Section 5.6: Systems with Three Variables

Systems with three variables (x, y, z) can be solved using substitution or elimination. The goal is to reduce the system to two equations with two variables, then solve as usual.

Example 1

Solve the system:

\( x + y + z = 6 \)

\( 2x - y + z = 3 \)

\( x + 2y - z = 4 \)

Step 1: Use first equation to express \( z = 6 - x - y \)

Step 2: Substitute into other equations:

Equation 2: \( 2x - y + (6 - x - y) = 3 \) → \( x - 2y + 6 = 3 \) → \( x - 2y = -3 \)

Equation 3: \( x + 2y - (6 - x - y) = 4 \) → \( x + 2y - 6 + x + y = 4 \) → \( 2x + 3y - 6 = 4 \) → \( 2x + 3y = 10 \)

Step 3: Solve the 2×2 system: \( x - 2y = -3 \), \( 2x + 3y = 10 \)

Multiply first by 2 → \( 2x - 4y = -6 \). Subtract from second → \( (2x + 3y) - (2x -4y) = 10 - (-6) \) → \( 7y = 16 \) → \( y = \frac{16}{7} \)

Step 4: \( x = -3 + 2y = -3 + \frac{32}{7} = \frac{11}{7} \)

Step 5: \( z = 6 - x - y = 6 - \frac{11}{7} - \frac{16}{7} = \frac{15}{7} \)

Solution: \( (x,y,z) = \left(\frac{11}{7}, \frac{16}{7}, \frac{15}{7}\right) \)

Practice Problems

  1. Solve: \( x + y + z = 9 \), \( 2x - y + z = 4 \), \( x + 2y - z = 5 \)
  2. Solve: \( x + 2y + z = 10 \), \( 3x - y + 2z = 7 \), \( 2x + y - z = 3 \)
  3. Solve: \( 2x + y - z = 1 \), \( x - y + 2z = 4 \), \( 3x + 2y + z = 9 \)
  4. Solve: \( x + y + z = 7 \), \( x - y + 2z = 8 \), \( 2x + y - z = 5 \)
  5. Solve: \( 3x + y + z = 12 \), \( x + 2y - z = 4 \), \( 2x - y + 3z = 10 \)
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