Simultaneous Equations: Elimination and Substitution — GCSE Maths

Simultaneous equations are two equations, each with two unknowns, that must both be true at the same time. At GCSE, the two main algebraic methods are elimination (adding or subtracting the equations to remove one unknown) and substitution (rearranging one equation and substituting into the other).

What is the elimination method?

Elimination works by making the coefficient of one variable identical in both equations, then adding or subtracting the equations to remove that variable entirely.

Step 1 — Label the equations (1) and (2).
Step 2 — If necessary, multiply one or both equations so that one variable has the same coefficient.
Step 3 — Add the equations if the equal coefficients have opposite signs; subtract if they have the same sign.
Step 4 — Solve the resulting single-variable equation.
Step 5 — Substitute back to find the second variable.
Step 6 — Check both values in the original equations.

How do you use elimination when the coefficients already match?

Worked example 1: solve 3x + 2y = 11 and 5x − 2y = 13

The y-coefficients are +2 and −2 (opposite signs), so ADD the equations.

(3x + 2y) + (5x − 2y) = 11 + 13
8x = 24
x = 3

Substitute x = 3 into equation (1):
3(3) + 2y = 11 → 9 + 2y = 11 → 2y = 2 → y = 1.

Answer: x = 3, y = 1

Check in (2): 5(3) − 2(1) = 15 − 2 = 13 ✓

How do you use elimination when you must scale first?

Worked example 2: solve 2x + 3y = 16 and 5x + 2y = 19

The coefficients of neither variable match. Multiply equation (1) by 5 and equation (2) by 2 to make x-coefficients equal:

(1) × 5: 10x + 15y = 80
(2) × 2: 10x + 4y = 38

Subtract (2) × 2 from (1) × 5 (same sign, so subtract):
(10x + 15y) − (10x + 4y) = 80 − 38
11y = 42
y = 42/11

Hmm — integer answers are more common in GCSE. Let me try targeting y instead: multiply (1) by 2 and (2) by 3.

(1) × 2: 4x + 6y = 32
(2) × 3: 15x + 6y = 57

Subtract: (15x + 6y) − (4x + 6y) = 57 − 32
11x = 25 — still not integer.

With this example the answer is non-integer; let us check with substitution.

Worked example 3 (cleaner): solve 2x + 5y = 24 and 3x − y = 5

Multiply equation (2) by 5 to match y-coefficients:
(2) × 5: 15x − 5y = 25

Add to equation (1):
(2x + 5y) + (15x − 5y) = 24 + 25
17x = 49 — still non-integer.

Let me choose a clean example: solve 4x + y = 14 and 3x + 2y = 13.

Multiply (1) by 2: 8x + 2y = 28.
Subtract (2): (8x + 2y) − (3x + 2y) = 28 − 13 → 5x = 15 → x = 3.
Substitute into (1): 4(3) + y = 14 → y = 2. y = 2.

Check in (2): 3(3) + 2(2) = 9 + 4 = 13 ✓ Answer: x = 3, y = 2

What is the substitution method?

Substitution works best when one equation has a variable with coefficient 1 (or −1), making rearrangement easy.

Step 1 — Rearrange one equation to express one variable in terms of the other.
Step 2 — Substitute this expression into the second equation.
Step 3 — Solve the resulting equation.
Step 4 — Substitute back to find the first variable.
Step 5 — Check.

Worked example 4: solve y = 2x − 1 and 3x + 4y = 18

Equation (1) is already rearranged: y = 2x − 1.
Substitute into equation (2):
3x + 4(2x − 1) = 18
3x + 8x − 4 = 18
11x = 22
x = 2

Substitute into (1): y = 2(2) − 1 = 3.

Answer: x = 2, y = 3

Check in (2): 3(2) + 4(3) = 6 + 12 = 18 ✓

Which method should you choose?

How do you check your answer?

Always substitute both values back into both original equations — not just one. A value that satisfies only one equation is not a solution to the simultaneous system. This takes ten seconds and earns full accuracy marks.

Frequently asked questions

What does it mean if the result of elimination gives 0 = 0?

This means the two equations represent the same line — they have infinitely many solutions. Every point on the line satisfies both equations simultaneously. This situation does not typically appear in straightforward GCSE simultaneous equation questions, but may appear in proof or "show that" style problems.

What happens if the result is 0 = 5 (a contradiction)?

The equations represent two parallel lines that never intersect — there is no solution. Again, this is a possible GCSE Higher question rather than a standard solution-finding question. Recognising the contradiction and stating "no solution" is the correct response.

Do I always have to multiply both equations when scaling?

No. If multiplying just one equation makes the coefficients equal, that is sufficient. For example, if the equations have y-coefficients of 3 and 6, multiply only the first equation by 2. Only multiply both when no single multiplier creates a match.

How do simultaneous equations appear in GCSE exam questions?

They appear as pure algebra ("solve simultaneously"), as word problems ("two numbers add to 17 and differ by 5 — find them"), and inside geometry questions ("find the intersection point of two lines"). In all cases, the algebraic method is the same — identify the pair of equations and solve using elimination or substitution.

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