Solve 3×3 System of Equations: Worked Example

A 3×3 system of equations means three equations with three unknowns. Students usually find these hard for one reason: there are too many moving parts. With a 2×2 system you juggle two equations; with a 3×3 system the working can spread across the page and it is easy to lose track of which equation is which, or to make a small sign error in step four that ruins step nine.

The method that keeps everything under control is elimination: you deliberately destroy one variable at a time until the system collapses into a single equation you can solve. Then you climb back up, finding each variable in turn. Nothing here is clever or tricky. It is bookkeeping done carefully.

Here is the system we will solve:

(1) x + y + z = 6
(2) 2x − y + z = 3
(3) x + 2y − z = 2

Our target answer, which we will earn honestly and then verify, is x = 1, y = 2, z = 3.

Worked solution

The plan: combine pairs of equations to kill off one variable, leaving a 2×2 system. Then kill off a second variable, leaving one equation. Then substitute backwards.

Step 1: Pick the variable to eliminate first.

Look at the y-coefficients: +1 in equation (1), −1 in equation (2), +2 in equation (3). The +1 and −1 in equations (1) and (2) are begging to cancel. Add (1) + (2):

x-terms: x + 2x = 3x
y-terms: y + (−y) = 0 ✓ (y is gone)
z-terms: z + z = 2z
right sides: 6 + 3 = 9

So we get a brand-new equation, call it (4):

(4) 3x + 2z = 9

Step 2: Make a second new equation, also without y.

Use equation (3), which has 2y. Multiply equation (1) by 2 so its y-term becomes 2y, then subtract equation (3):

2 × (1): 2x + 2y + 2z = 12
Subtract (3): (2x + 2y + 2z) − (x + 2y − z) = 12 − 2
x-terms: 2x − x = x
y-terms: 2y − 2y = 0 ✓ (y is gone again)
z-terms: 2z − (−z) = 2z + z = 3z
right side: 12 − 2 = 10

New equation (5):

(5) x + 3z = 10

Notice what happened: equations (4) and (5) are a 2×2 system in x and z only. We have shrunk the problem.

Step 3: Solve the 2×2 system.

From (5), express x in terms of z:

x = 10 − 3z

Substitute this into (4):

3(10 − 3z) + 2z = 9
30 − 9z + 2z = 9
30 − 7z = 9
−7z = 9 − 30 = −21
z = (−21) ÷ (−7) = 3 ✓

So z = 3.

Step 4: Back-substitute to find x.

Put z = 3 into x = 10 − 3z:

x = 10 − 3(3) = 10 − 9 = 1 ✓

So x = 1.

Step 5: Back-substitute to find y.

Use the simplest original equation, (1): x + y + z = 6.

1 + y + 3 = 6
y + 4 = 6
y = 2 ✓

So y = 2. The solution is the ordered triple (1, 2, 3).

Step 6: Check (never skip).

Plug (1, 2, 3) into all three original equations. Every single one must balance, or something went wrong upstream.

Equation (1): 1 + 2 + 3 = 6 ✓
Equation (2): 2(1) − 2 + 3 = 2 − 2 + 3 = 3 ✓
Equation (3): 1 + 2(2) − 3 = 1 + 4 − 3 = 2 ✓

All three check out. The answer is confirmed: x = 1, y = 2, z = 3.

Why this method works

Each time you add or subtract whole equations, you are doing the same legal move to both sides, so the solution set never changes. You are not guessing; you are rewriting the system into an equivalent but simpler one. When y disappears from (4) and (5), it is not lost, it is just temporarily out of the picture. Back-substitution restores it, because the original equations still tie all three variables together. Elimination works because equality is preserved at every step: if two equations are both true, their sum is true too.

Common mistakes

Sign slips when subtracting. In Step 2, subtracting equation (3) turns −z into +z. Forgetting to flip every sign in the subtracted equation is the number one error in 3×3 systems. Write the subtraction out term by term.
Reusing an equation pair that does not eliminate anything. If you combine (1) and (2) to kill y, your second combination must also kill y (not x). Both new equations must be missing the same variable, or you have not actually shrunk the system.
Arithmetic in the final substitution. Students often solve z correctly and then write x = 10 − 3(3) = 10 − 6 = 4. Slow down on the last two lines; most 3×3 errors happen after the hard part is already done.

FAQ

Q: Can I eliminate a different variable first? A: Yes. We chose y because +1 and −1 cancelled for free. You could eliminate x or z first instead; the final triple will be the same (1, 2, 3). Pick whichever variable gives you the easiest cancellation and the working stays short.

Q: What if adding two equations does not cancel anything? A: Multiply one equation first, exactly as we did in Step 2 (multiplying (1) by 2 to make the y-terms match). You can always force a cancellation by scaling. The only requirement is that you multiply every term, including the right-hand side.

Q: How do I know my answer is right? A: Substitute the triple into all three original equations, not just one or two. A wrong triple can accidentally satisfy one equation. If all three balance, as (1, 2, 3) does here, the solution is certain.

Solve it with steps on our system of equations calculator.