Systems of Three Linear Equations
A free College Algebra lesson from the “Systems, Matrices, and Linear Programming” unit, with a worked example and practice problems including step-by-step solutions.
A 3x3 linear system has three variables (typically x, y, z) and three equations. Solve by eliminating one variable from two pairs of equations to reduce to a 2x2 system, then solve that and back-substitute. If the elimination process produces a contradiction (like 0 = 5), the system has no solution. If it produces 0 = 0, there are infinitely many solutions.
What you'll learn
- Solve a 3x3 system of linear equations by elimination
- Recognize inconsistent (no solution) and dependent (infinitely many) cases
- Verify a solution by substituting back into all three equations
Worked example
Problem. Solve x + y + z = 6; 2x - y + z = 3; x + 2y - z = 5. Enter z.
- Add eq1 and eq2: 3x + 2z = 9. Add eq1 and eq3: 2x + 3y + 0z = 11 (z cancels).
- From eq1: y = 6 - x - z. Substitute and solve. After arithmetic: x = 1, y = 2, z = 3.
- Check eq2: 2(1) - 2 + 3 = 3. ✓
Answer: 3
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. If a 3x3 linear system reduces to 0 = 5, how many solutions are there?
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- Contradiction means no solution.
Answer: 0
2. If a 3x3 system reduces to 0 = 0, how many solutions are there?
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- A true identity means dependent system.
Answer: infinitely many
3. How many equations are needed for a unique solution in 3 unknowns?
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- Need at least as many independent equations as unknowns.
Answer: 3
4. Solve x + y + z = 6; x - y + z = 2; x + y - z = 0. Enter z.
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- Add eq1 + eq3: 2x + 2y = 6 -> x + y = 3.
- Subtract eq2 from eq1: 2y = 4 -> y = 2.
- Then x = 1 and z = 6 - 1 - 2 = 3.
Answer: 3
5. Same system. Enter x.
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- From the work above: x = 1.
Answer: 1
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