In the equation $x = 5y + 4z$, the variable $x$ is the *subject* of the equation. This means it is expressed in terms of the other variables. To rearrange an equation so that another variable becomes the subject, perform the same operations on both sides of the equals sign so that eventually this variable is by itself on the left hand side. Performing the same operations on both sides makes sure that the left hand side is always equal to the right hand side. Operations which we can use include:

- Adding or subtracting a quantity
- Multiplying or dividing by a quantity
- Raising to any power (except the power zero)
- Taking logarithms or exponentiating

**Note**: We don't raise both sides of an equation to the power of zero because anything raised the power of zero is equal to $1$ and so this would give us $1=1$.

Rearrange $x=5y+4z$ to make $z$ the subject.

First write the equation as $5y+4z=x$ so that $z$ appears on the left hand side.

Next subtract $5y$ from both sides. \[\Rightarrow 4z = x-5y\] Finally, divide by $4$ on both sides to get $z$ on its own. \[\Rightarrow z = \frac{1}{4}(x-5y)\]

Rearrange $6 + 3y = \dfrac{1}{3x} +2 $ to make $x$ the subject.

Begin by collecting any like terms; here we can collect the constants by subtracting $2$ from both sides. \[\Rightarrow 4 + 3y = \frac{1}{3x}\] Then multiply by $3x$ and divide by $4+3y$. An equivalent operation is to take the reciprocal of both sides. \[\Rightarrow 3x = \frac{1}{4+3y}\] Finally, divide by $3$ to get $x$ on its own. \[\Rightarrow x = \frac{1}{3(4+3y)}\]

Rearrange $\displaystyle{\sqrt{xyz-3} = 5}$ to make $z$ the subject.

Begin by squaring both sides. \begin{align} \Rightarrow xyz-3 &= 5^2 \\ \Rightarrow xyz -3 &= 25 \end{align} Add $3$ to both sides to give \[xyz = 28\] Divide by $xy$ on both sides to get $z$ on its own. This gives \[z = \frac{28}{xy}\]

Rearrange $x = 2 e^{4y^2}$ to make $y$ the subject.

Begin by dividing through by $2$. \[\Rightarrow \frac{x}{2} = e^{4y^2}\] Take natural logarithms of both sides to get rid of the exponential. \[\Rightarrow \ln{\left(\frac{x}{2}\right)} = 4y^2\] Divde through by $4$. \[\Rightarrow \frac{1}{4}\ln{\left(\frac{x}{2}\right)} = y^2\] Finally, take the square roots of both sides. \[\Rightarrow y = \pm\sqrt{\frac{1}{4}\ln{\left(\frac{x}{2}\right)} }\]

**Note:** There are two solutions as taking the square root gives two possible values.

Prof. Robin Johnson rearranges the equation $x=y+\dfrac{1}{u}$ to make $u$ the subject.

Prof. Robin Johnson rearranges the equation $\sqrt{u-v}=1+\dfrac{1}{2}u$ to make $v$ the subject.

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

- Formulae and transposition including work on rearranging equations.

Test yourself: Numbas test on rearranging equations

Test yourself: Numbas test on Algebraic Manipulation

- Transposition of formulae workbook at
**math**centre.