Algebra questions appear throughout the ACT Math section and represent one of the highest-leverage areas to study. This guide covers every algebra topic the ACT tests — from basic substitution to intermediate topics like quadratics, functions, and logarithms — with worked examples for each one.

ACT Algebra: Topic frequency at a glance

The 60-question ACT Math section tests algebra heavily across two difficulty tiers:

TopicTierFrequencyPriority
SubstitutionElementaryMost commonHigh
Simplifying expressionsElementaryVery commonHigh
Writing equationsElementaryCommonHigh
Linear equationsElementaryCommonHigh
InequalitiesElementaryCommon (more than binomials)High
Multiplying binomialsElementaryLess frequentMedium
Quadratic equationsIntermediate~3 per testHigh
Systems of equationsIntermediateA few per testHigh
FunctionsIntermediateOccasionalMedium
MatricesIntermediateRareLow
LogarithmsIntermediateRareLow
01

To Algebra or Not to Algebra?

Choosing the right method for each problem

There's no single "right" method for ACT algebra questions. Every problem can typically be approached in at least two ways — and the best approach depends on your math comfort level and the time you have.

Faster
🧮

Use Algebra

Set up and solve an equation directly. More conceptually demanding but usually faster. Best when you can quickly identify the right equation.

Safer
🔢

Plug In Numbers

Start with the middle answer choice (C) and test it. If it doesn't work, eliminate half the remaining options based on whether you need a larger or smaller value.

Bonus
⚡

Spot the Shortcut

Look for a pattern or insight that makes the equation unnecessary. Can't be taught directly — but reviewing worked examples builds this instinct.

Worked Example — All Three Methods Coin Flip Problem

A man flipped a coin 162 times. The coin landed heads 62 more times than tails. How many times did it land heads? (A) 100   (B) 104   (C) 108   (D) 112   (E) 116

Method 1 — Plug In: Start with C (108). Tails = 162 – 108 = 54. Difference = 108 – 54 = 54, not 62. Need more heads → try D (112). Tails = 50. Difference = 62. ✓

Method 2 — Algebra:

x + (x – 62) = 162 2x – 62 = 162 2x = 224 x = 112

Method 3 — Shortcut: Remove the 62 surplus from 162. The remaining 100 split evenly → 50 + 50. Then 50 + 62 = 112.

Answer: D (112)

The Bottom Line

Don't lock in to one method. If algebra is flowing, use it — it's faster. If you can't set up the equation quickly, plug in. The only goal is the right answer in the least time.

02

Substitution

Elementary Algebra — Highest frequency topic
Elementary Algebra

Substitution questions give you an algebraic expression and the value of a variable (or a whole sub-expression) and ask you to calculate the result. These are the most straightforward algebra questions on the ACT — don't overthink them.

Example 1 — Direct SubstitutionSubstitution

If 2y + 8x = 11, what is the value of 3(2y + 8x)?

The entire expression 2y + 8x = 11. Substitute directly:

3(11) = 33

Answer: 33

Example 2 — Math Before SubstitutionSubstitution

If 3x – 7 = 8, then 23 – 3x = ?

First solve for 3x (not x individually):

3x – 7 = 8 → 3x = 15

Now substitute into the expression:

23 – 15 = 8

Answer: 8

Example 3 — Math After SubstitutionSubstitution

If a + b = 7 and b = 3, then 4a = ?

Substitute b = 3 to find a:

a + 3 = 7 → a = 4

Then: 4a = 4 × 4 = 16

03

Simplifying Algebraic Expressions

Factoring, unfactoring, and combining like terms
Elementary Algebra

There are two main techniques for simplifying algebraic expressions: factoring/unfactoring and combining like terms.

Factoring and Unfactoring

Factoring means finding a common factor across all terms and dividing it out. Unfactoring (distributing) is the reverse — you take a factored expression and multiply it out.

6y + 8x = 2(3y + 4x)   ← factoring out 2
8b + 24 = 8(b + 3)
3(x + y) + 4(x + y) = 7(x + y)
(2x + y)/x = 2 + y/x

Combining Like Terms

Two terms can be combined if and only if they share the same variable AND the same exponent. The variable itself does not change when you combine.

Can Combine

x² + 8x² = 9x²    y¹³ + 754y¹³ = 755y¹³    m³ + m³ = 2m³

Cannot Combine

x⁴ + x² (different exponents)    y² + x² (different variables)

04

Writing Expressions and Equations

Translating word problems into algebra
Elementary Algebra

Some ACT problems describe a situation in words and ask you to build or evaluate an algebraic expression. The key is identifying exactly what information matters — and ignoring what doesn't.

Worked ExampleWriting Equations

Mary poured g cups of water into a bucket, leaving f total cups. She then removed (g – 3) cups. How many cups remain?

We don't need to know the pre-pour amount. Start with f (total after pouring) and subtract what was removed:

f – (g – 3) = f – g + 3

Answer: f – g + 3

05

Solving Linear Equations

Isolation and shortcuts
Elementary Algebra

The standard method is to isolate the variable on one side of the equation. But on the ACT, taking a moment to look for shortcuts can save significant time.

Standard vs. ShortcutLinear Equations

If 6p + 2 = 20, then 6p – 3 = ?

Standard method: Solve for p: 6p = 18 → p = 3. Then 6(3) – 3 = 18 – 3 = 15.

Shortcut: Both expressions contain 6p. From the first equation, 6p = 18. Substitute directly into the second: 18 – 3 = 15. No need to solve for p at all.

Shortcut Strategy

When two equations share the same variable with the same coefficient (like 6p and 6p), you don't need to solve for the variable itself — just solve for the entire term and substitute.

06

Multiplying Binomials — FOIL

First · Outer · Inner · Last
Elementary Algebra

A binomial is any two-term expression joined by + or −, like (x + 2) or (y – 11). To multiply two binomials, use FOIL:

F
First
O
Outer
I
Inner
L
Last
(x + 2)(x + 3)
= x·x + x·3 + 2·x + 2·3
= x² + 3x + 2x + 6
= x² + 5x + 6
⚠️
Watch the signs When a negative term is involved, carry the sign through carefully. (x + 2)(x – 3) = x² + 2x – 3x – 6 = x² – x – 6. The –3 makes the outer and last products negative.

Three identities to memorize for the ACT:

Most important — difference of squares
(x + y)(x – y) = x² – y²
Perfect square (sum)
(x + y)² = x² + 2xy + y²
Perfect square (difference)
(x – y)² = x² – 2xy + y²
07

Inequalities

One critical rule difference from regular equations
Elementary Algebra

Solving inequalities follows all the same rules as solving regular equations — with one critical exception:

The Golden Rule of Inequalities

When you multiply or divide both sides by a negative number, you must flip the inequality sign.

If x > y, then –x < –y.

Example: If 2x + 6 ≥ y and you multiply by –2, the result is –4x – 12 ≤ –2y.

08

Quadratic Equations

Intermediate Algebra — ~3 questions per test
Intermediate Algebra

Quadratic equations are the most tested intermediate algebra topic — expect about 3 per test, making them a third of all intermediate algebra questions. Master these and you're most of the way to handling intermediate algebra.

Standard Form

Every quadratic on the ACT appears in the form: ax² + bx + c = 0 (where a ≠ 0)
If you see ax² + c = bx, subtract bx to get standard form: ax² – bx + c = 0

Solving by Factoring

Most ACT quadratics can be solved by factoring — essentially the reverse of FOIL. Find two numbers that multiply to c and add to b.

Factoring ExampleQuadratic Equations

Solve: x² + 9x + 18 = 0

Find factors of 18 that add to 9: 1×18, 2×9, 3×6 ✓ (3+6=9)

(x + 3)(x + 6) = 0 x + 3 = 0 → x = –3 x + 6 = 0 → x = –6

Solutions: x = –3 or x = –6

The Quadratic Formula (backup method)

When factoring isn't obvious, use the quadratic formula. This is rare on the ACT but good to have in reserve:

x = (–b ± √(b² – 4ac)) / 2a
09

Systems of Equations

Intermediate Algebra — Appears several times per test
Intermediate Algebra

These problems give you two equations and ask for the value of a variable or expression. The standard approach: substitution — solve one equation for one variable, then plug that value into the other equation.

Worked ExampleSystems of Equations

If 3x + 4y = 32 and 2y – x = 6, then x – y = ?

Step 1: Solve the simpler equation for x:

2y – x = 6 → x = 2y – 6

Step 2: Substitute into the first equation:

3(2y – 6) + 4y = 32 6y – 18 + 4y = 32 10y = 50 → y = 5

Step 3: Find x:

x = 2(5) – 6 = 4

Step 4: Answer the actual question:

x – y = 4 – 5 = –1

Answer: –1

⚠️
3 Common Systems of Equations Mistakes 1) Solve for x, not 2x — lowest form only.   2) Apply the distributive property carefully: 3(2y – 6) = 6y – 18, not 6y – 6.   3) Always answer what the question asks — here it's x – y, not just x.

Infinite solutions — a special case

A system has infinitely many solutions when both equations describe the same line (i.e., they're equivalent). Convert both to y = mx + b form and match the slopes. If the coefficients are proportional, the system has infinite solutions.

10

Functions & Relationships

f(x) notation, compound functions, and variation
Intermediate Algebra

f(x) = ax + b is just another way to write y = ax + b. On the ACT, treat f(x) exactly as you'd treat y — plug in the given x value and evaluate.

Compound Functions

A compound function like f(g(x)) is evaluated from the inside out. First evaluate the inner function, then plug that result into the outer function.

Compound Function ExampleFunctions

If h(x) = x² + 2x and j(x) = |x/4 + 2|, what is j(h(4))?

Inside first: h(4) = 4² + 2(4) = 16 + 8 = 24

Outside next: j(24) = |24/4 + 2| = |6 + 2| = |8| = 8

Direct and Inverse Variation

Direct Variation

w = kt²

As t increases, w increases. The variables move in the same direction.

Inverse Variation

w = k/t²

As t increases, w decreases. The variables move in opposite directions.

11

Matrices

Intermediate Algebra — Rare, but straightforward
Intermediate Algebra

Matrix problems appear rarely on the ACT — and when they do, they only test addition and subtraction. To add or subtract two matrices, simply add or subtract the corresponding entries in each position.

Matrix Addition ExampleMatrices

A = [2, 0 / 3, –5]    B = [–4, 1 / 6, 3]

A + B: Add position by position:

Row 1: 2 + (–4) = –2, 0 + 1 = 1 Row 2: 3 + 6 = 9, (–5) + 3 = –2

Result: [–2, 1 / 9, –2]

12

Logarithms

Intermediate Algebra — Rare, one key rule
Intermediate Algebra

Logarithms appear very rarely on the ACT, but you should know the one conversion rule — it's all you need:

The Only Logarithm Rule You Need for the ACT

The exponential equation x = aᵇ is equivalent to the logarithmic equation log_a(x) = b.

Example: If you see log_x(16) = 4, convert it to x⁴ = 16 → x = 2. Work with the exponent form — it's always easier to solve.

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Frequently Asked Questions

Common ACT Math algebra questions answered
How many algebra questions are on the ACT Math test?
The ACT Math test has 60 questions total. Elementary algebra is the most common category. Intermediate algebra accounts for approximately 9 questions — less than one-sixth of the math score. However, quadratics and systems of equations alone represent about 5–6 of those 9, so focusing there pays the biggest dividend.
Should I use algebra or plug in numbers on the ACT?
Both are valid — the right choice depends on the problem and your math comfort level. Use algebra when you can set up the equation quickly. Plug in when the equation setup isn't clear. The plug-in method is especially useful when you're stuck: always start with answer choice C (the middle value), then eliminate up or down based on whether you need a larger or smaller result.
What is the FOIL method and when do I use it?
FOIL stands for First, Outer, Inner, Last — a systematic method for multiplying two binomials. Multiply the First terms of each, then the Outer pair, then the Inner pair, then the Last terms, and add all four products together. Use it any time you need to expand an expression like (x + 3)(x – 2).
What quadratic equations should I memorize for the ACT?
Three identities appear frequently: (1) Difference of squares: (x+y)(x–y) = x²–y² — this is the most important and most tested. (2) Perfect square sum: (x+y)² = x²+2xy+y². (3) Perfect square difference: (x–y)² = x²–2xy+y². Recognizing these patterns lets you skip the full FOIL process and solve much faster.
What is the quadratic formula and when do I need it?
The quadratic formula is x = (–b ± √(b²–4ac)) / 2a. You use it when a quadratic equation cannot be solved by factoring — meaning you can't find two integers that multiply to c and add to b. On the ACT this is rare; most quadratics are designed to factor cleanly. Know the formula as a backup, but don't reach for it first.
How do I solve a system of two equations on the ACT?
Use the substitution method: (1) Pick the simpler equation and solve for one variable in its simplest form (solve for x, not 2x). (2) Substitute that expression into the second equation. (3) Solve for the remaining variable. (4) Plug back in to find the first variable. (5) Re-read the question — it often asks for x–y or some other expression, not just x or y individually.
Do I need to know matrices and logarithms for the ACT?
Yes, but only the basics. Matrix questions on the ACT only involve adding or subtracting — add corresponding entries position by position. Logarithm questions only require one conversion: log_a(x) = b is the same as x = aᵇ. Convert to the exponential form and solve from there. Neither topic appears often, but either can show up, so 10 minutes of review on each is worth it.

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