AQA AS Level Mathematics Paper 2, June 2025: Question 6

Mark scheme

Show the mark scheme

Substitutes to obtain a quadratic 1.1a M1 x = 2 2(x2 − 4x + 1) − 11

equation, must have = 0, 2

in terms of x or y only. 0 = 4x −9x −9

PI by two correct values of x or

two correct values of y.

x = 3, x = −

3 When x = 3

Obtains x = 3, x = −

4 y = 2 3(2 ) − 4 3( ) + 1 = 7

Or 1.1b A1

Obtains y = 7, y = 3

8 When x = −

CSO 2

33 41

y =2 − −4 − +1 =

Substitutes at least one of their 1.1a M1 4 4 8

x or y values into either equation

to obtain y or x values

Obtains

x = 3, x = −

41 1.1b A1

y = 7, y =

ACF CSO

Do not ISW

Question 6 Total 4

How to answer it

Solving Linear and Quadratic Simultaneous Equations

📌 What this question tests

This question assesses your ability to solve a pair of simultaneous equations where one is linear and the other is non-linear (quadratic). It tests algebraic substitution, expansion and rearranging into standard quadratic form ( ax² + bx + c = 0 ), factorisation or use of the quadratic formula, and finding corresponding paired values of both variables.

Question 6: Complete Solution & Examiner Breakdown

Total Marks: 4 | Recommended Time: 4–5 minutes

📐 Step-by-Step Solution

Given the system of equations:
(1) x = 2y - 11
(2) y = 2x² - 4x + 1

1 Substitute (2) into (1):
Substitute the expression for y directly into equation (1):
x = 2(2x² - 4x + 1) - 11

2 Expand and set to zero:
x = 4x² - 8x + 2 - 11
x = 4x² - 8x - 9
Subtract x from both sides:
4x² - 9x - 9 = 0

3 Solve the quadratic equation:
Factorising: we need two numbers that multiply to 4 × (-9) = -36 and add to -9 . These are -12 and +3 .
4x² - 12x + 3x - 9 = 0
4x(x - 3) + 3(x - 3) = 0
(4x + 3)(x - 3) = 0
So, x = 3 or x = -3/4 (or -0.75 )

4 Find corresponding y-values:
Rearrange (1) for easy evaluation: 2y = x + 11 ⇒ y = (x + 11) / 2
• When x = 3 :
y = (3 + 11) / 2 = 14 / 2 = 7
• When x = -3/4 :
y = (-3/4 + 11) / 2 = (41/4) / 2 = 41/8 (or 5.125 )

✅ Correct Answer & Marks

Paired solutions are:

Pair 1: x = 3, y = 7
Pair 2: x = -3/4, y = 41/8
(Decimal equivalents x = -0.75, y = 5.125 are fully accepted)

Mark Scheme Breakdown:
• M1 (1.1a): Valid substitution to obtain a quadratic equation in one variable only, equated to 0 (e.g. 4x² - 9x - 9 = 0 ).
• A1 (1.1b): Correct values for the first variable: x = 3, x = -3/4 (or y = 7, y = 41/8 ).
• M1 (1.1a): Substituting at least one obtained value back into an equation to find the corresponding second variable.
• A1 (1.1b): Both pairs of values completely correct and clearly paired (CSO - Correct Solution Only).

💡 Key Knowledge

Path of least resistance: Since y is already isolated in the quadratic ( y = 2x² - 4x + 1 ), substituting it into the linear equation avoids having to square a binomial.
Standard quadratic form: You cannot solve a quadratic equation until all terms are on one side and equal to zero: ax² + bx + c = 0 .
Exact values: Unless specified otherwise, keep solutions as exact fractions ( -3/4 , 41/8 ) or terminating decimals to prevent rounding errors.

🧠 Exam Technique & "Fully Justify"

Show intermediate working: The command phrase "Fully justify your answer" means you must show the formed quadratic and your method of solving it (factorisation or quadratic formula). Do not just write calculator outputs.
Pairing values: Clearly state which y goes with which x . Writing coordinates like (3, 7) and (-3/4, 41/8) is an excellent way to avoid ambiguity.
Double check: Substitute your solutions into equation (2) to verify: 2(3)² - 4(3) + 1 = 18 - 12 + 1 = 7 . Perfectly verified!

❌ Common Errors & Traps

Sign slip when rearranging: Forgetting to subtract x from -8x , mistakenly obtaining -7x instead of -9x .
Expansion bracket error: Multiplying only the first term by 2, e.g. 2(2x² - 4x + 1) = 4x² - 4x + 1 .
Stopping halfway: Finding x = 3 and x = -3/4 and forgetting to calculate the corresponding y values, dropping 2 out of the 4 marks.
Mismatched pairs: Pairing the larger x with the wrong y value.

Topics

Pure Mathematics · B: Algebra and functions