### 2 Rearranging a formula

In the formula for the area of a circle, $A=\pi {r}^{2}$ , we say that $A$ is the subject of the formula. A variable is the subject of the formula if it appears by itself on one side of the formula, usually the left-hand side, and nowhere else in the formula . If we are asked to transpose the formula for $r$ , or solve for $r$ , then we have to make $r$ the subject of the formula. When transposing a formula whatever is done to one side is done to the other . There are five rules that must be adhered to.

##### Key Point 22
Rearranging a formula

You may carry out the following operations

$\bullet$ add the same quantity to both sides of the formula

$\bullet$ subtract the same quantity from both sides of the formula

$\bullet$ multiply both sides of the formula by the same quantity

$\bullet$ divide both sides of the formula by the same quantity

$\bullet$ take a ‘function’ of both sides of the formula: for example,

find the reciprocal of both sides (i.e. invert).

##### Example 64

Transpose the formula $p=5t-17$ for $t$ .

##### Solution

We must obtain $t$ on its own on the left-hand side. We do this in stages by using one or more of the five rules in Key Point 22. For example, by adding 17 to both sides of $p=5t-17$ we find

Dividing both sides by 5 we obtain $t$ on its own:

$\phantom{\rule{2em}{0ex}}\frac{p+17}{5}=t$

so that $t=\frac{p+17}{5}$ .

##### Example 65

Transpose the formula $\sqrt{2q}=p$ for $q$ .

##### Solution

First we square both sides to remove the square root. Note that ${\left(\sqrt{2q}\right)}^{2}=2q$ . This gives

$\phantom{\rule{2em}{0ex}}2q={p}^{2}$

Second we divide both sides by $2$ to get $q=\frac{{p}^{2}}{2}$ .

Note that in general by squaring both sides of an equation may introduce extra solutions not valid for the original equation. In Example 65 if $p=2$ then $q=2$ is the only solution. However, if we transform to $q=\frac{{p}^{2}}{2}$ , then if $q=2$ , $p$ can be $+2$ or $-2$ .

Transpose the formula $v=\sqrt{{t}^{2}+w}$ for $w$ .

You must obtain $w$ on its own on the left-hand side. Do this in several stages.

First square both sides to remove the square root:

${v}^{2}={t}^{2}+w$

Then, subtract ${t}^{2}$ from both sides to obtain an expression for $w$ :

${v}^{2}-{t}^{2}=w$

Finally, write down the formula for $w$ :

$w={v}^{2}-{t}^{2}$

##### Example 66

Transpose $x=\frac{1}{y}$ for $y$ .

##### Solution

We must try to obtain an expression for $y$ . Multiplying both sides by $y$ has the effect of removing this fraction:

Divide both sides by $x$ to leaves $y$ on its own, $y=\frac{1}{x}$ .

Alternatively: simply invert both sides of the equation $x=\frac{1}{y}$ to get $\frac{1}{x}=y$ .

##### Example 67

Make $R$ the subject of the formula

$\phantom{\rule{2em}{0ex}}\frac{2}{R}=\frac{3}{x+y}$

##### Solution

In the given form $R$ appears in a fraction. Inverting both sides gives

$\phantom{\rule{2em}{0ex}}\frac{R}{2}=\frac{x+y}{3}$

Thus multiplying both sides by 2 gives

$\phantom{\rule{2em}{0ex}}R=\frac{2\left(x+y\right)}{3}$

Make $R$ the subject of the formula $\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$ .

1. Add the two terms on the right:

$\phantom{\rule{2em}{0ex}}\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}=\frac{{R}_{2}+{R}_{1}}{{R}_{1}{R}_{2}}$

2. Write down the complete formula:

$\phantom{\rule{2em}{0ex}}\frac{1}{R}=\frac{{R}_{2}+{R}_{1}}{{R}_{1}{R}_{2}}$

3. Now invert both sides:

$\phantom{\rule{2em}{0ex}}R=\frac{{R}_{1}{R}_{2}}{{R}_{2}+{R}_{1}}$