1 Cancelling common factors

Consider the fraction $\frac{10}{35}$ . To simplify it we can factorise the numerator and the denominator and then cancel any common factors. Common factors are those factors which occur in both the numerator and the denominator. Thus

$\frac{10}{35}=\frac{\mathrm{5̸}\times 2}{7\times \mathrm{5̸}}=\frac{2}{7}$

Note that the common factor 5 has been cancelled. It is important to remember that only common factors can be cancelled. The fractions $\frac{10}{35}$ and $\frac{2}{7}$ have identical values - they are equivalent fractions - but $\frac{2}{7}$ is in a simpler form than $\frac{10}{35}$ .

We apply the same process when simplifying algebraic fractions.

Example 49

Simplify, if possible,

1. $\frac{yx}{2x}$ ,
2. $\frac{x}{xy}$ ,
3. $\frac{x}{x+y}$

Solution

1. In the expression $\frac{yx}{2x}$ , $x$ is a factor common to both numerator and denominator. This common factor can be cancelled to give

   $\frac{y\mathrm{x̸}}{2\mathrm{x̸}}=\frac{y}{2}$

2. Note that $\frac{x}{xy}$ can be written $\frac{1x}{xy}$ . The common factor of $x$ can be cancelled to give

   $\frac{1\mathrm{x̸}}{\mathrm{x̸}y}=\frac{1}{y}$

3. In the expression $\frac{x}{x+y}$ notice that an $x$ appears in both numerator and denominator. However $x$ is not a common factor. Recall that factors of an expression are multiplied together whereas in the denominator $x$ is added to $y$ . This expression cannot be simplified.

Task!

Task!

Task!

Example 50

Simplify $\frac{3x+6}{6x+12}$ .

Solution

First we factorise the numerator and the denominator to see if there are any common factors.

$\frac{3x+6}{6x+12}=\frac{3\left(x+2\right)}{6\left(x+2\right)}=\frac{3}{6}=\frac{1}{2}$

The factors $x+2$ and 3 have been cancelled.

Task!

Example 51

Show that the algebraic fraction $\frac{3}{x+1}$ and $\frac{3\left(x+4\right)}{x^2+5x+4}$ are equivalent.

Solution

The denominator, $x^2+5x+4$ , can be factorised as $\left(x+1\right)\left(x+4\right)$ so that

$\frac{3\left(x+4\right)}{x^2+5x+4}=\frac{3\left(x+4\right)}{\left(x+1\right)\left(x+4\right)}$

Note that $\left(x+4\right)$ is a factor common to both the numerator and the denominator and can be cancelled to leave $\frac{3}{x+1}$ . Thus $\frac{3}{x+1}$ and $\frac{3\left(x+4\right)}{x^2+5x+4}$ are equivalent fractions.

Task!

Example 52

Simplify $\frac{6\left(4-8x\right)\left(x-2\right)}{1-2x}$

Solution

The factor $4-8x$ can be factorised to $4\left(1-2x\right)$ . Thus

$\frac{6\left(4-8x\right)\left(x-2\right)}{1-2x}=\frac{\left(6\right)\left(4\right)\left(1-2x\right)\left(x-2\right)}{\left(1-2x\right)}=24\left(x-2\right)$

Task!

Interactive Exercises