National 5 Maths
Algebraic Fractions

Note that the reason for \(x\) not being allowed to be 0, 2 or -1 is so that none of the factors of the denominator can possibly be zero. Division by zero is impossible.

In this example, both the numerator and denominator are already fully factorised. So we do not have to do any more factorising ourselves.

There are common factors in both the numerator and denominator of \(x\) and \(\left(x+1\right)\). These can cancel.

So we get the following working:

$$ \begin{eqnarray} \small\frac{{x^2}(x+1)}{x(x-2)(x+1)}\ &=&\small\ \frac{{x^\cancel{2}}\cancel{(x+1)}}{\cancel{x}(x-2)\cancel{(x+1)}}\\[8pt] &=&\small\ \frac{x}{x-2} \end{eqnarray} $$

Video solution by Clelland Maths 

In this example, we need to factorise both the numerator and denominator to see if there are any common factors that we can cancel.

$$ \begin{eqnarray} \small\frac{x^2+3x-4}{x^2-x}\ &=&\small\ \frac{(x-1)(x+4)}{x(x-1)}\\[8pt] &=&\small\ \frac{\cancel{(x-1)}(x+4)}{x\cancel{(x-1)}}\\[8pt] &=&\small\ \frac{x+4}{x} \end{eqnarray} $$

Video solution by Clelland Maths 

Just like adding numerical fractions, we need a common denominator.

The lowest common multiple of 4 and 6 is 12.

$$ \begin{eqnarray} \small\frac{3a}{4}+\frac{5a}{6}\ &=&\small\ \frac{9a}{12}+\frac{10a}{12}\\[8pt] &=&\small\ \frac{19a}{12} \end{eqnarray} $$

Video solution by Clelland Maths 

Just like subtracting numerical fractions, we need a common denominator.

With numerical fractions, we would use the lowest common multiple of the denominators. But in this example, we don't know the value of \(x\), so we don't know if it's a multiple of 5.

Because of this, we just have to multiply the denominators and use \(5x\) as the common denominator.

$$ \begin{eqnarray} \small\frac{3}{x}-\frac{2x}{5}\ &=&\small\ \frac{15}{5x}-\frac{2x^2}{5x}\\[8pt] &=&\small\ \frac{15-2x^2}{5x} \end{eqnarray} $$

Video solution by Clelland Maths 

Just like multiplying numerical fractions, we first try to cancel any common factors.

How do we find common factors? By factorising, of course!

$$ \begin{eqnarray} &\ &\small (x+3) \times \frac{x^2+7}{x^2-9} \\[12pt] &=&\small\ \frac{x+3}{1} \times \frac{x^2+7}{(x-3)(x+3)} \\[12pt] &=&\small\ \frac{\cancel{x+3}}{1} \times \frac{x^2+7}{(x-3)(\cancel{x+3})} \\[12pt] &=&\small\ \frac{x^2+7}{x-3} \end{eqnarray} $$

Video solution by Clelland Maths 

Just like dividing numerical fractions, we multiply by the reciprocal.

$$ \begin{eqnarray} \small\frac{p}{2t^2} \div \frac{p^2}{4t}\ &=&\ \small\frac{p}{2t^2}\times \frac{4t}{p^2}\\[10pt] &=&\small\ \frac{\cancel{p}}{\cancelto{1}{2}t^\cancel{2}}\times \frac{\cancelto{2}{4}\cancel{t}}{p^\cancel{2}}\\[10pt] &=&\small\ \frac{2}{pt} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small\frac{7}{x+5}-\frac{3}{x}\ &=&\small\ \frac{7x}{x(x+5)}-\frac{3(x+5)}{x(x+5)}\\[10pt] &=&\small\ \frac{7x-3(x+5)}{x(x+5)}\\[10pt] &=&\small\ \frac{7x-3x-15}{x(x+5)}\\[10pt] &=&\small\ \frac{4x-15}{x(x+5)} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small\frac{5t}{s} \div \frac{t}{2s^2}\ &=&\small\ \frac{5t}{s}\times \frac{2s^2}{t}\\[10pt] &=&\small\ \frac{5\cancel{t}}{\cancel{s}}\times \frac{2s\cancel{^2}}{\cancel{t}}\\[10pt] &=&\ 10s \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &\ &\small\frac{3}{x-2}+\frac{5}{x+1}\ \\[10pt] &=&\small\ \frac{3(x+1)}{(x-2)(x+1)}+\frac{5(x-2)}{(x+1)(x-2)}\\[10pt] &=&\small\ \frac{3(x+1)+5(x-2)}{(x+1)(x-2)}\\[10pt] &=&\small\ \frac{3x+3+5x-10}{(x+1)(x-2)}\\[10pt] &=&\small\ \frac{8x-7}{(x+1)(x-2)} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &\ &\small\frac{4}{x+2}-\frac{3}{x-4}\ \\[10pt] &=&\small\ \frac{4(x-4)}{(x-4)(x+2)}-\frac{3(x+2)}{(x-4)(x+2)}\\[10pt] &=&\small\ \frac{4(x-4)-3(x+2)}{(x-4)(x+2)}\\[10pt] &=&\small\ \frac{4x-16-3x-6}{(x-4)(x+2)}\\[10pt] &=&\small\ \frac{x-22}{(x-4)(x+2)} \end{eqnarray} $$

Video solution by Clelland Maths 

This numerator has a common factor of \(x.\) The denominator is a trinomial. So we factorise as follows:

$$ \begin{eqnarray} \small\frac{x^2-4x}{x^2+x-20}&=&\small\frac{x(x-4)}{(x-4)(x+5)}\\[10pt] &=&\small\ \frac{x}{x+5} \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{3}{a^2}-\frac{2}{a}\ &=&\small\ \frac{3}{a^2}-\frac{2a}{a^2}\\[10pt] &=&\small\ \frac{3-2a}{a^2} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} 4x^2-25 &=&(2x)^2-5^2\\[10pt] &=& (2x-5)(2x+5) \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{4x^2-25}{2x^2-x-10} &=&\small\ \frac{(2x-5)(2x+5)}{(x+2)(2x-5)}\\[10pt] &=&\small\ \frac{2x+5}{x+2} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &\ &\small\frac{n}{n^2-4} \div \frac{3}{n-2}\\[10pt] &=&\small\ \frac{n}{(n-2)(n+2)}\times \frac{n-2}{3}\\[10pt] &=&\small\ \frac{n}{\cancel{(n-2)}(n+2)}\times \frac{\cancel{n-2}}{3}\\[10pt] &=&\small\ \frac{n}{3(n+2)} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &\ &\small\frac{4}{x-2}-\frac{3}{x+5}\ \\[10pt] &=&\small\ \frac{4(x+5)}{(x-2)(x+5)}-\frac{3(x-2)}{(x+5)(x-2)}\\[10pt] &=&\small\ \frac{4(x+5)-3(x-2)}{(x-2)(x+5)}\\[10pt] &=&\small\ \frac{4x+20-3x+6}{(x-2)(x+5)}\\[10pt] &=&\small\ \frac{x+26}{(x-2)(x+5)} \end{eqnarray} $$

Video solution by Clelland Maths 

We need to remove a common factor on the top and factorise a difference of two squares on the bottom.

$$ \begin{eqnarray} \small\frac{2ab+6a}{b^2-9}\ &=&\small\ \frac{2a(b+3)}{(b-3)(b+3)}\\[10pt] &=&\small\ \frac{2a}{b-3} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small\frac{7}{x-3}-\frac{2}{x}\ &=&\small\ \frac{7x}{x(x-3)}-\frac{2(x-3)}{x(x-3)}\\[10pt] &=&\small\ \frac{7x-2(x-3)}{x(x-3)}\\[10pt] &=&\small\ \frac{7x-2x+6}{x(x-3)}\\[10pt] &=&\small\ \frac{5x+6}{x(x-3)} \end{eqnarray} $$

Video solution by Clelland Maths 

This numerator is a difference of two squares. The denominator is a trinomial. So we factorise as follows:

$$ \begin{eqnarray} \small\frac{x^2-16}{x^2+x-20}&=&\small\frac{(x+4)(x-4)}{(x-4)(x+5)}\\[10pt] &=&\small\ \frac{x+4}{x+5} \end{eqnarray} $$

Video solution by Clelland Maths 

(a)  This is just a simple common factor.

$$ \begin{eqnarray} y^2-6y &=&y(y-6)\\[10pt] \end{eqnarray} $$

(b)  Here we have the expression from (a) as the numerator and a trinomial as the denominator.

$$ \begin{eqnarray} \small\frac{y^2-6y}{y^2-3y-18} &=&\small\ \frac{y(y-6)}{(y-6)(y+3)}\\[10pt] &=&\small\ \frac{y}{y+3} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &\ &\small\frac{2}{x+5}+\frac{3}{x-4}\ \\[10pt] &=&\small\ \frac{2(x-4)}{(x+5)(x-4)}+\frac{3(x+5)}{(x-4)(x+5)}\\[10pt] &=&\small\ \frac{2(x-4)+3(x+5)}{(x+5)(x-4)}\\[10pt] &=&\small\ \frac{2x-8+3x+15}{(x+5)(x-4)}\\[10pt] &=&\small\ \frac{5x+7}{(x+5)(x-4)} \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small\frac{5}{x-1}-\frac{4}{x}\ &=&\small\ \frac{5x}{x(x-1)}-\frac{4(x-1)}{x(x-1)}\\[10pt] &=&\small\ \frac{5x-4(x-1)}{x(x-1)}\\[10pt] &=&\small\ \frac{5x-4x+4}{x(x-1)}\\[10pt] &=&\small\ \frac{x+4}{x(x-1)} \end{eqnarray} $$