ACT Math Inequalities: 16 Practice Problems with Step-by-Step Explanations
Inequalities behave almost exactly like equations, with one rule that makes all the difference: whenever you multiply or divide both sides by a negative number, the inequality sign flips direction. Forgetting this single rule is the single most common mistake on this entire chapter, and the ACT writes wrong-answer choices specifically to catch it. Beyond that rule, solving a linear inequality follows the same steps as solving a linear equation — isolate the variable using inverse operations.
Compound inequalities (like 4 < 3x−5 < 13) are solved by applying the same operation to all three parts at once, keeping the parts in the same relative order. Quadratic inequalities require extra care: an inequality like x2<6 becomes −√6<x<√6 (a range), while x2>6 becomes two separate rays (x<−√6 or x>√6) — squaring or taking a square root doesn't work the same way it does with a straightforward linear equation, since x2 is never negative, which occasionally makes a quadratic inequality have no solution at all, or all real numbers as its solution.
For "which must be true" problems involving inequalities with unknown-but-constrained variables, plugging in a few different concrete numbers that satisfy the given constraints (and trying to break each answer choice) is often faster and more reliable than pure symbolic reasoning — just make sure to test more than one set of numbers before concluding a statement always holds. Work through the 16 problems below, including two that reference a number line graph, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Solving a linear inequality
- Problem 2 — Solving a linear inequality
- Problem 3 — Graphing a solution set
- Problem 4 — Greatest integer solution
- Problem 5 — Solving a linear inequality
- Problem 6 — Clearing fractions in an inequality
- Problem 7 — Isolating a variable
- Problem 8 — Graphing a compound inequality
- Problem 9 — Maximizing an expression
- Problem 10 — Solving a compound inequality
- Problem 11 — Quadratic inequality
- Problem 12 — Testing constrained values
- Problem 13 — Testing constrained values
- Problem 14 — Quadratic inequality with no solution
- Problem 15 — Quadratic inequality
- Problem 16 — Setting up an inequality
Practice Problems
Which of the following inequalities describes the solution set for 4x−2 ≥ 10x−14?
- A) x≤2
- B) x≥2
- C) x≥8/3
- D) x≤−1
Move variable terms to one side and constants to the other: −2+14≤10x−4x, so 12≤6x.
Divide by 6 (positive, no sign flip needed): 2≥x, which is the same as x≤2.
The solution set to 3x+4 > −11 is all real values of x such that:
- A) x > −15
- B) x < −15
- C) x > −5
- D) x < −5
Subtract 4 from both sides: 3x > −15.
Divide by 3 (positive, no sign flip needed): x > −5.
Which of the following graphs shows the solution set for 5−3x ≥ −2x+11?
- A) Closed dot at 0, ray extending right
- B) Closed dot at −6, ray extending right
- C) Closed dot at −6, ray extending left
- D) Closed dot at −16, ray extending left
Move variable terms to one side and constants to the other: 5−11≥−2x+3x, so −6≥x, which is the same as x≤−6.
This is graphed as a closed dot at −6 (since the inequality includes equality) with a ray extending left toward negative infinity.
What is the greatest integer solution to −3x+4 > 19.4?
- A) −3
- B) −4
- C) −5
- D) −6
Subtract 4 from both sides: −3x > 15.4.
Divide by −3 and flip the inequality (dividing by a negative): x < −5.1̄3. Since −5 is greater than −5.1̄3, it does not satisfy the inequality. The greatest integer that does is −6.
Which of the following is the solution set for (2/5)x+6 ≤ x+10?
- A) x≤−20/3
- B) x≥−20/3
- C) x≥4/7
- D) x≤20/3
Move variable terms to one side and constants to the other: 6−10≤x−(2/5)x, so −4≤(3/5)x.
Multiply both sides by 5/3 (positive, no sign flip needed): −(4)(5/3)≤x, so x≥−20/3.
The solution set to (1/3)x < (3/2)x + 11/6 is all real values of x such that:
- A) x > −11/7
- B) x < −11/7
- C) x > 11/7
- D) x > 2
Multiply every term by 6 to clear the fractions: 2x < 9x + 11.
Move variable terms to one side: 2x−9x < 11, so −7x < 11. Divide by −7 and flip the inequality: x > −11/7.
Which of the following inequalities is equivalent to 5x+9y ≤ 2x−6?
- A) x≥−3y−2
- B) x≤−3y−2
- C) x≥3y−2
- D) x≤3y−2
Move all x terms to one side and everything else to the other: 5x−2x≤−6−9y, so 3x≤−6−9y.
Divide by 3 (positive, no sign flip needed): x≤−2−3y, which is the same as x≤−3y−2.
Which of the following number line graphs shows the solution set to the inequality 4 < 3x−5 < 13?
- A) Closed dots at 3 and 6, ray extending right past 6
- B) Open dots at 3 and 6, bold segment between
- C) Open dots at 3 and 6, rays extending outward in both directions
- D) Open dots at 3 and 6, rays extending outward with additional shading
Add 5 to all three parts of the compound inequality: 9 < 3x < 18.
Divide all three parts by 3 (positive, no sign flip needed): 3 < x < 6. Since the inequality is strict (no equal sign), this is graphed with open dots at both 3 and 6, with a bold segment connecting them.
If x ≥ −14 and y ≤ 19, what is the greatest possible value of 2y−x?
- A) 24
- B) 33
- C) 38
- D) 52
To maximize 2y−x, maximize 2y (using the largest allowed y, 19) and maximize −x (which means minimizing x, using the smallest allowed x, −14).
2(19)−(−14) = 38+14 = 52.
Which of the following is the solution statement for the inequality shown below?
−3 < −2x+1 < 11
- A) −5 < x < 2
- B) −3 < x < 11
- C) −2 < x < 5
- D) −2 < x
Subtract 1 from all three parts: −4 < −2x < 10.
Divide all three parts by −2 and flip both inequality signs: 2 > x > −5, which is the same as −5 < x < 2.
Which of the following inequalities is equivalent to (x+2)2 ≤ 9?
- A) −5≤x≤1
- B) −3≤x≤3
- C) −1≤x≤3
- D) −1≤x≤1
Take the square root of both sides, remembering that this creates a compound inequality: −3≤x+2≤3 (since √9=3).
Subtract 2 from all three parts: −5≤x≤1.
If 0 > y > −1 > x, then which of the following CANNOT be true?
- A) x2 > y2
- B) x3 > y3
- C) −1/y > −1/x
- D) 0 < xy < 1
Since y is between −1 and 0, and x is less than −1, it's always true that x < y (x is further left on the number line).
Cubing preserves order for all real numbers (unlike squaring), so x < y always implies x3 < y3. This means x3 > y3 can never be true under these constraints. Testing x=−2, y=−0.5 confirms: x3=−8 is not greater than y3=−0.125.
If x and y are real numbers such that x > 1 and y < −1, then which of the following inequalities must be true?
- A) x/y < −1
- B) x2−3 > y2−3
- C) (1/5)x+1 > (1/5)y+1
- D) 3x−2 > 3y−2
Option C simplifies to x > y (dividing both sides by the positive constant 1/5 preserves the inequality). Since x > 1 and y < −1, x is always greater than y — this holds for every valid choice of x and y.
The other options fail for specific values: for example, with x=1.1 and y=−10, option A gives x/y≈−0.11 (not <−1), and option B gives x2≈3<y2≈100 (a contradiction).
The inequality x2−2 < −3 is true for the values of x in which of the following sets?
- A) {−1,1}
- B) {x | −1<x<1}
- C) {x | 0<x<1}
- D) The empty set
Add 2 to both sides: x2 < −1.
Since x2 is always non-negative for any real number x, it can never be less than −1. There is no real solution, so the solution set is the empty set.
Which of the following inequalities is the solution set for 10−3x2 > −8?
- A) x < √6
- B) x > √6
- C) −√6 < x < √6
- D) x < −√6 and x > √6
Subtract 10 from both sides: −3x2 > −18.
Divide by −3 and flip the inequality: x2 < 6. Taking the square root creates a range (not two separate rays, since this is a "less than" quadratic inequality): −√6 < x < √6.
Trey has $71.50 to spend at the golf store. He buys a putter for $45 and wants to buy golf balls that are $1.50 each. Which of the following inequalities, when solved, gives g, the number of golf balls, that Trey can buy?
- A) 1.5g+45 ≤ 71.50
- B) 1.5g−45 ≤ 71.50
- C) g+45 ≤ 71.50
- D) g−45 ≤ 71.50
The total cost is the putter's fixed price plus the cost per golf ball times the number of golf balls: 45 + 1.5g.
This total must be at most Trey's budget of $71.50: 1.5g+45 ≤ 71.50.
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