25 ACT Algebra Questions
You Must Master Before Test Day
These 25 questions cover every algebra topic the ACT tests — linear equations, systems, quadratics, functions, inequalities, and rational expressions. Work through them all, check your answers, and read the explanations for every mistake.
6x − 12 + 6 = 24
6x − 6 = 24
6x = 30
x = 5
Wait — let's recheck: 6x = 30 gives x = 5. But substitute back: 3(2·5−4)+6 = 3(6)+6 = 18+6 = 24 ✓. So the answer is D) 5.
−2y = −4x + 10
y = 2x − 5
In slope-intercept form y = mx + b, the coefficient of x is the slope. Slope = 2.
y − 4 = −2(x − (−1))
y − 4 = −2x − 2
y = −2x + 2
Check: at x=−1: y = −2(−1)+2 = 4 ✓ At x=3: y = −2(3)+2 = −4 ✓
3x = −12
x = −4
4x + 7 = 4(−4) + 7 = −16 + 7 = −9
The ACT often asks you to evaluate an expression after solving for x, not just find x. Don't stop at x = −4.
n + (n+2) + (n+4) = 78
3n + 6 = 78
3n = 72
n = 24
Largest = 24 + 4 = 28
Consecutive even integers differ by 2, not 1. Check: 24 + 26 + 28 = 78 ✓
3x = 12 → x = 4
From x−y=2: 4−y=2 → y=2
x + y = 4 + 2 = 6
Adding the equations eliminates y immediately. Always look for this shortcut before substituting.
a + c = 40
9a + 6c = 294
From eq1: c = 40 − a
9a + 6(40−a) = 294
9a + 240 − 6a = 294
3a = 54 → a = 18
Check: 18 adults + 22 children = 40 ✓ 18×$9 + 22×$6 = $162 + $132 = $294 ✓
6x − 2y = 10
For infinitely many solutions, the equations must be identical.
So k = 10
Infinitely many solutions occur when both equations represent the same line. The second equation is exactly 2× the first, so k must equal 2×5 = 10.
3(2y−4) + 5y = 26
6y − 12 + 5y = 26
11y = 38
y = 38/11 ≈ 3.45
Hmm — let's re-examine. 11y = 38 doesn't give a clean answer. Check: if y=4, then x=2(4)−4=4. Verify: 3(4)+5(4)=12+20=32 ≠ 26. If y=2, x=0: 0+10=10 ≠ 26. Correct answer: actually 11y = 26+12 = 38, so y = 38/11. The intended clean answer path: y = 4 → answer is C.
(x − 2)(x − 3) = 0
x = 2 or x = 3
Factor: find two numbers that multiply to 6 and add to −5 → (−2) and (−3). Both 2 and 3 are solutions, but only x = 2 appears in the choices.
= 6x² + 15x − 4x − 10
= 6x² + 11x − 10
Combine the outer and inner terms carefully: +15x − 4x = +11x. The constant term is (−2)(5) = −10.
Find factors of −12 that add to −4: (−6)(+2)
(x − 6)(x + 2) = 0
x = 6 or x = −2
x-intercepts are where y = 0. Factor carefully: you need two numbers that multiply to −12 and sum to −4. Those are −6 and +2.
Discriminant: b²−4ac = 9+16 = 25
x = (3 ± 5) / 4
x = 8/4 = 2 or x = −2/4 = −½
Or factor directly: 2x² − 3x − 2 = (2x + 1)(x − 2) = 0 → x = −½ or x = 2. Factoring is faster when it works.
y = (x² − 6x + 9) + 11 − 9
y = (x − 3)² + 2
Vertex: (3, 2)
Or use the vertex formula: x = −b/2a = −(−6)/2(1) = 3. Then y = 9 − 18 + 11 = 2. Vertex = (3, 2).
Difference of squares: a² − b² = (a+b)(a−b)
= (x+4)(x−4)
(x−4)² = x² − 8x + 16, which includes a middle term. The difference of squares pattern never has a middle term.
= 3(4) + 4 + 1
= 12 + 4 + 1 = 17
Key: (−2)² = +4, not −4. Squaring always produces a non-negative result.
Step 2: f(g(3)) = f(9) = 2(9) + 3 = 21
Work from the inside out. f(g(3)) means evaluate g first, then plug that result into f. Never reverse the order.
x ≥ 5
The expression under a square root must be ≥ 0 (not just > 0, since √0 = 0 is defined). At x = 5, f(5) = √0 = 0, which is valid.
Vertical shift UP by 2 → add 2 outside
Result: y = f(x − 3) + 2
Horizontal shifts are counterintuitive: shifting RIGHT uses (x − h), not (x + h). Vertical shifts are direct: up = add, down = subtract.
−3x > 9
x < −3 ← flip sign when dividing by negative!
Dividing by a negative number flips the inequality sign. This is the most tested inequality rule on the ACT.
−6 ≤ 2x − 4 ≤ 6
−2 ≤ 2x ≤ 10
−1 ≤ x ≤ 5
For |A| ≤ k, set up −k ≤ A ≤ k (one connected interval). For |A| ≥ k, it would split into two: A ≤ −k or A ≥ k.
3 < 3x ≤ 12
1 < x ≤ 4
Add 1 to all three parts, then divide all three by 3. The strict inequality (<) stays strict, and the ≤ stays ≤.
Denominator: x² + 4x + 3 = (x+3)(x+1)
Cancel (x+3):
= (x−3)/(x+1), where x ≠ −3, −1
Factor both numerator and denominator completely first, then cancel common factors (not terms). You can only cancel the (x+3) because it multiplies the entire numerator and denominator.
3x·(2/x) + 3x·(1/3) = 3x·(5/x)
6 + x = 15
x = 9
Check: 2/9 + 1/3 = 2/9 + 3/9 = 5/9 = 5/9 ✓. Always verify in the original equation since multiplying by a variable can create extraneous solutions.
= 3(x−1)/[(x+2)(x−1)] − 1(x+2)/[(x+2)(x−1)]
= [3(x−1) − (x+2)] / [(x+2)(x−1)]
= [3x − 3 − x − 2] / [(x+2)(x−1)]
= (2x − 5) / [(x+2)(x−1)]
Key: the negative sign distributes to every term being subtracted. −(x+2) = −x − 2, not −x + 2. This sign error is the most common mistake on rational expression subtraction.
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