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14 Mar 2026 · Revise Right Team

Solving Linear Equations with Three Unknowns Revision Notes | GCSE AQA Higher Further Maths

Solving Linear Equations with Three Unknowns: Learn how to solve linear equations involving three unknowns using algebraic methods. We will cover…

Solving Linear Equations with Three Unknowns is part of Algebraic solution of linear equations in three unknowns in GCSE AQA Higher Further Maths.

4 min

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24

Lessons in this topic

GCSE AQA Higher

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Reviewed by Revise Right

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What this lesson covers

Solving Linear Equations with Three Unknowns: Learn how to solve linear equations involving three unknowns using algebraic methods. We will cover…

  • Solve a system of three linear equations in three unknowns using elimination
  • Apply systematic elimination to reduce a three-variable system to a two-variable system
  • Verify solutions by substituting values back into all three original equations
  • Identify and avoid common errors when working with three simultaneous equations

Key ideas to keep in view

Use this lesson to stay inside Algebraic solution of linear equations in three unknowns while connecting the detail back to the wider Algebra topic.

  • Algebra
  • Algebraic solution of linear equations in three unknowns
  • GCSE AQA Higher Further Maths

Revision notes

Read the core explanations from Solving Linear Equations with Three Unknowns before testing yourself from memory.

1. Why Three Unknowns?

You already know how to solve two simultaneous equations with two unknowns — you eliminate one variable and solve for the other. Three unknowns follow exactly the same logic, extended by one extra step.

Think of it as peeling an onion. You start with three unknowns ($x$, $y$, $z$), eliminate one to get two unknowns, then eliminate another to get one unknown — and then work back up. Each layer reveals the next.

To find a unique solution for three unknowns, you need exactly three independent equations. One equation gives you a plane in 3D space; three equations (when consistent) intersect at a single point — that point is your solution.

2. The Method: Systematic Elimination

The algebraic method for three unknowns has a clear structure. Follow these four stages every time and you will not lose your way.

  • Stage 1 — Label your equations (1), (2), (3) so you can track which pair you are working with.
  • Stage 2 — Choose one variable to eliminate first. Use pairs of equations to remove it, producing two new equations in only two unknowns.
  • Stage 3 — Solve the resulting two-variable system using elimination or substitution.
  • Stage 4 — Back-substitute the found values into one of the original equations to find the third unknown, then verify in all three equations.

3. Fully Worked Example

Theory is clear in the abstract, but exam questions are concrete. Here is a complete walkthrough of a three-unknown system from start to verified answer.

Worked example: Solving a System of Three Linear Equations

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

  1. Equations: (1) $x + y + z = 6$, (2) $2x - y + z = 3$, (3) $x + 2y - z = 2$. Choose to eliminate $z$ first. — Labelling prevents confusion. Eliminating $z$ is natural here because its coefficients are already $+1$, $+1$, $-1$ — straightforward to match without scaling. [M1 for a valid elimination strategy]
  2. Add (1) and (3): $(x + y + z) + (x + 2y - z) = 6 + 2$ $2x + 3y = 8$... (4) — The $z$ terms cancel: $+z + (-z) = 0$. Equation (4) now has only $x$ and $y$. [M1 for correct addition producing a two-variable equation]
  3. Add (2) and (3): $(2x - y + z) + (x + 2y - z) = 3 + 2$ $3x + y = 5$... (5) — Again $z$ cancels. Now we have two equations, (4) and (5), in $x$ and $y$ only. [M1 for a second correct two-variable equation]
  4. (4): $2x + 3y = 8$ (5): $3x + y = 5$ Multiply (5) by 3: $9x + 3y = 15$... (6) Subtract (4) from (6): $(9x + 3y) - (2x + 3y) = 15 - 8$ $7x = 7$ $x = 1$ — Multiplying (5) by 3 makes the $y$-coefficients equal so they cancel on subtraction. [M1 for correct elimination; A1 for $x = 1$]
  5. $3(1) + y = 5$ $3 + y = 5$ $y = 2$ — Substituting $x = 1$ into the simpler two-variable equation (5) is the quickest route to $y$. [A1 for $y = 2$]
  6. $1 + 2 + z = 6$ $3 + z = 6$ $z = 3$ — Back-substitution into the original equation (1) gives $z$. [A1 for $z = 3$]
  7. (1): $1 + 2 + 3 = 6$ ✓ (2): $2(1) - 2 + 3 = 2 - 2 + 3 = 3$ ✓ (3): $1 + 2(2) - 3 = 1 + 4 - 3 = 2$ ✓ — All three equations are satisfied. Verification is the only way to be certain — and it takes under 30 seconds. Never skip it.

Answer: $x = 1$, $y = 2$, $z = 3$

Notice how the method is entirely mechanical once you choose which variable to eliminate first. The choice does not affect the final answer — only the arithmetic difficulty of the intermediate steps.

4. A Second Worked Example — Scaling Required

The previous example had convenient coefficients. Exam questions often require you to scale one or both equations before the coefficients match. Here is how that looks in practice.

Worked example: Three-Unknown System with Scaling

Solve the system: $2x + y - z = 1$... (1) $x - y + 2z = 7$... (2) $3x + 2y + z = 10$... (3)

  1. Coefficients of $y$: (1) has $+1$, (2) has $-1$, (3) has $+2$. Add (1) and (2) directly to eliminate $y$: $(2x + y - z) + (x - y + 2z) = 1 + 7$ $3x + z = 8$... (4) — $+y$ and $-y$ cancel without any scaling needed here. [M1 for correct elimination]
  2. Multiply (2) by 2: $2x - 2y + 4z = 14$... (2') Add (3) and (2'): $(3x + 2y + z) + (2x - 2y + 4z) = 10 + 14$ $5x + 5z = 24$... (5) — Scaling (2) by 2 makes the $y$-coefficients $-2$ and $+2$, which cancel on addition. [M1 for correct scaling and elimination]
  3. (4): $3x + z = 8 \Rightarrow z = 8 - 3x$ Substitute into (5): $5x + 5(8 - 3x) = 24$ $5x + 40 - 15x = 24$ $-10x = -16$ $x = \frac{16}{10} = \frac{8}{5}$ — Substitution is efficient here because (4) expresses $z$ in terms of $x$ directly. Keep the fraction exact — do not round mid-calculation. [M1 for correct substitution; A1 for $x = \frac{8}{5}$]
  4. $z = 8 - 3 \times \frac{8}{5} = 8 - \frac{24}{5} = \frac{40}{5} - \frac{24}{5} = \frac{16}{5}$ — Working with exact fractions avoids rounding errors that would cause the verification to fail. [A1 for $z = \frac{16}{5}$]
  5. $2 \times \frac{8}{5} + y - \frac{16}{5} = 1$ $\frac{16}{5} + y - \frac{16}{5} = 1$ $y = 1$ — Back-substituting into equation (1) gives $y = 1$ cleanly. [A1 for $y = 1$]
  6. (2): $\frac{8}{5} - 1 + 2 \times \frac{16}{5} = \frac{8}{5} - 1 + \frac{32}{5} = \frac{40}{5} - 1 = 8 - 1 = 7$ ✓ (3): $3 \times \frac{8}{5} + 2(1) + \frac{16}{5} = \frac{24}{5} + 2 + \frac{16}{5} = \frac{40}{5} + 2 = 8 + 2 = 10$ ✓ — Both remaining equations check out. The solution is confirmed.

Answer: $x = \frac{8}{5}$, $y = 1$, $z = \frac{16}{5}$

5. Guided Practice

Now it is your turn. Work through these problems using the four-stage method. The hints guide you if you get stuck, and the solutions let you check every step.

A note on the practice set: the third problem above illustrates that not every system produces neat integer answers. Fractions are perfectly valid solutions — keep them exact throughout and verify carefully.

6. Visualising the Structure

It helps to see the elimination process as a funnel: three unknowns narrow to two, then to one, and back-substitution fills the funnel back up. The visualisation below maps that structure.

7. Exam Technique and Common Mistakes

Three-unknown systems reward careful, organised working. The mistakes that cost marks are almost always avoidable — they come from rushing or skipping the verification step.

8. Exam-Style Question

This question is structured in the style you will see in an exam. Work through it fully before checking the ordering block below, which maps the correct sequence of steps.

Solve the system: $3x - y + 2z = 11$... (1), $x + 2y - z = 3$... (2), $2x + y + 3z = 16$... (3). Show all working clearly.

Worked example: Exam-Style Question: Full Solution

Solve: $3x - y + 2z = 11$... (1), $x + 2y - z = 3$... (2), $2x + y + 3z = 16$... (3)

  1. Add (1) and (3): $(3x - y + 2z) + (2x + y + 3z) = 11 + 16$ $5x + 5z = 27$... (4) — $-y$ and $+y$ cancel directly. [M1]
  2. Multiply (2) by 1 and (3) by 2: $2 \times (3): 4x + 2y + 6z = 32$... (3') Subtract (2) from (3'): $(4x + 2y + 6z) - (x + 2y - z) = 32 - 3$ $3x + 7z = 29$... (5) — $2y$ terms cancel. [M1 for correct scaling and subtraction]
  3. (4): $5x + 5z = 27 \Rightarrow x + z = \frac{27}{5}$... (4') Multiply (4') by 7: $7x + 7z = \frac{189}{5}$... (4'') Subtract (5) from (4''): $7x + 7z - 3x - 7z = \frac{189}{5} - 29$ $4x = \frac{189}{5} - \frac{145}{5} = \frac{44}{5}$ $x = \frac{44}{20} = \frac{11}{5}$ — Keeping fractions exact is essential here. [M1 for correct elimination; A1 for $x = \frac{11}{5}$]
  4. $z = \frac{27}{5} - x = \frac{27}{5} - \frac{11}{5} = \frac{16}{5}$ — [A1 for $z = \frac{16}{5}$]
  5. $\frac{11}{5} + 2y - \frac{16}{5} = 3$ $\frac{11 - 16}{5} + 2y = 3$ $-\frac{5}{5} + 2y = 3$ $-1 + 2y = 3$ $2y = 4$ $y = 2$ — [A1 for $y = 2$]
  6. (1): $3 \times \frac{11}{5} - 2 + 2 \times \frac{16}{5} = \frac{33}{5} - 2 + \frac{32}{5} = \frac{65}{5} - 2 = 13 - 2 = 11$ ✓ (2): $\frac{11}{5} + 4 - \frac{16}{5} = \frac{-5}{5} + 4 = -1 + 4 = 3$ ✓ (3): $2 \times \frac{11}{5} + 2 + 3 \times \frac{16}{5} = \frac{22}{5} + 2 + \frac{48}{5} = \frac{70}{5} + 2 = 14 + 2 = 16$ ✓ — All three equations satisfied. The solution is confirmed.

Answer: $x = \frac{11}{5}$, $y = 2$, $z = \frac{16}{5}$

This example reinforces the key lesson: non-integer answers are not a sign of an error. Work with exact fractions, verify in all three equations, and you will always know whether your solution is correct.

Frequently asked questions

What should I know about Solving Linear Equations with Three Unknowns?

To find a unique solution for three unknowns, you need exactly three independent equations. One equation gives you a plane in 3D space; three equations (when consistent) intersect at a single point — that point is your solution.

What should I learn from this Further Maths lesson?

By the end of this lesson, you should be able to Solve a system of three linear equations in three unknowns using elimination; Apply systematic elimination to reduce a three-variable system to a two-variable system; Verify solutions by substituting values back into all three original equations; Identify and avoid common errors when working with three simultaneous equations.

How should I revise Solving Linear Equations with Three Unknowns?

Read each section, then close the notes and explain the idea from memory. Check your explanation against the page and finish with an exam question from Algebra.

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