Addition & Subtraction with Algebraic Fractions (SQA National 5 Maths): Revision Note

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 $\left(x+2\right)\left(x+5\right)$

    • 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 common denominator, it is not necessarily the lowest common denominator

    • E.g. The LCD for the fractions $\frac{1}{x}$ and $\frac{1}{2x}$ is $2x$ (not $2x^2$) 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 $\frac{1}{\left(x+2\right)\left(x-1\right)}$ is $\left(x+2\right)\left(x-1\right)$

    • The LCD of $\frac{1}{x+1}$ and $\frac{1}{{\left(x+1\right)}^2}$ is ${\left(x+1\right)}^2$

    • The LCD of $\frac{1}{\left(x+3\right)\left(x-1\right)}$ and $\frac{1}{\left(x+4\right)\left(x-1\right)}$ is $\left(x+3\right)\left(x-1\right)\left(x+4\right)$

• 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. $\begin{array}{rcl}\frac{x}{x-4}+\frac{1}{x+2}& =& \frac{x\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}+\frac{\left(x-4\right)}{\left(x-4\right)\left(x+2\right)}\end{array}$

• 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 with minus signs if subtracting

  • E.g. $\begin{array}{rcl}& & \frac{x\left(x+2\right)+\left(x-4\right)}{\left(x-4\right)\left(x+2\right)}\end{array}=\frac{x^2+2x+x-4}{\left(x-4\right)\left(x+2\right)}=\frac{x^2+3x-4}{\left(x-4\right)\left(x+2\right)}$

• STEP 4

  Check at the end to see if the top factorises and the fraction can be simplified

  • E.g. $\frac{\left(x+4\right)\left(x-1\right)}{\left(x-4\right)\left(x+2\right)}$, the top factorises but there are no common factors so it is in its most simple form