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