Adding & Subtracting Algebraic Fractions (Edexcel IGCSE Maths A (Modular))

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Naomi C

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Maths

Adding & Subtracting Algebraic Fractions

How do I add (or subtract) two algebraic fractions?

The rules for adding and subtracting algebraic fractions are the same as they are for fractions with numbers
STEP 1
Find the lowest common denominator (LCD)
- Sometimes the LCD can be found by multiplying the denominators together
  - E.g. The LCD for the fractions $\frac{1}{x + 2}$ and $\frac{1}{x + 5}$ is $(x + 2) (x + 5)$
  - Similarly, with numbers, the LCD of $\frac{1}{2}$ and $\frac{1}{5}$ is 2 × 5 = 10
- Although multiplying the denominators will always give you a multiple, it is not necessarily the lowest multiple
  - E.g. The LCD for the fractions $\frac{1}{x}$ and $\frac{1}{2 x}$ is $2 x$ (not $2 x^{2}$ ) as both terms already include an $x$
  - Similarly, with numbers, the LCD of $\frac{1}{2}$ and $\frac{1}{4}$ is just 4, not 2 × 4 = 8
- Other examples include:
  - The LCD of $\frac{1}{x + 2}$ and $\frac{1}{(x + 2) (x - 1)}$ is $(x + 2) (x - 1)$
  - The LCD of $\frac{1}{x + 1}$ and $\frac{1}{{(x + 1)}^{2}}$ is ${(x + 1)}^{2}$
  - The LCD of $\frac{1}{(x + 3) (x - 1)}$ and $\frac{1}{(x + 4) (x - 1)}$ is $(x + 3) (x - 1) (x + 4)$
STEP 2
Write each fraction over the lowest common denominator
Multiply the numerator of each fraction by the same amount as the denominator
- E.g. $\begin{array}{rcl} \frac{x}{x - 4} + \frac{1}{x + 2} & = & \frac{x (x + 2)}{(x - 4) (x + 2)} + \frac{(x - 4)}{(x - 4) (x + 2)} \end{array}$
STEP 3
Write as a single fraction over the lowest common denominator and simplify the numerator
- Do this by adding or subtracting the numerators
- Take particular care if subtracting
- E.g. $\begin{array}{rcl} \frac{x (x + 2) + (x - 4)}{(x - 4) (x + 2)} \end{array} = \frac{x^{2} + 2 x + x - 4}{(x - 4) (x + 2)} = \frac{x^{2} + 3 x - 4}{(x - 4) (x + 2)}$
STEP 4
Check at the end to see if the top factorises and the fraction can be simplified
- E.g. $\frac{(x + 4) (x - 1)}{(x - 4) (x + 2)}$ , the top factorises but there are no common factors so it is in its most simple form

Exam Tip

Leaving the top and bottom of your answer in factorised form will help you see if anything cancels at the end

Worked Example

(a) Express $\frac{x}{x + 4} - \frac{3}{x - 1}$ as a single fraction.

The lowest common denominator is $(x + 4) (x - 1)$
Write each fraction over this common denominator, remember to multiply the top of the fractions too

$\frac{x (x - 1)}{(x + 4) (x - 1)} - \frac{3 (x + 4)}{(x - 1) (x + 4)}$

Combine the fractions, as they now have the same denominator

$\frac{x (x - 1) - 3 (x + 4)}{(x + 4) (x - 1)}$

Simplify the numerator
Be careful expanding with the negative signs

$\frac{x^{2} - x - 3 x - 12}{(x + 4) (x - 1)} = \frac{x^{2} - 4 x - 12}{(x + 4) (x - 1)}$

Factorise the top

$\frac{(x + 2) (x - 6)}{(x + 4) (x - 1)}$

There are no terms which would cancel here, so this is the final answer

$\frac{(x + 2) (x - 6)}{(x + 4) (x - 1)}$

(b) Express $\frac{x - 4}{2 (x - 3)} - \frac{x - 1}{2 x}$ as a single fraction.

The lowest common denominator is $2 x (x - 3)$
(You could also use $4 x (x - 3)$ but this wouldn't be the lowest common denominator)

Write each fraction over this common denominator, remember to multiply the top of the fractions too

$\frac{x (x - 4)}{2 x (x - 3)} - \frac{(x - 1) (x - 3)}{2 x (x - 3)}$

Combine the fractions, as they now have the same denominator

$\frac{x (x - 4) - (x - 1) (x - 3)}{2 x (x - 3)}$

Simplify the numerator
Be careful expanding with negative signs

$\frac{(x^{2} - 4 x) - (x^{2} - 4 x + 3)}{2 x (x - 3)} = \frac{x^{2} - 4 x - x^{2} + 4 x - 3}{2 x (x - 3)} = \frac{- 3}{2 x (x - 3)}$

There is nothing else that can be factorised on the numerator, so this is the final answer

$\frac{- 3}{2 x (x - 3)}$

There are other accepted answers, e.g. $\frac{3}{2 x (3 - x)}$

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