AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 14

Equating Coefficients in an Algebraic Identity

📋 What this question tests

This question tests your ability to expand single brackets, understand the identity symbol ( ≡ ), equate corresponding coefficients of like terms (the x -terms and the constant terms), and find integer solutions subject to specific constraints (positive integers: 1, 2, 3...).

Question 14

Given a(9x + 2) ≡ 45x + 3b + c , find one possible set of values for positive integers a , b , and c .

Total Marks: 3 • AQA GCSE Mathematics (Higher Tier)

📐 Step-by-Step Calculation

Expand the left-hand side:
a(9x + 2) = 9ax + 2a
Set up the identity:
9ax + 2a ≡ 45x + (3b + c)
Equate the x-coefficients:
9a = 45 → a = 45 ÷ 9 = 5 [1 mark]
Equate constant terms:
2a = 3b + c
Since a = 5 , substitute into the equation:
2(5) = 3b + c ⇒ 3b + c = 10 [1 mark]
Pick positive integer values for b and c:
Remember positive integers must be whole numbers ≥ 1.
• If b = 1 : 3(1) + c = 10 ⇒ c = 7
• If b = 2 : 3(2) + c = 10 ⇒ c = 4
• If b = 3 : 3(3) + c = 10 ⇒ c = 1

✅ Acceptable Answers (Choose Any ONE Set)

Any one of the following complete sets earns all 3 marks:

Set 1: a = 5, b = 1, c = 7
Set 2: a = 5, b = 2, c = 4
Set 3: a = 5, b = 3, c = 1

Mark Breakdown:
• B1: Finding a = 5 (or finding values where 3b + c = 2 × a )
• B2: Setting up the correct linear equation 3b + c = 10
• B3: Fully correct valid integer set for a , b , and c

💡 Key Knowledge

Identity ( ≡ ): Means the expression is true for all values of x . Therefore, the coefficients of matching powers of x on both sides must be identical.
Positive Integers: The set {1, 2, 3, 4, ...} . Note that zero ( 0 ) and negative numbers are not positive integers!
Constant Terms: Any term without an x attached belongs together: 3b + c forms the entire constant term on the RHS.

🧠 Exam Technique & Examiner Tips

Highlight keywords: Box the words "positive integers" and "one possible set" so you don't lose simple marks on condition traps.
Work systematically: Once you determine 3b + c = 10 , test smallest positive values of b starting at b = 1 .
Check your result: Substitute all three numbers back into both sides to verify that both expressions simplify to the exact same polynomial:
5(9x + 2) = 45x + 10
45x + 3(2) + 4 = 45x + 10 ✓

❌ Common Errors to Avoid

Using zero: Choosing b = 0 , c = 10 . Zero is not a positive integer! This scores only 2 marks (B2).
Using negative values: Choosing b = 4 , c = -2 . While 3(4) + (-2) = 10 is true, -2 is negative, violating the positive integer rule (caps at B2).
Using fractions: Giving non-integer values like b = 5/3, c = 5 (caps at B2).
Expanding incorrectly: Forgetting to multiply the bracket's second term: writing 9ax + 2 instead of 9ax + 2a .

AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 14

Question

Mark scheme

How to answer it