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GCSE substitution practice and worked examples

Replace letters with their given values to evaluate expressions and formulae. Pay special attention to negative values and powers.

Foundation and Higher · Useful for students building fluency and applying the method in GCSE questions.

Core pathway. Core GCSE methods are useful across Foundation and Higher. Start with the core examples and questions.

Higher extension. No questions in this selected set are labelled Higher extension. These core methods also support Higher revision.

3 worked examples and 8 original practice questions · Allow 25–40 minutes · Read online or print. Higher extensions are labelled. No account or email needed.

Learn the methodTry the questions

Printing includes the method, examples and space for working, followed by a separate answer section.

A useful approach

  1. Write the formula before inserting any numbers.
  2. Use brackets around substituted negative values; calculate powers before multiplication and addition.
  3. Check the size, sign and units of the result against the original formula.

A mistake to watch for

Squaring a negative number in brackets gives a positive result. Multiplying a number by two is different from squaring it.

Learn each method, then practise it

Read each line of working and explain why it follows from the previous line.

Replacing positive and negative values

Put each value in place of its letter. Bracket negative values so the signs remain clear.

Worked example 1

Substitution · Core GCSE skill · No calculator · 2 marks

E = 3a − 2b

Work out the value of E when a = −4 and b = 5.

  1. Substitute the values: E = 3 × (-4) - 2 × 5.
  2. E = -12 - 10 = -22.
Answer: -22
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Now practise this method: Question 1 · Question 2 · Question 3

Powers and order of operations

Calculate powers before multiplication and addition. A substituted negative number is squared as a whole when it replaces x in x².

Worked example 2

Substitution · Core GCSE skill · Calculator allowed · 2 marks

Work out the value of  x2 + 7x  when x = 6.

  1. x2 = 6 × 6 = 36.
  2. 7x = 7 × 6 = 42.
  3. 36 + 42 = 78.
Answer: 78
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Now practise this method: Question 4 · Question 5

Use a formula backwards

When the output is known, substitute that value and solve the resulting equation for the unknown input. Apply inverse operations to both sides, then check in the formula.

Worked example 3

Substitution · Core GCSE skill · No calculator · 4 marks

F = 5v + 20

  1. Work out the value of F when v = 6.
  2. Work out the value of v when F = 95.
  1. a) F = 5 × 6 + 20 = 30 + 20 = 50.
  2. b) Substitute F = 95: 95 = 5v + 20.
  3. b) Subtract 20: 5v = 75, so v = 15.
  4. Check: 5 × 15 + 20 = 95.
Answer: (a) 50   (b) 15
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Now practise this method: Question 6 · Question 7 · Question 8

Your practice questions

Write your working on paper. Marks indicate how much working to show; these questions are self-marked and do not change saved practice results. Use squared paper for drawing questions.

Question 6

Substitution · Core GCSE skill · Calculator allowed · 4 marks

A tool hire company uses this rule to work out the cost of hiring a floor sander.

Total cost = £45 × number of days + £60 deposit

  1. Meera hires the sander for 7 days.
    Work out her total cost.
  2. Liam's total cost is £555.
    Work out the number of days Liam hires the sander for.

Check answer and working for question 6

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Worked answers: Substitution

Compare the reasoning as well as the final answer. Another correct method is valid. If a step is unclear, revisit an example before trying a similar question.

Answer 1

Show answer and working for question 1
  1. Substitute the values: T = 4 × 5 + 3 × 2.
  2. T = 20 + 6 = 26.
Answer: 26
Back to question 1

Answer 2

Show answer and working for question 2
  1. Substitute t = 3.4: v = 5 × 3.4 - 8.
  2. v = 17 - 8 = 9.
Answer: 9
Back to question 2

Answer 3

Show answer and working for question 3
  1. a) 4 × 2 = 8 and the letters are c and d, so 8cd.
  2. b) Substitute into 8cd: 8 × 3 × 1.5
  3. b) 8 × 3 = 24, then 24 × 1.5 = 36.
Answer: (a) 8cd   (b) 36
Back to question 3

Answer 4

Show answer and working for question 4
  1. a) y = 35 ÷ 5 = 7.
  2. b) Square first: y2 = 49.
  3. b) Then multiply by 2: 2 × 49 = 98.
Answer: (a) 7   (b) 98
Back to question 4

Answer 5

Show answer and working for question 5
  1. a) 2 × (-1) = -2 and 3 × 4 = 12, so -2 - 12 = -14.
  2. b) a2 = (-3) × (-3) = 9, so 4a2 = 36.
  3. b) 36 + (-7) = 29.
Answer: (a) -14   (b) 29
Back to question 5

Answer 6

Show answer and working for question 6
  1. a) 45 × 7 = 315, then 315 + 60 = 375. Total cost £375.
  2. b) Take off the deposit: 555 - 60 = 495.
  3. b) Divide by the daily rate: 495 ÷ 45 = 11 days.
  4. Check: 45 × 11 + 60 = 555.
Answer: (a) £375   (b) 11
Back to question 6

Answer 7

Show answer and working for question 7
  1. a) v = 7 + 9.8 × 5 = 7 + 49 = 56.
  2. b) Substitute: 87 = 12 + 5t.
  3. b) Subtract 12: 5t = 75, so t = 15.
  4. Check: 12 + 5 × 15 = 87.
Answer: (a) 56   (b) 15
Back to question 7

Answer 8

Show answer and working for question 8
  1. a) m = 3 × (-5) - 2 × 4 = -15 - 8 = -23.
  2. b) Substitute: 19 = 3n - 2 × (-2) = 3n + 4.
  3. b) Subtract 4: 3n = 15, so n = 5.
  4. Check: 3 × 5 - 2 × (-2) = 15 + 4 = 19.
Answer: (a) -23   (b) 5
Back to question 8

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