Rearranging a formula means isolating one variable (the subject) on its own side of the equals sign, using the same inverse-operation logic as solving an equation: whatever you do to one side, do to the other.

Work from the "outside in" — undo whatever is structurally farthest from the subject first. A reliable priority order:

1. Clear any fractions first, by multiplying both sides by the denominator.
2. Expand brackets if the subject is trapped inside one.
3. Apply inverse operations (addition/subtraction, then multiplication/division, then powers/roots) working outward to inward.

Example: Make $r$ the subject of $V=\dfrac{4}{3}\pi r^3$.

1. Clear the fraction: $3V = 4\pi r^3$.
2. Isolate $r^3$: $r^3 = \dfrac{3V}{4\pi}$.
3. Undo the cube: $r = \sqrt[3]{\dfrac{3V}{4\pi}}$.

If the subject is multiplied by a bracket from outside (e.g. $P=2(l+w)$, making $w$ the subject), divide first: $\dfrac{P}{2}=l+w$, then subtract $l$. If the subject is trapped inside a bracket that's being expanded elsewhere, you must expand the bracket first to free it up for collection.

When $x$ appears in two terms, simple division won't work — you must factorise. For $p=\dfrac{2-ax}{x-b}$, making $x$ the subject:

1. Clear the fraction: $p(x-b)=2-ax$.
2. Expand: $px - pb = 2-ax$.
3. Collect all $x$-terms on one side: $px+ax = 2+pb$.
4. Factorise: $x(p+a)=2+pb$.
5. Divide by the bracket: $x=\dfrac{2+pb}{p+a}$.

Worked Example

Make $t$ the subject of $v=u+at$.

1. Subtract $u$ from both sides: $v-u=at$.
2. Divide both sides by $a$: $t=\dfrac{v-u}{a}$.

Check: substitute back — $u+a\left(\dfrac{v-u}{a}\right)=u+(v-u)=v$. ✓

Treat the power or root exactly like any other operation, but make sure it's the last thing undone, since it typically sits closest to the subject.

Example: Make $b$ the subject of Pythagoras' theorem, $a^2+b^2=c^2$.

1. Subtract $a^2$ from both sides (the term farthest from $b$): $b^2=c^2-a^2$.
2. Undo the square by square-rooting the entire right-hand side: $b=\sqrt{c^2-a^2}$.

This mirrors the sphere-volume example above: fractions and additive terms are cleared first, and the power/root is always the final step, acting on everything that remains on the other side — never on just one part of it.