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Linear Equations in One Variable

Linear equations in one variable are a fundamental topic in algebra and a key part of the ACT Math test. These problems require students to solve for a single unknown by simplifying expressions, working with fractions, and applying inverse operations step by step.This section of linear equations in one variable practice problems is designed to help you build accuracy and speed when solving multi-step algebra questions. Many ACT questions include fractions, parentheses, and require careful organization of terms before finding the final solution.Mastering this topic will improve your overall performance in algebra and help you solve exam questions more efficiently under time pressure.

A typical problem asks what value must be added to both the numerator and the denominator of $\dfrac{2}{7}$ to obtain $\dfrac{3}{4}$. We write: \begin{gather*} \dfrac{2+x}{7+x}=\dfrac{3}{4}, \end{gather*} Cross multiplying yields \begin{align*} 4(2+x) & = 3(7+x) \implies \\ 8+4x & = 21+3x \implies \\ x & = 13. \end{align*}

1. What value of $x$ satisfies the equation $-5(3 - 2x)= -2(5 - x)$? 2

$ \dfrac{-25}{8} $
$\dfrac{5}{8} $
$\dfrac{5}{6} $
$ \dfrac{25}{12} $

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B

2. If $\dfrac{x}{3}+\dfrac{x}{5}=\dfrac{4}{7}$, what is the value of $x$? 3

$\dfrac{14}{15}$
$ \dfrac{15}{14} $
$\dfrac{12}{11}$
$1$

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B

3. If $\dfrac{1}{x}+\dfrac{1}{3x}=5$, what is the value of $x$? 3

$\dfrac{2}{15}$
$\dfrac{8}{25}$
$10$
$ \dfrac{4}{15}$

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D

4. If $\dfrac{2x}{5}+\dfrac{3x}{11}=\dfrac{3}{4}$, what is the value of $x$? 3

$\dfrac{155}{148}$
$\dfrac{165}{148}$
$\dfrac{40}{37}$
$\dfrac{41}{37}$

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B

5.What number must be added to both the numerator and denominator of $\dfrac{1}{3}$ to get $\dfrac{3}{4}$? 4

$4$
$5$
$6$
$ 7 $

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B

6. What number must be added to both the numerator and denominator of $\dfrac{a}{b}$ to get $\dfrac{3}{4}$? 5

$3b-2a$
$3b+4a$
$3b-a$
$3b-4a$

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D

7. Suppose the sum of 4 consecutive integers is $a$. In terms of $a$, what is the sum of the largest 2 of those integers? 4

$\dfrac{a}{2}+4 $
$\dfrac{a}{3}+3$
$\dfrac{a}{3}+5 $
$ \dfrac{a}{2}+2 $

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D

Answer key: 1.B, 2.B, 3.D, 4.B, 5.B, 6.D, 7.D..

Solutions

1. (B) $-15+10x= -10+2x\implies 8x=5\implies x=\dfrac{5}{8}$.

2. (B) $\dfrac{x}{3}+\dfrac{x}{5}=\dfrac{4}{7} \implies \dfrac{(5+3) x}{15}=\dfrac{4}{7} \implies \dfrac{8x}{15}=\dfrac{4}{7} \implies x=\dfrac{4}{7}\times\dfrac{15}{8}=\dfrac{15}{14}$.

3. (D) $\dfrac{1}{x}+\dfrac{1}{3x}=5 \implies \dfrac{3}{3x}+\dfrac{1}{3x}=5 \implies \dfrac{4}{3x}=5 \implies 15x=4 \implies x=\dfrac{4}{15}$.

4. (B) $\dfrac{2x}{5}+\dfrac{3x}{11}=\dfrac{3}{4} \implies $ $\dfrac{22x+15x}{55}=\dfrac{3}{4} \implies 148x=165 \implies x=\dfrac{165}{148}$.

5. (B) Convert the word problem into an equation, cross multiply, and then solve. $\dfrac{1+x}{3+x}=\dfrac{3}{4}$. Cross-multiplying yields $$ 4(1+x)=3(3+x) \implies 4+4x=9+3x \implies x=5. $$ This problem can also be solved fairly easily by substituting in the answer choices and seeing which one works.

6.(D) Convert the word problem into an equation, cross multiply, and then solve. $\dfrac{a+x}{b+x}=\dfrac{3}{4}$. Cross-multiplying yields $$ 4(a+x) = 3(b+x) \implies 4a+4x=3b+3x \implies x=3b-4a. $$

7.(D) Let the first integer be $x$; then the four consecutive integers are $x, x+1, x+2, \text{and } x+3$. Summing them gives $4x+6=a \implies x=\dfrac{a}{4}-\dfrac{3}{2}$. The sum of the largest two integers is $$ (x+2)+(x +3)=2x+5=2\left(\dfrac{a}{4}-\dfrac{3}{2}\right) +5=\dfrac{a}{2}-3+5=\dfrac{a}{2}+2. $$

Difficulty Key

1 Easiest
2 – 4 Intermediate
5 Most Difficult

Why Linear Equations in One Variable Are Important on the ACT

Linear equations in one variable are one of the most commonly tested topics on the ACT Math exam because they measure your understanding of algebraic manipulation and problem-solving skills.

These problems test your ability to simplify expressions, handle fractions, and isolate variables efficiently.

Key Skills You Need

To solve linear equations in one variable successfully, you must:

Simplify both sides of the equation
Use least common denominator when dealing with fractions
Properly distribute negative signs
Isolate the variable step by step
Check your final answer when possible

Common Mistakes Students Make

Many students lose easy marks due to:

Incorrect handling of fractions
Skipping distribution steps
Sign errors when moving terms
Not simplifying before solving

ACT Exam Strategy

ACT linear equations problems are designed to test accuracy under time pressure.

The fastest method is:

Clear fractions first
Expand all expressions
Group like terms
Solve for the variable
Double-check quickly if time allows

Extra Practice Tip

Most linear equation problems on the ACT should take less than 60 seconds if the correct strategy is used.