$$
y = 3 – x
y – 3x = 5
Use inverse operations to isolate one variable on one side of the equation.
y = 3 – x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
y – 3x = 5
3 – x- 3x = 5
-4x = 5 – 3
-4x = 2
4x = -2
2x = -1
x = $$\frac{-1}{2}$$
Substitute the value of the variable in one of the equations and solve.
y – 3x = 5
y – 3($$\frac{-1}{2}$$) = 5
y +  $$\frac{3}{2}$$ = 5
y = 5 – $$\frac{3}{2}$$
y = $$\frac{10 – 3}{2}$$
y = $$\frac{7}{2}$$
Therefore, x = $$\frac{-1}{2}$$, y = $$\frac{7}{2}$$

b. -4x + 9y = 9
x = -6 + 3y
x = ___, y = ____
Answer: x =9, y = 5
-4x + 9y = 9
x = -6 + 3y
Use inverse operations to isolate one variable on one side of the equation.
x = -6 + 3y
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-4x + 9y = 9
-4(-6 + 3y) + 9y = 9
24 – 12y +9y = 9
– 12y +9y = 9 – 24
-3y = -15
3y = 15
y = 5
Substitute the value of the variable in one of the equations and solve.
x = -6 + 3y
x = -6 + 3(5)
x = -6 + 15
x = 9
Therefore, x =9, y = 5

Question 4.
a. 3g + f = 15
g + 2 = 10
g = ___, f = ____
Answer: g = 8, f =-9
3g + f = 15
g + 2 = 10
Use inverse operations to isolate one variable on one side of the equation.
g + 2 = 10
g = 10-2
g =8
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
3g + f = 15
3(8) + f = 15
24 + f = 15
f = 15 – 24
f = -9
Therefore, g = 8, f =-9

b.
3b + 5t = 17
2b + t = 9
b = ___, t = ____
Answer: b = 4, t = 1
3b + 5t = 17
2b + t = 9
Use inverse operations to isolate one variable on one side of the equation.
3b + 5t = 17
3b + 5t -3b= 17 -3b
5t = 17 -3b
t =  $$\frac{17}{5}$$ – $$\frac{3}{5}$$b
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
2b + $$\frac{17}{5}$$ – $$\frac{3}{5}$$b = 9
2b  – $$\frac{3}{5}$$b = 9 – $$\frac{17}{5}$$
$$\frac{10-3}{5}$$b = $$\frac{45 -17}{5}$$
$$\frac{7}{5}$$b = $$\frac{28}{5}$$
7b = 28
b = 4
Substitute the value of the variable in one of the equations and solve.
t =  $$\frac{17}{5}$$ – $$\frac{3}{5}$$b
t =  $$\frac{17}{5}$$ – $$\frac{3}{5}$$4
t =  $$\frac{17}{5}$$ – $$\frac{12}{5}$$
t = $$\frac{17 – 12}{5}$$
t = $$\frac{5}{5}$$
t = 1
Therefore, b = 4, t = 1

Question 5.
a.
y = 4 – 2x
y = -5 + 4x
x = ___, y = ____
y = 4 – 2x
y = -5 + 4x
Use inverse operations to isolate one variable on one side of the equation.
y = -5 + 4x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
y = 4 – 2x
-5 + 4x = 4 – 2x
4x +2x = 4+5
6x = 9
2x = 3
x = $$\frac{3}{2}$$
Substitute the value of the variable in one of the equations and solve.
y = -5 + 4x
y = -5 + 4{$$\frac{3}{2}$$}
y = -5 + $$\frac{12}{2}$$
y = $$\frac{-10 + 12}{2}$$
y = $$\frac{2}{2}$$
y = 1
Therefore, x = $$\frac{3}{2}$$, y = 1

b.
5b + 2e = 32
6b + 6e = 42
h = ____, e = ____
Answer: b = 6, e = 1
5b + 2e = 32
6b + 6e = 42
Use inverse operations to isolate one variable on one side of the equation.
5b + 2e = 32
5b + 2e – 5b= 32 -5b
2e = 32 – 5b
e = 16 – $$\frac{5}{2}$$b
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
6b + 6e = 42
6b + 6{16 – $$\frac{5}{2}$$b} = 42
6b + 96 – $$\frac{30}{2}$$b = 42
6b + 96 – 15b = 42
-9b = 42 – 96
-9b = -54
9b = 54
b = 6
Substitute the value of the variable in one of the equations and solve.
e = 16 – $$\frac{5}{2}$$b
e = 16 – $$\frac{5}{2}$$6
e = 16 – 15
e = 1
Therefore, b = 6, e = 1

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