Simultaneous Equations — Substitution Method

The substitution method is the second key technique for solving simultaneous equations, and it truly comes into its own when one equation is already solved for a single variable — or when you have a linear and a non-linear equation paired together. On Foundation papers you will meet purely linear pairs, while Higher tier extends this to systems involving a quadratic or a circle equation alongside a linear one. AQA, Edexcel, and OCR all examine this topic. On this page you will learn the step-by-step process for substitution with both linear and non-linear pairs, see worked examples at different difficulty levels, and practise with exam-style questions. Combined with the elimination method, substitution completes your toolkit for two-variable problems.

What Is the Substitution Method?

The substitution method involves rearranging one equation to make one variable the subject, then substituting that expression into the other equation. This reduces the system from two equations in two unknowns to a single equation in one unknown.

This approach is especially useful when:

• One equation is already in the form y = ... or x = ...
• The system contains a non-linear equation (Higher tier), because elimination does not work with x² or y² terms.

Step-by-Step Method

Linear pairs

1. Pick the equation where a variable is easiest to isolate. For example, if one equation is x + 2y = 10, rearrange to x = 10 − 2y.
2. Substitute this expression into the other equation. Replace every x with (10 − 2y).
3. Solve the resulting equation for y.
4. Substitute the value of y back into the rearranged equation to find x.
5. Check both values in the equation you did not use for substitution.

Linear + non-linear pairs (Higher tier)

1. Rearrange the linear equation to make y (or x) the subject.
2. Substitute into the non-linear equation (e.g., x² + y² = 25 or y = x² − 3x).
3. Expand, simplify, and rearrange to form a quadratic equation set equal to zero.
4. Solve the quadratic by factorising or using the quadratic formula.
5. Find the corresponding values of the other variable by substituting each solution back into the linear equation.
6. Write each solution as a pair: (x₁, y₁) and (x₂, y₂).

Worked Example 1 — Foundation Level (Linear Pair)

Question: Solve simultaneously: y = 3x − 1 and 2x + y = 9.

Working:

Step 1: The first equation already gives y in terms of x: y = 3x − 1.

Step 2: Substitute into the second equation.

2x + (3x − 1) = 9

Step 3: Simplify and solve.

5x − 1 = 9

5x = 10

x = 2

Step 4: Substitute x = 2 into y = 3x − 1.

y = 3(2) − 1 = 5

Check using 2x + y = 9: 2(2) + 5 = 9 ✓

Answer: x = 2, y = 5

Worked Example 2 — Higher Level (Linear + Quadratic)

Question: Solve simultaneously: y = 2x + 1 and x² + y² = 13.

Working:

Step 1: Substitute y = 2x + 1 into x² + y² = 13.

x² + (2x + 1)² = 13

Step 2: Expand (2x + 1)² = 4x² + 4x + 1.

x² + 4x² + 4x + 1 = 13

Step 3: Simplify.

5x² + 4x + 1 = 13

5x² + 4x − 12 = 0

Step 4: Factorise. Look for two numbers that multiply to 5 × (−12) = −60 and add to 4: (10, −6).

5x² + 10x − 6x − 12 = 0

5x(x + 2) − 6(x + 2) = 0

(5x − 6)(x + 2) = 0

x = 6/5 or x = −2.

Step 5: Find corresponding y values.

If x = 6/5: y = 2(6/5) + 1 = 12/5 + 1 = 17/5.

If x = −2: y = 2(−2) + 1 = −3.

Check for x = −2, y = −3: (−2)² + (−3)² = 4 + 9 = 13 ✓

Answer: x = 6/5, y = 17/5 or x = −2, y = −3

Common Mistakes

• Substituting into the same equation you rearranged. This leads to a tautology like y = y, which tells you nothing. Always substitute into the other equation.
• Forgetting to square the entire expression. When substituting y = 2x + 1 into y², you must expand (2x + 1)², not write 2x² + 1. The cross term (4x) is essential.
• Missing one solution in non-linear systems. A linear-quadratic system typically has two solutions. If you find only one, check your factorising or quadratic formula work.
• Not pairing x and y values correctly. Each x value has a specific y value that goes with it. Write your solutions as ordered pairs to avoid confusion.
• Arithmetic errors during expansion. Write out the expansion of (2x + 1)² fully, term by term: (2x)(2x) + (2x)(1) + (1)(2x) + (1)(1) = 4x² + 4x + 1.

Exam Tips

1. If one equation is already y = ... or x = ..., use substitution. It is faster than elimination in this case.
2. For linear + non-linear systems, substitution is the only option. Elimination cannot remove a variable from an x² or y² term. This is a Higher-only question type.
3. Present both solutions clearly as pairs. Write: "x = 2, y = 5" and "x = −1, y = 3" on separate lines or as coordinate pairs.
4. If your quadratic does not factorise, use the quadratic formula and give answers to an appropriate degree of accuracy. See Solving Quadratic Equations — Quadratic Formula for guidance.

Practice Questions

Q1 (Foundation): Solve: y = x + 4 and 3x + 2y = 23.

Q2 (Higher): Solve: y = x − 1 and x² + y² = 25.

Q3 (Higher): Solve: y = 3x + 2 and y = x² + 4x − 1.

Need to build confidence with substitution? Start revising with GCSEMathsAI — our AI tutor generates paired equations at your level and guides you through every substitution step.

Related Topics

• Simultaneous Equations — Elimination
• Solving Quadratic Equations by Factorising
• Solving Linear Equations
• Forming and Solving Equations

For a complete topic checklist, visit our GCSE Maths Topics Complete List.

Summary

• The substitution method rearranges one equation and inserts it into the other.
• It is ideal when one equation is already in the form y = ... or x = ...
• For linear pairs, substitution produces a simple linear equation to solve.
• For linear + non-linear pairs (Higher), substitution produces a quadratic to solve.
• Always pair your x and y solutions correctly.
• Check each solution pair in the equation you did not use for the substitution.
• Together with elimination, substitution gives you a complete set of tools for solving simultaneous equations at GCSE.