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: _____
B. Jaden’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.
Are You Ready?
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
Answer:
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
Answer:
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
Answer:
\(\frac{1}{3}\)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
Answer:
\(\frac{3}{4}\)(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
Answer:
3b + 4 = 5b – 8
3b – 5b = -8 – 4
-2b = -12
b = 6
Question 6.
-2(x + 6) = 4x
Answer:
-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
Answer:
\(\frac{1}{4}\)(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
Answer:
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
Answer: 6
Explanation:
3(2x + 1) = 6x + 3
Add 2 on both sides
3(2x + 1) + 2 = 6x + 3 + 2
3(2x + 1) + 2 = 6x + 5
Question 10.
-4n + 8 = 4n – 8n + infinitely many solutions
Answer:
-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.
Answer:
Let us use the equation
x + 3 = x + 3
x = x or 3 = 3
x + 3 = x + 3
x = 2
2 + 3 = 2 + 3
5 = 5
We can see that both sides of the equation are always the same with any value of x.