To solve three linear equations in x, y and z, eliminate the same variable from two different pairs, solve the resulting two equations, then substitute back.
This lesson is part of systems of equations. If two-unknown elimination still slips, read choosing substitution or elimination first.
What is the routine?
Number the equations (1), (2), (3) and follow the same steps each time.
- Choose a variable to eliminate, for example z.
- Combine (1) with (2) to remove z, and combine (1) with (3) to remove z.
- Solve the two new equations for x and y.
- Substitute x and y into one original equation to find z.
- Check all three originals.
Worked example 1: a clean reduction
Solve: x + y + z = 6 (1); 2x − y + z = 3 (2); x + 2y − z = 2 (3).
Step 1. Add (1) and (3) to remove z: 2x + 3y = 8 (4).
Step 2. Add (2) and (3) to remove z: 3x + y = 5 (5).
Step 3. From (5), y = 5 − 3x. Substitute into (4): 2x + 3(5 − 3x) = 8, so 2x + 15 − 9x = 8, giving −7x = −7 and x = 1. Then y = 2.
Step 4. From (1): z = 6 − 1 − 2 = 3.
Check. (2): 2 − 2 + 3 = 3. (3): 1 + 4 − 3 = 2. The solution is x = 1, y = 2, z = 3.
Worked example 2: a smarter first move
Solve: 3x + y − z = 8 (A); x − 2y + z = −3 (B); 2x + y + 2z = 9 (C).
Add (A) and (B): 4x − y = 5 (D). For the second pair, take (C) − 2 × (B): (2x + y + 2z) − (2x − 4y + 2z) = 5y = 15, so y = 3 straight away.
Then (D) gives 4x − 3 = 5, so x = 2. From (A): 6 + 3 − z = 8, so z = 1. Check (B): 2 − 6 + 1 = −3. Check (C): 4 + 3 + 2 = 9.
Look for a combination that removes two variables at once, as C − 2 × B did here. It saves a whole step.
The mistake that costs marks
The common slip is to eliminate z from one pair and x from another. The two new equations then have different unknowns, so you cannot solve them together.
| Step | Wrong | Right |
|---|---|---|
| Pair (1) and (2) | Remove z | Remove z |
| Pair (1) and (3) | Remove x | Remove z |
| Result | Equations in (x, y) and (y, z) | Both equations in (x, y) |
Write the eliminated variable beside each new equation, such as “(4) z removed”, so the pairing stays visible.
Check yourself
Solve: x + y + z = 9; 2x − y + z = 5; x + 2y − z = 4.
Answer
Label them (1), (2), (3). Add (1) and (3): 2x + 3y = 13 (4). Add (2) and (3): 3x + y = 9 (5).
From (5), y = 9 − 3x. Then 2x + 3(9 − 3x) = 13 gives 2x + 27 − 9x = 13, so −7x = −14 and x = 2, y = 3.
From (1): z = 9 − 2 − 3 = 4. Check (2): 4 − 3 + 4 = 5. Check (3): 2 + 6 − 4 = 4. The answer is x = 2, y = 3, z = 4.
What to study next
Next, see how a squared equation changes the method in solving one linear and one nonlinear equation. Then test the whole chapter with the systems of equations practice set.
If you want a teacher to work through three-unknown questions with you, see online one-to-one Additional Mathematics tuition.