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 = ____

Answer:

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