Rearranging formulaeIB MYP Maths Extended: Revision notes

A formula links quantities with letters, such as $v=u+at$. The subject is the letter on its own on one side, here $v$. Rearranging (changing the subject) rewrites the formula so that a different letter is on its own, such as $a=\frac{v-u}{t}$. The relationship does not change, only the way it is written. Rearranging first is useful when you need to use the same formula many times with different values.

Treat the formula like an equation: do the same to both sides, undoing operations with their inverses. Undo addition and subtraction first, then multiplication and division, then powers and roots, working in reverse order from the way the formula was built. To make $u$ the subject of $v=u+at$, subtract $at$: $u=v-at$. To make $a$ the subject, subtract $u$ and then divide by $t$: $a=\frac{v-u}{t}$.

For $A=\frac12(a+b)h$, multiply by 2 to clear the fraction: $2A=(a+b)h$. Then divide by $h$ to get $\frac{2A}{h}=a+b$, so $a=\frac{2A}{h}-b$ and, to make $h$ the subject, $h=\frac{2A}{a+b}$. When the subject is inside a bracket, you can divide by the other factor first, or expand and then move terms. Always carry the whole expression with each operation.

If the subject is squared, take the square root at the end. $V=\frac13\pi r^2h$: multiply by 3 to get $3V=\pi r^2h$, divide to get $r^2=\frac{3V}{\pi h}$, then $r=\sqrt{\frac{3V}{\pi h}}$. For $d=5t^2$: $t=\sqrt{\frac{d}{5}}$. If the subject is inside a square root, square both sides. From $y=\sqrt{x}+4$: $y-4=\sqrt{x}$, so $x=(y-4)^2$. In real situations such as lengths and times, use the positive root.

Rearrange first, then substitute. For $v=u+at$ with $v=25$, $u=7$ and $a=3$: $t=\frac{v-u}{a}=\frac{25-7}{3}=6$ seconds. Substitute into the original formula to check: $7+3\times6=25$. Include units in the answer. For a trapezium with $A=44$, $h=8$, $a=7$: $b=\frac{2A}{h}-a=\frac{88}{8}-7=4$ cm.