Key Idea: This topic solves three linear equations in three unknowns. On Paper 1 you eliminate by hand, on Paper 2 the GDC solves it. You also decide how many solutions there are.
Solving a system of three equations
- Eliminate — Remove the same variable from two different pairs of equations.
- Solve the 2×2 — Now two equations in two unknowns.
- Back-substitute — Find the third unknown.
- Check — Put x, y, z into the third equation.
| Last line of elimination | Number of solutions |
|---|---|
| Clean values | One unique solution |
| 0 = 0 | Infinitely many |
| 0 = a non-zero number | None |
A linear system has 0, 1 or infinitely many solutions, never exactly 2.
Equation 2 − equation 1 for x + y + z = 2 and x + 2y + 3z = 5? y + 2z = 3.
Elimination ends with 0 = 4. What does that mean? No solution.
Elimination gives y + 2z = 8 and y + 2z = k − 12. Which k works? k = 20, giving infinitely many solutions.
How do you check your answer? Substitute it into the equation you did not use last.