Rearranging Formulae

Simple Rearranging

What are formulae?

A formula (plural, formulae) is a mathematical relationship consisting of variables, constants and an equals sign
You will come across many formulae in your IGCSE course, including
- the formulae for areas and volumes of shapes
- equations of lines and curves
- the relationship between speed, distance and time
Some examples of formulae you should be familiar with are
- The equation of a straight line
  - $y = m x + c$
- The area of a trapezium
  - $Area = \frac{(a + b) h}{2}$
- Pythagoras' theorem
  - $a^{2} + b^{2} = c^{2}$
You will also be expected to rearrange formulae that you are not familiar with

How do I rearrange formulae where the subject appears only once?

Rearranging formulae can also be called changing the subject
- The subject is the variable that you want to find out, or get on its own on one side of the formula
The method for changing the subject is the same as the method used for solving linear equations
- STEP 1
  Remove any fractions or brackets
  - Remove fractions by multiplying both sides by anything on the denominator
  - Expand any brackets only if it helps to release the variable, if not it may be easier to leave the bracket there
- STEP 2
  Carry out inverse operations to isolate the variable you are trying to make the subject
  - This works in the same way as with linear equations, however you will create expressions rather than carry out calculations
- For example, to rearrange $A = \frac{(a + b) h}{2}$ so that $h$ is the subject
  - Multiply by 2

$2 A = (a + b) h$

- - Expanding the bracket will not help here as we would end up with the subject appearing twice, so instead divide by the whole expression $(a + b)$

$\frac{2 A}{a + b} = h$

- - You can now rewrite this with the subject ( $h$ ) on the left hand side

$h = \frac{2 A}{a + b}$

How do I rearrange formulae that include powers or roots?

If the formula contains a power of n, use the nth root to reverse this operation
- For example to make $x$ the subject of $y = a x^{5}$
  - Divide both sides by $a$ first

$\frac{y}{a} = x^{5}$

- - Then take the 5th root of both sides

$\sqrt[5]{\frac{y}{a}} = x$

If n is even then there will be two answers: a positive and a negative
- For example if $y = x^{2}$ then $x = \pm \sqrt{y}$
If the formula contains an nth root, reverse this operation by raising both sides to the power of n
- For example to make $a$ the subject of $m = \sqrt[3]{2 a b}$
  - Raise both sides to the power of 3 first

$m^{3} = 2 a b$

- - Divide both sides by $2 b$

$a = \frac{m^{3}}{2 b}$

Are there any common formulae to be aware of?

Formulae for accelerating objects are often used
- $v = u + a t$
- $v^{2} = u^{2} + 2 a s$
- $s = u t + \frac{1}{2} a t^{2}$
The letters mean the following:
- t stands for the amount of time something accelerates for (in seconds)
- u stands for its initial speed (in m/s) - the speed at the beginning
- v stands for its final speed (in m/s) - the speed after t seconds
- a stands for its acceleration (in m/s²) during in that time
- s stands for the distance covered in t seconds
You do not need to memorise these formulae, but you should know how to rearrange them
- You should also understand the difference between initial speed (u) and final speed (v)

Exam Tip

If you are unsure about the order in which you would carry out the inverse operations, try substituting numbers in and reverse the order that you would carry out the substitution

Worked example

Make $x$ the subject of $y = \sqrt{a x^{2} - b}$ .

Use inverse operations to isolate $x$ .

Square both sides.

$\begin{array}{rcl} y^{2} & = & {(\sqrt{a x^{2} - b})}^{2} \\ y^{2} & = & a x^{2} - b \end{array}$

Add $b$ to both sides.

$\begin{array}{rcl} y^{2} & = & a x^{2} - b \end{array}$

$\begin{array}{rcl} (+ b) (+ b) \end{array}$

$\begin{array}{rcl} y^{2} + b & = & a x^{2} \end{array}$

Divide both sides by $a$ .

$\begin{array}{rcl} y^{2} + b & = & a x^{2} \end{array}$

$\begin{array}{rcl} (\div a) & (\div a) \end{array}$

$\begin{array}{rcl} \frac{y^{2} + b}{a} & = & x^{2} \end{array}$

Square root both sides.

$\begin{array}{rcl} \begin{array}{rcl} \sqrt{\frac{y^{2} + b}{a}} & = & \sqrt{x^{2}} \end{array} \\ \begin{array}{rcl} \sqrt{\frac{y^{2} + b}{a}} & = & x \end{array} \end{array}$

The equation is fully correct as it is and will gain full marks, however the two sides can be swapped if preferred. Remember that when you square root you get two answers (a positive and a negative).

$x = \pm \sqrt{\frac{y^{2} + b}{a}}$

Subject Appears Twice

How do I rearrange formulae where the subject appears twice?

If the subject appears twice, you will need to factorise at some point
- Factorising means putting an expression into brackets, with the subject on the outside of the brackets
If the subject appears inside a set of brackets, you will need to expand these brackets before you can begin rearranging
If the subject appears on two sides of a formula, you will need to bring those terms to the same side before you can factorise

Worked example

Rearrange the formula $p = \frac{2 - a x}{x - b}$ to make $x$ the subject.

Get rid of the fraction by multiplying both sides by the expression on the denominator.

$p (x - b) = 2 - a x$

Expand the brackets on the left hand side to 'release' the $x$ .

$p x - p b = 2 - a x$

Bring the terms containing $x$ to one side of the equals sign and any other terms to the other side.

$\begin{array}{rcl} p x - p b & = & 2 - a x \end{array}$

$\begin{array}{rcl} (+ a x) (+ a x) \end{array}$

$\begin{array}{rcl} p x - p b + a x & = & 2 \end{array}$

$\begin{array}{rcl} (+ p b) (+ p b) \end{array}$

$\begin{array}{rcl} p x + a x & = & 2 + p b \end{array}$

Factorise the left-hand side to bring $x$ outside of the brackets, so that it appears only once.

$x (p + a) = 2 + p b$

Isolate $x$ by dividing by the whole expression $(p + a)$ .

$x = \frac{2 + p b}{p + a}$

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