
Solving Equations with Variables on Both Sides
Authored by Anthony Clark
Mathematics
8th Grade
CCSS covered

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19 questions
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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;
Addition Property of equality
Division Property of Equality
Subtraction Property of Equality
Combine Like Terms
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
Mathematical Equivalence
ON
Tags
CCSS.8.EE.C.7B
6.
MATCH QUESTION
1 min • 5 pts
Describe what was done in each solving step!
Combine Like Terms
Step 5
Subtract 10 from both sides
Step 3
Distribute
Step 4
Divide by -10 on both sides
Step 2
Subtract 4x from both sides
Step 1
7.
MATH RESPONSE QUESTION
1 min • 1 pt
Mathematical Equivalence
ON
Tags
CCSS.8.EE.C.7B
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