College Algebra / Unit 9 Alg 498 worked example, 60 to 90 seconds
Solving a Three-Variable System
The idea
Solving a three-variable system means pairing equations to eliminate the same variable twice, solving the resulting two-variable system, and back-substituting up the ladder — with all three originals as the final check.
In practice
Solve x + y + z = 6, x − y + z = 2, x + y − z = 0. Subtract the second from the first: 2y = 4, so y = 2 — one variable already. Subtract the third from the first: 2z = 6, so z = 3. Back-substitute both into the first: x + 2 + 3 = 6, so x = 1. The triple (1, 2, 3) now faces all three originals: 1 + 2 + 3 = 6, 1 − 2 + 3 = 2, 1 + 2 − 3 = 0 — three receipts. Most systems need a full elimination pair before anything falls out this fast; the ladder is the same either way, down to two, then one, then back up.
Where people slip
Eliminate the same variable in both pairings; mismatched eliminations rebuild a three-variable mess.
Your place on the trail
Every lesson rests on earlier ones and opens later ones. Tap any to jump.
Builds on
- Nothing. This is a starting point.
Unlocks
- Nothing yet depends on this.
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