# Rearranging Formulas Worksheets, Questions and Revision

GCSE 4 - 5GCSE 6 - 7KS3AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022

## 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.

Level 4-5 GCSE KS3 ## 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.

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

Level 4-5 GCSE KS3 ## 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}

Level 4-5 GCSE KS3 Level 4-5 GCSE KS3 ## 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,

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

Level 4-5 GCSE KS3 ## 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,

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

Level 4-5 GCSE KS3 Level 6-7 GCSE ## 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

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

Level 6-7 GCSE ## Example Questions

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,

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

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

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

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,

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

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,

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

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,

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

## Related Topics

Level 1-3GCSEKS3

Level 4-5GCSEKS3

#### Expanding Brackets

Level 1-3Level 4-5GCSEKS3

#### Factorising

Level 4-5Level 6-7GCSE

#### Surds

Level 6-7Level 8-9GCSE

## Worksheet and Example Questions

### (NEW) Rearranging Formulas Exam Solutions - MME

Level 4-5 GCSENewOfficial MME

Level 4-5 GCSE

Level 6-7 GCSE

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