ACT Math Quiz: Equivalent Expressions

What this quiz covers

This quiz focuses on Equivalent Expressions, 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. $5x - 4$
2. $5x - 12$ (correct answer)
3. $6x - 12$
4. $x - 12$

Explanation: The correct answer is B (5x − 12). Distribute the 3 across the parentheses: 3(x − 4) = 3x − 12. Then combine like terms with 2x: 3x − 12 + 2x = 5x − 12. A (5x − 4) results from distributing 3 to x but not to −4, keeping −4 instead of computing 3 × (−4) = −12. C (6x − 12) comes from incorrectly treating the 2x as adding to the coefficient 3 rather than to 3x. D (x − 12) results from subtracting 2x rather than adding it. Distribution errors are extremely common — always multiply the outside factor by every term inside the parentheses.

Question 2

1. $7x^3y^6$
2. $12x^3y^8$
3. $12x^4y^6$ (correct answer)
4. $7x^4y^6$

Explanation: The correct answer is C (12x⁴y⁶). Multiply the coefficients: 3 × 4 = 12. Apply the product rule for exponents (add exponents of like bases): x³ × x¹ = x^(3+1) = x⁴; y² × y⁴ = y^(2+4) = y⁶. Result: 12x⁴y⁶. A (7x³y⁶) adds coefficients (3+4=7) instead of multiplying and doesn't add the x exponent. B (12x³y⁸) correctly multiplies coefficients but treats x³ as fixed and incorrectly adds the y exponents (possibly counting y⁴ twice). D (7x⁴y⁶) adds coefficients but correctly adds the exponents. Key rule: multiply coefficients, add exponents of like bases.

Question 3

1. $9x - 2$
2. $9x - 8$
3. $9x + 2$ (correct answer)
4. $5x + 2$

Explanation: Simplifying means combining like terms, and each term travels with the sign in front of it. The $x$ terms give $7x + 2x = 9x$, and the constants give $-3 + 5 = 2$, so the expression simplifies to $9x + 2$. The answer $9x - 2$ subtracts the constants as $3 - 5$ instead of adding $-3$ and $5$; $9x - 8$ treats the $+5$ as $-5$ and computes $-3 - 5$; and $5x + 2$ subtracts the variable terms as $7x - 2x$ even though the $2x$ is being added. Before combining, rewrite the expression as a sum such as $7x + (-3) + 2x + 5$ so every sign is attached to its own term and cannot drift.

Question 4

1. $10$
2. $20x$ (correct answer)
3. $10x$
4. $20$

Explanation: Order of operations says to simplify inside the parentheses before multiplying, and the variable outside stays attached through the multiplication. Inside, $2 + 3 = 5$, so the expression becomes $4x \cdot 5 = 20x$; distributing gives the same result, since $4x(2) + 4x(3) = 8x + 12x = 20x$. The value $20$ carries out the multiplication correctly but drops the variable, which cannot disappear because nothing cancels it. Both $10x$ and $10$ come from multiplying the 5 by the 2 inside the parentheses instead of by the 4 outside, and $10$ additionally loses the $x$. Whenever a coefficient with a variable multiplies a numerical sum, collapse the sum first and keep the variable on the product.

Question 5

1. $6x+15$
2. $21x$ (correct answer)
3. $7x$
4. $21$

Explanation: Simplify inside the parentheses first, then multiply, keeping the variable attached to the product. Since $2 + 5 = 7$, the expression becomes $3x \cdot 7 = 21x$, and distributing agrees, because $3x(2) + 3x(5) = 6x + 15x = 21x$. The value $21$ does the multiplication correctly but loses the $x$, which nothing in the problem cancels. The expression $6x+15$ distributes but drops the variable from the second product, since $3x(5)$ is $15x$, not $15$, and $7x$ comes from adding inside the parentheses and then forgetting to multiply by the 3. When a term like $3x$ multiplies a numerical sum, every piece of the product keeps the variable, so a variable-free term signals a lost factor.

Question 6

1. $10m - 2$
2. $10m + 2$
3. $10m - 8$ (correct answer)
4. $10m + 10$

Explanation: To simplify this expression, we need to apply the distributive property to both terms. First, distribute: 6(m - 2) = 6m - 12 and 4(m + 1) = 4m + 4. The expression becomes 6m - 12 + 4m + 4. Combining like terms: (6m + 4m) + (-12 + 4) = 10m - 8. Choice B incorrectly has +2 instead of -8 as the constant term.

Question 7

1. $7x - 6x + 12$
2. $x + 12$ (correct answer)
3. $13x - 12$
4. $x - 12$

Explanation: To simplify this expression, we need to apply the distributive property. First, distribute the -3 to both terms in the parentheses: -3(2x - 4) = -6x + 12. The expression becomes 7x - 6x + 12. Combining like terms: (7x - 6x) + 12 = x + 12. Choice C incorrectly adds the coefficients of x terms instead of subtracting.

Question 8

1. $6x^6$
2. $6x^5$
3. $5x^2$
4. $6x^3$ (correct answer)

Explanation: Multiply the coefficients and add the exponents when multiplying powers with the same base. 2x² × 3x = (2 × 3)(x² × x) = 6x³. The exponents add: x² × x¹ = x²⁺¹ = x³. Choice A incorrectly multiplies the exponents instead of adding them.

Question 9

1. $2a + 2b$
2. $2a + 5b$
3. $6a + 14b$
4. $2a + 14b$ (correct answer)

Explanation: This is a distribution and simplification question testing the distributive property with negatives. Choice D (2a + 14b) is correct — distribute the 4: 4a + 8b. Distribute the −2 across (a − 3b): −2a + 6b. Note: −2 × (−3b) = +6b, not −6b. Combine like terms: (4a − 2a) + (8b + 6b) = 2a + 14b. Choice A (2a + 2b) results from treating the second distribution as −2(a − 3b) = −2a − 6b (wrong sign on 3b), giving 8b − 6b = 2b. Choice B (2a + 5b) is an arithmetic error in combining the b terms, possibly computing 8b − 3b. Choice C (6a + 14b) adds 4a + 2a = 6a instead of subtracting, getting the sign wrong on the a-coefficient of the second term. Pro tip: When distributing a negative number, BOTH terms inside the parentheses change sign. Write out −2(a − 3b) = −2a + 6b before combining anything.

Question 10

1. $(x + 5)(x + 1)$
2. $(x - 1)(x + 5)$
3. $(x - 5)(x - 1)$
4. $(x - 5)(x + 1)$ (correct answer)

Explanation: To factor this quadratic expression, find two numbers that multiply to -5 (the constant term) and add to -4 (the coefficient of x). The numbers -5 and 1 satisfy both conditions: (-5) × 1 = -5 and (-5) + 1 = -4. Therefore, x² - 4x - 5 = (x - 5)(x + 1). You can verify by expanding: (x - 5)(x + 1) = x² + x - 5x - 5 = x² - 4x - 5.

Question 11

1. $x^2 - 4x$ (correct answer)
2. $x^2 - 10x$
3. $x^2 + 4x$
4. $x^2 - 7$

Explanation: First distribute x through the parentheses: x(x - 7) = x² - 7x. Then add 3x to get: x² - 7x + 3x = x² - 4x. Combine like terms: -7x + 3x = -4x, while the x² term remains unchanged. Choice B incorrectly shows -10x instead of -4x.

Question 12

1. $x^2 - x - 6$ (correct answer)
2. $x^2 + 5x - 6$
3. $x^2 - 6$
4. $x^2 - x + 6$

Explanation: To expand this expression, we need to apply the FOIL method. First: x × x = x². Outer: x × 2 = 2x. Inner: (-3) × x = -3x. Last: (-3) × 2 = -6. Combining all terms: x² + 2x - 3x - 6 = x² - x - 6. Choice B incorrectly has +5x instead of -x as the middle term.

Question 13

1. $6x + 4x$
2. $14x$ (correct answer)
3. $7x$
4. $14$

Explanation: To simplify this expression, we need to first simplify inside the parentheses, then apply the distributive property. Inside the parentheses: 3 + 4 = 7. The expression becomes 2x(7) = 14x. We can also think of this as distributing: 2x(3 + 4) = 2x(3) + 2x(4) = 6x + 8x = 14x. Choice A incorrectly leaves the expression in distributed form without combining like terms.

Question 14

1. $(x + 2)(x + 3)$ (correct answer)
2. $(x + 1)(x + 6)$
3. $(x + 3)(x - 2)$
4. $(x - 1)(x - 6)$

Explanation: To factor this quadratic expression, we need to find two numbers that multiply to $6$ and add to $5$. The numbers $2$ and $3$ satisfy these conditions: $2 \times 3 = 6$ and $2 + 3 = 5$. Therefore, $x^2 + 5x + 6 = (x + 2)(x + 3)$. We can verify by expanding: $(x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$. Choice B incorrectly uses factors $1$ and $6$, which would give $x^2 + 7x + 6$.

Question 15

1. $4y + 12$
2. $2y + 12$ (correct answer)
3. $4y + 6$
4. $2y + 6$

Explanation: First, apply the distributive property to expand 4(y + 3), which gives 4y + 12. Then combine this with -2y to get 4y + 12 - 2y. Combine like terms by grouping the y terms: (4y - 2y) + 12 = 2y + 12. The simplified form is 2y + 12.

Question 16

1. $2z + 7$ (correct answer)
2. $2z - 3$
3. $4z + 7$
4. $4z + 3$

Explanation: Remove parentheses and combine like terms. Group the z terms: 3z - z = 2z. Group the constants: 2 + 5 = 7. Therefore, (3z + 2) + (5 - z) = 3z + 2 + 5 - z = 2z + 7. Choice C incorrectly shows 4z instead of 2z for the coefficient.

Question 17

1. $x^2(1 + 6)$
2. $(x + 3)^2$
3. $x(x + 6)$ (correct answer)
4. $6(x + 1)$

Explanation: Factor out the greatest common factor from both terms. Both x² and 6x contain the factor x: x² = x(x) and 6x = x(6). Therefore, x² + 6x = x(x + 6). Choice B represents a perfect square trinomial, which this expression is not.

Question 18

1. $12(k - 8)$
2. $2(6k - 8)$
3. $4(3k + 2)$
4. $4(3k - 2)$ (correct answer)

Explanation: Find the greatest common factor of 12k and 8. Both terms share a factor of 4: 12k = 4(3k) and 8 = 4(2). Therefore, 12k - 8 = 4(3k - 2). Choice B factors out 2 instead of 4, while choice C has the wrong sign inside the parentheses.

Question 19

1. $x + 12$ (correct answer)
2. $5x + 12$
3. $x + 4$
4. $x + 8$

Explanation: Apply the distributive property first: 3(x + 4) = 3x + 12. Then subtract 2x from this result: 3x + 12 - 2x = x + 12. Combine like terms: 3x - 2x = x, and the constant remains 12. Choice B incorrectly keeps the 3x term without subtracting 2x.

Question 20

1. $L = 2S - na$
2. $L = \frac{2S - n}{a}$
3. $L = \frac{S}{2n} - a$
4. $L = \frac{2S}{n} - a$ (correct answer)

Explanation: Solving literal equations (equations with multiple variables) requires the same algebraic steps as solving for x, just with more symbols. Starting with $S = \frac{n}{2}(a + L)$, first clear the fraction by multiplying both sides by 2: $2S = n(a + L)$. Divide by n: $\frac{2S}{n} = a + L$. Finally, subtract a: $L = \frac{2S}{n} - a$. Strategy: Work systematically and keep track of which variable you're solving for—treat all other variables as if they were numbers.