Mathematics · AQA 8300 §A1

Expressions and Formulae

Spec: AQA 8300 §A1 ⭐⭐ ⏱ 40 mins Boards: AQA · Edexcel · OCR Grade 9

Use algebraic vocabulary correctly: term, expression, equation, formula, identity
Collect like terms and simplify expressions including index notation
Substitute values into expressions and formulae accurately
Form expressions and formulae from word descriptions and geometric contexts
Change the subject of a formula including with powers and square roots

🔑 Core Concepts

Algebraic Vocabulary

Before manipulating algebra, you must understand precisely what each word means. Examiners award marks for using the correct term — and lose marks for confusion between an equation and an identity.

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DEFINITION — Term

A term is a single algebraic unit: a number, a variable, or the product of numbers and variables. Examples: $5x$, $-3y^2$, $7$. Terms are separated by $+$ or $-$ signs.

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DEFINITION — Expression

An expression is a collection of terms combined by $+$, $-$, $\times$, $\div$. It has no equals sign. Example: $3x^2 - 5x + 2$.

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DEFINITION — Equation

An equation has an equals sign and is true only for specific values of the variable. Example: $2x + 3 = 11$ is true only when $x = 4$.

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DEFINITION — Formula

A formula shows a relationship between two or more quantities. It can be rearranged and applied to find one quantity from others. Example: $A = \pi r^2$ gives the area of a circle.

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DEFINITION — Identity (≡)

An identity is true for all values of the variable. The symbol $\equiv$ means "is identically equal to". Example: $3(x+2) \equiv 3x + 6$ holds for every value of $x$.

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COMMON MISTAKE — $=$ vs $\equiv$

Writing $3(x+2) = 3x+6$ uses $=$ when $\equiv$ is required. Using $=$ implies it is only true for certain values. If a question asks you to show an identity, always use $\equiv$.

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EXAM TIP — Spotting Identities

If the examiner asks "Prove that $A \equiv B$", expand and simplify both sides independently and show they are equal. An identity needs no specific value — it must work for all values.

Collecting Like Terms

Like terms share the exact same variable part (including powers). You can only add or subtract like terms — just as you can only add apples to apples. The coefficients change; the variable part stays the same.

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DEFINITION — Like Terms

Terms whose variable parts are identical. For example: $3x^2$ and $-7x^2$ are like terms (both have $x^2$), but $3x^2$ and $3x$ are not like terms.

✏️

WORKED EXAMPLE — Simplifying

Simplify $5a^2b - 3ab^2 + 2a^2b + ab^2$.

Group like terms: $(5a^2b + 2a^2b) + (-3ab^2 + ab^2)$
$= 7a^2b - 2ab^2$
Note: $a^2b$ and $ab^2$ are not like terms because the powers on $a$ and $b$ differ.

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EXAM TIP — Index Terms

$x^2$ and $x^3$ are different terms. So are $xy$ and $x^2y$. Always compare the full variable part, including all indices, before combining terms.

Index Notation in Algebra

Indices (powers) appear constantly in algebra. Knowing the laws prevents the most common errors in simplification.

Index Laws (Algebra)

$$a^m \times a^n = a^{m+n}$$

$$a^m \div a^n = a^{m-n}$$

$$(a^m)^n = a^{mn}$$

$$a^0 = 1 \quad (a \neq 0)$$

$$a^{-n} = \dfrac{1}{a^n}$$

$a$ = base $m, n$ = indices (powers)

✗

COMMON MISTAKE — Multiplying Coefficients and Powers

$3x^2 \times 4x^3 \neq 12x^6$. The coefficients multiply ($3 \times 4 = 12$) but the indices add: $x^2 \times x^3 = x^5$. Correct answer: $12x^5$.

Substitution into Expressions and Formulae

Substitution means replacing a letter with a number. The golden rule: write the substituted expression fully before evaluating, and always use brackets when substituting negative values.

🧠

MEMORY TRICK — Brackets for Negatives

When substituting a negative number, always wrap it in brackets: if $x = -3$, then $x^2 = (-3)^2 = 9$, not $-3^2 = -9$. The bracket removes ambiguity about whether the negative is inside the power.

✏️

WORKED EXAMPLE — Multi-variable Substitution

Find the value of $p = \dfrac{2m^2 - 3n}{m + n}$ when $m = -2$ and $n = 5$.

Numerator: $2(-2)^2 - 3(5) = 2(4) - 15 = 8 - 15 = -7$
Denominator: $(-2) + 5 = 3$
$p = \dfrac{-7}{3} = -2\tfrac{1}{3}$

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EXAM TIP — Show Every Step

In a substitution question, write the formula, write with values substituted (in brackets), then evaluate step by step. This earns method marks even if you make an arithmetic error.

Forming Expressions and Formulae

Many exam questions describe a real-world or geometric situation in words and ask you to write an algebraic expression or formula. The skill is translating language into algebra precisely.

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DEFINITION — Forming an Expression

Assign a variable to the unknown quantity. Translate each operation described in words using the correct algebraic operation. Do not include an equals sign (that would make it an equation).

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WORKED EXAMPLE — Geometric Context

A rectangle has length $(3x + 1)$ cm and width $(x - 2)$ cm. Write an expression for the perimeter.

Perimeter $= 2l + 2w = 2(3x+1) + 2(x-2) = 6x + 2 + 2x - 4 = 8x - 2$ cm.

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EXAM TIP — State Your Variable

If forming an expression, always define your variable first: "Let $n$ represent the number of items." This earns marks and avoids ambiguity.

Changing the Subject of a Formula

Changing the subject means rearranging a formula so a different variable is isolated on the left-hand side. The strategy is always to reverse the operations in reverse order, using inverse operations — like undoing knots from the outside in.

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DEFINITION — Subject of a Formula

The subject is the variable on its own on one side of the formula. In $v = u + at$, the subject is $v$. To make $a$ the subject, rearrange to $a = \dfrac{v - u}{t}$.

Strategy: Changing the Subject

$$\text{1. Identify the target variable.}$$

$$\text{2. Isolate the term containing it.}$$

$$\text{3. Undo surrounding operations (reverse order, inverse operations).}$$

$$\text{4. If squared: take square root. If square-rooted: square both sides.}$$

$$\text{5. If the variable appears twice: factorise it out first.}$$

✏️

WORKED EXAMPLE — Powers

Make $r$ the subject of $A = \pi r^2$.

Divide both sides by $\pi$: $\dfrac{A}{\pi} = r^2$
Take square root: $r = \sqrt{\dfrac{A}{\pi}}$
Note: since $r$ is a length, we take the positive square root only.

✏️

WORKED EXAMPLE — Square Roots

Make $l$ the subject of $T = 2\pi\sqrt{\dfrac{l}{g}}$.

Step 1 — Divide both sides by $2\pi$: $\dfrac{T}{2\pi} = \sqrt{\dfrac{l}{g}}$
Step 2 — Square both sides: $\dfrac{T^2}{4\pi^2} = \dfrac{l}{g}$
Step 3 — Multiply both sides by $g$: $l = \dfrac{gT^2}{4\pi^2}$

⚠️

IMPORTANT — Variable Appearing Twice

When the target variable appears in two places, you must factorise. Example: make $x$ the subject of $ax + b = cx + d$.
$ax - cx = d - b$
$x(a - c) = d - b$
$x = \dfrac{d - b}{a - c}$

✗

COMMON MISTAKE — Forgetting ± When Taking Square Roots

When you take the square root of both sides in a rearrangement, the answer is technically $\pm\sqrt{...}$. In context (e.g. $r$ is a radius), only the positive root applies — but always consider which root is physically meaningful.

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EXAM TIP — Multi-Step Rearrangements

In Grade 9 questions, rearrangements will involve fractions, powers, roots, and sometimes the variable appearing twice. Work one step at a time, showing each line. Never try to do two operations in one line — that is where errors happen.

🗺️ Visual Notes

Expressions
& Formulae

Vocabulary

Term — single unit
Expression — no = sign
Equation — specific values
Identity ≡ — all values

Simplifying

Collect like terms only
Same variable part required
Index laws: add when multiply
Coefficients multiply separately

Substitution

Replace letters with numbers
Brackets around negatives
Work numerator/denominator separately
Show every step for marks

Forming

Define variable first
Translate words → operations
Geometric contexts common
No = sign in expression

Changing Subject

Isolate target variable
Inverse operations, reverse order
Square/square-root carefully
Two occurrences → factorise

Index Notation

Multiply → add powers
Divide → subtract powers
Power of power → multiply
$a^0 = 1$, $a^{-n} = 1/a^n$

Comparing Algebraic Types

Type	Has = sign?	True for…	Symbol	Example
Expression	No	N/A — not a statement	—	$3x^2 - 5x + 2$
Equation	Yes	Specific value(s)	$=$	$2x + 3 = 11$
Formula	Yes	All valid inputs	$=$	$A = \pi r^2$
Identity	Yes	All values of variable	$\equiv$	$x^2 - 1 \equiv (x+1)(x-1)$

Rearranging a Formula — Decision Flow

Identify target variable

→

Does it appear twice?
YES → Collect on one side & factorise

→

Isolate the term with target variable
(move everything else to opposite side)

→

Is the variable squared?
YES → take $\sqrt{\ }$ ; Is it under $\sqrt{}$? → square both sides

→

Divide by any remaining coefficient
Target variable is now the subject

✏️ Worked Examples

Grade 4–5

Simplify $4x^2 + 3x - 2x^2 - 7x + 5$ and then substitute $x = -2$ to find the value.

Identify like terms

Group terms with $x^2$: $4x^2 - 2x^2$
Group terms with $x$: $3x - 7x$
Constant: $+5$

Collect like terms

$4x^2 - 2x^2 = 2x^2$
$3x - 7x = -4x$
Simplified expression: $2x^2 - 4x + 5$

Substitute $x = -2$ with brackets

$2(-2)^2 - 4(-2) + 5$

Evaluate each term

$2(4) + 8 + 5 = 8 + 8 + 5 = 21$

Simplified: $2x^2 - 4x + 5$. When $x = -2$: value $= 21$.

Grade 6–7

Make $v$ the subject of $E = \dfrac{1}{2}mv^2$, then find $v$ when $E = 900$ J and $m = 8$ kg.

Multiply both sides by 2

$2E = mv^2$

Divide both sides by $m$

$\dfrac{2E}{m} = v^2$

Take the positive square root (speed is positive)

$$v = \sqrt{\dfrac{2E}{m}}$$

Substitute values

$v = \sqrt{\dfrac{2 \times 900}{8}} = \sqrt{\dfrac{1800}{8}} = \sqrt{225} = 15$

$v = \sqrt{\dfrac{2E}{m}}$. When $E = 900$, $m = 8$: $v = 15$ m/s.

Grade 9

Make $x$ the subject of $\dfrac{ax + b}{cx - d} = k$, where $a, b, c, d, k$ are constants and $cx \neq d$.

Multiply both sides by $(cx - d)$ to clear the fraction

$ax + b = k(cx - d)$

Expand the right-hand side

$ax + b = kcx - kd$

Collect all $x$-terms on the left

$ax - kcx = -kd - b$

Factorise the left-hand side — $x$ appears twice

$x(a - kc) = -kd - b$

Divide both sides by $(a - kc)$, provided $a \neq kc$

$$x = \dfrac{-kd - b}{a - kc} = \dfrac{-(kd + b)}{a - kc}$$

$x = \dfrac{-(kd + b)}{a - kc}$. The key Grade 9 skill here is recognising that $x$ appears on both sides after expanding, so you must factorise before dividing.

❓ Exam Questions

Q1 1 mark

State whether $5(x + 2) = 5x + 10$ is an equation, an expression, or an identity. Give a reason.

Mark scheme:
Identity ✓ — because $5(x + 2) \equiv 5x + 10$ holds for all values of $x$ (it is simply expanding the bracket). [1 mark: correct term with valid reason]

Q2 2 marks

Simplify $3x^2y + 5xy^2 - x^2y + 2xy^2 - 4xy$.

Working:
$x^2y$ terms: $3x^2y - x^2y = 2x^2y$ ✓
$xy^2$ terms: $5xy^2 + 2xy^2 = 7xy^2$ ✓
$xy$ term: $-4xy$ (no like terms to combine)
Answer: $2x^2y + 7xy^2 - 4xy$
[1 mark for any two pairs collected correctly; 1 mark for fully correct final expression]

Q3 3 marks

The formula for the surface area of a closed cylinder is $S = 2\pi r^2 + 2\pi r h$. Make $h$ the subject.

Working:
$S - 2\pi r^2 = 2\pi r h$ ✓ (subtract $2\pi r^2$)
$\dfrac{S - 2\pi r^2}{2\pi r} = h$ ✓ (divide by $2\pi r$)
Answer: $h = \dfrac{S - 2\pi r^2}{2\pi r}$ ✓
[1M subtract $2\pi r^2$; 1M divide by $2\pi r$; 1M correct final form]

Q4 4 marks

A trapezium has parallel sides of length $a$ cm and $b$ cm, and height $h$ cm. The area formula is $A = \frac{1}{2}(a + b)h$.

(a) Make $b$ the subject. [2 marks]
(b) Given $A = 60$ cm², $a = 8$ cm, $h = 6$ cm, find the value of $b$. [2 marks]

(a) Rearranging:
$2A = (a + b)h$ ✓
$\dfrac{2A}{h} = a + b$
$b = \dfrac{2A}{h} - a$ ✓

(b) Substituting:
$b = \dfrac{2(60)}{6} - 8 = 20 - 8 = 12$ ✓✓
Answer: $b = 12$ cm
[Part (a): 1M multiply by 2 and divide by $h$; 1M isolate $b$. Part (b): 1M correct substitution into rearranged formula; 1M correct value]

Q5 4 marks

Given $s = ut + \dfrac{1}{2}at^2$, make $a$ the subject. [2 marks]
Hence find $a$ when $s = 84$ m, $u = 6$ m/s, $t = 4$ s. [2 marks]

Rearranging:
$s - ut = \dfrac{1}{2}at^2$ ✓
$a = \dfrac{2(s - ut)}{t^2}$ ✓

Substituting:
$a = \dfrac{2(84 - 6 \times 4)}{4^2} = \dfrac{2(84 - 24)}{16} = \dfrac{2 \times 60}{16} = \dfrac{120}{16} = 7.5$ ✓✓
Answer: $a = 7.5$ m/s²
[1M subtract $ut$; 1M multiply by 2 and divide by $t^2$; 1M correct substitution; 1M correct value]

Q6 6 marks

Make $r$ the subject of $V = \dfrac{4}{3}\pi r^3$ [2 marks], then make $t$ the subject of $p = \dfrac{mt}{r + t}$ [4 marks].

Part 1 — Make $r$ subject of $V = \frac{4}{3}\pi r^3$:
$3V = 4\pi r^3$ ✓
$r^3 = \dfrac{3V}{4\pi}$ ✓
$r = \sqrt[3]{\dfrac{3V}{4\pi}}$

Part 2 — Make $t$ subject of $p = \dfrac{mt}{r+t}$:
$p(r + t) = mt$ ✓ (multiply both sides by $(r+t)$)
$pr + pt = mt$ ✓ (expand)
$pr = mt - pt$ ✓ (collect $t$ terms)
$pr = t(m - p)$ ✓ (factorise)
$t = \dfrac{pr}{m - p}$ ✓
[Part 1: 2 marks. Part 2: 1M multiply to clear fraction; 1M expand; 1M factorise; 1M divide to isolate $t$]

⭐ Grade 9 Model Answers

Full Model Answer: Q6 (Part 2) — $t = \dfrac{pr}{m-p}$ from $p = \dfrac{mt}{r+t}$

This question tests three Grade 9 skills simultaneously: clearing a fraction, recognising that the target variable appears on both sides, and factorising to isolate it. Here is what a full-marks answer looks like.

Grade 9 — Full Marks

Make $t$ the subject of $p = \dfrac{mt}{r + t}$

Multiply both sides by $(r + t)$ — earns M1

$p(r + t) = mt$
Why this earns marks: you have cleared the fraction. Never try to cross-multiply unless it is truly a single fraction on each side.

Expand the bracket — earns M1

$pr + pt = mt$
Why: expanding reveals that $t$ now appears on both sides, which is the key difficulty of this question.

Collect all $t$-terms on one side — earns M1

$pr = mt - pt$
Why: subtracting $pt$ from both sides gathers the two $t$-terms together so they can be factorised.

Factorise $t$ from the right-hand side — earns M1

$pr = t(m - p)$
Why: this is the crucial step. A common error is to write $t(m-p)$ incorrectly as $t \cdot m - p$ — you must factor out $t$ from every term.

Divide both sides by $(m - p)$ — earns final A1

$$t = \dfrac{pr}{m - p}$$
Note: valid only when $m \neq p$. In an exam, you do not need to state this unless asked.

$t = \dfrac{pr}{m - p}$ [6/6 marks]. Key moves: clear fraction → expand → collect → factorise → divide.

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Why Grade 9 students score full marks here

They recognise within the first line of working that $t$ will appear on both sides once the fraction is cleared. This tells them immediately that factorisation will be needed at the end — so they plan their working accordingly rather than being surprised mid-solution.

📋 Revision Sheet

Key Definitions

Term	Meaning
Term	Single algebraic unit (e.g. $5x$)
Expression	Terms combined; no $=$ sign
Equation	True for specific values only
Formula	Relationship between quantities
Identity $\equiv$	True for ALL values
Subject	Variable alone on LHS

Essential Formulae

Index Laws: $a^m \times a^n = a^{m+n}$

$a^m \div a^n = a^{m-n}$

$(a^m)^n = a^{mn}$

$a^0 = 1$; $a^{-n} = \dfrac{1}{a^n}$

SUVAT: $v = u + at$; $s = ut + \frac{1}{2}at^2$

Area of circle: $A = \pi r^2$; Sphere: $V = \frac{4}{3}\pi r^3$

Pendulum: $T = 2\pi\sqrt{\dfrac{l}{g}}$

Memory Hooks

BELT: Brackets first, Expand, Like terms, Tidy — order for simplifying
$\equiv$ = Eternal: identity is true forever (all values)
BIRDCAGE: Brackets on negatives when substituting
FOLD: Factorise Out when variable appears twice on Left & right-hand sides when Dividing
Reverse onion: peel off outer operations first when rearranging

Exam Tips

Show every substitution step — method marks available
Write $\equiv$ not $=$ when showing an identity
For subject changes: write each new line clearly
If target variable appears twice, you must factorise
When taking square roots, consider if $\pm$ applies in context
Define your variable if forming an expression from words
Check your answer by substituting back into the original formula

🔄 Flashcards

Click any card to reveal the answer. Use these for rapid recall practice.

✗ Common Mistakes

✗

MISTAKE 1 — Using $=$ instead of $\equiv$ for identities

What students write: $a^2 - b^2 = (a+b)(a-b)$
Why marks are lost: The equals sign implies this is only true for specific values. Examiners marking "prove the identity" questions specifically check for $\equiv$.
How to avoid: Whenever a result holds for all values of a variable, use $\equiv$. Ask yourself: "Is this true no matter what I substitute?" If yes, use $\equiv$.

✗

MISTAKE 2 — Treating unlike terms as like terms

What students write: $3x^2 + 5x = 8x^3$
Why marks are lost: $x^2$ and $x$ are different powers of $x$; they cannot be added. Students confuse "they both have $x$" with "they are the same term".
How to avoid: Write out the variable part fully: $x^2$ means $x \times x$; $x$ means $x \times 1$. They are structurally different. Like terms must match exactly.

✗

MISTAKE 3 — Omitting brackets when substituting negatives

What students write: $x = -3$, so $x^2 = -3^2 = -9$ (WRONG)
Why marks are lost: $-3^2 = -(3^2) = -9$ due to order of operations. The correct answer is $(-3)^2 = 9$.
How to avoid: Always write brackets: $(-3)^2$. This is a strict rule — drill it until automatic.

✗

MISTAKE 4 — Adding indices when they should be multiplied (power of a power)

What students write: $(x^3)^4 = x^7$ (WRONG — added $3+4$)
Why marks are lost: Power of a power requires multiplication of indices: $(x^3)^4 = x^{3 \times 4} = x^{12}$.
How to avoid: Distinguish the two rules clearly. $x^3 \times x^4 = x^7$ (same base, multiply — ADD indices). $(x^3)^4 = x^{12}$ (power of a power — MULTIPLY indices).

✗

MISTAKE 5 — Forgetting to factorise when target variable appears twice

What students write: From $ax = bx + c$, writing $x = \frac{c}{a-b}$ without the factorisation step shown.
Why marks are lost: The method mark is specifically for $x(a-b) = c$. Writing the answer without the factorisation step loses a mark even if the final answer is correct.
How to avoid: Always show $x(\ldots) = \ldots$ explicitly as a separate line. The examiner needs to see the factorisation.

✗

MISTAKE 6 — Incorrectly rearranging square roots (squaring only one term)

What students write: From $\sqrt{x + 1} = k$, writing $x + 1 = k^2 + 1$ (WRONG — only squared $k$, not both sides)
Why marks are lost: When squaring both sides, you must square the entire side, not individual terms. $(\sqrt{x+1})^2 = x + 1$; $k^2$ is the result of squaring the right-hand side.
How to avoid: Think of each side as a single object in brackets: $(\sqrt{x+1})^2 = (k)^2$ gives $x + 1 = k^2$.

✅ Final Checklist

Click each item when you feel confident. Track your progress below.

I can define: term, expression, equation, formula, identity
I know when to use $\equiv$ versus $=$
I can identify like terms including with multiple variables and indices
I can collect like terms in expressions with three or more different term types
I know all five index laws and can apply them to algebraic expressions
I can substitute positive and negative values (with brackets) into expressions
I can substitute values into complex formulae with fractions and powers
I can form an expression from a word description, defining my variable first
I can form expressions and formulae from geometric contexts (area, perimeter)
I can change the subject of a formula using inverse operations
I can rearrange formulae involving squares (e.g. $A = \pi r^2 \Rightarrow r = \sqrt{A/\pi}$)
I can rearrange formulae involving square roots (e.g. $T = 2\pi\sqrt{l/g}$)
I can rearrange when the target variable appears on both sides (factorise method)
I can rearrange formulae involving fractions by clearing the denominator first
I can verify a rearrangement by substituting numbers back into the original formula

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