Linear Equations and Inequalities — ACT Math

Introduction

Linear equations are the backbone of algebra — and the ACT tests them in sneaky ways, from absolute value traps to inequalities that flip when you divide by a negative.

Linear equations and inequalities account for 8-12 questions per test across the easy, medium, and hard range. Absolute value and inequality problems appear as late as question 50.

By the end of this lesson you will be able to:

Solve one-variable linear equations and inequalities, including those with fractions
Solve absolute value equations and identify when they produce two solutions
Translate word problems into linear equations and solve

You'll solve |3x − 7| = 11 and explain why you must check both solutions in the original problem context.

The Concept

The Core Rule

To solve a linear equation, isolate the variable using inverse operations. For inequalities, the same rules apply EXCEPT: multiplying or dividing both sides by a negative number reverses the inequality sign. For |ax + b| = c (c > 0), split into two equations: ax + b = c and ax + b = −c.

How the ACT tests this

Hiding a linear equation inside a word problem about ages, money, or geometry
Testing whether students remember to flip the inequality sign when dividing by a negative
Presenting an absolute value equation and asking how many solutions exist

Solving Linear Equations

Step 1: Distribute and clear fractions (multiply through by LCD). Step 2: Combine like terms on each side. Step 3: Move variables to one side, constants to the other. Step 4: Divide by the coefficient.

Clearing fractions: multiply every term by the LCD before solving
Variables on both sides: subtract the smaller variable term from both sides
Check your answer by substituting back into the original equation

Inequalities

Solve exactly like equations with one critical exception: flip the inequality sign when multiplying or dividing by a negative number. Graph solutions on a number line: open circle for < or >, closed circle for ≤ or ≥.

Compound inequalities: a < x < b means both a < x AND x < b
Negative coefficient: −2x > 6 → x < −3 (flip the sign!)
'No solution' vs 'all real numbers': check if the inequality is ever true or always true

Absolute Value

|x| = distance from zero on the number line. |ax + b| = c gives two equations when c > 0: ax + b = c OR ax + b = −c. When c < 0, there is no solution. When c = 0, there is exactly one solution.

Always set up both cases and solve each separately
Verify both solutions in the original equation
|ax + b| < c becomes −c < ax + b < c (a bounded interval)

Your strategy

Clear fractions first by multiplying every term by the LCD
For inequalities, highlight the negative coefficient as a reminder to flip the sign
For absolute value, write 'Case 1:' and 'Case 2:' to stay organized
Backsolve: for word problems, plug each answer choice into the problem statement

Worked Examples

Easy Example 1 Adding 3 To The Coefficient Instead Of Isolating The Variable First

If 4x − 3 = 17, what is the value of x?

A. 3
B. 4
C. 5 (Correct answer)
D. 6
E. 7

Step 1

Add 3 to both sides: 4x = 20

Step 2

Divide by 4: x = 5

Step 3

Check: 4(5) − 3 = 17 ✓

Correct answer: C

Why C is correct

Correct: 4(5) − 3 = 17

Why other options are wrong

A: 4(3) − 3 = 9, not 17

B: 4(4) − 3 = 13, not 17

D: 4(6) − 3 = 21, not 17

E: 4(7) − 3 = 25, not 17

⚠ Trap: Adding 3 to the coefficient instead of isolating the variable first

Medium Example 2 Forgetting To Flip The Inequality Sign When Dividing By −3

If −3x + 5 > 14, which of the following describes all solutions for x?

A. x > 3
B. x < 3
C. x > −3
D. x < −3 (Correct answer)
E. x > −6

Step 1

Subtract 5 from both sides: −3x > 9

Step 2

Divide by −3 and flip the inequality: x < −3

Step 3

Answer: x < −3

Correct answer: D

Why D is correct

Correct: −3x > 9 → x < −3

Why other options are wrong

A: Forgot to flip: divided by −3 without reversing the sign

B: Divided correctly but used +3 instead of −3

C: Got magnitude right but wrong direction for the flip

E: Divided 9 by −3 incorrectly as −6

⚠ Trap: Forgetting to flip the inequality sign when dividing by −3

Hard Example 3 Setting Up |2x−5| = 3 (equality Only) Instead Of ≤ 3 (interval)

How many integer solutions does |2x − 5| ≤ 3 have?

A. 2
B. 3
C. 4 (Correct answer)
D. 5
E. Infinitely many

Step 1

Rewrite as compound inequality: −3 ≤ 2x − 5 ≤ 3

Step 2

Add 5 throughout: 2 ≤ 2x ≤ 8

Step 3

Divide by 2: 1 ≤ x ≤ 4

Step 4

Integer solutions: x = 1, 2, 3, 4 → 4 integers

Correct answer: C

Why C is correct

Correct: x ∈ {1, 2, 3, 4}

Why other options are wrong

A: Only counted the two endpoints 1 and 4

B: Missed one boundary value

D: Added an extra integer outside the range

E: The solution set is bounded, not infinite

⚠ Trap: Setting up |2x−5| = 3 (equality only) instead of ≤ 3 (interval)

Strategy Tips

Circle or underline the inequality sign in every problem — it's easy to lose track
When dividing by a negative, write a big arrow pointing at the sign as a reminder to flip
For absolute value, always check if c is negative first (no solution if c < 0)
Backsolving is extremely reliable for linear equations — test C first, then adjust
For word problems, define your variable in a sentence: 'Let x = the number of apples'

Common pitfalls

Not flipping the inequality sign when multiplying/dividing by a negative number

Solving only one case for absolute value equations

Mistaking the direction of an inequality after solving

Simple linear equations: 30 seconds. Inequality or absolute value: 60-90 seconds. If a word problem is taking over 2 minutes, backsolve using the answer choices.

Summary

Flip the inequality sign whenever you multiply or divide both sides by a negative
Absolute value equations always produce two cases (unless the right side is 0 or negative)
Backsolving (plugging in answer choices) is the fastest method when the equation looks complex

Solve |5x + 2| = 13. Write out both cases, solve each, and verify both answers by substituting back into the original equation.

Next: Systems of Equations and Polynomials All ACT Math lessons