ACT Math study guide · Chapter 7
Linear Equations and Systems
Solving a linear equation means isolating the variable: apply the same operation to both sides of the equation until the variable stands alone. This chapter applies that method to equations, inequalities, and systems of two equations.
Inequalities follow the same steps as equations, with one exception. Multiplying or dividing both sides by a negative number reverses the inequality symbol. Skipping this step produces an incorrect answer choice even when every other step is correct.
A system of two linear equations identifies the point where two lines intersect. Two methods solve a system. Substitution solves one equation for a variable and substitutes that expression into the other equation. Elimination adds or subtracts the two equations to remove a variable. The more efficient method depends on the form of the given equations.
Where this appears as a model: two competing costs described in words. A repair shop charges a $15 diagnostic fee plus $2 per part, while a second shop charges $10 plus $3 per part; the two costs are equal at 5 parts, found by setting 15+2x=10+3x.
▸ Solve for the requested quantity
A question often asks for an expression built from a variable, such as 2x + 3 or the sum of two variables, instead of the variable alone. Solving the equation correctly but reporting the value of the wrong quantity produces an incorrect answer even when every calculation is correct.
Worked Example Solve 5x − 3 = 2x + 12. Then find the value of 3x − 1.
Gather the x-terms on one side and the constants on the other.
5x − 3 = 2x + 12
3x = 15
x = 5
3x − 1 = 3(5) − 1 = 14
▸ Systems: selecting the more efficient method
When one equation already isolates a variable, as in y = 2x − 1, substitution requires a single step: replace that variable in the other equation and solve. When both equations appear in standard form, ax + by = c, elimination is typically faster, because adding or subtracting the equations can remove a variable directly.
Worked Example Solve the system y = 2x − 1 and 3x + y = 14. Find the value of x.
y is already isolated, so substitute directly into the second equation.
3x + (2x − 1) = 14
5x − 1 = 14
5x = 15
x = 3
A more advanced version of this skill presents the system as a word problem: two numbers, a total, and a ratio between them. Translating the given information into two equations is the primary skill tested. Solving the system is a direct application of the methods above.
▸ Practice
1. Solve for x: 4x − 7 = 2x + 9.
A) 8 B) −8 C) 1 D) 16
2. Solve P = 2l + 2w for w.
F) w = (P − 2l) ⁄ 2 G) w = P − l H) w = (P + 2l) ⁄ 2 J) w = P/2 − 2l
3. Solve −5x + 3 > 18.
A) x > −3 B) x < 3 C) x > 3 D) x < −3
4. Solve −1 ≤ 3x − 4 < 8.
F) −1 ≤ x < 4 G) 1 < x ≤ 4 H) 1 ≤ x < 4 J) −5/3 ≤ x < 4/3
5. For what value of k does 2x + ky = 6 and 4x + 8y = 12 have infinitely many solutions?
A) k = 2 B) k = 8 C) k = 4 D) k = 6
6. If y = 2x − 1 and 3x + y = 9, what is the value of x + y?
F) 5 G) 2 H) −1 J) 3
7. Solve for x: 8x − 3 = 5x + 18.
A) 5 B) 7 C) 9 D) 6
8. Solve the inequality −3x + 10 < 19.
F) x < −3 G) x > −3 H) x < 3 J) x > 3
9. Solve the system 2x − y = 5 and x + 3y = 13. Find x + y.
A) 7 B) 5 C) 9 D) 11
10. Solve y = mx + b for b.
F) y + mx G) mx − y H) y − mx J) −y − mx
11. For what value of k does the system kx + 2y = 6 and 3x + y = 9 have no solution?
A) 3 B) 9 C) 6 D) 2
12. Solve the compound inequality −8 < 3x + 1 < 13.
F) 3 < x < 4 G) −4 < x < −3 H) −4 < x < 3 J) −3 < x < 4
13. Solve the system x − 2y = −4 and 3x + y = 9. Find x − y.
A) 1 B) −1 C) −5 D) 5
14. For what value of k does the system x + 3y = 7 and 2x + ky = 14 have infinitely many solutions?
F) 3 G) 6 H) 9 J) 12
15. Solve for x: 9x + 5 = 4x + 35.
A) 6 B) 8 C) 5 D) 7
16. Solve the inequality 4x − 9 ≤ 15.
F) x ≤ 6 G) x ≥ 6 H) x < 7 J) x > 6
17. Solve the system 3x + y = 14 and x − 2y = 0. Find x + y.
A) 4 B) 6 C) 8 D) 10
18. Solve A = (1/2)bh for h.
F) 2b/A G) 2A/b H) A/2b J) bA/2
19. For what value of k does the system 3x + ky = 9 and 6x + 4y = 18 have infinitely many solutions?
A) 6 B) 4 C) 3 D) 2
20. Solve the compound inequality −5 < 2x + 3 ≤ 11.
F) −4 ≤ x < 4 G) −4 ≤ x ≤ 4 H) −4 < x < 4 J) −4 < x ≤ 4
▸ Answers and Explanations
1: A. Gather the x-terms: 4x − 2x = 9 + 7, so 2x = 16, giving x = 8. Choice C combines the constants with the wrong sign, 9 − 7 instead of 9 + 7.
2: F. Isolate the w-term: P − 2l = 2w. Divide the entire right side by 2: w = (P − 2l)/2. Choice G divides only the 2l term by 2 and leaves P undivided.
3: D. Subtract 3 from both sides: −5x > 15. Dividing by −5 reverses the inequality: x < −3. Choice A keeps the inequality direction unchanged, the result of not flipping the sign when dividing by a negative number.
4: H. Add 4 to every part: 3 ≤ 3x < 12. Divide every part by 3: 1 ≤ x < 4. Choice G reverses which boundary is strict, using < where ≤ belongs.
5: C. Doubling the first equation gives 4x + 2ky = 12. Matching this to 4x + 8y = 12 requires 2k = 8, so k = 4. Choice B copies the coefficient 8 directly from the second equation without accounting for the factor of 2 used to scale the first equation.
6: F. Substitute y = 2x − 1 into the second equation: 3x + (2x − 1) = 9, so 5x = 10, x = 2, y = 3, and x + y = 5. Choice G reports x alone instead of the requested sum x + y.
7: B. Gather the x-terms: 8x − 5x = 18 + 3, so 3x = 21, giving x = 7. Choice A combines the constants as 18 − 3 instead of 18 + 3.
8: G. Subtract 10 from both sides: −3x < 9. Dividing by −3 reverses the inequality: x > −3. Choice F keeps the inequality direction unchanged when dividing by a negative number.
9: A. From the first equation, y = 2x − 5. Substituting gives x + 3(2x − 5) = 13, so 7x = 28, x = 4, y = 3, and x + y = 7. Choice D computes 2x + y instead of the requested sum x + y.
10: H. Subtract mx from both sides: b = y − mx. Choice F adds mx to both sides instead of subtracting it.
11: C. For no solution, the coefficient ratios must match while the constant ratios differ: k/3 = 2/1, giving k = 6. Checking the constants, 6/9 does not equal 2/1, confirming no solution. Choice A copies the x-coefficient of the second equation directly instead of solving the ratio equation.
12: J. Subtract 1 from every part: −9 < 3x < 12. Divide every part by 3: −3 < x < 4. Choice F drops the negative sign on the left boundary after dividing.
13: B. From the first equation, x = 2y − 4. Substituting gives 3(2y − 4) + y = 9, so 7y = 21, y = 3, x = 2, and x − y = −1. Choice A reports y − x instead of the requested x − y.
14: G. For infinitely many solutions, both the coefficient ratio and the constant ratio must match: 2/1 = k/3 gives k = 6, and 14/7 = 2 matches the coefficient ratio 2/1. Choice F copies the y-coefficient of the first equation directly instead of solving for k.
15: A. Gather the x-terms: 9x − 4x = 35 − 5, so 5x = 30, giving x = 6. Choice B combines the constants as 35 + 5 instead of 35 − 5.
16: F. Add 9 to both sides: 4x ≤ 24. Divide by 4: x ≤ 6. Choice G reverses the inequality direction even though both sides were divided by a positive number.
17: B. From the second equation, x = 2y. Substituting gives 3(2y) + y = 14, so 7y = 14, y = 2, x = 4, and x + y = 6. Choice A reports x alone instead of the requested sum x + y.
18: G. Multiply both sides by 2, then divide by b: h = 2A/b. Choice F inverts the position of A and b during the final division.
19: D. For infinitely many solutions, the coefficient ratio must match: 3/6 = k/4, so k = 2. The constant ratio, 9/18 = 1/2, also matches. Choice A copies the x-coefficient of the second equation directly instead of solving the ratio equation.
20: J. Subtract 3 from every part: −8 < 2x ≤ 8. Divide every part by 2: −4 < x ≤ 4. Choice F swaps which boundary is strict.