National 5 Maths
Algebraic Fractions

Before we start this solution, let's understand why \(x\neq0,\ x\neq2,\ x\neq-1\small.\) These specific values of \(x\) must be excluded so that none of the factors in the denominator equal zero. Remember that division by zero is impossible. So the fact that \(x\neq 0,\,\) \(x\!-\!2\neq 0\) and \(x\!+\!1\neq 0\) means that we can divide by them when simplifying.

Now let's look at how to simplify the actual fraction.

In this example, both the numerator and denominator are already fully factorised, so we can skip that step.

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

So the working is as follows:

$$ \begin{eqnarray} && \!\!\!\!\!\!\frac{x^{\color{#0c0}\bcancel{\color{#444}2}} {\color{red}\cancel{\color{#444}(x+1)}}} {{\color{#0c0}\bcancel{\color{#444}x}}(x-2) {\color{red}\cancel{\color{#444}(x+1)}}}\\[8pt] &=&\ \frac{x}{x-2} \end{eqnarray} $$

Note 1: We remove the brackets around \((x\!-\!2)\) in the final answer, because \((x\!-\!2)\) is the only factor in the denominator. It is tidier to drop brackets whenever possible.

Note 2: If you aren't sure why \(\large\frac{x^2}{x}\) cancelled to \(x\) above, recall that \(x^2=xx\small.\)

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.

The numerator is a simple unitary trinomial and the denominator has a common factor of \(x\small.\)

$$ \begin{eqnarray} && \!\!\!\frac{x^2+3x-4}{x^2-x}\\[10pt] &=& \frac{\color{red}\cancel{\color{#444}(x-1)}\color{#444}(x+4)}{x\color{red}\cancel{\color{#444}(x-1)}}\\[8pt] &=& \frac{x+4}{x} \end{eqnarray} $$

Note: The worst mistake that students make with questions like this is to cancel the \(x^2\) terms top and bottom. This is completely wrong because \(x^2\) isn't a factor of either the top or bottom. The only way to simplify an algebraic fraction is to factorise both the numerator and denominator, and then to cancel common factors.

Video solution by Clelland Maths 

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

This example is quite simple because both denominators are just numbers, with no algebra letters.

So we just use the lowest common multiple of 4 and 6, which is 12.

$$ \begin{eqnarray} && \!\!\frac{\overset{\textcolor{red}{\small\times\normalsize 3}}{3a}}{\underset{\textcolor{red}{\small\times\normalsize 3}}{4}} + \frac{\overset{\textcolor{red}{\small\times\normalsize 2}}{5a}}{\underset{\textcolor{red}{\small\times\normalsize 2}}{6}} \\[10pt] &=& \frac{9a}{12}+\frac{10a}{12}\\[9pt] &=& \frac{19a}{12} \end{eqnarray} $$

Video solution by Clelland Maths 

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

With numerical fractions, we can 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.

$$ \begin{eqnarray} && \frac{{\overset{\textcolor{red}{\small\times\normalsize 5}}{3}}}{\underset{\textcolor{red}{\small\times\normalsize 5}}{x}}-\frac{\overset{\textcolor{red}{\small\times\normalsize x}}{2x}}{\underset{\textcolor{red}{\small\times\normalsize x}}{5}}\\[10pt] &=& \frac{15}{5x}-\frac{2x^2}{5x}\\[8pt] &=& \frac{15-2x^2}{5x} \end{eqnarray} $$

Video solution by Clelland Maths 

Note that the \((x\!+\!3)\) is only in the numerator. We can write it over \(1\small,\) or we can just move it into the numerator.

As always, we factorise fully, so that we can cancel any factors that are common to the numerator and denominator.

The \((x^2\!+\!7)\) in the numerator doesn't factorise, so we leave it alone.

But in the denominator, \((x^2\!-\!9)\) is a difference of two squares, so we must factorise it.

$$ \begin{eqnarray} &\ & (x+3) \times \frac{x^2+7}{x^2-9} \\[12pt] &=& \frac{x+3}{1} \times \frac{x^2+7}{(x-3)(x+3)} \\[12pt] &=& \frac{\color{red}\cancel{\color{#444}(x+3)}\color{#444}(x^2+7)}{(x-3)\color{red}\cancel{\color{#444}(x+3)}} \\[12pt] &=& \frac{x^2+7}{x-3} \end{eqnarray} $$

Note: We don't use the fact that \(x\neq \pm 3\) in our working. However, it's a vital part of the question. We need to know that \(x\) cannot be \(-3\) or \(3\small,\) so that the denominator is never zero.

Video solution by Clelland Maths 

To divide by a fraction, we multiply by the reciprocal ('keep, change, flip').

Then we cancel all of the common factors:

$$ \begin{eqnarray} && \frac{p}{2t^2} \div \frac{p^2}{4t}\\[10pt] &=& \frac{p}{2t^2}\times \frac{4t}{p^2}\\[10pt] &=& \frac{{\color{red}\bcancel{\color{#444}p}}}{\cancelto{1}{2}t^{\color{#0c0}\cancel{\color{#444}2}}}\times \frac{\cancelto{2}{4}{\color{#0c0}\cancel{\color{#444}t}}}{p^{\color{red}\bcancel{\color{#444}2}}}\\[10pt] &=& \frac{2}{pt} \end{eqnarray} $$

Video solution by Clelland Maths 

We use \(x(x\!+\!5)\) as the common denominator, like this:

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

Video solution by Clelland Maths 

Like example 6 above, we multiply by the reciprocal ('keep, change, flip').

$$ \begin{eqnarray} && \!\!\frac{5t}{s} \div \frac{t}{2s^2}\\[11pt] &=& \frac{5t}{s}\times \frac{2s^2}{t}\\[11pt] &=& \frac{5{\color{#0c0}\cancel{\color{#444}t}}}{{\color{red}\bcancel{\color{#444}s}}}\times \frac{2s^{\color{red}\bcancel{\color{#444}2}}}{{\color{#0c0}\cancel{\color{#444}t}}}\\[11pt] &=& \frac{10s}{1}\\[11pt] &=& 10s \end{eqnarray} $$

Video solution by Clelland Maths 

Like example 7 above, we multiply the denominators and use that as our common denominator.

Then we expand and simplify the numerator, leaving the denominator in factorised form.

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

Video solution by Clelland Maths 

This question is similar to example 7 and example 9 above.

We multiply the denominators and use that as our common denominator.

Then we expand and simplify the numerator, leaving the denominator in factorised form.

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

Video solution by Clelland Maths 

This numerator has a common factor of \(x.\) The denominator is a quadratic trinomial.

So we factorise each of them as follows, and cancel the common factor:

$$ \begin{eqnarray} && \!\!\frac{x^2-4x}{x^2+x-20}\\[10pt] &=& \frac{{\color{#444}x}\color{red}\cancel{\color{#444}(x-4)}}{\color{red}\cancel{\color{#444}(x-4)}\color{#444}(x+5)}\\[10pt] &=& \frac{x}{x+5} \end{eqnarray} $$

When subtracting two algebraic fractions, we need a common denominator.

We usually just multiply the two denominators and use that as our common denominator.

However, in this question, using \(a^3\) would be unnecessarily complicated. Because \(a\) is a factor of \(a^2\small,\) it is better to just use \(a^2\small.\) So the first fraction can stay as it is, and we also avoid having to simplify at the end.

$$ \begin{eqnarray} && \!\!\frac{3}{a^2}-\frac{{\overset{\textcolor{red}{\small\times\normalsize a}}{2}}}{{\underset{\textcolor{red}{\small\times\normalsize a}}{a}}}\\[11pt] &=& \frac{3}{a^2}-\frac{2a}{a^2}\\[11pt] &=& \frac{3-2a}{a^2} \end{eqnarray} $$

Video solution by Clelland Maths 

\(4x^2-25\) is a difference of two squares, so it factorises like this:

$$ \begin{eqnarray} && 4x^2-25\\[8pt] &=& (2x)^2-5^2\\[8pt] &=& (2x-5)(2x+5) \end{eqnarray} $$

Note: You don't need to show the middle step above. We only included it to make the method clearer

Now we can start work on the algebraic fraction. We already know how the numerator factorises.

The denominator is a non-unitary quadratic, so it factorises into a double bracket.

$$ \begin{eqnarray} && \!\!\frac{4x^2-25}{2x^2-x-10}\\[10pt] &=& \color{#444} \frac{\color{red}{\cancel{\color{#444}{(2x-5)}}} \color{#444}{(2x+5)}}{(x+2) \color{red}{\cancel{\color{#444}{(2x-5)}}}}\\[10pt] &=& \frac{2x+5}{x+2} \end{eqnarray} $$

Video solution by Clelland Maths 

To divide by a fraction, we multiply by its reciprocal ('keep, change, flip').

Then we factorise the difference of two squares and cancel the common factor.

$$ \begin{eqnarray} &\ & \!\!\frac{n}{n^2-4} \div \frac{3}{n-2}\\[10pt] &=& \frac{n}{(n-2)(n+2)}\times \frac{n-2}{3}\\[10pt] &=& \color{#444} \frac{n \color{red}{\cancel{\color{#444}{(n-2)}}}}{3 \color{red}{\cancel{\color{#444}{(n-2)}}} \color{#444}{(n+2)}}\\[10pt] &=& \frac{n}{3(n+2)} \end{eqnarray} $$

Note: Expanding the denominator to \(3n+6\) wouldn't lose any marks here, but in general, it is better to leave the denominator in factorised form.

Video solution by Clelland Maths 

We multiply the denominators and use that as the common denominator.

Then we expand and simplify the numerator. We leave the denominator in factorised form.

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

Video solution by Clelland Maths 

The terms in the numerator have a common factor of \(2a\small,\) so that gets factorised out.

The denominator is a difference of two squares.

$$ \begin{eqnarray} && \!\!\frac{2ab+6a}{b^2-9}\\[10pt] &=& \color{#444} \frac{2a \color{red}{\cancel{\color{#444}{(b+3)}}}}{(b-3) \color{red}{\cancel{\color{#444}{(b+3)}}}}\\[10pt] &=& \frac{2a}{b-3} \end{eqnarray} $$

Video solution by Clelland Maths 

We multiply and use \(x(x-3)\) as the common denominator.

Then we expand and simplify the numerator. We leave the denominator in factorised form.

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

Video solution by Clelland Maths 

This numerator is a difference of two squares. The denominator is a trinomial quadratic.

So we factorise as follows:

$$ \begin{eqnarray} && \!\!\frac{x^2-16}{x^2+x-20}\\[10pt] &=& \color{#444} \frac{(x+4) \color{red}{\cancel{\color{#444}{(x-4)}}}}{\color{red}{\cancel{\color{#444}{(x-4)}}} \color{#444}{(x+5)}}\\[10pt] &=& \frac{x+4}{x+5} \end{eqnarray} $$

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)  The numerator is the expression from the previous part. The denominator is a trinomial quadratic.

$$ \begin{eqnarray} && \!\!\frac{y^2-6y}{y^2-3y-18}\\[10pt] &=& \color{#444} \frac{y \color{red}{\cancel{\color{#444}{(y-6)}}}}{\color{red}{\cancel{\color{#444}{(y-6)}}} \color{#444}{(y+3)}}\\[10pt] &=& \frac{y}{y+3} \end{eqnarray} $$

Video solution by Clelland Maths 

We multiply the denominators and use that as our common denominator.

Then we expand and simplify the numerator. We leave the denominator in factorised form.

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

Video solution by Clelland Maths 

We multiply the denominators and use that as our common denominator.

Then we expand and simplify the numerator. We leave the denominator in factorised form.

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