Systems of Linear Equations
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.
A system of equations asks for values that make both equations true at the same time. On a graph, the solution is the point where the lines intersect. 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
- Understand solutions as intersection points
- Solve simple systems by substitution
- Solve simple systems by elimination
Worked example
Problem. Solve x + y = 10 and y = 4.
- Substitute y = 4 into x + y = 10.
- x + 4 = 10, so x = 6.
- The solution is (6, 4).
- Connect the result back to Systems of Linear Equations so the method and meaning are both clear.
Answer: (6, 4)
Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.
Practice problems
1. The graph solution to a system is the...
Choices: Intersection point · x-axis only · steepest line · largest y-intercept
Show solution
- Both equations are true where their lines meet.
Answer: Intersection point
2. Solve x + y = 8 and y = 3. Enter x.
Show solution
- Substitute y = 3.
- x + 3 = 8, so x = 5.
Answer: 5
3. Solve x = 2 and y = x + 5. Enter y.
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- Substitute x = 2.
- y = 2 + 5 = 7.
Answer: 7
4. Solve y = x + 1 and y = 5. Enter x.
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- Set x + 1 equal to 5.
- x = 4.
Answer: 4
5. Solve x + y = 12 and x = 7. Enter y.
Show solution
- Substitute x = 7.
- 7 + y = 12, so y = 5.
Answer: 5
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