## What you need to know

An algebraic fraction is exactly what it sounds like: a fraction where you’ll find terms involving algebra on the numerator (top) and/or the denominator (bottom).

Whilst the introduction of algebra naturally makes fractions more complicated, it is important to remember that the rules which you learned regarding adding, subtracting, multiplying, dividing, simplifying, and generally manipulating normal fractions (i.e., the ones with only numbers in them) have not changed. First and foremost, it is still true that if you multiply (or divide) the top and bottom of a fraction by the same thing, then the value of the fraction does not change. Unless that thing is zero, in which case you’ll end up in heaps of trouble. What’s different now is that rather than just numbers, we might be multiplying by $5a$ or dividing by $(x + 3)$.

Simplifying

When simplifying fractions, the aim is to find a common factor in the top and the bottom and then cancel them out, e.g. we get that $\dfrac{6}{8} = \dfrac{3}{4}$ by observing 6 and 8 have a common factor of 2, which we can then cancel from the top and bottom (in other words, we divide both numbers by it). Here, the aim is the same, but factors look a little different, and as the name suggests, we’re going to have to do some factorising to find them. Time to dust off all those factorising methods you learned.

Note: It may be the case that you have to do some expanding first before you can factorise. Bear this in mind when you’re having a go at the questions below.

Example: Simplify fully the fraction $\dfrac{a^2 + a - 6}{ab + 3b}$.

First, we will consider factorising the numerator. We see that is a quadratic, so observing that the numbers 3 and -2 multiply to make -6 and add to make 1, we get

$a^2 + a - 6 = (a + 3)(a - 2)$

Now, for the denominator:

$ab + 3b = b(a + 3)$

This might seem insignificant, but what we now have is a factor of $(a + 3)$ in both the numerator and the denominator, which means we can cancel it. This looks like:

$\dfrac{a^2 + a - 6}{ab + 3b}=\dfrac{(a + 3)(a - 2)}{b(a + 3)} = \dfrac{a - 2}{b}$

There is nothing else that the top and bottom have in common, so that must mean that we’re done.

When adding or subtracting algebraic fractions, the aim is still to find a common denominator. If you can, it might make your life a little easier to use the lowest common multiple of both the denominators in question as your common denominator. However, it isn’t always easy to know what this is so if you’re unsure, you can always safely use the product of both denominators instead (and to be honest, in algebraic fraction this often is the lowest common multiple anyway).

Example: Write $\dfrac{m}{m - 6} + \dfrac{m}{7}$ as one fraction. Simplify your answer fully.

Our choice of common denominator will be the product of both denominators, which is

$7 \times (m - 6) = 7(m - 6)$

We could expand the brackets, but it’s better to leave it factorised in case we can cancel anything later. So, our left-hand fraction is being multiplied by 7 on top and bottom, whilst our right-hand fraction is being multiplied by $(m - 6)$. So, we get

\begin{aligned}\dfrac{m}{m - 6}+\dfrac{m}{7} &= \dfrac{7m}{7(m - 6)}+\dfrac{m(m - 6)}{7(m - 6)} \\ &= \dfrac{7m + m(m - 6)}{7(m - 6)}\end{aligned}

We have written it as one fraction, now we must simplify it fully. First, we should expand the brackets and collect terms on the top, which we can then factorise to see if there are any common factors we can cancel.

Denominator:

$7m + m(m - 6) = 7m + m^2 - 6m = m^2 + m = m(m + 1)$

We cannot factorise this denominator further, so our fraction becomes

$\dfrac{m(m + 1)}{7(m - 6)}$

which shows no common factors in the top and bottom, so cannot be simplified further.

Multiplying & Dividing

Multiplying two fractions together remains a straightforward procedure even when algebra is involved: we simply multiply the numerators together and the denominators together (and then, as we have been doing, we simplify if possible). For division, the same rule of flipping the second fraction over and turning the division into a multiplication also remains.

Example: Simplify fully

$\dfrac{2x + 4}{3xy} \times \dfrac{x}{x + 2}$.

Multiply the numerators:

$(2x + 4) \times x = x(2x + 4)$

The bonus here is that we don’t have to do any tricky factorisation since it’s already factorised.

Multiply the denominators:

$3xy \times (x + 2) = 3xy(x + 2)$

So our fraction is:

$\dfrac{x(2x + 4)}{3xy(x + 2)}$

Firstly, we can cancel out the factor of $x$ on the top and bottom.

$\dfrac{2x + 4}{3y(x + 2)}$

Now, this might look like we have simplified it fully, but if we consider that $2x + 4 = 2(x + 2)$, then suddenly we have a common factor. We get:

$\dfrac{2(x + 2)}{3y(x + 2)} = \dfrac{2}{3y}$

After cancelling the $(x + 2)$ there are no more common factors, so we’re done.

## Example Questions

#### 1) Write the following algebraic fraction in its simplest form.$$\dfrac{2(8 - k) + 4(k - 1)}{k^2 - 36}$$

First, we’ll look at the numerator. Before we can factorise it, we must expand the brackets.

$2(8 - k) + 4(k - 1) = 16 - 2k + 4k - 4 = 2k + 12$

Then, the most we can do is take the 2 out as a factor, leaving us with

$2k + 12 = 2(k + 6)$

Now, the denominator is a special type of quadratic expression referred to as the difference of two squares. Therefore, recognising that $36 = 6^2$, we can immediately factorise to get

$k^2 - 36 = (k + 6)(k - 6)$

As there is a factor of $(k + 6)$ in both the numerator and denominator, these will cancel.

$\dfrac{2(k + 6)}{(k + 6)(k - 6)} = \dfrac{2}{k - 6}$

There are no more common factors, so we are done.

#### 2) Write the result of the division below in its simplest form.$\dfrac{7ab}{12}\div\dfrac{4a}{9b^2}$

Our first step when dividing any fractions should be to flip the second fraction over and turn the division into a multiplication.

$\dfrac{7ab}{12} \div \dfrac{4a}{9b^2} = \dfrac{7ab}{12} \times \dfrac{9b^2}{4a}$

Now, simply do the multiplication. Firstly, the numerators:

$7ab \times 9b^2 = 7 \times 9 \times ab \times b^2 = 63ab^3$

Next, the denominators:

$12 \times 4a = 48a$

This means the result of our division is:

$\dfrac{63ab^3}{48a}$

Firstly, there is clearly a factor of $a$ that we can cancel. Secondly, we can also take out a factor of 3 from 63 and 48, meaning we can cancel a factor of 3 as well. The fully simplified fraction is then as follows.

$\dfrac{3a \times 21b^3}{3a \times 16} = \dfrac{21b^3}{16}$

#### 3) Work out $\dfrac{z + 2}{z - 1} - \dfrac{z}{z + 5}$ and write your answer in its simplest form.

To do this subtraction, we need to find a common denominator. For this, we will use the product of the two denominators, which is $(z - 1)(z + 5)$.

Now, for the left-hand fraction to have this denominator, it must be multiplied by $(z + 5)$ on top and bottom.

$\dfrac{z + 2}{z - 1} = \dfrac{(z + 2)(z + 5)}{(z - 1)(z + 5)}$

For the right-hand fraction have this denominator, it must be multiplied by $(z - 1)$ on top and bottom.

$\dfrac{z}{z + 5} = \dfrac{z(z - 1)}{(z - 1)(z + 5)}$

Then, the subtraction is:

\begin{aligned}\dfrac{z + 2}{z - 1} - \dfrac{z}{z + 5} &= \dfrac{(z + 2)(z + 5)}{(z - 1)(z + 5)} - \dfrac{z(z - 1)}{(z - 1)(z + 5)} \\ &= \dfrac{(z + 2)(z + 5) - z(z - 1)}{(z - 1)(z + 5)}\end{aligned}

Expanding the denominator, we get:

$(z + 2)(z + 5) - z(z - 1) = z^2 + 7z + 10 - z^2 + z = 8z + 10$

The furthest we can factorise this is to take the 2 out as a factor and get $8z + 10 = 2(4z + 5)$. Considering the denominator is $(z - 1)(z + 5)$, we can see that there are no common factors, so this is as simple as it gets, meaning our final answer is:

$\dfrac{2(4z + 5)}{(z - 1)(z + 5)}$

## Algebraic Fractions Teaching Resources

Teaching Resources From Around The Web
Algebraic Fractions Subtract (RAG)
6-7
Algebraic Fractions Multily and Divide (RAG)
6-7

Algebraic fraction resources for the new GCSE Maths 9-1 curriculum can be accessed above. You may be a GCSE Maths tutor or a teacher in a secondary school, these resources should be useful either way. The algebraic fraction worksheets can be used as starters, refreshers and homework and the online tests can be used to determine how a student is progressing.