✓ Revise Right course
To substitute into a formula, replace each letter with its value; to rearrange, do the same to both sides to isolate the new subject. 10 questions answered.
Formulae and Expressions: Use and Manipulation is part of Use and manipulation of formulae and expressions in GCSE AQA Higher Further Maths.
3 min
Reading time
24
Lessons in this topic
GCSE AQA Higher
Pathway
Reviewed by Revise Right
Content review
To substitute into a formula, replace each letter with its value; to rearrange, do the same to both sides to isolate the new subject. 10 questions answered.
Use this lesson to stay inside Use and manipulation of formulae and expressions while connecting the detail back to the wider Algebra topic.
Read the core explanations from Formulae and Expressions: Use and Manipulation before testing yourself from memory.
Every time you calculate speed, area, or interest, you are using a formula. The ability to substitute into, rearrange, expand, and factorise algebraic expressions is one of the most frequently tested skills across all GCSE Maths papers.
Think of a formula as a machine: you feed in known values, and it produces the answer you need. Rearranging the formula is like rewiring the machine so a different quantity comes out. Both skills appear on Paper 1 (non-calculator) and Papers 2–3 (calculator), so you need to be confident with the algebra regardless of whether a calculator is available.
Substitution means replacing each letter in a formula with its given numerical value, then evaluating the result. The key discipline is to write the substitution out explicitly before you calculate — this earns method marks even if arithmetic slips occur later.
A common source of errors is negative values. When substituting a negative number, always wrap it in brackets: write $(-3)^2$, not $-3^2$. Without brackets, $-3^2 = -9$, but $(-3)^2 = 9$ — two very different results.
Worked example: Substituting into a Formula
The formula for the area of a trapezium is $A = \frac{1}{2}(a + b)h$. Find $A$ when $a = 5$, $b = 9$, and $h = 4$.
Answer: $A = 28 \text{ (square units)}$
Rearranging a formula means isolating a different variable — changing the subject. The method mirrors solving an equation: whatever you do to one side, you do to the other. Work outward from the variable you want, undoing operations in reverse order.
Worked example: Rearranging a Formula — One Step
Make $u$ the subject of $v = u + at$.
Answer: $u = v - at$
Worked example: Rearranging a Formula — Multiple Steps
Make $r$ the subject of $A = \pi r^2$.
Answer: $r = \sqrt{\frac{A}{\pi}}$
Expanding brackets removes the bracket by multiplying every term inside by the term outside. For a single bracket, multiply each term inside by the factor outside. For double brackets, every term in the first bracket multiplies every term in the second.
Worked example: Expanding Double Brackets
Expand and simplify $(2x + 3)(x - 5)$.
Answer: $2x^2 - 7x - 15$
Factorising is the reverse of expanding — you rewrite an expression as a product of factors. There are three main types you need at GCSE: taking out a common factor, factorising quadratics of the form $x^2 + bx + c$, and factorising quadratics of the form $ax^2 + bx + c$.
Always check for a common factor first. If every term shares a factor, take it outside a bracket before attempting any other method. This keeps the numbers smaller and reduces errors in later steps.
Worked example: Factorising — Common Factor
Factorise $6x^2 - 9x$ fully.
Answer: $3x(2x - 3)$
Worked example: Factorising a Quadratic — $x^2 + bx + c$
Factorise $x^2 + 7x + 12$.
Answer: $(x + 3)(x + 4)$
Worked example: Factorising a Quadratic — $ax^2 + bx + c$
Factorise $2x^2 + 7x + 3$.
Answer: $(2x + 1)(x + 3)$
Exam questions on formulae and expressions often chain several skills together. You might need to substitute, then rearrange, or expand and then factorise. Recognising which skill is needed — and in which order — is what separates a confident answer from a confused one.
The most reliable way to lose marks on these topics is through sign errors and skipped steps. Examiners award method marks for correct working even when the final answer is wrong — but only if the working is visible. Write every substitution and every algebraic step on a separate line.
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.
Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
To expand double brackets, multiply every term in the first bracket by every term in the second, multiplying the First, Outer, Inner and Last terms. For (2x + 3)(x − 5), this gives 2x² − 10x + 3x − 15 = 2x² − 7x − 15.
To substitute into a formula, write the formula, replace each letter with its value, then evaluate using the order of operations. For the trapezium area A = ½(a + b)h with a = 5, b = 9 and h = 4, A = ½ × 14 × 4 = 28 square units.
Substitution means replacing variables in a formula or expression with specific numerical values to calculate a result. Writing the substitution out before calculating earns method marks even if arithmetic slips occur later.
To rearrange a formula, isolate the new subject by doing the same to both sides, undoing operations in reverse order. For example, to make u the subject of v = u + at, subtract at from both sides to get u = v − at.
To make r the subject of A = πr², undo the operations in reverse order: divide both sides by π to get A/π = r², then take the square root of both sides to get r = √(A/π). Since r is a length, take the positive root.
Negative numbers must go in brackets when substituting so the sign is squared too. Without brackets, −3² = −9, but (−3)² = 9, which are two very different results.
To make u the subject of v = u + at, subtract at from both sides to get v − at = u, then write the subject on the left: u = v − at. The method mirrors solving an equation, doing the same to both sides.
(2x + 3)(x − 5) expands to 2x² − 7x − 15. The First, Outer, Inner and Last products are 2x², −10x, 3x and −15, and −10x + 3x = −7x.
x² + 7x + 12 factorises to (x + 3)(x + 4), because 3 and 4 multiply to 12 and add to 7. Expanding (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 is a check.
A formula with a squared term is rearranged by undoing the operations in reverse order, with a square root last. For A = πr², divide both sides by π to get A/π = r², then take the square root: r = √(A/π).