MA3 Algebra

Algebra on the ACT includes linear equations and inequalities, systems of equations, and quadratics, the backbone of the Math section. Strong algebra means you can set up and solve for unknowns quickly and read what a problem is really asking. A concrete tip: for systems of two equations, substitution works well when one variable is already alone, and elimination works well when you can add or subtract to cancel a variable, so pick the method that fits the setup. This page covers solving equations, handling inequalities, systems, and quadratics with clear steps, plus free interactive practice on the problem types the ACT repeats. You'll drill isolating variables, flipping inequality signs correctly, choosing substitution or elimination, and factoring or using the quadratic formula. Practice a set, review each solution, and your setup speed and accuracy climb together.

Algebra is the largest single slice of ACT Math. If you can solve a linear equation without stalling, read a word problem into an equation, and factor a quadratic on sight, you have covered most of what the section asks of you — and you have freed up the time you need for the handful of harder questions.

The clock is the real test. Mathematics is 45 questions in 50 minutes — about 66 seconds per question, on average. A question you grind through for two minutes has spent the time budgeted for two questions. That is the trade you are managing all section long, and the four answer choices are usually the cheapest way out of it.

What actually shows up

Linear equations and inequalities in one variable, including the sign flip.
Systems of two equations, by substitution or elimination — plus the two special cases, no solution and infinitely many.
Lines: slope from two points, slope-intercept form, parallel and perpendicular slopes, distance and midpoint.
Quadratics: factoring out a GCF, difference of squares, trinomials, the quadratic formula, and what the discriminant tells you.
Rational and radical expressions, where the answer you compute is not always the answer you keep.
Exponent rules and scientific notation.
Word problems — mixtures, rates, costs, ages, consecutive integers — that turn into a linear equation or a system.

Four choices are information

You are never asked to produce an answer out of thin air: the ACT has no student-produced-response items, so one of four printed options is always correct. That changes the arithmetic of how you should work.

Test the choices when the algebra is messier than the substitution. "Which value of x satisfies…" with four integers in front of you is often a ten-second problem: substitute and see.
Work backwards from the answers in word problems with an awkward set-up. If the choices are 6, 8, 12 and 15 guests, try one and see whether the two costs come out equal.
Pick numbers when the choices are expressions rather than values. Let x = 2, run it through the stem and through each choice, and keep the choice that matches.
Solve directly when the equation is short. Backsolving a two-step equation is slower than just solving it.

None of these is a trick; they are all just cheaper routes to the same place. Choose the cheap route when the direct one looks long, and notice which one you are on before a minute has gone.

Three traps that cost real points

Extraneous roots. Squaring both sides of a radical equation, or multiplying both sides of a rational equation by an expression containing the variable, can manufacture solutions the original equation never had. √(x + 9) = x − 3 squares to x² − 7x = 0, giving x = 0 and x = 7 — but x = 0 makes the right-hand side −3, and a square root is never negative. Only x = 7 survives. The same goes for any value that makes a denominator zero: it is disqualified no matter how cleanly it came out of the algebra. Substitute your answer back into the equation as printed.

The inequality sign flip. Multiplying or dividing both sides of an inequality by a negative number reverses the sign: from −5x > 20 you get x < −4, not x > −4. Adding and subtracting never flip it, and dividing by a positive number never flips it. The wrong-direction version of the right boundary is on the answer sheet every time.

Losing a solution by dividing by a variable. Given 3x² = 12x, dividing both sides by x gives x = 4 and quietly throws away x = 0, which also satisfies the equation. Move everything to one side and factor instead: 3x(x − 4) = 0, so x = 0 or x = 4. Dividing by a variable is only safe when you already know that variable is not zero.

The trap in practice. "What are all the solutions of 2x² = 10x?" Divide by x and you get the single answer x = 5, which will be sitting there as a choice. Factor instead — 2x(x − 5) = 0 — and you get x = 0 and x = 5. When a question asks for all the values, the choice listing only one of them is usually the bait.

How to spend the 66 seconds

Read the last line first in a word problem. Knowing whether it wants the width or the length, the student tickets or the adult tickets, stops you solving the right system and reporting the wrong variable — the most common way to lose a question you actually did correctly.
Write the equation before you compute anything. Most word-problem errors are set-up errors, not arithmetic errors.
Move on and come back. There is no penalty for guessing, so an unanswered question is strictly worse than a guessed one. Mark it, answer everything you find easy, and return with whatever time is left.
Use the calculator you are allowed. Calculators are permitted on the Mathematics section, but no question requires one. If you are reaching for it on every item, the set-up is doing too little of the work.

Worked examples

Solve: 5x + 3 = 2x + 18

Get the variables on one side: subtract 2x from both sides → 3x + 3 = 18.
Get the numbers on the other side: subtract 3 from both sides → 3x = 15.
Divide both sides by 3 → x = 5.
Check: 5(5) + 3 = 28 and 2(5) + 18 = 28. Both sides match, so x = 5 is correct.

A line passes through (2, 3) and (6, 11). Write its equation in slope-intercept form.

Find the slope: m = (11 − 3) / (6 − 2) = 8/4 = 2.
Use y = mx + b with one point. Using (2, 3): 3 = 2(2) + b.
Solve for b: 3 = 4 + b, so b = −1.
The equation is y = 2x − 1. Check with the other point: 2(6) − 1 = 11. ✓

Solve the inequality: 7 − 3x ≥ 22

Subtract 7 from both sides: −3x ≥ 15.
Divide both sides by −3. Dividing by a negative FLIPS the inequality sign.
x ≤ −5.
Check with x = −6: 7 − 3(−6) = 25, and 25 ≥ 22 ✓.

Common mistakes to avoid

Forgetting to flip the inequality sign when you multiply or divide by a negative number. This is the single most common algebra error on the SAT.
Mixing up rise and run in the slope formula. Slope is change in y over change in x, not the other way around.
Distributing a negative incorrectly: −2(x − 5) is −2x + 10, not −2x − 10.
Answering the wrong question. If the problem asks for 3x and you solved x = 4, the answer is 12, not 4. Always re-read what was asked.

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Frequently asked questions

How do I solve a system of equations on the ACT?

Use substitution or elimination. Substitution: solve one equation for a variable and plug it into the other. Elimination: add or subtract the equations to cancel a variable. Pick whichever is less work, if a variable already stands alone, substitute; if coefficients line up to cancel, eliminate. Then solve for one variable and back-substitute for the other.

When do I flip the inequality sign on the ACT?

Flip the inequality sign whenever you multiply or divide both sides by a negative number. For example, -2x < 6 becomes x > -3 after dividing by -2. Adding or subtracting never flips it. This single rule is a common ACT trap, so watch every time a negative coefficient is involved.

How do I solve a quadratic equation on the ACT?

Set the equation equal to zero, then factor if possible and set each factor to zero. If it doesn't factor cleanly, use the quadratic formula: x equals negative b plus or minus the square root of (b squared minus 4ac), all over 2a. Factoring is faster when it works, so try it first, then fall back on the formula.

How do I set up a word problem in algebra on the ACT?

Name the unknown with a variable, then translate each sentence into an equation. Words like 'is' mean equals, 'more than' means add, 'per' signals a rate. Write what you know, solve, and check that the answer fits the question actually asked, sometimes the question wants a total or a difference, not the variable you solved for.