**Rearranging Formulas**

A formula is a way of giving a mathematical relationship between different things, expressed using algebra. For example, the relationship between **speed**, **distance** and **time** can be captured by the simple formula,

s=\dfrac{{d}}{{t}}.

You will need to be able to answer **5 key question types** when rearranging formulas, these are shown below.

Make sure you are happy with the following topics before continuing.

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## 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: I**f 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 letter 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}

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### Example Questions

**Question 1:** Rearrange the formula F=\dfrac{mv}{t} to make m the subject. **[2 marks]**

In order to make \textcolor{blue}{m} the subject, we will multiply both sides by t and get:

Ft=\textcolor{blue}{m}v

Then, if we divide both sides of the equation by v, and rewrite it so that m is on the left hand side, we get,

\textcolor{blue}{m}=\dfrac{Ft}{v}

**Question 2:** Rearrange the formula A=\dfrac{1}{2}(a+b)h to make a the subject. **[2 marks]**

Firstly, we will multiply both sides of the equation by 2 to get rid of the fraction:

2A=(\textcolor{blue}{a}+b)h

Then, if we divide both sides by h, we get

\dfrac{2A}{h}=\textcolor{blue}{a}+b

Finally, subtracting b from both sides, we get

\textcolor{blue}{a}=\dfrac{2A}{h}-b

**Question 3:** Rearrange the formula F=\dfrac{kq}{r^2} to make r the subject. **[3 marks]**

**Step 1:** Multiplying both sides by r^2, we get

F\textcolor{blue}{r}^2=kq

**Step 2:** Next, divide both sides by F to get

\textcolor{blue}{r}^2=\dfrac{kq}{F}

**Step 3:** Finally, square rooting both sides gives us,

\textcolor{blue}{r}=\sqrt{\dfrac{kq}{F}}

**Question 4:** Rearrange the formula \dfrac{x}{x+c}=\dfrac{a}{b} to make x the subject. **[3 marks]**

Multiplying both sides of the equation by x+c, we get

\textcolor{blue}{x}=\dfrac{a(\textcolor{blue}{x}+c)}{b}

Next, multiply both sides by b and expanding the right-hand side, to get

b\textcolor{blue}{x}=a\textcolor{blue}{x}+ac

Subtracting ax from both sides and factorising,

b\textcolor{blue}{x}-a\textcolor{blue}{x}=\textcolor{blue}{x}(b-a)=ac

Finally, dividing both sides by (b-a) gives us,

\textcolor{blue}{x}=\dfrac{ac}{b-a}

**Question 5:** Rearrange the formula a=\dfrac{3-2b}{b-4} to make b the subject. **[3 marks]**

Multiplying both sides of the equation by b-4, we get

a(\textcolor{blue}{b}-4)=(3-2\textcolor{blue}{b})

Next, expanding the left-hand side, to get

a\textcolor{blue}{b}-4a=3-2\textcolor{blue}{b}

Collecting all the terms with a factor of b on one side of the equation,

a\textcolor{blue}{b}+2\textcolor{blue}{b}=3+4a

Factorising b out of the left-hand side and dividing by the terms left we find,

\textcolor{blue}{b}=\dfrac{3+4a}{a+2}

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