Formulae, Rearranging and Functions

Rearranging (or changing the subject of) a formula means isolating a different variable on one side of the equals sign. The same inverse operations used to solve equations apply here.

Golden rule: whatever operation is done to one side must be done to the other.

Single-step examples:

$v=u+at$ — make $u$ the subject: $u=v-at$

$A=\frac{1}{2}bh$ — make $h$ the subject: $h=\frac{2A}{b}$

$v^2=u^2+2as$ — make $s$ the subject: $s=\frac{v^2-u^2}{2a}$

Worked example — make $r$ the subject of $A=\pi r^2$:

$A=\pi r^2⟹r^2=\frac{A}{\pi }⟹r=\sqrt{\frac{A}{\pi }}$

(Take the positive root since $r$ is a length.) ✓

When the subject appears more than once, or is inside a fraction or root, more steps are required.

Worked example — make $x$ the subject of $\frac{x+3}{x-1}=k$:

$x+3=k(x-1)=kx-k$

$x-kx=-k-3$

$x(1-k)=-k-3$

$x=\frac{-k-3}{1-k}=\frac{k+3}{k-1}$

Worked example — make $x$ the subject of $y=\frac{2x-1}{3}$:

$3y=2x-1⟹2x=3y+1⟹x=\frac{3y+1}{2}$

Key moves:

• Multiply through by denominator first
• Expand brackets
• Collect all terms containing the subject on one side
• Factorise the subject out if it appears more than once
• Divide by the coefficient

An equation is true only for specific values of the variable — it can be solved.

An identity is true for all values of the variable — it cannot be "solved" for a specific value. Identities are written with $\equiv$ (the identity symbol).

Example	Type	Why
$3x+1=7$	Equation	True only when $x=2$
$2(x+3)\equiv 2x+6$	Identity	True for all values of $x$
$x^2-4\equiv (x+2)(x-2)$	Identity	True for all values of $x$
$(x+1{)}^2=x^2+3$	Equation	True only for specific $x$ values ($x^2+2x+1=x^2+3⟹x=1$)

Worked example (Higher) — show that $(n+1{)}^2-n^2\equiv 2n+1$.

$(n+1{)}^2-n^2=n^2+2n+1-n^2=2n+1\equiv 2n+1$ ✓

This shows the difference between consecutive squares is always an odd number.

Algebraic argument — to show two expressions are equivalent, expand and simplify both sides independently until they match. Do not work "across" the equals/identity sign.

A function is a rule that maps each input to exactly one output. At GCSE:

$f(x)=3x-1\text{means "multiply the input by 3, then subtract 1"}$

Evaluating functions — substitute the given value for $x$:

$f(4)=3(4)-1=11$

$f(-2)=3(-2)-1=-7$

$f(x+1)=3(x+1)-1=3x+2$

Common functions:

$g(x)=x^2+2x$: $g(3)=9+6=15$; $g(-1)=1-2=-1$

Finding $x$ given $f(x)$: set up an equation.

$f(x)=20⟹3x-1=20⟹x=7$ ✓

The inverse function $f^{-1}(x)$ reverses the effect of $f(x)$: if $f(a)=b$ then $f^{-1}(b)=a$.

Finding $f^{-1}(x)$: write $y=f(x)$, rearrange to make $x$ the subject, then write $f^{-1}(x)$ replacing $y$ with $x$.

Worked example — find $f^{-1}(x)$ for $f(x)=3x-1$:

$y=3x-1⟹3x=y+1⟹x=\frac{y+1}{3}$

$f^{-1}(x)=\frac{x+1}{3}$

Check: $f(f^{-1}(x))=3\cdot \frac{x+1}{3}-1=(x+1)-1=x$ ✓ (applying $f$ then $f^{-1}$ returns $x$)

A composite function $fg(x)$ applies $g$ first, then $f$ to the result.

$fg(x)=f(g(x))$

Right to left — the function closer to $x$ is applied first.

Worked example — given $f(x)=2x+1$ and $g(x)=x^2-3$, find $fg(x)$ and $gf(x)$.

$fg(x)=f(g(x))=f(x^2-3)=2(x^2-3)+1=2x^2-5$

$gf(x)=g(f(x))=g(2x+1)=(2x+1{)}^2-3=4x^2+4x+1-3=4x^2+4x-2$

Note: $fg(x)\ne gf(x)$ in general — the order matters.

$fg(2)=2(4)-5=3$ ✓; check: $g(2)=1$, $f(1)=3$ ✓

1. Rearranging — multiplying out too late

When the subject is inside a fraction and also outside (e.g. $\frac{x+a}{x+b}=k$), multiply through by the denominator immediately before any other step. Attempting to cancel terms without doing this leads to errors.

2. Confusing equation and identity

An equation has specific solutions; an identity is universally true. Writing $=$ when $\equiv$ is required loses marks in "show that" questions.

3. $fg(x)$ — applying $f$ first instead of $g$

In $fg(x)$, the function written closest to $x$ is applied first. So $fg(x)=f(g(x))$ — evaluate $g$ first, then substitute into $f$.

4. Inverse function — taking the reciprocal

$f^{-1}(x)\ne \frac{1}{f(x)}$. The inverse function reverses the rule; $\frac{1}{f(x)}$ is the reciprocal. These are different operations with different results.

Mistake	Correction
"Make $r$ the subject of $C=2\pi r$: $r=2\pi C$"	Divide both sides by $2\pi$: $r=\frac{C}{2\pi }$
"$fg(3)$ means $f(3)\times g(3)$"	$fg(3)=f(g(3))$ — apply $g$ to 3 first, then apply $f$ to the result
"The inverse of $f(x)=5x$ is $f^{-1}(x)=\frac{1}{5x}$"	$f^{-1}(x)=\frac{x}{5}$ (reverse: divide by 5) — not the reciprocal