3 Methods for Solving Simultaneous Equations in IGCSE Maths

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Simultaneous equations appear on virtually every IGCSE Mathematics 0580 paper. You will face them on Paper 2, Paper 4, or both. The good news is that there are only three methods you need to know, and once you have practised each one, you can tackle any question the exam throws at you.

This guide covers all three methods with full worked examples. By the end, you should know when to use each method and how to execute it accurately.

Method 1: Elimination

Elimination is the most common method and the one you will use most often in the exam. The idea is to add or subtract the two equations to eliminate one of the variables.

Worked Example 1

Solve simultaneously:

• 2x + 3y = 12 … (1)
• 4x + 3y = 18 … (2)

Step 1: Look at the coefficients. Both equations have 3y, so we can eliminate y by subtracting.

Step 2: Subtract equation (1) from equation (2):

(4x + 3y) - (2x + 3y) = 18 - 12

4x - 2x + 3y - 3y = 6

2x = 6

x = 3

Step 3: Substitute x = 3 back into equation (1):

2(3) + 3y = 12

6 + 3y = 12

3y = 6

y = 2

Solution: x = 3, y = 2

Step 4 (always do this): Check in equation (2): 4(3) + 3(2) = 12 + 6 = 18. Correct.

Worked Example 2: When Coefficients Don’t Match

Solve simultaneously:

• 3x + 2y = 16 … (1)
• 5x - 3y = 14 … (2)

Step 1: Neither x nor y has matching coefficients, so we need to create matching coefficients. Multiply equation (1) by 3 and equation (2) by 2 to make the y coefficients match:

9x + 6y = 48 … (3)

10x - 6y = 28 … (4)

Step 2: Add equations (3) and (4) — we add because the y terms have opposite signs:

19x = 76

x = 4

Step 3: Substitute x = 4 into equation (1):

3(4) + 2y = 16

12 + 2y = 16

2y = 4

y = 2

Solution: x = 4, y = 2

Check in equation (2): 5(4) - 3(2) = 20 - 6 = 14. Correct.

Exam tip: Always label your equations with numbers. This makes your working clearer and earns method marks even if you make an arithmetic error later.

Method 2: Substitution

Substitution works by rearranging one equation to express one variable in terms of the other, then substituting into the second equation. This method is especially useful when one equation is already solved for a variable, or when one equation is non-linear.

Worked Example 3

Solve simultaneously:

• y = 2x + 1 … (1)
• 3x + 2y = 16 … (2)

Step 1: Equation (1) already gives y in terms of x. Substitute this into equation (2):

3x + 2(2x + 1) = 16

Step 2: Expand and solve:

3x + 4x + 2 = 16

7x + 2 = 16

7x = 14

x = 2

Step 3: Substitute x = 2 into equation (1):

y = 2(2) + 1 = 5

Solution: x = 2, y = 5

Worked Example 4: Linear and Quadratic (Extended Tier)

On the Extended paper, you may be asked to solve a pair of simultaneous equations where one is linear and one is quadratic. Substitution is the required method here.

Solve simultaneously:

• y = x + 3 … (1)
• x² + y² = 29 … (2)

Step 1: Substitute equation (1) into equation (2):

x² + (x + 3)² = 29

Step 2: Expand:

x² + x² + 6x + 9 = 29

2x² + 6x + 9 = 29

2x² + 6x - 20 = 0

x² + 3x - 10 = 0

Step 3: Factorise:

(x + 5)(x - 2) = 0

x = -5 or x = 2

Step 4: Find corresponding y values using equation (1):

When x = -5: y = -5 + 3 = -2

When x = 2: y = 2 + 3 = 5

Solution: x = -5, y = -2 or x = 2, y = 5

Exam tip: When solving linear-quadratic simultaneous equations, always remember there are usually two pairs of solutions. Do not stop after finding only one.

Method 3: Graphical

The graphical method involves plotting both equations on a graph and reading off the coordinates of the point(s) where they intersect. This method is less accurate than algebraic methods but is sometimes specifically requested in exam questions.

Worked Example 5

Solve graphically:

• y = 2x - 1
• y = -x + 5

Step 1: Create a table of values for each equation.

For y = 2x - 1:

x	0	1	2	3
y	-1	1	3	5

For y = -x + 5:

x	0	1	2	3
y	5	4	3	2

Step 2: Plot both lines on the same axes. Use a sharp pencil and ruler for straight lines.

Step 3: Read the coordinates of the intersection point. The lines cross at (2, 3).

Solution: x = 2, y = 3

Verify algebraically: 2(2) - 1 = 3 and -(2) + 5 = 3. Both give y = 3. Correct.

Exam tip: When the exam asks you to solve simultaneous equations graphically, they will provide a grid. Plot your points carefully, use a ruler, and read the intersection coordinates as accurately as possible. If the intersection does not land exactly on a grid point, give your answer to one decimal place.

Which Method Should You Use?

In the exam, the question often tells you which method to use. If it does not, follow this guide:

• Both equations are linear and neither variable is already isolated: Use elimination.
• One equation already gives one variable in terms of the other: Use substitution.
• One equation is linear and one is quadratic or circular: Use substitution.
• The question asks you to solve graphically or provides a grid: Use the graphical method.

Regardless of method, always check your answer by substituting both values back into the equation you did not use to find them.

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