Changing the subject of a formula

The subject of a formula is a letter on its own on one side of an equal sign. Look at the following examples:

\({b}\) is the subject of the formula here:

\({b=}\frac{ac}{d}\)

\({f}\) is the subject of the formula here:

\({6c~+~d}={f}\)

Quite often in mathematics we want to change which letter is the subject of the formula, the method for doing this is very similar to how we solve equations.

Example

Make \({t}\) the subject of the formula \(\frac{6t}{p}={9}\)

Solution

We need to isolate \({t}\) so that it is on its own on one side of the formula. We can do this by dividing both sides of the expression by 6 and then multiplying both sides by \({p}\).

Dividing both sides of the equation by 6:

\(\frac{t}{p}={1.5}\)

Multiplying both sides of the equation by \({p}\):

\({t}={1.5p}\)

\({t}\) is now the subject of the formula.

Example

Make \({s}\) the subject of the equation \({3s~+~9}={t}\)

Solution

We must be very careful here as we have to ensure that when we divide by 3 we divide every term by 3.

First we subtract 9 from both sides:

\(\text{3s}=~{t~–~9}\)

Then we divide by 3:

\({s}=\frac{t~–~9}{3}\)

We could also write this as:

\({s}=\frac{t}{3}{–3}\)

You should be able to see that both of the above equations are just two different ways of writing the same answer.

Example

Make \({k}\) the subject of the formula \(\frac{7}{k}~=~{28p}\)

Solution

This is one of the most difficult rearrange questions that you can encounter. It is made difficult by the fact that the letter which we want to be the subject of the formula is in the denominator of the fraction \(\frac{7}{k}\). In order to solve this problem, our first step is going to be to multiply both sides of the equation by \({k}\) which will restore it to the top line of the equation. In other words, this will remove it from the denominator.

Multiplying both sides by \({k}\):

\({7}={28pk}\)

Dividing both sides by \({28p}\):

\(\frac{7}{28p}={k}\)

This is not our final answer as we can simplify the fraction \(\frac{7}{28}\):

\(\frac{1}{4p}={k}\)

How would you solve the equation \({z^3}\) = 27? When solving an equation our aim, as when simplifying, is to get "\({z}=\)" on one side of the equation. In order to turn \({z^3}\) into \({z}\) we need to take the cube root. Taking the cube root of both sides gives \({z}\) = 3.

Similarly, if we had the equation \(\sqrt{P}={4}\) we would have to square both sides of the equation to recover "\({P}=\)". This would give \({P}\) = 16.

The same logic must also be applied when we are changing the subject of a formula.

Example

Make \({T}\) the subject of \({w}{T}^{4}={E}\).

Solution

First divide both sides of the equation by \({w}\):

\(\text{T}^{4}=\frac{E}{w}\)

Then find the fourth root of both sides:

\({T}=\sqrt[4]{(\frac{E}{w})}\)

Question

Make \({q}\) the subject of \(4(\sqrt[3]{q})={3t}\)

Equations and formulae – WJECChanging the subject of a formula