Simultaneous equations appear on virtually every GCSE Maths paper across AQA, Edexcel and OCR — and they carry multiple marks. The good news is that once you master the method, every question follows the same pattern. This guide covers all three solving methods and every type you will meet at GCSE.
What Are Simultaneous Equations?
Simultaneous equations are two (or more) equations that are both true at the same time, for the same values of x and y. Your job is to find the values of x and y that satisfy both equations at once.
Example:
- Equation 1: 2x + y = 10
- Equation 2: x + y = 6
Both equations are true when x = 4 and y = 2. (Check: 2(4) + 2 = 10 ✓ and 4 + 2 = 6 ✓)
Method 1: Elimination (Foundation and Higher)
Elimination is the most commonly used method at GCSE. You combine the two equations to eliminate one variable, then solve for the other.
When to use it
Use elimination when the same variable appears in both equations and the coefficients can be made equal by multiplying.
Step-by-step method
Step 1: Make the coefficients of one variable equal (multiply one or both equations if needed).
Step 2: Add or subtract the equations to eliminate that variable.
- If the matching coefficients have the same sign → subtract
- If the matching coefficients have opposite signs → add
Step 3: Solve the resulting single-variable equation.
Step 4: Substitute back into one of the original equations to find the other variable.
Step 5: Check your answer satisfies both original equations.
Worked Example 1 — Simple elimination
Solve: 3x + 2y = 16 and 5x + 2y = 24
The y-coefficients are both 2. Same sign → subtract.
(5x + 2y) − (3x + 2y) = 24 − 16
2x = 8 → x = 4
Substitute into equation 1: 3(4) + 2y = 16 → 12 + 2y = 16 → 2y = 4 → y = 2
Check: 3(4) + 2(2) = 16 ✓ and 5(4) + 2(2) = 24 ✓
Worked Example 2 — Elimination requiring multiplication
Solve: 2x + 3y = 13 and 4x − y = 5
The coefficients do not match. Multiply equation 2 by 3 to match the y-coefficients:
Equation 1: 2x + 3y = 13
Equation 2 × 3: 12x − 3y = 15
The y-coefficients are +3 and −3 → add the equations:
(2x + 3y) + (12x − 3y) = 13 + 15
14x = 28 → x = 2
Substitute: 2(2) + 3y = 13 → 4 + 3y = 13 → 3y = 9 → y = 3
Check: 2(2) + 3(3) = 13 ✓ and 4(2) − 3 = 5 ✓
Worked Example 3 — Both equations need multiplying
Solve: 3x + 4y = 22 and 5x + 3y = 24
To eliminate y: multiply equation 1 by 3 and equation 2 by 4.
Equation 1 × 3: 9x + 12y = 66
Equation 2 × 4: 20x + 12y = 96
Both y-coefficients are +12 → subtract:
(20x + 12y) − (9x + 12y) = 96 − 66
11x = 30 → x = 30/11 (non-integer answers are fine — keep the fraction)
Substitute: 3(30/11) + 4y = 22 → 90/11 + 4y = 22 → 4y = 22 − 90/11 = 242/11 − 90/11 = 152/11 → y = 38/11
Method 2: Substitution (Foundation and Higher)
Substitution involves rearranging one equation to make one variable the subject, then substituting that expression into the other equation.
When to use it
Substitution is most useful when one equation has a variable with coefficient 1 (making rearrangement clean), and it is essential for Higher tier simultaneous equations involving a quadratic.
Step-by-step method
Step 1: Choose the simpler equation and rearrange to express one variable in terms of the other (e.g. x = 3 − 2y).
Step 2: Substitute this expression into the other equation.
Step 3: Expand and simplify to get a single-variable equation. Solve it.
Step 4: Substitute back to find the other variable.
Step 5: Check in both original equations.
Worked Example 4 — Substitution (linear)
Solve: y = 2x − 1 and 3x + 2y = 13
Equation 1 is already in the form y = ... so substitute directly:
3x + 2(2x − 1) = 13
3x + 4x − 2 = 13
7x = 15 → x = 15/7
Substitute back: y = 2(15/7) − 1 = 30/7 − 7/7 = 23/7
Method 3: Graphical Method (Foundation and Higher)
You can solve simultaneous equations by drawing both equations as straight lines on a graph and reading off their intersection point.
When to use it
The graphical method is asked explicitly on some GCSE papers (usually with a pre-drawn grid). It gives approximate answers and is less precise than algebraic methods.
Method
- Choose 3 x-values for each equation and calculate the corresponding y-values.
- Plot both lines on the same axes.
- Read the coordinates of the intersection point — this is the solution (x, y).
Worked Example 5 — Graphical
Solve graphically: y = x + 2 and y = 4 − x
For y = x + 2: when x = 0, y = 2; when x = 2, y = 4; when x = −1, y = 1. For y = 4 − x: when x = 0, y = 4; when x = 2, y = 2; when x = 4, y = 0.
Plot both lines. They intersect at (1, 3).
Check: y = 1 + 2 = 3 ✓ and y = 4 − 1 = 3 ✓
Higher Tier: One Linear and One Quadratic
At Higher tier, you may be asked to solve a pair of simultaneous equations where one is linear (y = mx + c type) and one is quadratic (containing x²). You always use substitution for this type.
Step-by-step method
Step 1: Rearrange the linear equation to express y (or x) in terms of the other variable.
Step 2: Substitute into the quadratic equation to form a quadratic in one variable.
Step 3: Rearrange into the form ax² + bx + c = 0.
Step 4: Solve by factorising, completing the square, or the quadratic formula.
Step 5: For each value of x, substitute back into the linear equation to find the corresponding y value. Present answers as coordinate pairs.
Worked Example 6 — Linear and quadratic
Solve: y = x + 3 and x² + y² = 29
Substitute y = x + 3 into x² + y² = 29:
x² + (x + 3)² = 29
x² + x² + 6x + 9 = 29
2x² + 6x + 9 − 29 = 0
2x² + 6x − 20 = 0
x² + 3x − 10 = 0 (divide by 2)
(x + 5)(x − 2) = 0
x = −5 or x = 2
For x = −5: y = −5 + 3 = −2 → point (−5, −2)
For x = 2: y = 2 + 3 = 5 → point (2, 5)
Solutions: (−5, −2) and (2, 5)
Geometric interpretation: This is where the line y = x + 3 intersects the circle x² + y² = 29. Two intersection points means two solutions.
Worked Example 7 — Linear and quadratic (another type)
Solve: y = 2x − 1 and y = x² − 3x + 5
Substitute: x² − 3x + 5 = 2x − 1
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2 or x = 3
For x = 2: y = 2(2) − 1 = 3 → (2, 3)
For x = 3: y = 2(3) − 1 = 5 → (3, 5)
Simultaneous Equations in Word Problems
GCSE papers often present simultaneous equations as written problems. Your job is to form the equations yourself, solve them, and give a contextual answer.
Worked Example 8 — Word problem
Three adult tickets and two child tickets cost £43. One adult ticket and four child tickets cost £31. Find the cost of each type of ticket.
Let a = adult ticket price, c = child ticket price.
Equation 1: 3a + 2c = 43
Equation 2: a + 4c = 31
Multiply equation 2 by 3: 3a + 12c = 93
Subtract equation 1: 10c = 50 → c = 5
Substitute: a + 4(5) = 31 → a = 11 → a = 11
Adult ticket: £11, child ticket: £5
Exam Tips for Simultaneous Equations
- Always show full working — method marks are available even if you make an arithmetic error. A bare answer earns no marks on multi-step questions.
- Always substitute back to find both variables — a question asking you to "solve" means finding both x and y. Stopping at x loses marks.
- Always check your answer in both original equations — this takes 30 seconds and confirms correctness.
- For quadratic simultaneous equations — remember to find both pairs of solutions and present them clearly as coordinate pairs.
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