Rearranging a formula

2 Rearranging a formula

In the formula for the area of a circle, $A = π r^{2}$ , 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

$∙$ add the same quantity to both sides of the formula

$∙$ subtract the same quantity from both sides of the formula

$∙$ multiply both sides of the formula by the same quantity

$∙$ divide both sides of the formula by the same quantity

$∙$ 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 $p = 5 t - 17$ 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 $p = 5 t - 17$ we find

\begin{array}{rcl} p + 17 & = & 5 t - 17 + 17 \\ so that p + 17 & = & 5 t \end{array}

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

$\frac{p + 17}{5} = t$

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

Example 65

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

Solution

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

$2 q = p^{2}$

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 $- 2$ .

Task!

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:

\begin{array}{rcl} Multiply both sides of x & = & \frac{1}{y} by y to get \\ y x & = & y \times \frac{1}{y} \\ so that y x & = & 1 \end{array}

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

$\frac{2}{R} = \frac{3}{x + y}$

Solution

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

$\frac{R}{2} = \frac{x + y}{3}$

Thus multiplying both sides by 2 gives

$R = \frac{2 (x + y)}{3}$

Task!

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

Add the two terms on the right:

$\frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{R_{2} + R_{1}}{R_{1} R_{2}}$
Write down the complete formula:

$\frac{1}{R} = \frac{R_{2} + R_{1}}{R_{1} R_{2}}$
Now invert both sides:

$R = \frac{R_{1} R_{2}}{R_{2} + R_{1}}$

2 Rearranging a formula

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