Knock it down to two equations: Three equations in x, y, z. The plan: eliminate one variable (add or subtract equations) to leave two equations in two unknowns. Solve those, then back-substitute for the third.
IB-style question — solve the system
Solve the system:
x + y + z = 6, x − y + z = 2, 2x + y − z = 1.
Step by step
- Subtract equation 2 from equation 1 to remove x and z.
- Put y = 2 into equations 1 and 3 to get two equations in x and z.
- Add those two to eliminate z.
- Back-substitute: x + z = 4 gives z.
Final answer
x = 1, y = 2, z = 3.
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Scale first if needed; GDC on Paper 2: Sometimes you must multiply an equation before two will cancel. On Paper 2, skip all that: the GDC's simultaneous-equation solver (or a matrix) gives x, y, z in seconds.
IB-style question — eliminate with scaling
Solve:
x + y + z = 4, 2x + y − z = 0, x − y + 2z = 9.
Step by step
- Subtract equation 2 from equation 1 (removes y).
- Add equations 1 and 3 (removes y again).
- Solve the 2×2: from −x + 2z = 4, x = 2z − 4; substitute.
- Back-substitute: x = 2(3) − 4 = 2, then y = 4 − x − z.
Final answer
x = 2, y = −1, z = 3.
On the GDC (Paper 2): TI-84: PlySmlt2 ▸ Simultaneous Eqn Solver ▸ 3 equations, 3 unknowns ▸ type the coefficients ▸ SOLVE. (Or enter the coefficient matrix and use rref.)