Solving Equations with Variables on Both Sides

Solving Equations with Variables on Both Sides

8th Grade

19 Qs

quiz-placeholder

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Solving Equations with Variables on Both Sides

Solving Equations with Variables on Both Sides

Assessment

Passage

Mathematics

8th Grade

Practice Problem

Hard

CCSS
8.EE.C.7B, HSA.REI.A.1, 7.EE.B.4A

Standards-aligned

Created by

Anthony Clark

FREE Resource

19 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

1 min • 5 pts

If I wanted to start to simplify this equation what would I do in order to get variables on one side and constants on the other?

Move the 2x to the other side of the equation by adding it to both sides.

Move the 6 to the other side of the equation by adding it to both side.

Move the 2x to the other side of the equation by subtracting it from both sides.

Move the 6 to the other side of the equation by subtracting it from both sides.

Answer explanation

If we move the 2x, then all terms with a variable will be on the right, and all constants will be on the left. Since the 2x is positive on the left, we want to do the opposite and subtract it from both sides.

2.

MULTIPLE CHOICE QUESTION

1 min • 5 pts

What is the answer? 3x + 2 = -5x + 10

x = 1

x = -1

x = 6

x = -6

Answer explanation

If we add 5x to both sides, we get;

8x + 2 = 10.

Then we need to subtract 2 from both sides to get;

8x = 8.

Finally we divide both sides by what is attached to the x, which in this case is 8.

8x/8 = x and 8/8 = 1 so x=1

3.

REORDER QUESTION

1 min • 5 pts

Reorder the following based on which properties I used in solving the following equation;

Combine Like Terms

Addition Property of equality

Division Property of Equality

Subtraction Property of Equality

Distributive Property

Answer explanation

To get rid of the parentheses w use the distributive property.

Then to simplify each side, we combine like terms.

Next the variable terms were all moved to the right side of the equation. The 4x was subtracted from both sides, hence the subtraction property of equality.

Next, we needed to get the constants on the opposite side of the equation, so we added 5 to both sides, hence the addition property of equality.

Finally we used the division prperty and divided both sides by what was attached to the x, which in this case was a 3. We used the division property of equality to get an answer of 1/3 = x.

Tags

CCSS.HSA.REI.A.1

4.

MULTIPLE CHOICE QUESTION

1 min • 5 pts

Solve for x.

-3(2x + 5) = -9x

x = 1

x = 5

x = -1

x = -5

Answer explanation

-3(2x + 5) = -9x

-6x - 15 = -9x distribute -3 to the 2x and 5.

-15 = -3x add 6x to both sides

5 = x Divide both sides by -3.

Tags

CCSS.8.EE.C.7B

5.

MATH RESPONSE QUESTION

1 min • 1 pt

Media Image

Mathematical Equivalence

ON

Tags

CCSS.8.EE.C.7B

6.

MATCH QUESTION

1 min • 5 pts

Media Image

Describe what was done in each solving step!

Distribute

Step 2

Combine Like Terms

Step 1

Subtract 4x from both sides

Step 4

Subtract 10 from both sides

Step 5

Divide by -10 on both sides

Step 3

7.

MATH RESPONSE QUESTION

1 min • 1 pt

Media Image

Mathematical Equivalence

ON

Tags

CCSS.8.EE.C.7B

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