Pre-Algebra Quiz: One Step Equations

7 questions · exam conditions

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One Step EquationsQuestion 1 of 7

The equation $-4x = 20$ has solution $x = -5$ . If this equation is modified by multiplying both sides by $-1$ , then adding the same constant $k$ to both sides, the resulting equation becomes $4x + k = -20 + k$ . What value of $x$ satisfies this new equation?

x = 5

x = -5

x = \frac{-20 + k}{4}

x = \frac{-20}{4 + k}

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Pre-Algebra Quiz

Pre-Algebra Quiz: One Step Equations

Practice One Step Equations in Pre-Algebra 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 One Step Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Pre-Algebra.

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

$x = 5$
$x = -5$ (correct answer)
$x = \frac{-20 + k}{4}$
$x = \frac{-20}{4 + k}$

Explanation: Starting with

-4x = 20

, multiply both sides by

-1

to get

4x = -20

. Adding

k

to both sides gives

4x + k = -20 + k

. Subtracting

k

from both sides:

4x = -20

, so

x = -5

. The value of

x

is the same as the original equation because adding the same constant to both sides doesn't change the solution. Choice A incorrectly changes the sign, Choice C treats

k

as if it's not eliminated, Choice D incorrectly puts

k

in the denominator.

Question 2

A student claims that the equation $\frac{x - 5}{3} = 7$ is a one-step equation because "you just multiply both sides by 3." After doing this operation, what additional step is required to solve for $x$ ?

Divide both sides by 5
Subtract 5 from both sides
Add 5 to both sides (correct answer)
No additional step is needed

Explanation: Multiplying both sides by 3 gives:

x - 5 = 21

. To isolate

x

, we need to add 5 to both sides:

x = 26

. The student was incorrect that this is a one-step equation - it requires two steps total. Choice A would divide by the wrong number, Choice B would subtract instead of add (making

x = 16

), Choice D ignores that

x - 5 = 21

is not solved for

x

Question 3

Consider the equation $|x| = 12$ . Jamie says this is equivalent to the one-step equation $x = 12$ . Taylor says it's equivalent to two one-step equations. Which statement best describes who is correct?

Jamie is correct because absolute value equations always have one solution
Taylor is correct because $|x| = 12$ means $x = 12$ or $x = -12$ (correct answer)
Both are correct because it depends on whether $x$ is positive or negative
Neither is correct because absolute value equations require two-step solving methods

Explanation: Taylor is correct. The equation

|x| = 12

means the distance from

x

to 0 is 12, which occurs when

x = 12

x = -12

. This gives us two one-step equations to solve:

x = 12

and

x = -12

. Choice A is wrong because absolute value equations typically have two solutions. Choice C misses that both solutions exist simultaneously. Choice D is incorrect because each individual case (

x = 12

x = -12

) is solved in one step.

Question 4

Two students solved the same one-step equation but got different answers. Student A got $x = -8$ and Student B got $x = 8$ . If both students used correct algebraic procedures, which type of equation were they most likely solving?

An equation of the form $ax = b$ where $a$ and $b$ have different signs
An equation of the form $x + a = b$ where Student A subtracted and Student B added
An equation where one student made a sign error in the arithmetic
An equation involving absolute value, such as $|x| = 8$ (correct answer)

Explanation: When you encounter a problem where two students get opposite answers using correct procedures, you need to think about what type of equation naturally produces multiple solutions or symmetric solutions. The key insight is recognizing that absolute value equations often have two solutions that are opposites of each other. When you solve

|x| = 8

, you're asking "what number has an absolute value of 8?" Since absolute value measures distance from zero, both

x = 8

and

x = -8

work perfectly. This is because

|8| = 8

and

|-8| = 8

. Both students used correct algebraic reasoning—they just found the two different valid solutions. Choice A is incorrect because equations of the form

ax = b

have exactly one solution:

x = \frac{b}{a}

. There's no way for two students to get different correct answers. Choice B is wrong because addition and subtraction are opposite operations. If one student added when they should have subtracted (or vice versa), they made an error in procedure, contradicting the problem's statement that both used correct methods. Choice C directly contradicts the given information that both students used correct procedures. A sign error would be an algebraic mistake, not a correct procedure. Remember this pattern: when a problem mentions two different correct answers that are opposites, immediately consider absolute value equations. Absolute value is one of the few concepts in pre-algebra where opposite solutions naturally occur from the same equation.

Question 5

A rectangle has area 84 square units and width $w$ . The length is 3 units more than twice the width. After setting up the equation $w(2w + 3) = 84$ , a student realizes this creates a quadratic equation. To avoid quadratic equations, what one-step equation could be used if the width were known to be 6 units?

$6(2w + 3) = 84$
$w(15) = 84$
$6L = 84$ (correct answer)
$L = 15$

Explanation: If

w = 6

, then the length

L = 2(6) + 3 = 15

. The area equation becomes

6L = 84

, which is a one-step equation that can be solved by dividing both sides by 6. Choice A still contains the variable

w

when

w

should be substituted with 6. Choice B uses the wrong variable and wrong length value. Choice D is already solved, not an equation to solve.

Question 6

The temperature in degrees Fahrenheit is 32 more than $\frac{9}{5}$ times the temperature in degrees Celsius. If the temperature is 68°F, which one-step equation correctly represents finding the Celsius temperature $C$ ?

$\frac{9}{5}C = 36$ (correct answer)
$\frac{9}{5}C = 68$
$C + 32 = 68$
$\frac{9}{5}C + 32 = 68$

Explanation: When you encounter word problems involving formulas, the key is translating the written description into mathematical symbols step by step. The problem states that Fahrenheit temperature equals "32 more than

\frac{9}{5}

times the Celsius temperature." This translates to:

F = \frac{9}{5}C + 32

. Since we know

F = 68

, we substitute:

68 = \frac{9}{5}C + 32

. To solve for

C

, we need to isolate the term with

C

. Subtracting 32 from both sides gives us:

68 - 32 = \frac{9}{5}C

, which simplifies to

36 = \frac{9}{5}C

\frac{9}{5}C = 36

. Looking at the wrong answers: Choice B (

\frac{9}{5}C = 68

) skips the subtraction step entirely, forgetting to account for the 32. Choice C (

C + 32 = 68

) incorrectly adds 32 to

C

instead of to

\frac{9}{5}C

, missing the fractional coefficient. Choice D (

\frac{9}{5}C + 32 = 68

) correctly sets up the initial equation but doesn't perform the algebraic manipulation needed to isolate the

C

term. The correct answer is A:

\frac{9}{5}C = 36

. Study tip: When working with temperature conversion problems, always write out the full formula first (

F = \frac{9}{5}C + 32

), then substitute known values and perform one algebraic step at a time. Don't try to jump directly to the final equation.

Question 7

The equation $0.25x = 15$ can be solved by multiplying both sides by 4. An equivalent approach is to rewrite 0.25 as a fraction first. If $0.25 = \frac{1}{4}$ , what operation should be applied to both sides of $\frac{1}{4}x = 15$ ?

Multiply both sides by 4 (correct answer)
Multiply both sides by $\frac{1}{4}$
Divide both sides by 4
Divide both sides by $\frac{1}{4}$

Explanation: When you encounter equations with fractions or decimals multiplied by a variable, your goal is to isolate the variable by "undoing" that multiplication. The key insight is understanding what operation cancels out the coefficient. Starting with

\frac{1}{4}x = 15

, you need to eliminate the

\frac{1}{4}

that's multiplying

x

. Since

\frac{1}{4} \times 4 = 1

, multiplying both sides by 4 gives you:

4 \cdot \frac{1}{4}x = 4 \cdot 15

1 \cdot x = 60

x = 60

This confirms that (A) Multiply both sides by 4 is correct. Let's see why the other choices don't work. (B) Multiply both sides by $\frac{1}{4}$ would give you

\frac{1}{4} \cdot \frac{1}{4}x = \frac{1}{4} \cdot 15

, which becomes

\frac{1}{16}x = \frac{15}{4}

— you've made the coefficient even smaller and more complicated. (C) Divide both sides by 4 is the same as multiplying by

\frac{1}{4}

, leading to the same problem as choice B. **(D) Divide both sides by

\frac{1}{4}

is actually correct mathematically (dividing by a fraction equals multiplying by its reciprocal), but the question asks what operation to apply, and "multiply by 4" is the more direct description. Study tip: When you have a fractional coefficient, multiply both sides by the denominator of that fraction. This is usually the quickest path to isolating your variable.