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Understanding One-Step Equations

Understanding One-Step Equations

Assessment

Interactive Video

Mathematics

5th - 7th Grade

Practice Problem

Hard

CCSS
6.EE.B.7, 7.NS.A.1C, 6.EE.A.2C

+1

Standards-aligned

Created by

Ethan Morris

FREE Resource

Standards-aligned

CCSS.6.EE.B.7
,
CCSS.7.NS.A.1C
,
CCSS.6.EE.A.2C
CCSS.6.EE.B.5
,
The video tutorial from Moomoomath explains how to solve a one-step equation, x + 8 = 5, using opposite operations. It begins with a quick demonstration and then provides a detailed, step-by-step explanation in slow motion. The tutorial emphasizes the importance of isolating the variable by subtracting 8 from both sides and checking the solution by substituting back into the original equation. The video concludes by verifying that the solution is correct.

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8 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in solving the equation x + 8 = 5?

Subtract 8 from both sides

Multiply both sides by 8

Add 8 to both sides

Divide both sides by 8

Tags

CCSS.6.EE.B.7

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What operation is used to isolate x in the equation x + 8 = 5?

Addition

Multiplication

Subtraction

Division

Tags

CCSS.6.EE.B.7

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When subtracting 8 from both sides of the equation x + 8 = 5, what is the result on the left side?

5

0

x

x + 8

Tags

CCSS.7.NS.A.1C

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the result of 5 - 8?

8

0

-3

3

Tags

CCSS.7.NS.A.1C

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the sign of the result when subtracting a larger number from a smaller number?

Negative

Positive

Undefined

Zero

Tags

CCSS.6.EE.B.7

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the purpose of drawing a line when solving equations?

To add numbers

To keep the work organized

To separate the equation

To multiply numbers

Tags

CCSS.6.EE.B.5

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you verify the solution x = -3 for the equation x + 8 = 5?

By dividing both sides by 3

By multiplying both sides by -3

By substituting -3 back into the equation

By adding 3 to both sides

Tags

CCSS.6.EE.A.2C

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