Linear equations are the backbone of Algebra 1. Before a student can tackle systems, functions, or quadratics, they need to be able to solve linear equations fluently and confidently — including the ones buried inside word problems. This post gives you 20 practice problems at four difficulty levels, each with a complete solution walkthrough.
Work through these in order, or jump to the level that matches where you are right now.
What Is a Linear Equation?
A linear equation is an equation where the variable appears to the first power only — no squares, no square roots, no absolute values. The solution is the value of the variable that makes the equation true. Linear equations can have one variable or two variables. In Algebra 1, most practice focuses on one-variable equations like 2x + 5 = 13, though two-variable equations like y = 3x − 4 appear frequently in graphing and functions units.
Level 1: One-Step Equations
Solve by applying one inverse operation to isolate the variable.
Problem 1: x + 7 = 15
Solution: Subtract 7 from both sides: x = 8. Check: 8 + 7 = 15 ✓
Problem 2: 4x = 28
Solution: Divide both sides by 4: x = 7. Check: 4(7) = 28 ✓
Problem 3: x − 9 = −2
Solution: Add 9 to both sides: x = 7. Check: 7 − 9 = −2 ✓
Problem 4: x/5 = −3
Solution: Multiply both sides by 5: x = −15. Check: −15/5 = −3 ✓
Problem 5: −6x = 42
Solution: Divide both sides by −6: x = −7. Check: −6(−7) = 42 ✓
Level 2: Two-Step Equations
Undo addition or subtraction first, then multiplication or division.
Problem 6: 2x + 3 = 11
Solution: Subtract 3: 2x = 8. Divide by 2: x = 4. Check: 2(4) + 3 = 11 ✓
Problem 7: 5x − 4 = 21
Solution: Add 4: 5x = 25. Divide by 5: x = 5. Check: 5(5) − 4 = 21 ✓
Problem 8: −3x + 8 = −1
Solution: Subtract 8: −3x = −9. Divide by −3: x = 3. Check: −3(3) + 8 = −1 ✓
Problem 9: x/4 + 6 = 10
Solution: Subtract 6: x/4 = 4. Multiply by 4: x = 16. Check: 16/4 + 6 = 10 ✓
Problem 10: 7 − 2x = −5
Solution: Subtract 7: −2x = −12. Divide by −2: x = 6. Check: 7 − 2(6) = −5 ✓
Level 3: Multi-Step Equations
Distribute, combine like terms, then solve.
Problem 11: 3(x + 4) = 21
Solution: Distribute: 3x + 12 = 21. Subtract 12: 3x = 9. Divide by 3: x = 3. Check: 3(3+4) = 21 ✓
Problem 12: 2x + 5 = x − 3
Solution: Subtract x: x + 5 = −3. Subtract 5: x = −8. Check: 2(−8)+5 = −11 = (−8)−3 ✓
Problem 13: 4(2x − 1) = 3x + 11
Solution: Distribute: 8x − 4 = 3x + 11. Subtract 3x: 5x − 4 = 11. Add 4: 5x = 15. Divide by 5: x = 3. Check: 4(6−1) = 20 = 9+11 ✓
Problem 14: 5x − (2x + 6) = 9
Solution: Distribute the minus: 5x − 2x − 6 = 9. Combine: 3x − 6 = 9. Add 6: 3x = 15. Divide: x = 5. Check: 5(5)−(10+6) = 25−16 = 9 ✓
Problem 15: 2(3x − 5) = 4(x + 1)
Solution: Expand both sides: 6x − 10 = 4x + 4. Subtract 4x: 2x − 10 = 4. Add 10: 2x = 14. Divide: x = 7. Check: 2(21−5) = 32 = 4(8) ✓
Level 4: Equations from Word Problems
Read carefully: define the variable, write the equation, solve, and check that the answer makes sense.
Problem 16: A number is 8 more than twice another number. If the larger number is 22, what is the smaller number?
Solution: Let x = the smaller number. Write: 2x + 8 = 22. Subtract 8: 2x = 14. Divide: x = 7. The smaller number is 7. Check: 2(7)+8 = 22 ✓
Problem 17: A plumber charges a $45 service fee plus $60 per hour. A customer’s bill was $225. How many hours did the plumber work?
Solution: Let h = hours. Write: 45 + 60h = 225. Subtract 45: 60h = 180. Divide: h = 3 hours. Check: 45 + 60(3) = 225 ✓
Problem 18: Three friends split the cost of a gift equally. Each person paid $18. Later they remembered they had a $9 coupon that should have reduced the total. What was the original price of the gift before the coupon?
Solution: Total paid = 3 × $18 = $54. Original price = 54 + 9 = $63. (No variable needed here, but you can also write: p − 9 = 54, so p = 63.) Check: 63 − 9 = 54 = 3 × 18 ✓
Problem 19: The length of a rectangle is 5 cm more than twice the width. The perimeter is 52 cm. Find the width and length.
Solution: Let w = width. Length = 2w + 5. Perimeter: 2w + 2(2w + 5) = 52. Expand: 2w + 4w + 10 = 52. Combine: 6w + 10 = 52. Subtract 10: 6w = 42. Divide: w = 7 cm. Length = 2(7)+5 = 19 cm. Check: 2(7)+2(19) = 14+38 = 52 ✓
Problem 20: Two cars start from the same point and drive in opposite directions. Car A travels at 55 mph and Car B at 65 mph. After how many hours will they be 300 miles apart?
Solution: Combined rate = 55 + 65 = 120 mph. Let t = time. Write: 120t = 300. Divide: t = 2.5 hours. Check: 55(2.5)+65(2.5) = 137.5+162.5 = 300 ✓
Key Strategies for Solving Linear Equations
- Work in reverse order of operations: undo addition/subtraction first, then multiplication/division.
- Distribute before combining like terms when you see parentheses.
- Move all variable terms to one side when the variable appears on both sides of the equation.
- Always check your answer by substituting back into the original equation.
- In word problems, define your variable clearly and re-read the problem after solving to confirm the answer makes real-world sense.
Build Your Skills with ViewMath Algebra 1 Workbooks
These 20 problems are a strong warm-up, but building real fluency requires hundreds of varied practice problems across all Algebra 1 topics. ViewMath offers complete Algebra 1 study guides, workbooks, and practice test books designed to take students from basic equations all the way through quadratics, functions, and statistics. You can browse the collection in the sidebar.
