Systems by Substitution
A free Algebra I lesson from the “Systems of Linear Equations and Inequalities” unit, with a worked example and practice problems including step-by-step solutions.
Substitution works best when one equation already tells you what a variable equals. Replace that variable in the other equation, solve, then substitute back. In Systems of Linear Equations and Inequalities, students need more than a memorized rule: they need to recognize the structure, select a method, carry out the algebra cleanly, and interpret the answer in a graph, table, equation, or real context. The expanded practice now mixes skill fluency, transfer questions, and cumulative review so the lesson builds durable Algebra I readiness.
What you'll learn
- Substitute an isolated expression
- Solve for one variable first
- Use the first solution to find the second variable
Worked example
Problem. Solve y = 2x + 1 and x + y = 10.
- Substitute 2x + 1 for y in x + y = 10.
- x + 2x + 1 = 10, so 3x = 9 and x = 3.
- y = 2(3) + 1 = 7, so the solution is (3, 7).
- Connect the result back to Systems by Substitution so the method and meaning are both clear.
Answer: (3, 7)
Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.
Practice problems
1. Solve y = x + 2 and x + y = 8. Enter x.
Show solution
- Substitute x + 2 for y.
- x + x + 2 = 8, so x = 3.
Answer: 3
2. For y = x + 2 and x + y = 8, enter y.
Show solution
- Use x = 3.
- y = 3 + 2 = 5.
Answer: 5
3. Which equation is easiest to substitute into 3x + y = 11?
Choices: y = 2x - 1 · 3x + 2y = 14 · x + y = 8 · 4x - y = 2
Show solution
- y is already isolated.
Answer: y = 2x - 1
4. Solve y = 4 and 2x + y = 14. Enter x.
Show solution
- Substitute y = 4.
- 2x + 4 = 14, so x = 5.
Answer: 5
5. Solve x = y + 1 and x + y = 9. Enter y.
Show solution
- Substitute y + 1 for x.
- y + 1 + y = 9, so 2y = 8 and y = 4.
Answer: 4
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