ACT Algebra Practice: Equations, Inequalities, and Word Problems

Algebra is the backbone of ACT math. This lesson focuses on the equation-solving skills that appear most often: linear equations, systems of equations, inequalities, and word problems. Every question below is an original PrepSolution item written in the enhanced four-choice format, graduated from easy to hard. Each answer was verified with two independent methods, and every wrong choice is explained so you can see the exact trap it tests. For the bigger math picture — topics, pacing, and a full worked set — start with our enhanced ACT math walkthrough. For formulas, see the ACT math formulas guide.

Linear equations

A linear equation in one variable has one solution unless the variable cancels to a true or false statement. The ACT usually tests distribution, combining like terms, and careful isolation. Work in stages: expand parentheses, collect variable terms on one side and constants on the other, then divide.

Systems of equations

A system of two linear equations asks for the pair (x, y) that satisfies both equations at once. Substitution works well when one variable is already isolated or has a coefficient of 1. Elimination works well when the coefficients line up to cancel. Both methods should give the same answer; using both on the same system is one of the fastest ways to catch an arithmetic error.

Inequalities

Inequalities are solved like equations, with one extra rule: when you multiply or divide both sides by a negative number, reverse the inequality sign. Adding or subtracting any number, positive or negative, never flips the sign. A common ACT trap asks you to divide by −2 or −3 late in the problem, right when you are most likely to forget the flip.

Word problems into equations

The hardest part of a word problem is usually the setup, not the solving. Read the question sentence once to identify what you are solving for, then reread to extract numbers and relationships. Translate phrases directly: "is" becomes =, "of" often becomes multiplication, "more than" becomes addition, "less than" becomes subtraction, and "per" or "out of" signals a ratio or rate.

Graduated practice set

The ten questions below are grouped by difficulty. Each includes the question, a primary solution, an independent check, a diagnosis of every wrong choice, and a note on the trap the question tests.

Easy

Question 1: Multi-step linear equation

If 2(x − 3) + 5 = 3x − 2, what is the value of x?
A) −1
B) 1
C) 3
D) 7

Solution: Distribute the 2 to get 2x − 6 + 5 = 3x − 2, which simplifies to 2x − 1 = 3x − 2. Subtract 2x from both sides: −1 = x − 2. Add 2 to both sides: x = 1.

Check another way: Substitute x = 1 into the original equation. Left side: 2(1 − 3) + 5 = 2(−2) + 5 = −4 + 5 = 1. Right side: 3(1) − 2 = 1. Both sides match.

• A) −1 fails from subtracting 2 instead of adding 2 in the final isolation step.
• C) 3 fails from treating 2(x − 3) as 2x − 3 and then combining constants incorrectly.
• D) 7 fails from adding the constants −6 and 5 incorrectly as −11, then misisolating the variable.

Trap this question tests: distribution sign errors and combining constants after the parentheses are cleared.

Question 2: Word problem — fixed fee plus rate

A plumber charges a $50 service fee plus $40 per hour. If a repair bill is $210, how many hours did the repair take?
A) 3
B) 4
C) 5.25
D) 6.5

Solution: Let h be the number of hours. The total bill is 50 + 40h = 210. Subtract 50: 40h = 160. Divide by 40: h = 4.

Check another way: Work backward from 4 hours. The service fee is $50 and 4 hours at $40 per hour is $160, for a total of $210.

• A) 3 fails from dividing the total bill by $70 (service fee plus one hour) instead of setting up the equation.
• C) 5.25 fails from dividing $210 by $40 and ignoring the service fee.
• D) 6.5 fails from adding the service fee to the total before dividing by the hourly rate: (210 + 50) ÷ 40.

Trap this question tests: whether you subtract the fixed fee before dividing by the rate.

Medium

Question 3: System of equations — substitution and elimination

A theater sells adult tickets for $12 and student tickets for $8. If 100 tickets are sold for a total of $1,040, how many adult tickets were sold?
A) 40
B) 60
C) 80
D) 90

Solution by substitution: Let a be adult tickets and s be student tickets. Then a + s = 100 and 12a + 8s = 1,040. From the first equation, s = 100 − a. Substitute into the second: 12a + 8(100 − a) = 1,040. That gives 12a + 800 − 8a = 1,040, so 4a = 240 and a = 60.

Solution by elimination: Multiply a + s = 100 by 8 to get 8a + 8s = 800. Subtract this from 12a + 8s = 1,040: 4a = 240, so a = 60.

Check another way: If 60 adult tickets and 40 student tickets were sold, then 60 + 40 = 100 tickets and 60 × $12 + 40 × $8 = $720 + $320 = $1,040.

• A) 40 fails because 40 is the number of student tickets, not adult tickets.
• C) 80 fails from averaging the two ticket counts without using the prices.
• D) 90 fails from swapping the two prices: 90 × $8 + 10 × $12 = $840, not $1,040.

Trap this question tests: solving for the requested variable rather than its complement, and keeping the ticket prices paired with the correct groups.

Question 4: Inequality with sign flip

If −3x + 7 ≥ 16, which of the following describes all possible values of x?
A) x ≤ −3
B) x ≥ −3
C) x ≤ 3
D) x ≥ 3

Solution: Subtract 7 from both sides: −3x ≥ 9. Divide by −3 and reverse the inequality: x ≤ −3.

Check another way: Test x = −4 in the original inequality. −3(−4) + 7 = 12 + 7 = 19, and 19 ≥ 16 is true. Test x = 0: −3(0) + 7 = 7, and 7 ≥ 16 is false. Only values less than or equal to −3 work.

• B) x ≥ −3 fails from forgetting to reverse the inequality when dividing by −3.
• C) x ≤ 3 fails from dividing 9 by 3 without the negative sign and without flipping.
• D) x ≥ 3 fails from dividing 9 by 3 without the negative sign and without flipping.

Trap this question tests: reversing the inequality symbol when dividing by a negative number.

Question 5: Word problem — percent markup

A store sells a jacket for $84 after a 40% markup from the wholesale price. What was the wholesale price?
A) $50.40
B) $56.00
C) $60.00
D) $117.60

Solution: A 40% markup means the selling price is 140% of the wholesale price. Let w be the wholesale price. Then 1.40w = 84, so w = 84 ÷ 1.40 = 60.

Check another way: 40% of $60 is $24. Adding the markup to the wholesale price gives $60 + $24 = $84, which matches the selling price.

• A) $50.40 fails from taking 40% of the selling price instead of the wholesale price.
• B) $56.00 fails from dividing the selling price by 1.5 instead of 1.4.
• D) $117.60 fails from adding 40% of the selling price to the selling price.

Trap this question tests: recognizing that the percent is taken from the original (wholesale) price, not the final price.

Question 6: Quadratic solved by factoring

If x² − 7x = −12, what are the values of x?
A) 3 and 4
B) −3 and −4
C) 2 and 6
D) −2 and 6

Solution: Add 12 to both sides to set the equation equal to zero: x² − 7x + 12 = 0. Factor: (x − 3)(x − 4) = 0. So x = 3 or x = 4.

Check another way: Use the sum and product of roots. For x² − 7x + 12 = 0, the sum of the roots is 7 and the product is 12. The pair 3 and 4 satisfies both conditions.

• B) −3 and −4 fails from factoring without moving −12 to the left, producing x(x − 7) = −12 and guessing signs.
• C) 2 and 6 fails from finding numbers that add to 8 instead of checking the middle term −7x.
• D) −2 and 6 fails from signs that do not multiply to +12 or add to −7.

Trap this question tests: setting a quadratic equal to zero before factoring, and matching both the sum and product of the roots.

Hard

Question 7: Word problem — mixture

A chemist has 6 liters of a 40% acid solution. How many liters of a 10% acid solution must be added to produce a 25% acid solution?
A) 3
B) 4
C) 5
D) 6

Solution: Let x be the liters of 10% solution added. The amount of pure acid from the 40% solution is 0.40(6) = 2.4 liters. The amount from the 10% solution is 0.10x. The total volume after mixing is 6 + x liters, and it should be 25% acid. So 2.4 + 0.10x = 0.25(6 + x). Expand: 2.4 + 0.10x = 1.5 + 0.25x. Subtract 0.10x and 1.5: 0.9 = 0.15x. Divide: x = 6.

Check another way: Adding 6 liters of 10% solution to 6 liters of 40% solution gives 12 liters total. Total acid is 2.4 + 0.6 = 3.0 liters, and 3.0 ÷ 12 = 0.25 = 25%.

• A) 3 fails from averaging the two percentages (40 + 10) ÷ 2 without tracking the actual acid amounts.
• B) 4 fails from setting up the equation 0.40(6) + 0.10x = 0.25(6) and ignoring the added volume x.
• C) 5 fails from a decimal-shift error, such as using 0.025 instead of 0.25 for the final concentration.

Trap this question tests: keeping track of total volume after mixing, not just the amount of pure substance.

Question 8: Linear equation with fractions

If (2x + 1)/3 = (x − 4)/2 + 2, what is the value of x?
A) −2
B) 2
C) 7
D) 14

Solution: Multiply every term by 6, the least common denominator, to clear fractions: 6 · (2x + 1)/3 = 6 · (x − 4)/2 + 6 · 2. This gives 2(2x + 1) = 3(x − 4) + 12. Expand: 4x + 2 = 3x − 12 + 12. Simplify: 4x + 2 = 3x. Subtract 3x: x + 2 = 0, so x = −2.

Check another way: Substitute x = −2. Left side: (2(−2) + 1)/3 = (−4 + 1)/3 = −3/3 = −1. Right side: (−2 − 4)/2 + 2 = −6/2 + 2 = −3 + 2 = −1. Both sides match.

• B) 2 fails from distributing 3 to x − 4 as 3x − 4 instead of 3x − 12.
• C) 7 fails from adding 2 to both sides instead of subtracting 3x when isolating the variable.
• D) 14 fails from multiplying only the left side by 6 and ignoring the right side.

Trap this question tests: clearing fractions by multiplying every term by the LCD, then distributing negatives and constants correctly.

Question 9: Inequality with distribution and sign flip

If 5 − 2(3x + 4) > 7x − 1, which of the following describes all possible values of x?
A) x < −2/13
B) x > −2/13
C) x < −2
D) x > −2

Solution: Distribute the −2: 5 − 6x − 8 > 7x − 1. Combine constants: −6x − 3 > 7x − 1. Add 6x to both sides: −3 > 13x − 1. Add 1: −2 > 13x. Divide by 13: −2/13 > x, which is the same as x < −2/13.

Check another way: Test x = −1, which is less than −2/13. Left side: 5 − 2(3(−1) + 4) = 5 − 2(1) = 3. Right side: 7(−1) − 1 = −8. Since 3 > −8, x = −1 satisfies the inequality. Test x = 0: left side is 5 − 8 = −3, right side is −1, and −3 > −1 is false.

• B) x > −2/13 fails from treating the final division by 13 as if it required flipping the sign.
• C) x < −2 fails from combining 5 and −8 as −3, then adding 3 to both sides instead of subtracting 7x.
• D) x > −2 fails from both a sign flip at the wrong step and dropping the fraction.

Trap this question tests: distributing a negative through parentheses and keeping the inequality direction straight across multiple moves.

Question 10: Word problem — rate and time

A car travels 300 miles. If the car had traveled 10 miles per hour faster, the trip would have taken 1 hour less. What was the car's actual speed, in miles per hour?
A) 40
B) 50
C) 60
D) 80

Solution: Let r be the actual speed. The actual time is 300/r hours, and the faster time is 300/(r + 10) hours. The difference is 1 hour: 300/r − 300/(r + 10) = 1. Multiply by r(r + 10): 300(r + 10) − 300r = r(r + 10). Simplify: 3,000 = r² + 10r. Rearrange: r² + 10r − 3,000 = 0. Factor: (r + 60)(r − 50) = 0. Since speed is positive, r = 50.

Check another way: At 50 mph, the 300-mile trip takes 6 hours. At 60 mph, it takes 5 hours. The difference is exactly 1 hour.

• A) 40 fails because 300/40 = 7.5 hours and 300/50 = 6 hours, a difference of 1.5 hours, not 1.
• C) 60 fails from choosing the faster speed instead of the actual speed.
• D) 80 fails from dividing the distance by a guessed time without setting up the rate equation.

Trap this question tests: building a rational equation from the relationship time = distance/rate and rejecting the extraneous negative root.

How to use this set

Treat these ten questions as a diagnostic, not a score prediction. A short set cannot estimate your ACT scaled score — only a full official form with its own scoring key can do that. After you finish, classify every miss as knowledge, reading, or pacing, then drill the highest-frequency type.

When you are ready for full-section pacing, follow the plan in our ACT study plan and take timed official math sections with their matching scoring keys.