Grade 8 Math: Solving by Substitution Method

Grade 8 Math: Solving by Substitution Method

Assessment

Interactive Video

Mathematics

8th Grade

Easy

Created by

Emma Peterson

Used 5+ times

FREE Resource

The video tutorial covers solving linear equations using the substitution method. It begins with an introduction to the concept, followed by two detailed examples demonstrating how to solve systems of equations by substituting one equation into another. The tutorial concludes with a motivational message encouraging students to continue learning.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in solving a system of equations using the substitution method?

Combine like terms

Isolate one variable in one of the equations

Multiply both equations by a common factor

Add the equations together

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the example given, what is the value of x when solving the system of equations 3x + 5y = 26 and y = 2x?

3

5

4

2

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

After finding the value of x, how do you find the value of y in the substitution method?

Divide the value of x by 2

Add the value of x to the second equation

Multiply the value of x by 2

Substitute x back into the original equation

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the ordered pair solution for the system of equations 3x + 5y = 26 and y = 2x?

(2, 4)

(3, 6)

(1, 2)

(4, 8)

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the second example, what is the first step to isolate y in the equation -3x + y = 1?

Subtract 3x from both sides

Add 3x to both sides

Divide both sides by 3

Multiply both sides by -1

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of x in the second example after isolating y and substituting into the second equation?

2

1

3

4

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you verify the solution of a system of equations?

By dividing the equations by a common factor

By multiplying the equations by a common factor

By substituting the values back into the original equations

By adding the equations together

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