Solve linear equations and inequalities in one variable, including multi-step and variables-on-both-sides cases, and remember to flip the inequality when multiplying or dividing by a negative (Algebra).

What this topic is asking

Solving linear equations and inequalities is the bedrock of ACT Algebra: a one-variable relationship that you simplify and rearrange to isolate the variable. The procedures are short, but the ACT plants reliable traps, especially distributing carefully and flipping the inequality when you multiply or divide by a negative.

Solving a linear equation

The method is always the same sequence of inverse operations.

Variables on both sides

When the variable appears on both sides, gather it on whichever side keeps the coefficient positive, then finish. For $5x - 4 = 2x + 8$, subtract $2x$ to get $3x - 4 = 8$, add 4 to get $3x = 12$, and divide to get $x = 4$. Moving the smaller variable term avoids a negative coefficient and reduces sign errors.

Special cases: no solution and all solutions

Sometimes the variable cancels entirely:

• If you reach a false statement (such as $7 = 2$), the equation has no solution.
• If you reach a true statement (such as $4 = 4$ or $0 = 0$), every real number is a solution.

For example, $2(x + 3) = 2x + 5$ becomes $2x + 6 = 2x + 5$, then $6 = 5$, which is false, so there is no solution. Recognising these saves time and prevents you from inventing a value.

Solving linear inequalities

Inequalities solve like equations with one extra rule.

For $-3x \le 12$, divide by $-3$ and flip: $x \ge -4$. A useful check is to test a value from your solution set in the original inequality; if it fails, you probably forgot to flip.

Compound inequalities

The ACT sometimes uses a compound inequality such as $-3 < 2x + 1 \le 7$, which means both $-3 < 2x + 1$ and $2x + 1 \le 7$ hold at once. Solve all three parts together: subtract 1 throughout to get $-4 < 2x \le 6$, then divide by 2 to get $-2 < x \le 3$. Whatever you do to the middle, do to both ends, and apply the flip rule to both comparisons if you divide by a negative. The solution is the set of values between the two bounds, here all $x$ greater than $-2$ and at most 3.

Why the checks matter

Linear work is fast, so the points are won or lost on care: distributing across every term, clearing fractions correctly, choosing the side that keeps coefficients positive, and flipping the inequality at the right moment. A quick substitution check on an equation, or a test value on an inequality, catches almost every slip in a couple of seconds, which is well worth it on a timed test.

Try this

Q1. Solve $4x + 7 = 2x - 5$. [1 point]

• Cue. $2x = -12$, so $x = -6$.

Q2. Solve $-5x + 2 \ge 17$. [1 point]

• Cue. $-5x \ge 15$; divide by $-5$ and flip: $x \le -3$.