Solving three equations in three unknowns

Covers AS91269

Why three equations

To pin down three unknowns ( $x$ , $y$ , $z$ ) you need three independent equations.
Each equation now represents a plane in 3-D. A unique solution is the single point where all three planes meet.

The strategy: reduce 3 down to 2

The plan is to eliminate the same variable twice to get a smaller system:
- Pick a variable to remove (choose the easiest — one that already has matching or opposite coefficients).
- Combine equation 1 with equation 2 to eliminate it → a new equation in the other two unknowns.
- Combine equation 1 with equation 3 to eliminate the same variable → a second new equation.
- You now have two equations in two unknowns — solve them as before.
- Back-substitute into any original equation to find the third variable.

Worked ExampleThree equations, three unknowns

Solve the system $x + y + z = 6 \quad (1)$ $2x - y + z = 3 \quad (2)$ $x + 2y - z = 2 \quad (3)$

Step 1 — Eliminate $z$ from two pairs

$(1) - (2)$ : $(x - 2x) + (y + y) + (z - z) = 6 - 3$ , so

$-x + 2y = 3 \quad (A)$

$(1) + (3)$ : $(x + x) + (y + 2y) + (z - z) = 6 + 2$ , so

$2x + 3y = 8 \quad (B)$

Step 2 — Solve the 2×2 system (A) and (B)

From (A): $x = 2y - 3$ . Substitute into (B):

$2(2y - 3) + 3y = 8 \;\Rightarrow\; 4y - 6 + 3y = 8 \;\Rightarrow\; 7y = 14 \;\Rightarrow\; y = 2$

Then $x = 2(2) - 3 = 1$ .

Step 3 — Back-substitute for $z$

Using (1): $1 + 2 + z = 6 \Rightarrow z = 3$ .

$\textbf{Solution: } x = 1, \; y = 2, \; z = 3$

The same three outcomes

Just like two-variable systems, three-variable systems can have one solution (planes meet at a point), no solution (no common point — e.g. planes forming a triangular "tube"), or infinitely many (planes share a whole line).
The algebra signals it the same way: a variable-free false line ( $0 = 4$ ) means no solution; a variable-free true line ( $0 = 0$ ) means infinitely many.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Given $x + y + z = 6$ , $y + z = 5$ and $z = 3$ , find $x$ , $y$ and $z$ .

Merit

Solve $x + y + z = 6$ , $x - y + z = 2$ , $x + y - z = 0$ .

Excellence

Solve $x + 2y + z = 7$ , $2x - y + 3z = 12$ , $3x + y - z = 4$ , showing your elimination clearly.

Solving three equations in three unknowns

Covers AS91269

Why three equations

To pin down three unknowns ( $x$ , $y$ , $z$ ) you need three independent equations.
Each equation now represents a plane in 3-D. A unique solution is the single point where all three planes meet.

The strategy: reduce 3 down to 2

The plan is to eliminate the same variable twice to get a smaller system:
- Pick a variable to remove (choose the easiest — one that already has matching or opposite coefficients).
- Combine equation 1 with equation 2 to eliminate it → a new equation in the other two unknowns.
- Combine equation 1 with equation 3 to eliminate the same variable → a second new equation.
- You now have two equations in two unknowns — solve them as before.
- Back-substitute into any original equation to find the third variable.

Worked ExampleThree equations, three unknowns

Solve the system $x + y + z = 6 \quad (1)$ $2x - y + z = 3 \quad (2)$ $x + 2y - z = 2 \quad (3)$

Step 1 — Eliminate $z$ from two pairs

$(1) - (2)$ : $(x - 2x) + (y + y) + (z - z) = 6 - 3$ , so

$-x + 2y = 3 \quad (A)$

$(1) + (3)$ : $(x + x) + (y + 2y) + (z - z) = 6 + 2$ , so

$2x + 3y = 8 \quad (B)$

Step 2 — Solve the 2×2 system (A) and (B)

From (A): $x = 2y - 3$ . Substitute into (B):

$2(2y - 3) + 3y = 8 \;\Rightarrow\; 4y - 6 + 3y = 8 \;\Rightarrow\; 7y = 14 \;\Rightarrow\; y = 2$

Then $x = 2(2) - 3 = 1$ .

Step 3 — Back-substitute for $z$

Using (1): $1 + 2 + z = 6 \Rightarrow z = 3$ .

$\textbf{Solution: } x = 1, \; y = 2, \; z = 3$

The same three outcomes

Just like two-variable systems, three-variable systems can have one solution (planes meet at a point), no solution (no common point — e.g. planes forming a triangular "tube"), or infinitely many (planes share a whole line).
The algebra signals it the same way: a variable-free false line ( $0 = 4$ ) means no solution; a variable-free true line ( $0 = 0$ ) means infinitely many.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Given $x + y + z = 6$ , $y + z = 5$ and $z = 3$ , find $x$ , $y$ and $z$ .

Merit

Solve $x + y + z = 6$ , $x - y + z = 2$ , $x + y - z = 0$ .

Excellence

Solve $x + 2y + z = 7$ , $2x - y + 3z = 12$ , $3x + y - z = 4$ , showing your elimination clearly.

Solving three equations in three unknowns

Why three equations

The strategy: reduce 3 down to 2

Step 1 — Eliminate zzz from two pairs

Step 2 — Solve the 2×2 system (A) and (B)

Step 3 — Back-substitute for zzz

The same three outcomes

Test yourself

Practice by grade

Solving three equations in three unknowns

Why three equations

The strategy: reduce 3 down to 2

Step 1 — Eliminate zzz from two pairs

Step 2 — Solve the 2×2 system (A) and (B)

Step 3 — Back-substitute for zzz

The same three outcomes

Test yourself

Practice by grade

Step 1 — Eliminate $z$ from two pairs

Step 3 — Back-substitute for $z$

Step 1 — Eliminate $z$ from two pairs

Step 3 — Back-substitute for $z$