ACT Math: Algebra (Drill 2)

Question 1. If 5y + 12 = 3y − 4, what is the value of y?

• A) −8 ✓
• B) −4
• C) 4
• D) 8

Explanation: Subtract 3y from both sides: 2y + 12 = −4. Subtract 12 from both sides: 2y = −16. Divide by 2: y = −8. Choice B results from a sign error when subtracting 12: treating −4 + 12 = 8 instead of −4 − 12 = −16, then dividing by 2 to get −4 rather than −8. Choice C results from adding 12 and 4 and dividing by 2, ignoring the sign: 16 ÷ 2 = 4 (but missing the negative). Choice D results from computing 12 + 4 = 16 and then dividing by 2 without applying any negatives.

Question 2. Which of the following is equivalent to (2x − 3)(x + 5)?

• A) 2x² + 7x − 15 ✓
• B) 2x² − 7x − 15
• C) 2x² + 10x − 15
• D) 2x² − 15

Explanation: Use FOIL: (2x)(x) + (2x)(5) + (−3)(x) + (−3)(5) = 2x² + 10x − 3x − 15 = 2x² + 7x − 15. Choice B comes from sign errors in the middle terms: computing (2x)(5) as −10x and (−3)(x) as +3x, giving −10x + 3x = −7x. Choice C omits the inner term (−3)(x) entirely, keeping only +10x. Choice D drops both middle terms, leaving only the first and last: 2x² − 15.

Question 3. What are all values of x that satisfy |2x − 6| = 10?

• A) x = 8 only
• B) x = −2 only
• C) x = 8 or x = −2 ✓
• D) x = 2 or x = −8

Explanation: An absolute value equation |A| = b splits into two cases: A = b and A = −b. Case 1: 2x − 6 = 10 → 2x = 16 → x = 8. Case 2: 2x − 6 = −10 → 2x = −4 → x = −2. Both values satisfy the original equation: |2(8) − 6| = |10| = 10 ✓ and |2(−2) − 6| = |−10| = 10 ✓. Choice A and B each give only one of the two solutions. Choice D results from solving |2x| = 10 − 6 = 4, incorrectly moving the 6 to the right side before removing the absolute value signs.

Question 4. At a school store, notebooks cost $3 each and pens cost $1.50 each. Marco buys a total of 10 items and spends exactly $21. How many notebooks did Marco buy?

• A) 4 ✓
• B) 5
• C) 6
• D) 7

Explanation: Let n = number of notebooks and p = number of pens. Set up the system: n + p = 10 and 3n + 1.5p = 21. From the first equation, p = 10 − n. Substitute: 3n + 1.5(10 − n) = 21 → 3n + 15 − 1.5n = 21 → 1.5n = 6 → n = 4. Check: 4 notebooks + 6 pens = 10 items; 4(3) + 6(1.5) = 12 + 9 = $21 ✓. Choice B results from a sign error when distributing: treating 1.5(10 − n) as 15 + 1.5n. Choice C results from dividing $21 by $3.50 (average price) without using the system. Choice D results from computing 21 ÷ 3 = 7 and ignoring the total item count.

Question 5. What is the solution to √2x − 3 = x − 3, where x is a real number?

• A) x = 2 only
• B) x = 6 only ✓
• C) x = 2 or x = 6
• D) x = 3 only

Explanation: Square both sides: 2x − 3 = (x − 3)² = x² − 6x + 9. Rearrange: 0 = x² − 8x + 12 = (x − 2)(x − 6). So x = 2 or x = 6. Always check for extraneous solutions in radical equations. Check x = 2: √2(2) − 3 = √1 = 1, but x − 3 = 2 − 3 = −1. Since 1 ≠ −1, x = 2 is extraneous. Check x = 6: √2(6) − 3 = √9 = 3, and x − 3 = 6 − 3 = 3 ✓. The only valid solution is x = 6. Choice A results from choosing the extraneous solution. Choice C results from keeping both algebraic solutions without checking for extraneous ones. Choice D results from solving √2x − 3 = 0 by setting the radicand equal to zero: 2x − 3 = 0 → x = 1.5, then rounding to 3.