Systems of equations are one of the most versatile topics in Algebra 1. They model everything from mixing two solutions to comparing pricing plans — and they appear on nearly every major algebra test from STAAR to CAASPP EOC. The challenge is that students have two reliable methods to choose from — substitution and elimination — and knowing when to use which one is as important as knowing how to use either.
This post gives you 15 practice problems with complete, step-by-step solutions. The first five use substitution, the next five use elimination, and the final five are word problems where you choose the method.
When to Use Substitution vs. Elimination
Use substitution when: one equation already has a variable isolated (like y = 2x + 3), or when isolating a variable requires only one easy step.
Use elimination when: both equations are in standard form (Ax + By = C) and the coefficients of one variable can be made equal (or opposite) with simple multiplication.
Both methods always work. Choosing the easier one for a given system just saves time.
Part 1: Substitution Method (5 Problems)
Problem 1:
y = 2x − 3
3x + y = 12
Solution: Substitute y = 2x − 3 into the second equation:
3x + (2x − 3) = 12 → 5x − 3 = 12 → 5x = 15 → x = 3.
Back-substitute: y = 2(3) − 3 = 3. Solution: (3, 3).
Check: 3(3) + 3 = 12 ✓
Problem 2:
x = 4y + 1
2x − 3y = 8
Solution: Substitute x = 4y + 1:
2(4y + 1) − 3y = 8 → 8y + 2 − 3y = 8 → 5y = 6 → y = 6/5.
x = 4(6/5) + 1 = 24/5 + 5/5 = 29/5. Solution: (29/5, 6/5).
Problem 3:
y = −x + 5
y = 3x − 7
Solution: Set the two right sides equal:
−x + 5 = 3x − 7 → 12 = 4x → x = 3.
y = −3 + 5 = 2. Solution: (3, 2).
Problem 4:
2x + y = 10
y = x − 1
Solution: Substitute y = x − 1:
2x + (x − 1) = 10 → 3x − 1 = 10 → 3x = 11 → x = 11/3.
y = 11/3 − 1 = 8/3. Solution: (11/3, 8/3).
Problem 5:
x + 2y = 6
x = 10 − 4y
Solution: Substitute x = 10 − 4y:
(10 − 4y) + 2y = 6 → 10 − 2y = 6 → 2y = 4 → y = 2.
x = 10 − 4(2) = 2. Solution: (2, 2).
Part 2: Elimination Method (5 Problems)
Problem 6:
2x + 3y = 12
2x − y = 4
Solution: Subtract the second equation from the first to eliminate 2x:
(2x + 3y) − (2x − y) = 12 − 4 → 4y = 8 → y = 2.
Back-substitute: 2x + 3(2) = 12 → 2x = 6 → x = 3. Solution: (3, 2).
Problem 7:
3x + 2y = 14
−3x + 5y = 21
Solution: Add the equations to eliminate 3x:
7y = 35 → y = 5.
3x + 2(5) = 14 → 3x = 4 → x = 4/3. Solution: (4/3, 5).
Problem 8:
4x + y = 9
2x + 3y = 7
Solution: Multiply the second equation by 2: 4x + 6y = 14. Subtract the first:
(4x + 6y) − (4x + y) = 14 − 9 → 5y = 5 → y = 1.
4x + 1 = 9 → 4x = 8 → x = 2. Solution: (2, 1).
Problem 9:
5x − 2y = 3
3x + 4y = 17
Solution: Multiply the first equation by 2: 10x − 4y = 6. Add to second:
13x = 23 → x = 23/13.
5(23/13) − 2y = 3 → 115/13 − 2y = 3 → 2y = 115/13 − 39/13 = 76/13 → y = 38/13. Solution: (23/13, 38/13).
Problem 10:
2x + 5y = 20
−4x − 5y = −30
Solution: Add the equations:
−2x = −10 → x = 5.
2(5) + 5y = 20 → 5y = 10 → y = 2. Solution: (5, 2).
Part 3: Word Problems — Choose Your Method (5 Problems)
Problem 11: Two numbers have a sum of 38 and a difference of 12. Find both numbers.
Solution: Let x + y = 38 and x − y = 12. Add: 2x = 50 → x = 25. Then y = 38 − 25 = 13. Numbers: 25 and 13.
Problem 12: A movie theater sells adult tickets for $12 and child tickets for $7. A group of 8 people paid $76 total. How many adults and how many children were in the group?
Solution: Let a = adults, c = children. a + c = 8 and 12a + 7c = 76. From first: a = 8 − c. Substitute: 12(8 − c) + 7c = 76 → 96 − 12c + 7c = 76 → −5c = −20 → c = 4. Then a = 4. 4 adults and 4 children.
Problem 13: Company A charges $30 per hour plus a $50 setup fee. Company B charges $45 per hour with no setup fee. After how many hours will the total cost be the same?
Solution: 30h + 50 = 45h → 50 = 15h → h = 10/3 ≈ 3.3 hours. (After about 3 hours 20 minutes, costs are equal.)
Problem 14: A piggy bank contains $3.85 in dimes and nickels. There are 50 coins in total. How many of each coin are there?
Solution: Let d = dimes, n = nickels. d + n = 50 and 0.10d + 0.05n = 3.85 (multiply by 100: 10d + 5n = 385). From first: n = 50 − d. Substitute: 10d + 5(50 − d) = 385 → 5d + 250 = 385 → 5d = 135 → d = 27. Then n = 23. 27 dimes and 23 nickels.
Problem 15: Two cyclists start 120 miles apart and ride toward each other. Cyclist A rides at 18 mph and Cyclist B rides at 22 mph. How many hours until they meet?
Solution: Combined closing speed = 18 + 22 = 40 mph. Time = 120 ÷ 40 = 3 hours.
Special Cases: No Solution and Infinitely Many Solutions
Not every system has one solution. Remember:
- No solution: The equations represent parallel lines. Mathematically, you end up with a contradiction like 0 = 5. In context, the two conditions can never be met simultaneously.
- Infinitely many solutions: The equations represent the same line. Mathematically, you end up with a tautology like 0 = 0. In context, every point on the line is a solution.
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