AQA GCSE Mathematics: Algebra — Revision Notes

Condensed for the final weeks. For the full explanation, use the Algebra study guide.

Notation, vocabulary and manipulation (3.2.1)

Vocabulary: expression, equation, formula, inequality, term, factor, and identity (true for every value, unlike an equation which is only true for specific values — tested directly). Identities and the equation/identity distinction are Foundation content (A3, A6); only algebraic proof is Higher.

Notation: ab, 3y, a², 1/a in place of the longer expressions. Simplest form always expected — leaving “2×x + 3×x” instead of “5x” loses marks even with correct method.

Manipulation: collecting like terms, expanding brackets, factorising, expanding two binomials, factorising x²+bx+c (including the difference of two squares); Higher tier — expanding products of more than two binomials, factorising quadratics ax²+bx+c, algebraic fractions (A4); rearranging a formula to change the subject.

Functions (A7): interpreting simple expressions as functions with inputs and outputs is Foundation; (Higher) function notation f(x), inverse f⁻¹(x) and composite fg(x).

Graphs (3.2.2)

y = mx + c    -- m = gradient, c = y-intercept

Common trap: given 2y = 4x + 6, the gradient is 2, not 4 — rearrange to y = mx + c form first.

Function type	Shape
Linear	Straight line
Quadratic	Parabola — roots, intercepts, turning point
Cubic	S-curve
Reciprocal	Two curved branches
Exponential (Higher)	Curve growing/decaying
Trigonometric (Higher)	Periodic wave

Linear, quadratic, simple cubic and reciprocal graphs are Foundation; exponential and trigonometric graphs are Higher only (A12, A14).

Straight lines (A9): using y = mx + c for parallel lines and finding the equation of a line through two points, or through one point with a given gradient, is Foundation; perpendicular lines are Higher.

Higher tier also: gradients/areas under graphs (kinematics, financial contexts), equation of a circle centred at the origin, tangent to a circle.

Solving equations and inequalities (3.2.3)

Method	When
Factorising (Foundation for x² + bx + c)	Quadratic has clean integer roots
Completing the square (Higher)	No clean roots, or need the turning point
Quadratic formula (Higher)	Always works, any quadratic
Graph	Approximate solutions, any case
Iteration (Higher)	Numerical approximation

Simultaneous equations: linear/linear (Foundation); linear/quadratic (Higher) — substitute the linear equation into the quadratic.

Inequalities: linear (all tiers), quadratic (Higher) — represent on a number line; Higher tier also uses set notation.

Practise solving the SAME equation both algebraically and graphically — the specification deliberately tests both, since graphical solutions are approximate and algebraic ones exact.

Sequences (3.2.4)

Sequence type	Rule
Arithmetic	Constant common difference
Triangular/square/cube	Named number patterns
Fibonacci-type	Each term = sum of previous two
Quadratic	Second differences constant; finding its nth term is Higher
Geometric	Constant common ratio (rational at Foundation; a surd ratio is Higher)

Recognising and using Fibonacci-type, quadratic and simple geometric sequences is Foundation (A24); other sequences, a surd common ratio and the nth term of a quadratic sequence (A25) are Higher.

nth term of a linear sequence: the coefficient of n is always the common difference — check quickly against the first few terms rather than memorising per question.

Worked example: nth term

Find the nth term of 5, 8, 11, 14, …

Common difference = 3, so coefficient of n is 3
3n gives: 3, 6, 9, 12 -- each 2 less than the actual sequence
So nth term = 3n + 2
Check: n=1 -> 5, n=2 -> 8 (correct)

Worked example: solving a linear-quadratic simultaneous pair (Higher)

Solve y = x + 1 and y = x² − 1 simultaneously.

x + 1 = x^2 - 1
0 = x^2 - x - 2
0 = (x - 2)(x + 1)
x = 2 or x = -1

Substitute back: x=2 -> y=3;  x=-1 -> y=0
Solutions: (2, 3) and (-1, 0)

How the four sub-sections connect

Notation/manipulation (3.2.1) is the toolkit everything else depends on. Graphs (3.2.2) is manipulation made visual — y = mx + c is a straight-line graph; a quadratic’s roots are the solutions from factorising/the formula. Solving (3.2.3) is manipulation used to find an answer, testable algebraically and graphically. Sequences (3.2.4) use the same algebraic skills to build and rearrange nth-term expressions.

Worked example: quadratic sequence nth term (Higher)

Find the nth term of 3, 8, 15, 24, 35, …

First differences:  5, 7, 9, 11        (not constant -- not linear)
Second differences: 2, 2, 2            (constant -- quadratic sequence)

Coefficient of n^2 = second difference / 2 = 2 / 2 = 1
So the sequence starts from n^2: 1, 4, 9, 16, 25
Compare to actual: 3, 8, 15, 24, 35
Difference each time: +2, +4, +6, +8, +10 -- this is 2n

So nth term = n^2 + 2n
Check: n=1 -> 1+2=3 (correct); n=2 -> 4+4=8 (correct)

Finding the second differences first, using them to identify the n² coefficient, and then finding the remaining linear part by comparison is the standard, reliable method for quadratic nth-term questions – attempting to spot the pattern directly, without this structured approach, is where most errors happen under exam pressure.

Key terms

Equation — true only for specific values of the unknown. Identity — true for every value. Gradient — rate of change; the m in y = mx + c. nth term — a formula generating any term in a sequence from its position.

Common mistakes

• Leaving an answer not in simplest form.
• Confusing an equation with an identity.
• Misreading gradient from a rearranged equation (2y = 4x + 6 → gradient is 2, not 4).
• Forgetting which content is Higher-tier-only — wastes revision time (Foundation) or leaves gaps (Higher).
• Memorising nth-term formulas per question instead of deriving from the common difference.

Quick self-test

1. Simplify 4a × 3b, and rearrange 2y = 6x − 8 to identify the gradient and y-intercept.
2. Solve x² − x − 6 = 0 by factorising.
3. Find the nth term of 7, 11, 15, 19.
4. (Higher) Solve the simultaneous pair y = 2x + 2 and y = x² − 1.
5. List three pieces of Higher-tier-only Algebra content from memory.

Answers: 1. 4a × 3b = 12ab; 2y = 6x − 8 gives y = 3x − 4, so the gradient is 3 and the y-intercept is −4. 2. (x − 3)(x + 2) = 0, so x = 3 or x = −2. 3. The common difference is 4, and 4n gives 4, 8, 12, 16, each 3 less than the sequence, so the nth term is 4n + 3. 4. 2x + 2 = x² − 1 gives x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0 and x = 3 or x = −1; substituting back gives the solutions (3, 8) and (−1, 0). 5. Any three of: factorising ax² + bx + c; solving quadratics by completing the square or the quadratic formula; perpendicular lines; exponential and trigonometric graphs; inverse and composite functions; algebraic proof; the equation of a circle and its tangent; linear/quadratic simultaneous equations; iteration; quadratic inequalities; the nth term of a quadratic sequence.

Related resources

• Algebra study guide
• Algebra practice questions

Official syllabus

AQA, GCSE Mathematics (8300) Specification, for first teaching 2015, subject content section 3.2 Algebra, https://www.aqa.org.uk/subjects/mathematics/gcse/mathematics-8300/specification/subject-content/3.2-algebra, fetched and verified in full 2026-09-02.