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Rearrange further equations


Previously, you've looked at how to change the subject of an equation for linear equations, and also more challenging equations involving powers and quadratics. We are going to take this further, looking at a wide variety of types of equations, including radicals, logs, and other skills.


The subject of an equation is the variable that is by itself on one side of the equals sign and usually it is at the start of the formula.

For example, in the formula $A=pb+y$A=pb+y, $A$A is the subject because it is by itself on the left hand side of the equal sign.

When we're changing the subject of a formula, we often have more than one variable, but we use a process similar to when we solve equations. 

  • Group any like terms
  • Simplify using the inverse of addition or subtraction.
  • Simplify further by using the inverse of multiplication or division.
  • Other inverse operations might be needed like square and square root



Question 1

Make $r$r the subject of $v=\sqrt{\frac{mr}{\pi}}$v=mrπ

Think: In this question, $r$r has been multiplied by $m$m, divided by $\pi$π, then square rooted.

Reversing this, what would be the first operation to do to both sides? 

The first operation will be to square both sides, in order to reverse the square root.

Then we multiply by $\pi$π, and divide by $m$m, finally isolating $r$r.


$v$v $=$= $\sqrt{\frac{mr}{\pi}}$mrπ square both sides
$v^2$v2 $=$= $\left(\sqrt{\frac{mr}{\pi}}\right)^2$(mrπ)2 cancel the square and square root
$v^2$v2 $=$= $\frac{mr}{\pi}$mrπ multiply both sides by $\pi$π
$v^2\times\pi$v2×π $=$= $\frac{mr}{\pi}\times\pi$mrπ×π cancel the $\pi$π's
$\pi v^2$πv2 $=$= $mr$mr divide both sides by $m$m
$\pi v^2\div m$πv2÷m $=$= $mr\div m$mr÷m cancel the $m$m's
$\frac{\pi v^2}{m}$πv2m $=$= $r$r rewrite the equation with $r$r on the left hand side
$r$r $=$= $\frac{\pi v^2}{m}$πv2m  


Question 2

The volume of a sphere is given by the formula $V=\frac{4}{3}\pi r^3$V=43πr3. Make the radius, $r$r, the subject of the equation.

a) Make the radius, $r$r, the subject of the equation.

Think: The radius $r$r is cubed, and multiplied by $4\pi$4π, then divided by $3$3.

What is the first step in reversing these operations? 

The first step is to multiply both sides by $3$3.

Then we divide by $4\pi$4π, and then take the cube root of both sides.


$V$V $=$= $\frac{4}{3}\pi r^3$43πr3
$V\times3$V×3 $=$= $\frac{4}{3}\pi r^3\times3$43πr3×3
$3V$3V $=$= $4\pi r^3$4πr3
$3V\div4\pi$3V÷4π $=$= $4\pi r^3\div4\pi$4πr3÷4π
$r^3$r3 $=$= $\frac{3V}{4\pi}$3V4π
$\sqrt[3]{r^3}$3r3 $=$= $\sqrt[3]{\frac{3V}{4\pi}}$33V4π
$r$r $=$= $\sqrt[3]{\frac{3V}{4\pi}}$33V4π

b) If the volume of the sphere is $20\pi$20π cm3, what is the exact radius of the sphere? Leave your answer in exact radical form.

Think: One of the main reasons we want to be able to change the subject of a formula is so that we can do these types of substitutions.


$r$r $=$= $\sqrt[3]{\frac{3V}{4\pi}}$33V4π
  $=$= $\sqrt[3]{\frac{3\times20\pi}{4\pi}}$33×20π4π
  $=$= $\sqrt[3]{\frac{60\pi}{4\pi}}$360π4π
  $=$= $\sqrt[3]{15}$315

In this case, the question has requested we leave the questions in exact radical form, so we have our answer, and do not need to use our calculator to find a value.

Question 3

The equation for the population at time $t$t is given by $P=Me^{4t}$P=Me4t. Make $t$t the subject of the equation.

Think: We want to make $e^{4t}$e4t the subject of the equation, and then use the definition of a logarithm to solve for $t$t.

Our first step is to divide both sides of the equation by $M$M

Then, we can make the power $4t$4t the subject . We can use the fact that an equation, such as $a^x=y$ax=y, which is in exponential form, can be written in log form, making the power $x$x the subject : $x=\log_ay$x=logay

So in this case, we can make $4t$4t the subject by taking the log of $\frac{P}{M}$PM.

We finish by dividing both sides by $4$4.


$P$P $=$= $Me^{4t}$Me4t
$e^{4t}$e4t $=$= $\frac{P}{M}$PM
$4t$4t $=$= $\ln\left(\frac{P}{M}\right)$ln(PM)
$t$t $=$= $\frac{\ln\left(\frac{P}{M}\right)}{4}$ln(PM)4

Worked Examples


Solve $\frac{V_1}{V_2}=\frac{P_2}{P_1}$V1V2=P2P1 for $P_1$P1.


Rearrange the expression $p=\frac{q\sqrt{k}}{r}$p=qkr to make $k$k the subject.


If $p=\sqrt{\frac{5m^2+4}{2m^2-3}q}$p=5m2+42m23q, express $m$m in terms of $p$p and $q$q.

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