ACT Math: Equations
Equations is a foundational Algebra topic on the ACT, and solving them accurately and efficiently underlies nearly everything else in the Algebra and Functions domains. This free, complete lesson covers one-step and two-step equations, multi-step equations with variables on both sides, literal equations (solving a formula for a specific variable), and absolute value equations, along with real-world word problems that translate into first-degree equations. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.
1. One-Step and Two-Step Equations
Solve an equation by performing the same operation to both sides until the variable is isolated. A one-step equation needs a single operation; a two-step equation needs two, typically undoing addition/subtraction before multiplication/division.
Worked Example: Solve 3x − 7 = 14.
3x = 14 + 7 = 21
x = 21 ÷ 3 = 7
1. Solve: x + 9 = 23
Explanation: Subtract 9 from both sides: x = 23 − 9 = 14.
2. Solve: 5x = 45
Explanation: Divide both sides by 5: x = 45 ÷ 5 = 9.
3. Solve: x4 = 12
Explanation: Multiply both sides by 4: x = 12 × 4 = 48.
4. Solve: 2x + 5 = 17
Explanation: Subtract 5: 2x = 12. Divide by 2: x = 6.
5. Solve: 4x − 9 = 15
Explanation: Add 9: 4x = 24. Divide by 4: x = 6.
2. Multi-Step Equations
Multi-step equations may need distributing first, and often have the variable on both sides. Move all variable terms to one side and all constants to the other before isolating the variable.
Worked Example: Solve 3(2x − 4) = 5x + 6.
6x − 12 = 5x + 6
6x − 5x = 6 + 12
x = 18
1. Solve: 2(x + 3) = 16
Explanation: Distribute: 2x + 6 = 16. Subtract 6: 2x = 10. Divide by 2: x = 5.
2. Solve: 5x − 2 = 2x + 13
Explanation: Subtract 2x from both sides: 3x − 2 = 13. Add 2: 3x = 15. Divide by 3: x = 5.
3. Solve: 4(x − 2) = 2(x + 6)
Explanation: Distribute: 4x − 8 = 2x + 12. Subtract 2x: 2x − 8 = 12. Add 8: 2x = 20. Divide by 2: x = 10.
4. Solve: 3x + 7 = 5x − 9
Explanation: Subtract 3x: 7 = 2x − 9. Add 9: 16 = 2x. Divide by 2: x = 8.
5. Solve: 2(3x − 1) + 4 = 3(x + 5)
Explanation: Distribute and combine: 6x − 2 + 4 = 6x + 2, and 3(x + 5) = 3x + 15. So 6x + 2 = 3x + 15. Subtract 3x: 3x + 2 = 15. Subtract 2: 3x = 13. Divide by 3: x = 133.
3. Literal Equations
A literal equation is a formula with multiple variables. Solving for one specific variable means isolating it, treating every other letter the same way you would treat a number.
Worked Example: Solve for h: A = 12bh, in terms of A and b.
2A = bh
h = 2Ab
1. Solve for r: C = 2πr, in terms of C.
Explanation: Divide both sides by 2π: r = C2π.
2. Solve for w: P = 2l + 2w, in terms of P and l.
Explanation: Subtract 2l: P − 2l = 2w. Divide by 2: w = P − 2l2.
3. Solve for x: y = mx + b, in terms of y, m, and b.
Explanation: Subtract b: y − b = mx. Divide by m: x = y − bm.
4. Solve for t: d = rt, in terms of d and r.
Explanation: Divide both sides by r: t = dr.
5. Solve for F: C = 59(F − 32), in terms of C.
Explanation: Multiply both sides by 95: 95C = F − 32. Add 32: F = 95C + 32.
4. Absolute Value Equations and Word Problems
Since |x| represents distance from 0, an absolute value equation like |expression| = k (for positive k) splits into two separate equations: expression = k, or expression = −k.
Worked Example: Solve |2x − 3| = 7.
2x − 3 = 7 or 2x − 3 = −7
x = 5 or x = −2
1. Solve: |x + 4| = 9
Explanation: x + 4 = 9 gives x = 5. x + 4 = −9 gives x = −13.
2. Solve: |3x − 6| = 12
Explanation: 3x − 6 = 12 gives x = 6. 3x − 6 = −12 gives x = −2.
3. Solve: 2|x − 1| = 10
Explanation: Divide by 2 first: |x − 1| = 5. Then x − 1 = 5 gives x = 6, and x − 1 = −5 gives x = −4.
4. A rectangle's length is 3 more than twice its width. If the perimeter is 36, what is the width?
Explanation: Let w = width, so length = 2w + 3. Perimeter: 2(w + 2w + 3) = 36, so 2(3w + 3) = 36, giving 6w + 6 = 36, 6w = 30, and w = 5.
5. Maria has $45 more than twice what Josh has. Together they have $315. How much does Josh have?
Explanation: Let j = Josh's amount, so Maria has 2j + 45. Together: j + 2j + 45 = 315, so 3j = 270, and j = 90.
Common Mistakes to Avoid
- Forgetting to distribute to every term inside parentheses. 3(2x − 4) means both 2x and −4 get multiplied by 3, not just the first term.
- Making a sign error when moving terms across the equals sign. Moving a term to the other side always flips its sign; track this carefully, especially with negative coefficients.
- Only finding one solution to an absolute value equation. |expression| = k almost always produces two separate equations to solve, missing the negative case loses half the valid solutions.
- Treating every letter in a literal equation like it should disappear. When solving for one variable in a formula, the other variables stay in the answer, they're being treated as known constants, not solved away.
- Setting up a word problem's equation from the wrong starting quantity. Carefully identify which quantity is being defined "in terms of" another, reversing this relationship is one of the most common word-problem errors.
Frequently Asked Questions
What's the general strategy for solving any multi-step equation?
Distribute first if needed, then combine like terms on each side separately, then move all variable terms to one side and all constants to the other, and finally divide to isolate the variable. Following the same order every time reduces careless errors.
How is a literal equation different from a regular equation?
A regular equation has one variable and a numeric answer. A literal equation has several variables, and "solving" it means isolating one variable in terms of all the others, the answer is an expression, not a single number.
Why does an absolute value equation usually have two solutions?
Since |x| equals both x and −x when x is negative, a statement like |expression| = k is true whenever the expression equals k or its opposite, −k. Both cases need to be solved and checked.
Where can I practice more problems like these?
The Equations quizzes in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.