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A2: Substitution

Foundation Higher AQAEdexcelOCREduqasCCEA

Substitute numerical values into formulae and expressions

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📋 Key Concepts

Definition: Substitution means replacing variables (letters) with their numerical values, then calculating the result.
Method:
  1. Replace each letter with its given value
  2. Calculate following BIDMAS/BODMAS rules
  3. Give your final answer

📝 Basic Substitution

Example 1

Find the value of 3x + 5 when x = 4

Solution:

3x + 5 = 3(4) + 5

= 12 + 5

= 17

Example 2

Find the value of 2a - 3b when a = 7 and b = 2

Solution:

2a - 3b = 2(7) - 3(2)

= 14 - 6

= 8

📝 Substitution with Powers

Remember: Powers are calculated before multiplication. x² means x × x, then multiply by any coefficient.
Example 3

Find the value of 2x² when x = 3

Solution:

2x² = 2 × 3²

= 2 × 9

= 18

Note: This is NOT (2×3)² = 36

Example 4

Find the value of x² + 3x - 2 when x = 5

Solution:

x² + 3x - 2 = 5² + 3(5) - 2

= 25 + 15 - 2

= 40 - 2

= 38

📝 Substitution with Brackets

Example 5

Find the value of 4(x - 2) when x = 7

Solution:

4(x - 2) = 4(7 - 2)

= 4 × 5

= 20

Example 6

Find the value of (a + b)² when a = 3 and b = 1

Solution:

(a + b)² = (3 + 1)²

= 4²

= 16

📝 Substitution in Formulae

Formula: A formula shows how quantities are related. Substitute values to find unknown quantities.
Example 7

The formula for the area of a circle is A = πr². Find A when r = 5 (use π = 3.14)

Solution:

A = πr² = 3.14 × 5²

= 3.14 × 25

= 78.5

Example 8

The formula for speed is s = dt. Find s when d = 120 and t = 3

Solution:

s = 1203 = 40

📝 Negative Numbers in Substitution

Warning: Be careful with negative signs, especially when substituting negative values.
Example 9

Find the value of 3x + 7 when x = -4

Solution:

3x + 7 = 3(-4) + 7

= -12 + 7

= -5

Example 10

Find the value of x² - 2x when x = -3

Solution:

x² - 2x = (-3)² - 2(-3)

= 9 + 6

= 15

❓ Practice Questions

Q1: Find the value of 5x - 2 when x = 3.

Q2: Find the value of 2a + 3b when a = 4 and b = 5.

Q3: Find the value of 3x² when x = 4.

Q4: Find the value of (m + 2)(n - 1) when m = 5 and n = 4.

Q5: Use the formula A = 12bh. Find A when b = 10 and h = 7.

Q6: Find the value of x² + 2x - 3 when x = -2.

✅ Answers

  1. 13
  2. 23
  3. 48
  4. 18
  5. 35
  6. -3

🎯 Exam Tips

🧠 Problem-Solving Strategies

Problem-Solving

When substituting, always put values in brackets to avoid sign errors. Use BIDMAS carefully — powers come before multiplication. Check whether you need (2x)² or 2x² as they give different results.
Multi-Step Problem

The formula for kinetic energy is E = ½mv². Find the energy when m = 8 and v = 5, then find the new energy if v doubles.

Solution:

Step 1: E = ½ × 8 × 5² = ½ × 8 × 25 = 4 × 25 = 100 J

Step 2: If v doubles, v = 10: E = ½ × 8 × 10² = ½ × 8 × 100 = 400 J

The energy is 4 times larger (because v is squared, doubling v gives 2² = 4 times the energy).

⚠️ Common Errors

Watch Out!

1. Wrong: 2x² when x = 3 gives (2×3)² = 36 Correct: 2 × 3² = 2 × 9 = 18

2. Wrong: (-3)² = -9 Correct: (-3)² = (-3)×(-3) = +9

3. Wrong: Substituting x = -2 into 3x + 1 gives 3-2+1 = 2 Correct: 3(-2) + 1 = -6 + 1 = -5 (use brackets!)

✍️ 6-Mark Exam Question

Extended Answer

6 marks: The volume of a cone is V = 13πr²h. (a) Find V when r = 5 and h = 12, giving your answer in terms of π. (b) Find V when r = 5, h = 12 and π = 3.14. (c) The radius doubles and the height stays the same. What happens to the volume?

(a) V = 13 × π × 5² × 12 = 13 × π × 25 × 12 = 3003π = 100π cm³

(b) V = 100 × 3.14 = 314 cm³

(c) New V = 13π(2r)²h = 13π × 4r² × h = 4 times original volume.

Mark scheme: (a) 2 marks for correct substitution and simplification. (b) 1 mark for numerical answer. (c) 3 marks — 1 for (2r)², 1 for showing 4r², 1 for stating 4 times.

📊 AO3: Reason & Interpret

Reasoning and Interpretation

The surface area of a sphere is A = 4πr². Earth has radius approximately 6400 km.

(a) Calculate the surface area of Earth in terms of π.

(b) If 70% of the surface is water, write an expression for the water surface area.

(c) Explain why your answer to (a) is only approximate.

Answers: (a) A = 4π(6400)² = 4π × 40,960,000 = 163,840,000π km². (b) 0.7 × 163,840,000π = 114,688,000π km². (c) The radius 6400 km is approximate — Earth is not a perfect sphere.

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