ACT Math Quiz: Inequalities And Absolute Value

What this quiz covers

This quiz focuses on Inequalities And Absolute Value, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $x \geq 5$
2. $x \geq 1$
3. $x \leq 1$ (correct answer)
4. $x \leq 5$

Explanation: Linear inequalities are solved exactly like equations, with one extra rule: the inequality symbol flips only when you multiply or divide by a negative number. Adding $3$ to both sides of $5x - 3 \leq 2$ gives $5x \leq 5$, and dividing both sides by the positive number $5$ leaves the symbol alone, so $x \leq 1$. The answer $x \geq 1$ has the correct boundary but flips the direction for no reason, since nothing negative was ever divided out, and $x \leq 5$ and $x \geq 5$ both stop at $5x \leq 5$ and forget to divide by the coefficient. After isolating the variable term, check whether the number you divided by was negative; if it was positive, the symbol you started with is the symbol you end with.

Question 2

1. $x > 7$
2. $x \leq -1$ or $x \geq 7$
3. $-1 \leq x \leq 7$ (correct answer)
4. $x < 7$

Explanation: An absolute value expression bounded above by a number describes an interval around a center, so $|x - 3| \leq 4$ means $x$ sits within $4$ units of $3$. Writing that as $-4 \leq x - 3 \leq 4$ and adding $3$ to all three parts gives $-1 \leq x \leq 7$. The answer $x \leq -1$ or $x \geq 7$ uses the same two boundary numbers but describes everything outside the interval, which is the solution to $|x - 3| \geq 4$, and both $x > 7$ and $x < 7$ report only one boundary and abandon the other end entirely. A correct solution to $|\text{expression}| \leq c$ always has two boundaries, so if your answer names only one endpoint, you have lost half the inequality.

Question 3

1. $-13$
2. $-3$ (correct answer)
3. $3$
4. $13$

Explanation: This is an absolute value question testing evaluation order. Choice B (−3) is correct — resolve each absolute value first: |−3| = 3, |−8| = 8, |2| = 2. Then apply the operations left to right: 3 − 8 + 2 = −3. Choice A (−13) ignores the absolute value symbols entirely, computing −3 − 8 + (−2) = −13 as if the bars weren't there. Choice C (3) likely comes from computing |−3 − (−8) + 2| = |7| = 7... or from incorrectly adding all values as positives: 3 + 8 + 2 = 13, then dividing or applying some other operation. Choice D (13) adds all three absolute values together — 3 + 8 + 2 = 13 — treating the minus sign between the first two as a plus sign. Pro tip: Absolute value bars are a grouping symbol — resolve them first to get the positive values, THEN carry out the arithmetic operations (subtraction, addition) between those values.

Question 4

1. $2\le x\le 7$
2. $2<x\le 7$ (correct answer)
3. $2\le x<7$
4. $2<x<7$

Explanation: The interval notation (2,7] means all values greater than 2 but not including 2, up to and including 7. The parenthesis at 2 means x>2 (not x≥2), and the bracket at 7 means x≤7. Combined, this gives 2<x≤7. Choice A incorrectly includes 2 with ≥, choice C incorrectly excludes 7 with <, and choice D excludes both endpoints. When converting between interval and inequality notation, parentheses mean strict inequalities while brackets mean inclusive inequalities.

Question 5

1. Number line with shading from $2$ to the right, open circle at $2$
2. Number line with shading from $-2$ to the left, closed circle at $-2$
3. Number line with shading from $2$ to the left, open circle at $2$ (correct answer)
4. Number line with shading from $-2$ to the right, closed circle at $-2$

Explanation: This is a linear inequality with a negative coefficient and graphical representation. Starting with -x + 5 > 3, subtract 5 from both sides to get -x > -2. Multiply by -1 (negative, so flip the inequality sign) to get x < 2. On a number line, this is represented by shading from 2 to the left with an open circle at 2. The solution matches choice C. Choice A would represent x > 2, which forgets the sign flip when dealing with the negative coefficient.

Question 6

1. $x > 3$ (correct answer)
2. $x < 3$
3. $x > \frac{9}{3}$
4. $x < \frac{9}{3}$

Explanation: This is a linear inequality requiring basic algebraic manipulation. Starting with 3x - 5 > 4, add 5 to both sides to get 3x > 9. Divide both sides by 3 (positive, so no sign flip) to get x > 3. The solution is x > 3, which matches choice A. Choices C and D are mathematically equivalent to A and B respectively since 9/3 = 3, but they present the answer in an unnecessarily complex form. Always simplify fractions when possible for clarity.

Question 7

1. Number line with shading from $2$ to the right, closed circle at $2$ (correct answer)
2. Number line with shading from $3$ to the left, open circle at $3$
3. Number line with shading from $2$ to the left, open circle at $2$
4. Number line with shading from $3$ to the right, closed circle at $3$

Explanation: This is a linear inequality requiring basic algebraic manipulation and graphical representation. Starting with 2x + 1 ≥ 5, subtract 1 from both sides to get 2x ≥ 4. Divide by 2 (positive, so no sign flip) to get x ≥ 2. On a number line, this is represented by shading from 2 to the right with a closed circle at 2 (since the inequality includes equality). The solution matches choice A. Choice C would represent x < 2, which is the opposite direction.

Question 8

1. $x \geq 1$ (correct answer)
2. $x \leq 1$
3. $x \leq -1$
4. $x \geq -1$

Explanation: This is a linear inequality with a negative coefficient requiring sign reversal. Starting with -4x + 5 ≤ 1, subtract 5 from both sides to get -4x ≤ -4. Divide by -4 (negative, so flip the inequality sign) to get x ≥ 1. The solution is x ≥ 1, which matches choice A. Choice B forgot to flip the inequality sign when dividing by a negative number, resulting in the incorrect x ≤ 1. This sign flip error is extremely common in inequality problems.

Question 9

1. $3n - 5 > 10$
2. $3n - 5 \geq 10$ (correct answer)
3. $5 - 3n \leq 10$
4. $3(n - 5) \geq 10$

Explanation: This is an inequality translation question testing mathematical language. Choice B (3n − 5 ≥ 10) is correct — "3 times a number n" → 3n. "5 less than" that quantity → subtract 5 → 3n − 5. "Is at least 10" → ≥ 10. Combined: 3n − 5 ≥ 10. Choice A (3n − 5 > 10) is close but uses strict inequality (>) instead of ≥. "At least" means "greater than OR equal to" — the ≥ symbol is required. Choice C (5 − 3n ≤ 10) reverses the subtraction — "5 less than 3n" means subtract 5 FROM 3n, which is 3n − 5, not 5 − 3n. These expressions are not equivalent. Choice D (3(n − 5) ≥ 10) misgroups the expression — it means "3 times the quantity (n minus 5)," which is different from "5 less than 3 times n." Pro tip: Build math phrases in the order they're stated. "5 less than 3n" means start with 3n, then subtract 5 — never flip the subtraction. And "at least" always translates to ≥, while "more than" translates to >.

Question 10

1. 1
2. 7
3. 11 (correct answer)
4. 15

Explanation: This is a max/min with constraints question testing strategic substitution. Choice C (11) is correct — to maximize x − y, use the largest possible x and the smallest possible y (most negative). Largest x = 6. Smallest y = −5. Maximum: 6 − (−5) = 6 + 5 = 11. Choice A (1) minimizes instead of maximizes: use smallest x = 2 and largest y = −1: 2 − (−1) = 3... or student computes 2 − 1 = 1, treating y as positive. Choice B (7) uses largest x = 6 but doesn't choose the smallest y: 6 − (−1) = 7, using y = −1 instead of y = −5. Choice D (15) overcounts: perhaps 6 + (−(−9)) = 15, or treats y as ranging to −9. Pro tip: To maximize a difference x − y, pull x as large as possible and pull y as small (most negative) as possible. Subtracting a very negative number gives a very large result: 6 − (−5) = 6 + 5 = 11. Don't be tripped by the negative signs on y's range — the smallest y in −5 ≤ y ≤ −1 is −5, not −1.

Question 11

1. $1 < x \le 7$ (correct answer)
2. $-4 < x \le 2$
3. $1 \le x < 7$
4. $2 < x \le 14$

Explanation: The correct answer is A (1 < x ≤ 7). Solve by adding 5 to all three parts of the compound inequality: −3 + 5 < 2x − 5 + 5 ≤ 9 + 5 → 2 < 2x ≤ 14. Divide all parts by 2: 1 < x ≤ 7. The open inequality (strict <) stays on the left; the closed inequality (≤) stays on the right. B (−4 < x ≤ 2) results from subtracting 5 instead of adding: left side becomes −8, divided by 2 = −4; right side becomes 4, divided by 2 = 2. C (1 ≤ x < 7) gets the correct bounds but swaps the open and closed inequality signs. D (2 < x ≤ 14) correctly adds 5 but stops without dividing by 2. Track open vs. closed endpoints carefully throughout all steps.

Question 12

1. $x > -4$
2. $x < -4$ (correct answer)
3. $x > 4$
4. $x < 4$

Explanation: The correct answer is B (x < −4). Solve the inequality step by step: −2x + 5 > 13 → −2x > 8 → x < −4. The critical step is flipping the inequality sign when dividing by a negative number. A (x > −4) correctly computes the boundary of −4 but fails to flip the inequality direction. C (x > 4) drops the negative sign on the coefficient of x, solving 2x > 8 instead. D (x < 4) also drops the negative on 2 but does flip the sign, arriving at 2x < 8 → x < 4. Pro tip: write a reminder to yourself — any time you divide or multiply both sides by a negative number, the inequality sign flips direction.

Question 13

1. $x \leq -3$ and $x \geq 3$
2. $x < -3$ or $x > 3$
3. $x = 0$
4. $-3 < x < 3$ (correct answer)

Explanation: This is a quadratic inequality that can be solved by factoring. The inequality x² - 9 < 0 can be factored as (x - 3)(x + 3) < 0. For a product to be negative, the factors must have opposite signs. This occurs when -3 < x < 3, where (x + 3) is positive and (x - 3) is negative. The solution is -3 < x < 3, which matches choice D. Choice B represents the solution to x² - 9 > 0, where the product is positive.

Question 14

1. $x \leq 2$ (correct answer)
2. $x \geq 2$
3. $x < 2$
4. $x > 2$

Explanation: This is a linear inequality requiring basic algebraic manipulation. Starting with 2x + 3 ≤ 7, subtract 3 from both sides to get 2x ≤ 4. Divide by 2 (positive, so no sign flip) to get x ≤ 2. The solution is x ≤ 2, which matches choice A. This is a straightforward problem with no negative coefficients requiring sign flips. Choice B would represent the opposite inequality direction.

Question 15

1. $x<-1$ or $x>7$
2. $-1<x<7$ (correct answer)
3. $-1\le x\le 7$
4. $-7<x<1$

Explanation: This absolute value inequality represents distance and splits into two cases. |x - 3| < 4 means the distance from x to 3 is less than 4. This gives us -4 < x - 3 < 4. Adding 3 to all parts: -1 < x < 7. The solution is all values strictly between -1 and 7. Choice A shows the exterior solution (for ≥), while choices C and D have incorrect boundaries. For |expression| < constant, always get the interior solution between two bounds.

Question 16

1. $x\ge -6$
2. $x\le -6$ (correct answer)
3. $x\ge 6$
4. $x\le 6$

Explanation: This inequality involves a negative fraction requiring sign flip awareness. Start with $-\dfrac{x}{3} \ge 2$, then multiply both sides by -3 and flip the inequality sign: $x \le -6$. The negative coefficient of x necessitates the sign flip when clearing the fraction. Choice A shows $x \ge -6$, which results from forgetting to flip the inequality sign. Choice C shows $x \ge 6$, combining both sign errors. Always flip inequality signs when multiplying or dividing by negative numbers.

Question 17

1. $x > 3$ (correct answer)
2. $x < 3$
3. $x > -3$
4. $x < -3$

Explanation: This linear inequality involves isolating x through standard operations with positive coefficients. Start with $7 + 3x > 16$, then subtract 7 from both sides: $3x > 9$. Divide both sides by 3 (positive number, so no sign flip): $x > 3$. The solution is $x > 3$, meaning all values greater than 3 satisfy the inequality. Choice B ($x < 3$) represents the opposite direction, which could result from incorrectly flipping the inequality sign. Since we're working with positive coefficients throughout, the inequality direction remains unchanged at each step.

Question 18

1. $x>3$
2. $x<3$ (correct answer)
3. $x\ge 3$
4. $x\le 3$

Explanation: This is a linear inequality with a negative coefficient. Start with -2x + 6 > 0, then subtract 6 from both sides: -2x > -6. Divide both sides by -2 and flip the inequality sign (critical when dividing by negative): x < 3. The solution is all real numbers less than 3. Choice A incorrectly shows x > 3, likely from forgetting to flip the inequality sign when dividing by a negative number. Always remember: dividing or multiplying by a negative flips the inequality.

Question 19

1. $[-3,7]$
2. $( -\infty,-7]\cup[3,\infty)$
3. $(-7,3)$
4. $[-7,3]$ (correct answer)

Explanation: An absolute value inequality using $\le$ becomes a compound inequality with the expression trapped between the negative and positive versions of the bound. From $|x+2|\le 5$ you get $-5 \le x + 2 \le 5$, and subtracting $2$ from all three parts gives $-7 \le x \le 3$, written $[-7,3]$ with square brackets because equality is allowed at both ends. The interval $[-3,7]$ comes from adding $2$ instead of subtracting it, undoing the $+2$ in the wrong direction; $(-7,3)$ has the right boundaries but uses parentheses, which excludes the two values where $|x+2|$ equals exactly $5$; and $( -\infty,-7]\cup[3,\infty)$ is the exterior solution belonging to $|x+2| \ge 5$. To undo an addition inside the absolute value, subtract the same amount from every part of the compound inequality, and let the $\le$ decide brackets over parentheses.

Question 20

1. $x\ge -7$
2. $x\ge 3$
3. $x\le -7$
4. $x\le 3$ (correct answer)

Explanation: Solving a linear inequality uses the same steps as solving an equation, with one extra rule: the direction of the inequality flips only when you multiply or divide by a negative number. Here you undo the subtraction by adding $5$ to both sides of $x-5\le -2$, and since $-2+5=3$, the solution is $x\le 3$, with the direction unchanged because adding a constant never flips it. The version $x\ge 3$ has the right boundary but flips the symbol for no reason, a mistake students make when they see the negative $-2$ and assume a flip is required. The versions $x\le -7$ and $x\ge -7$ come from subtracting $5$ instead of adding it, since $-2-5=-7$ moves the constant the wrong way. Decide the boundary number by undoing the operation on the variable, and change the inequality direction only when a negative multiplier or divisor is involved.