Simplifying Algebraic Fractions | Questions and Revision | MME

# Algebraic Fractions Worksheets, Questions and Revision

Level 6 Level 7

## What you need to know

### Algebraic Fractions

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).

It is important to remember that the rules which you learned regarding adding, subtracting, multiplying, dividing, simplifying fractions still apply to algebraic fractions. Therefore to confidently tackle algebraic fractions you need to be confident with the following topics:

### Take Note:

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 dividing 6 and 8 by 2. The aim of simplifying algebraic fractions is the same, but factors look a little different, and we’re going to have to do some factorising to find them. Make sure are really confident with factorising.

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 1: Simplifying Algebraic Fractions

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

When asked to simplify an algebraic fraction, this almost always involved factorising

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:

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

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:

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

### Example 2: Adding Algebraic Fractions

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

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

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

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

As they have the same denominator now, we can simply add the numerators to get

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

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, then factorise to see if there are any common factors we can cancel.

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

We cannot simplify this fraction any further so the final answer is

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

Note: Subtracting algebraic fractions uses the same method as shown here but instead of adding the numerators you subtract them.

### Example 3: Multiplying Algebraic Fractions

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:

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

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.

Note: Dividing algebraic fractions takes has the same method with one additional step of KFCKeep, Flip, Change if you are unsure about this then revise fractions

### Example Questions

We need to find a common denominator between all three fractions before we can do the addition and subtraction.

As 30 is the lowest common multiple of 2, 3 & 5, we will choose that as the common denominator.

Hence we can multiply each term such that,

\begin{aligned}\dfrac{1}{2x}+\dfrac{1}{3x}-\dfrac{1}{5x} &= \dfrac{1}{2x}\times\dfrac{15}{15}+\dfrac{1}{3x}\times\dfrac{10}{10}-\dfrac{1}{5x}\times\dfrac{6}{6} \\ \\ &= \dfrac{15}{30x}+\dfrac{10}{30x}-\dfrac{6}{30x} \\ \\ &= \dfrac{15+10-6}{30x} \\ \\ &=\dfrac{19}{30x}\end{aligned}

We need to find a common denominator between the fractions before we can do the addition, hence,

$\dfrac{8}{x}-\dfrac{1}{x-3}=\dfrac{8(x-3)}{x(x-3)}-\dfrac{1(x)}{x-3(x)}$

This can be simplified to,

$\dfrac{8x-24-x}{x(x-3)}=\dfrac{7x-24}{x(x-3)}$

There are no more common terms so this is fully simplified.

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,

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

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

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

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

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

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

Completing the multiplication,

$\dfrac{7ab}{12} \times \dfrac{9b^2}{4a}=\dfrac{63ab^3}{48a}$

There is a factor of $a$ that we can cancel and can also take out a factor of 3 from 63 and 48,

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

To do this subtraction, we need to find a common denominator.

Hence the left-hand fraction 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 we will multiply $(z - 1)$ 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 numerator, we get:

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

Considering the denominator is $(z - 1)(z + 5)$, we can see that there are no common factors, meaning our final answer is:

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

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