Middle School Math Quiz: Equations With Variables On Both Sides

10 questions · exam conditions

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Equations With Variables On Both SidesQuestion 1 of 10

When solving $5x - 3(x + 4) = 2x + 1$ , a student gets $5x - 3x - 12 = 2x + 1$ , then $2x - 12 = 2x + 1$ . What should the student conclude?

The equation has infinitely many solutions because both sides have the same coefficient for

x

The equation has no solution because

-12 \neq 1

after subtracting

2x

from both sides

The solution is

x = 0

because the coefficients of

x

are equal on both sides

The solution is

x = 6.5

because

\frac{1 + 12}{2} = 6.5

when solving for the difference

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Middle School Math Quiz

Middle School Math Quiz: Equations With Variables On Both Sides

Practice Equations With Variables On Both Sides in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Equations With Variables On Both Sides, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

When solving $5x - 3(x + 4) = 2x + 1$ , a student gets $5x - 3x - 12 = 2x + 1$ , then $2x - 12 = 2x + 1$ . What should the student conclude?

The equation has infinitely many solutions because both sides have the same coefficient for $x$
The equation has no solution because $-12 \neq 1$ after subtracting $2x$ from both sides (correct answer)
The solution is $x = 0$ because the coefficients of $x$ are equal on both sides
The solution is $x = 6.5$ because $\frac{1 + 12}{2} = 6.5$ when solving for the difference

Explanation: When you encounter an equation that simplifies to something like

2x - 12 = 2x + 1

, you're dealing with a special case where the variable terms are identical on both sides. The key is recognizing what happens when you isolate the variable terms from the constants. Starting from

2x - 12 = 2x + 1

, subtract

2x

from both sides to eliminate the variable terms. This gives you

-12 = 1

, which is clearly a false statement. When solving an equation leads to a false statement (like

-12 = 1

), the original equation has no solution. No value of

x

can make the equation true because the equation is inherently contradictory. Choice A is incorrect because having the same coefficient for

x

on both sides doesn't guarantee infinitely many solutions—you must also check what happens to the constant terms. Choice C is wrong because equal coefficients don't mean

x = 0

is the solution; you need to solve completely by eliminating the variable terms. Choice D makes an error by trying to manipulate the numbers

1

and

12

arithmetically, but this approach ignores the fundamental contradiction that

-12 \neq 1

. The correct answer is B because after properly eliminating the variable terms, you're left with the impossible statement

-12 = 1

. Remember this pattern: when solving linear equations, if you end up with identical variable terms on both sides but different constants, the equation has no solution. Always complete the solving process by isolating the constants to check for contradictions.

Question 2

A student solves $3(2x - 1) + 4 = 2(3x + 2) - 1$ by expanding both sides. After simplification, what conclusion should the student reach?

The equation has infinitely many solutions because it simplifies to an identity
The equation has no solution because it leads to a contradiction (correct answer)
The solution is $x = 0$ because all terms with $x$ cancel out
The solution is $x = 1$ because the constant terms sum to the coefficient

Explanation: Expanding the left side:

3(2x - 1) + 4 = 6x - 3 + 4 = 6x + 1

. Expanding the right side:

2(3x + 2) - 1 = 6x + 4 - 1 = 6x + 3

. So we have

6x + 1 = 6x + 3

. Subtracting

6x

from both sides gives

1 = 3

, which is false. This contradiction means the equation has no solution.

Question 3

Marcus is solving the equation $3x + 7 = 5x - 9$ . After moving all terms with $x$ to one side and all constants to the other side, what equation should he have before solving for $x$ ?

$-2x = -16$ (correct answer)
$2x = 16$
$8x = -2$
$-8x = 2$

Explanation: Starting with

3x + 7 = 5x - 9

, subtract

3x

from both sides:

7 = 2x - 9

. Then subtract

-9

from both sides (or add 9):

16 = 2x

. Finally, subtract

2x

from both sides to get

16 - 2x = 0

, or equivalently

-2x = -16

. Choice B incorrectly adds coefficients instead of subtracting. Choice C incorrectly adds all terms. Choice D has incorrect signs from moving terms incorrectly.

Question 4

For what value of $k$ will the equation $3x + k = 7x - 12$ have the solution $x = 4$ ?

$k = 4$ (correct answer)
$k = 16$
$k = -4$
$k = 28$

Explanation: If

x = 4

is the solution, then substituting into the equation:

3(4) + k = 7(4) - 12

. This gives us

12 + k = 28 - 12

, so

12 + k = 16

. Therefore

k = 4

. Choice B gives the value of the right side before subtracting 12. Choice C incorrectly subtracts instead of adding to isolate

k

. Choice D gives the value of

7x

when

x = 4

Question 5

The equation $ax + 5 = 3x + b$ has the solution $x = 2$ . If $a + b = 8$ , what are the values of $a$ and $b$ ?

$a = 4, b = 4$
$a = 1, b = 7$
$a = 3, b = 5$ (correct answer)
$a = 5, b = 3$

Explanation: When you encounter an equation with unknown coefficients and a given solution, you need to use the fact that the solution must satisfy the equation to find relationships between those unknowns. Since

x = 2

is the solution to

ax + 5 = 3x + b

, substituting

x = 2

into the equation must make it true:

a(2) + 5 = 3(2) + b

2a + 5 = 6 + b

2a - b = 1

Now you have two equations with two unknowns:

$2a - b = 1$ (from substituting the solution)
$a + b = 8$ (given condition)

Adding these equations eliminates

b

3a = 9

a = 3

Substituting back:

3 + b = 8

, so

b = 5

. Let's check why the other answers fail. Choice A (

a = 4, b = 4

) gives us

2(4) - 4 = 4 \neq 1

, violating our first equation. Choice B (

a = 1, b = 7

) produces

2(1) - 7 = -5 \neq 1

, also incorrect. Choice D (

a = 5, b = 3

) yields

2(5) - 3 = 7 \neq 1

, failing the substitution test. Only choice C satisfies both

2a - b = 1

and

a + b = 8

. Study tip: When solving systems involving given solutions, always substitute the solution into the original equation first to create a new relationship. This converts an abstract problem into a concrete system of linear equations you can solve using elimination or substitution.

Question 6

The equation $0.5x + 1.2 = 0.3x - 0.6$ can be solved by first converting to integers. After multiplying by an appropriate power of 10, what equation results?

$x + 2.4 = 0.6x - 1.2$
$50x + 120 = 30x - 60$
$5x + 1.2 = 3x - 0.6$
$5x + 12 = 3x - 6$ (correct answer)

Explanation: When you encounter linear equations with decimals, converting them to integers by multiplying by an appropriate power of 10 makes solving much easier and reduces calculation errors. To eliminate all decimals in

0.5x + 1.2 = 0.3x - 0.6

, you need to identify the decimal with the most places after the decimal point. Here, all decimals have one decimal place (0.5, 1.2, 0.3, 0.6), so multiply the entire equation by 10¹ = 10. Multiplying each term by 10:

10(0.5x) + 10(1.2) = 10(0.3x) - 10(0.6)

, which gives you

5x + 12 = 3x - 6

. This matches answer choice D. Let's examine why the other choices are incorrect. Choice A multiplies only the first term by 10 while leaving the other decimals unchanged—you must multiply every term in the equation. Choice B multiplies by 100 instead of 10, which unnecessarily creates larger numbers:

50x + 120 = 30x - 60

. While this equation is mathematically equivalent, the question asks for the result after multiplying by the "appropriate" power of 10, which is the smallest power that eliminates all decimals. Choice C multiplies only the x-terms by 10 but leaves the constants as decimals, failing to achieve the goal of converting to integers. Study tip: Always multiply the entire equation by the same power of 10. Find the decimal with the most places after the decimal point—that tells you which power of 10 to use. One decimal place means multiply by 10, two decimal places means multiply by 100, and so on.

Question 7

Consider the equation $\frac{x + 2}{4} - \frac{x - 1}{6} = 1$ . After finding a common denominator and clearing fractions, what is the coefficient of $x$ in the resulting equation?

$3$
$2$
$1$ (correct answer)
$-1$

Explanation: When you encounter an equation with fractions, your goal is to eliminate them by finding a common denominator and then clearing the fractions entirely. This transforms the equation into a simpler form where you can easily identify coefficients. Let's work through

\frac{x + 2}{4} - \frac{x - 1}{6} = 1

step by step. First, find the least common denominator of 4 and 6, which is 12. Rewrite each fraction:

\frac{x + 2}{4} = \frac{3(x + 2)}{12} = \frac{3x + 6}{12}

\frac{x - 1}{6} = \frac{2(x - 1)}{12} = \frac{2x - 2}{12}

Now your equation becomes:

\frac{3x + 6}{12} - \frac{2x - 2}{12} = 1

Combine the fractions:

\frac{(3x + 6) - (2x - 2)}{12} = 1

\frac{3x + 6 - 2x + 2}{12} = 1

\frac{x + 8}{12} = 1

To clear the fraction, multiply both sides by 12:

x + 8 = 12

The coefficient of

x

is 1, making C correct. Choice A (3) likely comes from mistakenly using just the coefficient from the first fraction. Choice B (2) might result from using the coefficient from the second fraction. Choice D (-1) could occur if you incorrectly handled the subtraction when combining fractions. Remember: when clearing fractions, always multiply the entire equation by the common denominator. The resulting linear equation will have integer coefficients that are much easier to work with than the original fractions.

Question 8

The equation $\frac{2x + 6}{3} = \frac{x - 1}{2} + 4$ can be solved by first clearing the fractions. What is the resulting equation after multiplying both sides by the least common denominator?

$2(2x + 6) = 3(x - 1) + 24$ (correct answer)
$2(2x + 6) = 3(x - 1) + 12$
$6(2x + 6) = 6(x - 1) + 4$
$3(2x + 6) = 2(x - 1) + 6$

Explanation: The LCD of 3 and 2 is 6. Multiplying both sides by 6:

6 \cdot \frac{2x + 6}{3} = 6 \cdot \left(\frac{x - 1}{2} + 4\right)

. This simplifies to

2(2x + 6) = 3(x - 1) + 24

because

6 \cdot 4 = 24

. Choice B incorrectly calculates

6 \cdot 4 = 12

. Choice C incorrectly keeps the 6 with the first fraction. Choice D uses wrong coefficients from incorrect LCD application.

Question 9

Lisa solves $2(x + 3) - x = 3x + 1$ and gets $x = 5$ . To check her answer, she substitutes back into the original equation. What should she get for the left side when $x = 5$ ?

$13$
$16$
$6$
$11$ (correct answer)

Explanation: When checking solutions to equations, you need to substitute your answer back into the original equation and evaluate both sides to see if they're equal. This verification step catches any algebraic errors you might have made. To find what Lisa gets for the left side when

x = 5

, substitute

5

into

2(x + 3) - x

2(5 + 3) - 5 = 2(8) - 5 = 16 - 5 = 11

So the left side equals

11

when

x = 5

. You can verify this is correct by checking the right side:

3(5) + 1 = 15 + 1 = 16

. Since

11 \neq 16

, Lisa's solution

x = 5

is actually incorrect, but the question only asks what she gets for the left side. Looking at the wrong answers: Choice A (

13

) might come from incorrectly calculating

2(8) - 5

13

instead of

11

. Choice B (

16

) is what you get if you evaluate the right side

3x + 1

instead of the left side, or if you forget to subtract the

x

term entirely. Choice C (

6

) could result from mistakenly calculating

2(3) - 0 = 6

by using just the coefficient inside the parentheses. Remember: when substituting into expressions with parentheses, always follow the order of operations carefully. First handle what's inside parentheses, then multiply, then subtract. Double-check by working through each step slowly rather than trying to do multiple operations at once.

Question 10

Sarah claims that the equation $4(x - 2) = 4x - 8$ has infinitely many solutions. Jake argues it has no solution. Who is correct and why?

Sarah is correct because the equation simplifies to $0 = 0$ , which is always true (correct answer)
Jake is correct because the equation simplifies to $0 = -16$ , which is never true
Sarah is correct because the equation simplifies to $x = 0$ , giving one specific solution
Jake is correct because the equation simplifies to $x = -2$ , contradicting the original equation

Explanation: Expanding the left side:

4(x - 2) = 4x - 8

. This gives us

4x - 8 = 4x - 8

. Subtracting

4x

from both sides yields

-8 = -8

, or

0 = 0

, which is always true regardless of the value of

x

. This means the equation is an identity with infinitely many solutions. Choice B incorrectly calculates the simplification. Choices C and D incorrectly assume there's a unique solution.