Systems with Three Variables

By the end of this lesson, you will be able to solve systems of three linear equations in three variables using elimination, identify consistent/inconsistent/dependent systems, and apply three-variable systems to real-world problems.

The Big Idea

A system of three equations in three unknowns represents three planes in 3D space. The solution is the point (or line or plane) where all three intersect. Algebraically, we use the same elimination approach as two-variable systems — but we eliminate one variable at a time to reduce to a two-variable system, then solve.

The Strategy

1. Label the equations (1), (2), (3)
2. Choose a variable to eliminate first
3. Use pairs of equations to eliminate that variable → two new equations in two unknowns
4. Solve the two-variable system (substitution or elimination)
5. Back-substitute to find the third variable
6. Check in all three original equations

Example 1 — Full Solution

System:

$(1); x + y + z = 6$

$(2); 2x - y + z = 3$

$(3); x + 2y - z = 2$

Step 1: Eliminate $z$ using (1) and (3):

$(1) + (3)$: $2x + 3y = 8$ → call this $(4)$

Step 2: Eliminate $z$ using (1) and (2):

$(1) - (2)$: $-x + 2y = 3$ → call this $(5)$

Step 3: Solve system $(4)$, $(5)$:

From $(5)$: $x = 2y - 3$

Substitute into $(4)$: $2(2y-3) + 3y = 8 \implies 7y = 14 \implies y = 2$

$x = 2(2) - 3 = 1$

Step 4: Back-substitute into $(1)$: $1 + 2 + z = 6 \implies z = 3$

Solution: $(1, 2, 3)$

Check in (2): $2(1) - 2 + 3 = 3$ ✓; Check in (3): $1 + 4 - 3 = 2$ ✓

Example 2 — No Solution (Inconsistent)

If the reduced two-variable system yields a contradiction (e.g., $0 = 5$), the three planes have no common intersection — no solution.

Example 3 — Infinitely Many Solutions (Dependent)

If the reduced system yields a tautology (e.g., $0 = 0$), two of the planes coincide or all three share a common line — infinitely many solutions, described parametrically.

Example 4 — Application: Polynomial Fitting