The subject is whichever variable is alone on one side — in $y=mx+c$, $y$ is the subject. To make $x$ the subject instead, you unwind the formula by applying inverse operations in the reverse order they were applied: since $y=5x+8$ was built by multiplying then adding, you undo it by subtracting then dividing: $x=\dfrac{y-8}{5}$.

Worked example: Make $a$ the subject of $v^2=u^2+2as$.

Subtract $u^2$ from both sides: $v^2-u^2=2as$. Divide both sides by $2s$: $a=\dfrac{v^2-u^2}{2s}$.

Worked example: Make $L$ the subject of $T=2\pi\sqrt{\dfrac Lg}$.

Square both entire sides first: $T^2=4\pi^2\left(\dfrac Lg\right)$. Multiply by $g$: $T^2g=4\pi^2L$. Divide by $4\pi^2$: $L=\dfrac{T^2g}{4\pi^2}$.

You cannot invert term-by-term ($\frac{1}{A+B}\neq\frac1A+\frac1B$). Instead: isolate the term with the subject, combine the other side into a single fraction, then invert both sides.

Make $u$ the subject of $\dfrac1f=\dfrac1u+\dfrac1v$:

1. Isolate: $\dfrac1f-\dfrac1v=\dfrac1u$.
2. Combine (common denominator $fv$): $\dfrac{v-f}{fv}=\dfrac1u$.
3. Invert both sides: $u=\dfrac{fv}{v-f}$.

This is the key Higher Tier skill. A single inverse operation can't free a variable that's trapped in two separate terms — you need factorisation to consolidate it into one term first. For example, in $A=2xy+xz$, dividing by $y$ to "free" the first $x$ gives $\dfrac Ay=2x+\dfrac{xz}y$, which hasn't actually isolated anything — the only tool that turns a sum of $x$-terms into a single $x$-term is factorising it out as a common factor.

The 4-step algorithm:

1. Clear any fractions and expand any brackets so every term is "free".
2. Collect every term containing the subject onto one side.
3. Factorise the subject out as a common factor.
4. Divide both sides by that bracket to isolate the subject.

Worked example: Make $x$ the subject of $y=\dfrac{2x+1}{x-3}$.

1. Clear the fraction: $y(x-3)=2x+1$, then expand: $xy-3y=2x+1$.
2. Collect $x$-terms on the left: $xy-2x=1+3y$.
3. Factorise: $x(y-2)=1+3y$.
4. Isolate: $x=\dfrac{1+3y}{y-2}$.

Make $k$ the subject of $t=\dfrac{k}{k-2}$ (an Edexcel-style 3-mark question).

1. Clear the fraction: $t(k-2)=k$.
2. Expand: $tk-2t=k$.
3. Collect $k$-terms on the left: $tk-k=2t$.
4. Factorise: $k(t-1)=2t$.
5. Isolate: $k=\dfrac{2t}{t-1}$.

• "Splitting the root" over addition/subtraction, e.g. writing $\sqrt{Z^2-X^2}=Z-X$ (test with a 3-4-5 triangle: $\sqrt{25-9}=4$, but $5-3=2$ — not equal).
• Inverting a reciprocal formula term-by-term instead of combining first.
• Freezing once the subject appears twice instead of factorising it out.
• Sign errors when moving subject-terms across the equals sign during the "collect" step.