Systems of three linear equations | Year 12

01 / LEARN

Understand the idea.

Eliminate systematically

A solution of three linear equations must satisfy all three equations simultaneously. Use substitution or elimination to reduce the system, solve the remaining variables, then check the result in every original equation to catch algebra or transcription errors.

A nonzero coefficient determinant guarantees a unique solution. A zero determinant only rules out uniqueness; it does not distinguish no solution from infinitely many solutions. Elimination must still check whether the reduced equations are consistent.

Read dependence geometrically

Each nontrivial linear equation in three variables represents a plane. Independent restrictions may reduce their common set to a point; dependent equations can leave a line or plane. Repeated information does not impose an extra geometric restriction.

During elimination, $0 = 0$ contributes no new condition, while $0 = c$ with $c \neq 0$ is impossible. For a consistent system with free variables, introduce parameters and express every solution, rather than choosing one convenient point.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Substitute the known third coordinate.
$z = 2 ⟹ x + y = 4$
The first equation becomes a two-variable condition.
2
Use equality of the first two coordinates.
$x - y = 0 ⟹ x = y, 2 x = 4$
Replacing $y$ by $x$ combines the remaining equations.
3
State the solution.
$x = 2, y = 2, z = 2$
The two-variable system fixes both remaining coordinates.
4
Check all original equations.
$2 + 2 + 2 = 6, 2 - 2 = 0, z = 2$
The triple satisfies every original restriction.

Answer

$(2, 2, 2)$

EXAMPLE 02

Worked example 2

Solve x+y+z=1, 2x+2y+2z=2, x-y=0.

Worked solution

1
Identify the dependent equation.
$2 x + 2 y + 2 z = 2 ⟺ x + y + z = 1$
Dividing the second equation by two reproduces the first.
2
Use the remaining independent equality.
$x - y = 0 ⟹ x = y = t, t \in ℝ$
One coordinate can remain free because only two independent restrictions remain.
3
Solve for the third coordinate.
$t + t + z = 1 ⟹ z = 1 - 2 t$
The first equation determines $z$ once $t$ is chosen.
4
Describe every common point.
$(x, y, z) = (t, t, 1 - 2 t) = (0, 0, 1) + t (1, 1, - 2), t \in ℝ$
The solution set is a line, and substitution satisfies all three equations for every real $t$ .

Answer

$(t, t, 1 - 2 t)$ for real t.

03 / YOUR TURN

Try it yourself.

Check your understanding

Can x+y+z=6 and 2x+2y+2z=13 hold together?

Show answer & reasoning

No.

Doubling $x + y + z = 6$ forces $2 x + 2 y + 2 z = 12$ . The same left side cannot equal $13$ , so the two equations are inconsistent.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Solve $x + y + z = 9$ , $x - y = 1$ , $z = 2$ .
2 marks
Solve $x + y = 3$ , $y + z = 5$ , $x + z = 4$ .
2 marks
Classify $x + y + z = 2$ , $2 x + 2 y + 2 z = 5$ , $z = 1$ .
2 marks
Parametrise $x + y = 2$ , $z = 1$ , $2 x + 2 y = 4$ .
2 marks
Solve $x + 2 y + z = 8$ , $y - z = 1$ , $x - z = 2$ .
2 marks
What does a unique solution of three independent plane equations represent?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $x + y + z = 9$ , $x - y = 1$ , $z = 2$ .
2 marks
Write $x = y + 1$ .
Then $2 y + 3 = 9$ , so $y = 3$ .
$(4, 3, 2)$
Solve $x + y = 3$ , $y + z = 5$ , $x + z = 4$ .
2 marks
Add the first two and subtract the third: $2 y = 4$ .
Back-substitute.
$(1, 2, 3)$
Classify $x + y + z = 2$ , $2 x + 2 y + 2 z = 5$ , $z = 1$ .
2 marks
Twice the first equation would require left side 4 in the second.
The second constant is 5, giving a contradiction.
No solution.
Parametrise $x + y = 2$ , $z = 1$ , $2 x + 2 y = 4$ .
2 marks
The last equation is redundant.
Choose $y = t$ and solve for $x$ .
$(2 - t, t, 1)$ , real $t$ .
Solve $x + 2 y + z = 8$ , $y - z = 1$ , $x - z = 2$ .
2 marks
Use $y = z + 1$ and $x = z + 2$ .
Substitution gives $4 z + 4 = 8$ .
$(3, 2, 1)$
What does a unique solution of three independent plane equations represent?
2 marks
Each equation imposes an independent restriction.
No free direction remains.
Exactly one common point of all three planes.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics vector algebra, vector/Cartesian equations and linear systems. Original explanations and tasks.

Official curriculum source ↗

In this lesson

Understand the idea.

Eliminate systematically

Read dependence geometrically

See the method in action.

Worked example 1

Substitute the known third coordinate.

Use equality of the first two coordinates.

State the solution.

Check all original equations.

Worked example 2

Identify the dependent equation.

Use the remaining independent equality.

Solve for the third coordinate.

Describe every common point.

Try it yourself.

Check your understanding

Keep practising