Formulas where Subject Appears Once (Cambridge (CIE) IGCSE International Maths)

Revision Note

Written by: Mark Curtis

Reviewed by: Dan Finlay

Simple Rearranging

What are formulas?

A formula is a rule, definition or relationship between different quantities, written in shorthand using letters (variables)
- They include an equals sign
Some examples 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
  - $a^{2} + b^{2} = c^{2}$

How do I rearrange formulas?

The letter (variable) that is on its own on one side is called the subject
- y is the subject of y = mx + c
To make a different letter the subject, we need to rearrange the formula
- This is also called changing the subject
The method is as follows:
- First, remove any fractions
  - Multiply both sides by the lowest common denominator
- Then use inverse (opposite) operations to get the variable on its own
  - This is similar to solving equations
For example, make $x$ the subject of $\frac{5 x + 6}{2} = y$
- First remove fractions
  - Multiply both sides by 2
    $5 x + 6 = 2 y$
- Then get $x$ on its own
  - Subtract 6 from both sides
    $5 x = 2 y - 6$
  - Divide both sides by 5
    $x = \frac{2 y - 6}{5}$
- There may be more than one correct way to write an answer
  - The following are acceptable alternative forms
    $x = \frac{2 y}{5} - \frac{6}{5}$
    $x = \frac{2 (y - 3)}{5}$
    $x = 0.4 (y - 3)$
    $x = 0.4 y - 1.2$

Should I expand brackets?

Expand brackets if it releases the variable you want from inside the brackets
- If not, you can leave them in
To make $x$ the subject of $3 (1 + x) = y$
- $x$ is inside the brackets, so expand
  - $3 + 3 x = y$
- Rearrange
  - $\begin{array}{rcl} 3 x & = & y - 3 \end{array}$
    $\begin{array}{rcl} x & = & \frac{y - 3}{3} \end{array}$

To make $x$ the subject of $(1 + k) x = y$
- $x$ is not inside the brackets, so you do not need to expand
- Instead, divide both sides by the bracket $(1 + k)$
  - $x = \frac{y}{1 + k}$

What if I get fractions in fractions?

Some rearrangements can lead to fractions in fractions
- $x = \frac{\frac{3}{t}}{2}$
Either rewrite with a divide sign, $\div$ , then use the method of dividing two fractions
- $x = \frac{3}{t} \div 2$
  $x = \frac{3}{t} \div \frac{2}{1} x = \frac{3}{t} \times \frac{1}{2} x = \frac{3}{2 t}$
Or multiply top and bottom by the the lowest common denominator of the two fractions and cancel
- $x = \frac{\frac{5}{y}}{\frac{t}{8}}$ becomes $x = \frac{\frac{5}{y} \times 8 y}{\frac{t}{8} \times 8 y} = \frac{40}{t y}$

What if I end up dividing by a negative?

Remember that $\frac{a}{- b}$ (minus below) is the same as $\frac{- a}{b}$ (minus above) and the same as $- \frac{a}{b}$ (minus outside)
- Though be careful, as $\frac{- a}{- b}$ is $\frac{a}{b}$
$- 2 x = y - 3$ becomes $x = \frac{y - 3}{- 2}$ (minus below)
- This is the same as $x = \frac{- (y - 3)}{2}$ (minus above) or $- \frac{y - 3}{2}$ (minus outside)
  - brackets are required for minus above
  - brackets are assumed for minus outside
- You can also expand the brackets
  $\frac{- (y - 3)}{2} = \frac{- y + 3}{2} = \frac{3 - y}{2}$

Examiner Tips and Tricks

Mark schemes will accept different forms of the same answer, as long as they are correct and fully simplified

Worked Example

Make $x$ the subject of the following.

(a) $4 m + 5 x = 3$

Get 5x on its own by subtracting 4m from both sides

$5 x = 3 - 4 m$

Get x on its own by dividing both sides by 5

$x = \frac{3 - 4 m}{5}$

(b) $3 t = \frac{2}{x}$

Remove fractions by multiplying both sides by the denominator, x

$3 t x = 2$

Get x on its own by dividing both sides by 3t

$x = \frac{2}{3 t}$

Remove fractions by multiplying both sides by the denominator, 2g

$2 g A = 9 (1 - 4 x)$

x is inside the brackets
Expand the brackets to release the x term

$2 g A = 9 - 36 x$

One way to get x on its own is by subtracting 9 then dividing by -36
Or you can first add 36x to both sides, to create positive 36x on the left

$2 g A + 36 x = 9$

Now get x on its own by subtracting 2gA then dividing by 36

$\begin{array}{rcl} 36 x & = & 9 - 2 g A \\ x & = & \frac{9 - 2 g A}{36} \end{array}$

$x = \frac{9 - 2 g A}{36}$

Other accepted forms of the answer are $\frac{2 g A - 9}{- 36}, \frac{- (2 g A - 9)}{36}, - \frac{2 g A - 9}{36}, \frac{1}{4} - \frac{g A}{18}$

Last updated: 24 May 2024

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