Rearranging Formulae | Rearranging Equations | MME

# Rearranging Formulae Worksheets, Questions and Revision

Level 6 Level 7

## What you need to know

### Rearranging Formulae

A formula is a mathematical relationship between different quantities that is expressed with algebra. For example, one formula for $\textcolor{red}{speed}$ is $\textcolor{yellow}{distance}$ divided by $\textcolor{green}{time}$ , which we express like

$\textcolor{red}{s}=\dfrac\textcolor{yellow}{{d}}\textcolor{green}{{t}}$.

In this case, we say $\textcolor{red}{s}$ $\textcolor{red}{(speed)}$ is the subject of the formula. We can change the subject of the formula, for example by multiplying both sides by $t$ makes $d$ the subject:

$\textcolor{red}{s}\textcolor{green}{t}=\textcolor{yellow}{d}$

Before tacking rearranging equations, being able to solve equations is a good place to start and get up to speed with the fundamentals.

### Take Note:

Rearranging formulae is all about understanding the key rules but also knowing what order to apply them in.

As a rule of thumb, it’s good to try to get rid of any algebraic fractions as soon as you can. This is to say that if the thing that you want to make the subject is part of a fraction, then it’s generally a good idea to multiply up by the denominator of that fraction as a first step.

If you see square terms in your formula, then it is usually best to leave these until the end.

Always remember the golden rules:

1. Whatever you do to one side of an equation you must do to the other.
2. To get rid of something off one side you must do the opposite.

### Example 1: Rearranging Equations

Rearrange the formula $F=\dfrac{mv}{t}$ to make $m$ the subject.

So, to kick off making $m$ the subject, we will multiply both sides by $t$ and get

$Ft=mv$

Then, if we divide both sides of the equation by $v$, we get

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

### Example 2: Rearranging Equations

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

$A=\pi r^2$

So, dividing both sides of the equation by $\pi$, we get

$\dfrac{A}{\pi}=r^2$

Now, $r$ is on its own on one side but it’s not technically the subject, since it is being squared. So, if we now square root both sides of the equation, we get

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

### Example 3: Rearranging Formulae when the Letter Appears Twice

Rearrange the formula $H=2R-gR$ to make $R$ the subject.

This is different to what we’ve seen so far in that there are two instances of $R$ – the thing we want to make the subject. In these examples we factorise the terms involving the subject.

We take out a factor of $R$ from them both terms i.e. we factorise, and get

$H=R(2-g)$

Now we divide by $(2-g)$ (and yes, we divide by the whole thing in one go, since it’s the whole that is being multiplied by $R$), to get

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

Thus, we have made $R$ the subject of the equation.

### Example Questions

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

$2A=(a+b)h$

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

$\dfrac{2A}{h}=a+b$

Finally, subtracting $b$ from both sides, we get

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

Thus, we have made $a$ the subject.

Multiplying both sides by $r^2$, we get

$Fr^2=kq$

Next, divide both sides by $F$ to get

$r^2=\dfrac{kq}{F}$

Finally, square rooting both sides gives us,

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

Thus making $r$ the subject of the formula.

Squaring both sides of the equation, we get

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

Next, multiply both sides by $2$ to get

$2d^2=3h$

Finally, dividing both sides by 3 gives us,

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

Thus making $h$ the subject of the formula.

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

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

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

$bx=ax+ac$

Subracting $ax$ from both sides and factorising,

$bx-ax=x(b-a)=ac$

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

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

Thus making $x$ the subject of the formula.

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

$a(b-4)=(3-2b)$

Next, expanding the left-hand side, to get

$ab-4a=3-2b$

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

$ab+2b=3+4a$

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

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

Thus making $b$ the subject of the formula.

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