# Algebraic Fractions

## Algebraic Fractions Revision

**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.

**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}

**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}

**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}

**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)}

**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.

**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})}

**Step 2:** add the fractions

\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)}

## Algebraic Fractions Example Questions

**Question 1:** Simplify, \dfrac{1}{2x}+\dfrac{1}{3x}-\dfrac{1}{5x}

**[4 marks]**

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}

**Question 2:** Simplify, \dfrac{8}{x}-\dfrac{1}{x-3}

**[3 marks]**

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.

**Question 3:** Simplify, \dfrac{2(8 - k) + 4(k - 1)}{k^2 - 36}

**[4 marks]**

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.

**Question 4:** Simplify, \dfrac{7ab}{12}\div\dfrac{4a}{9b^2}

**[3 marks]**

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}

**Question 5:** Simplify, \dfrac{z + 2}{z - 1} - \dfrac{z}{z + 5}

**[4 marks]**

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)}