Precalculus / Unit 8: Systems and Matrices
8.1 Systems in Three Variables
Notes
A solution of a 3-variable system is an ordered triple $(x, y, z)$ satisfying all three equations.
- Use elimination: combine two equations to remove one variable, repeat with another pair, then solve the resulting 2-variable system and back-substitute.
- If you reach a false statement like $0 = 5$, there is no solution. If you reach $0 = 0$ with a free variable, there are infinitely many solutions.
- Always check the triple in all three original equations.
- Example: $x + y + z = 6$, $x - y + z = 2$, $2x + y - z = 1$. Subtracting the first two gives $2y = 4$, so $y = 2$, and then $x = 1$, $z = 3$.
Practice
Solve: $x + y + z = 6$, $x - y + z = 2$, $2x + y - z = 1$.
Subtracting the first two equations gives $y = 2$; then $x + z = 4$ and $2x - z = -1$, so $x = 1$ and $z = 3$.
Solve: $x + y = 5$, $y + z = 7$, $x + z = 6$. What is $z$?
Adding all three gives $2(x + y + z) = 18$, so $x + y + z = 9$ and $z = 9 - 5 = 4$.
While eliminating variables you reach the statement $0 = 5$. What does this mean?
A false statement means the system is inconsistent.
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