# Into Math Grade 8 Module 7 Answer Key Systems of Linear Equations

We included HMH Into Math Grade 8 Answer Key PDF Module 7 Systems of Linear Equations to make students experts in learning maths.

## HMH Into Math Grade 8 Module 7 Answer Key Systems of Linear Equations

Wags Per Mile

Write expressions showing the cost of using each service for d days.
A. Nikki’s: _____
Find when the dog services will cost the same.
C. The dog services cost the same on day _____
A. Nikki’s: (10 + 15) = 25d
B. Jaden’s: (20 + 13)d = 33d

Turn and Talk
Describe the steps you used to solve Part C.

Complete these problems to review prior concepts and skills you will need for this module.

Identify Solutions of Equations

Use substitution to determine whether the given value of the variable is a solution of the equation.

Question 1.
6x – 2x = 10; x = 2
6x – 2x = 10
x = 2
4x = 10
4(2) = 10
8 ≠ 10
It is not a solution.

Question 2.
4n + 4 = 14; n = 2.5
4n + 4 = 14;
n = 2.5
4n = 14 – 4
4n = 10
4(2.5) = 10
10 = 10
Thus n = 2.5 is a solution.

Question 3.
$$\frac{1}{3}$$a – 1 = $$\frac{1}{4}$$a; a = 12
$$\frac{1}{3}$$a – 1 = $$\frac{1}{4}$$a;
a = 12
$$\frac{1}{3}$$a – $$\frac{1}{4}$$a = 1
$$\frac{1}{3}$$(12) – $$\frac{1}{4}$$(12) = 1
4 – 3 = 1
1 = 1
a = 12 is a solution.

Question 4.
$$\frac{3}{4}$$(t – 6) = 9; t = 20
$$\frac{3}{4}$$(t – 6) = 9;
t = 20
$$\frac{3}{4}$$(20 – 6) = 9
$$\frac{3}{4}$$ (14) = 9
14 × 3 = 9 × 4
42 ≠ 36
It is not a solution.

Solve Multi-Step Equations

Solve each equation.

Question 5.
3b + 4 = 5b – 8
3b + 4 = 5b – 8
3b – 5b = -8 – 4
-2b = -12
b = 6

Question 6.
-2(x + 6) = 4x
-2(x + 6) = 4x
-2x – 12 = 4x
-2x – 4x = 12
-6x = 12
x = -12/6
x = -2

Question 7.
$$\frac{1}{4}$$(s + 6) = -3
$$\frac{1}{4}$$(s + 6) = -3
$$\frac{1}{4}$$s + $$\frac{6}{4}$$ = -3
$$\frac{1}{4}$$s = -3 – $$\frac{3}{2}$$
$$\frac{1}{4}$$s = (-6 – 3)/2
$$\frac{1}{4}$$s = -9/2
s = -9/2 × 4
s = -9 × 2
s = -18

Question 8.
2m + 1.5 = 3m – 11.7
2m + 1.5 = 3m – 11.7
2m – 3m = -1.5 – 11.7
-m = -13.2
m = 13.2

Special Cases of Equations

Complete each equation by writing a number in the box so that the equation has the given number of solutions.

Question 9.
x + 5 = 3(2x + 1); no solution
Explanation:
3(2x + 1) = 6x + 3
3(2x + 1) + 2 = 6x + 3 + 2
3(2x + 1) + 2 = 6x + 5

Question 10.
-4n + 8 = 4n – 8n + infinitely many solutions
-4n + 8 = 4n – 8n + 8

Question 11.
Saskia is solving an equation in one variable and gets the equivalent equation 3 = 3. Assuming her work is correct, what can she conclude about the original equation? Explain.