 Simplifying Algebraic Fractions | Questions and Revision | MME

# Algebraic Fractions Worksheets, Questions and Revision

Level 6-7

## Algebraic Fractions

An algebraic fraction is exactly what it sounds like: a fraction where you’ll find terms involving algebra. It is important to remember that the rules which you learned regarding fractions still apply to algebraic fractions, so make sure you are happy with the following topics before continuing.

Level 6-7

## Key Skill 1: Simplifying Algebraic Fractions

This is the easiest type you will need to simplify, requiring only the need to cancel down like terms.

Example: Simplify the following $\dfrac{55x^4y^3}{15x^2y}$

Using our knowledge of indices rules we can cancel down as follows.

$\dfrac{\sout{55}x^{\sout{4}} y^{\sout{3}}}{\sout{15}\sout{x}^{\sout{2}}\sout{y}} = \dfrac{11x^2y^2}{3}$

Level 6-7
Level 6-7

## Key Skill 2: Simplifying Algebraic Fractions – Involving Quadratics

When the fractions involve a quadratic, you first need to factorise the quadratic in order to simplify.

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

Step 1: First, We need to factorise the numerator and the denominator of the fraction. (Revise factorising quadratics here)

First the numerator,

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

Now, for the denominator,

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

Step 2: cancel down our fraction, in this case we can cancel down $(a + 3)$ in both the numerator and the denominator

This looks like:

$\dfrac{(a+3)(a-2)}{b(a+3)}=\dfrac{\xcancel{(a + 3)}(a - 2)}{b \xcancel{(a + 3)}} = \dfrac{a - 2}{b}$

Level 6-7
Level 6-7

## Key Skill 3: Multiplying and Dividing Algebraic Fractions

Multiplying and dividing algebraic fractions is exactly the same as regular fractions.

• When multiplying, multiply the top by the top and the bottom by the bottom separately.
• When dividing, simply flip the second fraction, then multiply.

Example: Simplify the following $\dfrac{(3x+1)}{(x-1)} \div \dfrac{2x}{(x-1)}$

Step 1: Flip the second fraction upside down and change the $\div$ to a $\times$

$\dfrac{(3x+1)}{(x-1)} \div \dfrac{2x}{(x-1)}=\dfrac{(3x+1)}{(x-1)} \times \dfrac{(x-1)}{2x}$

Step 2: Cancel down the fraction if possible.

$\dfrac{(3x+1)\xcancel{(x-1)}}{2x\xcancel{(x-1)}} = \dfrac{3x+1}{2x}$

Level 6-7
Level 6-7

## Skill 4: Adding and Subtracting Algebraic Fractions

When adding and subtracting fractions you always need to find a common denominator, this is the same for algebraic fractions.

Example: Write $\dfrac{2}{x+2} + \dfrac{3}{2x+1}$ as a single fraction.

Step 1: We need to multiply each fraction by the denominator of the other fraction.

$\bigg(\dfrac{2}{x+2}\times\textcolor{red}{\dfrac{2x+1}{2x+1}}\bigg) + \bigg(\dfrac{3}{2x+1}\times \textcolor{blue}{\dfrac{x+2}{x+2}}\bigg) = \dfrac{2\textcolor{red}{(2x+1)}}{(x+2)\textcolor{red}{(2x+1)}} + \dfrac{3\textcolor{blue}{(x+2)}}{(2x+1)\textcolor{blue}{(x+2)}}$

Step 2: Multiply out the numerators if needed.

$\dfrac{2\textcolor{red}{(2x+1)}}{(x+2)\textcolor{red}{(2x+1)}} + \dfrac{3\textcolor{blue}{(x+2)}}{(2x+1)\textcolor{blue}{(x+2)}} = \dfrac{4x + 2}{(x+2)(2x+1)} + \dfrac{3x+6}{(2x+1)(x+2)}$

Step 3: Add (or subtract) the fractions

$\dfrac{4x + 2}{(x+2)(2x+1)} + \dfrac{3x+6}{(2x+1)(x+2)} = \dfrac{7x+8}{(2x+1)(x+2)}$

Level 8-9
Level 6-7

## Example 1: Multiplying Algebraic Fractions

Simplify the following algebraic fraction fully,

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

[4 marks]

Step 1: Multiply the top by the top and the bottom by the bottom

Multiply the numerators:

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

Multiply the denominators:

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

So our fraction is:

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

Step 2: Cancel down

We can cancel out the factor of $x$ on the top and bottom.

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

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:

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

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

Level 8-9

## Example 2: Adding Algebraic Fractions

Write $\dfrac{m}{m - 6} + \dfrac{m}{7}$ as one fraction in its simplest form.

[4 marks]

Step 1: We need to multiply each fraction by the denominator of the other fraction.

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

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

Step 3: Simplify where possible.

$\dfrac{7m + m(m - 6)}{7(m - 6)}=\dfrac{7m + m^2 - 6m}{7(m - 6)}=\dfrac{m^2 + m}{7(m - 6)}=\dfrac{m(m + 1)}{7(m - 6)}$

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

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

Level 8-9

### 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 $30x$ as the common denominator.

Hence we can multiply each term such that,

$\dfrac{1}{2x}+\dfrac{1}{3x}-\dfrac{1}{5x} = \bigg(\dfrac{1}{2x}\times\dfrac{15}{15}\bigg)+\bigg(\dfrac{1}{3x}\times\dfrac{10}{10}\bigg)-\bigg(\dfrac{1}{5x}\times\dfrac{6}{6}\bigg)$

\begin{aligned} &= \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:

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

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:

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

### Worksheets and Exam Questions

#### (NEW) Algebraic Fractions Exam Style Questions - MME

Level 6-7 New Official MME

Level 6-7

Level 1-3

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