Rearranging Formulas

Question type 1: Simple linear formula

The formula for the circumference of a circle is C=2\pi r. Here the circumference is the subject of the formula.

Making a different letter the subject of a formula means manipulating the formula to get that letter by itself on one side of the equation.

Example: For the equation C=2\pi \textcolor{blue}{r}, make \textcolor{blue}{r} the subject.

Step 1: Dividing both sides of the equation by 2\pi

\begin{aligned}(\textcolor{maroon}{\div 2\pi})\,\,\,\,\,\,\,\,\, C &= 2\pi \textcolor{blue}{r} \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{\div 2\pi}) \\ \dfrac{C}{2\pi} &= \dfrac{\cancel{2\pi} \textcolor{blue}{r}}{\cancel{2\pi}} \end{aligned}

Step 2: Rewrite the formula in its final form, typically having the subject on the left hand side.

\textcolor{blue}{r} = \dfrac{C}{2\pi}

Question type 2: Formulas involving fractions

Some questions will include making a letter that appears in a fraction the subject of the formula.

d=\dfrac{(u+v)t}{2}

This formula describes the relation that distance, d, initial velocity, u, final velocity v and time t given constant acceleration. We can manipulate this formula to make u the subject.

Example: For the equation d=\dfrac{(\textcolor{blue}{u}+v)t}{2}, make \textcolor{blue}{u} the subject.

Step 1: Multiplying both sides by 2,

\begin{aligned}(\textcolor{maroon}{\times 2})\,\,\,\,\,\,\,\,\, d & =\dfrac{(\textcolor{blue}{u}+v)t}{2} \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{\times 2}) \\ 2d & =(\textcolor{blue}{u}+v)t \end{aligned}

Step 2: Dividing both sides of the equation by t,

\begin{aligned}(\textcolor{maroon}{\div t})\,\,\,\,\,\,\,\,\, 2d & =(\textcolor{blue}{u}+v)t \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{\div t}) \\ \dfrac{2d}{t} & =\textcolor{blue}{u}+v\end{aligned}

Step 3: Finally subtracting v and rewriting so \textcolor{blue}u is on the left hand side,

\begin{aligned}(\textcolor{maroon}{-v})\,\,\,\,\,\,\,\,\, \dfrac{2d}{t} & =\textcolor{blue}{u}+v \,\,\,\,\,\,\,\,\, (\textcolor{maroon}{-v}) \\  \textcolor{blue}{u} & = \dfrac{2d}{t} -v\end{aligned}

Question type 3: Formulas involving squares

Sometimes the required subject can appear as a square. For example, if we are asked to rearrange  the formula for the area of a circle to make r the subject. Firstly, recall that the formula for the area of a circle is,

A=\pi \textcolor{blue}{r^2}

Example: For the equation A=\pi \textcolor{blue}{r^2}, make \textcolor{blue}{r} the subject.

Step 1: Dividing both sides of the equation by \pi, we get,

\dfrac{A}{\pi}=\textcolor{blue}{r^2}

Now, r is on its own on one side but it’s not technically the subject, since it is squared.

Step 2: If we now square root both sides of the equation, and since r is always positive, we get,

\textcolor{blue}{r}=\sqrt{\dfrac{A}{\pi}}

Question type 4: Formulas involving square roots

Conversely, some questions may have the subject appear in a square root.

Example: Rearrange the following formula d=\sqrt{\dfrac{3\textcolor{blue}{h}}{2}} to make \textcolor{blue}{h} the subject.

Step 1: Squaring both sides of the equation, we get,

d^2=\dfrac{3\textcolor{blue}{h}}{2}

Step 2: Next, multiply both sides by 2 to get,

2d^2=3\textcolor{blue}{h}

Step 3: Finally, dividing both sides by 3 gives us,

\textcolor{blue}{h}=\dfrac{2d^2}{3}

Question type 5: Formulas when the subject appears twice

In some instances the required subject will appear more than once in the given formula. In these examples we factorise the terms involving the subject.

Example: Rearrange the following formula H=2\textcolor{blue}{R}-g\textcolor{blue}{R} to make \textcolor{blue}{R} the subject.

Step 1: We take out a factor of R from both terms i.e. we factorise, and get

H=\textcolor{blue}{R}(2-g)

Step 2: Now we divide by (2-g), to get

\textcolor{blue}{R}=\dfrac{H}{2-g}

Note: If your exams are with the OCR exam board, then you will need to know how to do this if you are a foundation student!