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Algebraic terms Simplifying algebraic expressions Collecting like terms Laws of indices Expanding brackets Factorising Simplifying fractions Simplifying algebraic fractions Multiplying and dividing fractions Difference of two squaresThis topic is relevant for:

Here we will learn about multiplying and dividing algebraic fractions, including algebraic fractions with monomial and binomial numerators and denominators.

There are also multiplying and dividing algebraic fractions worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**Multiplying and dividing algebraic fractions** is the skill of multiplying and dividing two or more fractions that contain **algebraic terms.**

For example, \cfrac{2a}{5}, \cfrac{3-gh}{4f}, or \cfrac{1}{x-x^{2}} \, .

To do this we must combine our knowledge of multiplying and dividing fractions with our understanding of algebra.

**Step-by-step guide:** Multiplying and dividing fractions

**Step-by-step guide:** Algebraic terms

To **multiply with fractions,** we multiply the **numerators** together, and multiply the **denominators** together. This is the same for algebraic fractions, but we need to take extra care when multiplying algebraic terms or expressions.

For example, \cfrac{3x^3}{a} \times \cfrac{5x}{2b}=\cfrac{3x^3\times 5x}{a\times 2b}=\cfrac{15x^4}{2ab} \, .

To **divide with fractions,** we first write the **reciprocal** of the **dividing fraction** and then multiply the numerators together, and multiply the denominators together. This is the same for algebraic fractions, but we need to take extra care when multiplying algebraic terms or expressions.

For example, \cfrac{4b}{3} \div \cfrac{7a}{b}=\cfrac{4b}{3} \times \cfrac{b}{7a}=\cfrac{4b\times b}{3\times 7a}=\cfrac{4b^2}{21a} \, .

In order to multiply algebraic fractions:

**Multiply the numerators together and multiply the denominators together.****Simplify the fraction if possible.**

Get your free multiplying algebraic fractions worksheet of 20+ algebraic fractions questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free multiplying algebraic fractions worksheet of 20+ algebraic fractions questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEWrite as a single fraction in its simplest form, \cfrac{2}{a}\times\cfrac{5}{b} \, .

**Multiply the numerators together and multiply the denominators together.**

\cfrac{2\times5}{a\times b}=\cfrac{10}{ab}

2**Simplify the fraction if possible.**

The fraction \cfrac{10}{ab} cannot be simplified as the numerator and the denominator do not have any common factors.

The final answer is \cfrac{10}{ab}.

Write as a single fraction in its simplest form, \cfrac{4x}{5}\times\cfrac{x+7}{8} \, .

**Multiply the numerators together and multiply the denominators together.**

\cfrac{4x\times(x+7)}{5\times8}=\cfrac{4x(x+7)}{40}

Note that you can leave algebraic products in factorised form. However, to help with step 2 it is best to check that they are **fully** factorised.

The terms in the bracket here do not share a common factor other than 1, and therefore the expression is fully factorised.

**Simplify the fraction if possible.**

\cfrac{4x(x+7)}{40}

The numerator and the denominator have a common factor of 4 so we divide both the numerator and the denominator by 4 to simplify the fraction.

\cfrac{4x(x+7)\div4}{40\div4}=\cfrac{x(x+7)}{10}

The final answer is \cfrac{x(x+7)}{10}.

If you expand the brackets then another way of writing the final answer is \cfrac{x^2+7x}{10}.

Write as a single fraction in its simplest form, \cfrac{3x+9}{4x}\times\cfrac{5}{2x+6} \, .

**Multiply the numerators together and multiply the denominators together.**

\cfrac{(3x+9)\times5}{4x\times(2x+6)}=\cfrac{5(3x+9)}{4x(2x+6)}

Note that you can leave algebraic products in factorised form. However, to help with step 2 it is best to check that they are **fully** factorised.

In this case the terms in the brackets of the numerator both have a factor of 3, and the terms in the brackets of the denominator both have a factor of 2. Therefore we can write the fraction like this.

\cfrac{5(3x+9)}{4x(2x+6)}=\cfrac{5\times3(x+3)}{4x\times2(x+3)}=\cfrac{15(x+3)}{8x(x+3)}

**Simplify the fraction if possible.**

\cfrac{15(x+3)}{8x(x+3)}

We can simplify this fraction as the numerator and the denominator share the binomial factor (x+3).

\cfrac{15\cancel{(x+3)}}{8x\cancel{(x+3)}}=\cfrac{15}{8x}

The final answer is \cfrac{15}{8x}.

In order to divide algebraic fractions:

**Find the reciprocal of the dividing fraction and rewrite the question with multiplication instead of division.****Multiply the numerators together and multiply the denominators together.****Simplify the fraction if possible.**

Write as a single fraction in its simplest form, \cfrac{10}{3c}\div\cfrac{5}{3d} \, .

**Find the reciprocal of the dividing fraction and rewrite the question with multiplication instead of division.**

\cfrac{10}{3c}\div\cfrac{5}{3d}=\cfrac{10}{3c}\times\cfrac{3d}{5}

**Multiply the numerators together and multiply the denominators together.**

\cfrac{10}{3c}\times\cfrac{3d}{5}=\cfrac{10\times 3d}{3c\times5}=\cfrac{30d}{15c}

**Simplify the fraction if possible.**

\cfrac{30d}{15c}

Both the numerator and the denominator have a common factor of 15 so we divide both the numerator and the denominator by 15 to simplify the fraction.

\cfrac{30d\div15}{15c\div15}=\cfrac{2d}{c}

The final answer is \cfrac{2d}{c}.

Write as a single fraction in its simplest form, \cfrac{3x}{4}\div\cfrac{x+1}{8} \, .

\cfrac{3x}{4}\div\cfrac{x+1}{8}=\cfrac{3x}{4}\times\cfrac{8}{x+1}

**Multiply the numerators together and multiply the denominators together.**

\cfrac{3x}{4}\times\cfrac{8}{x+1}=\cfrac{3x\times 8}{4\times(x+1)}=\cfrac{24x}{4(x+1)}

Note that you can leave algebraic products in factorised form. However, to help with step 2 it is best to check that they are **fully** factorised.

The terms in the bracket here do not share a common factor other than 1, and therefore the expression is fully factorised.

**Simplify the fraction if possible.**

\cfrac{24x}{4(x+1)}

Both the numerator and the denominator have a common factor of 4 so we divide both the numerator and the denominator by 4 to simplify the fraction.

\cfrac{24x}{4(x+1)}=\cfrac{24x\div4}{4(x+1)\div4}=\cfrac{6x}{x+1}

The final answer is \cfrac{6x}{x+1}.

Write as a single fraction in its simplest form, \cfrac{x^{2}-9}{x+1}\div\cfrac{x+3}{2} \, .

\cfrac{x^{2}-9}{x+1}\div\cfrac{x+3}{2}=\cfrac{x^{2}-9}{x+1}\times\cfrac{2}{x+3}

**Multiply the numerators together and multiply the denominators together.**

\cfrac{x^{2}-9}{x+1}\times\cfrac{2}{x+3}=\cfrac{(x^{2}-9)\times2}{(x+1)\times(x+3)}=\cfrac{2(x^{2}-9)}{(x+1)(x+3)}

Note that you can leave algebraic products in factorised form. However, to help with step 2 it is best to check that they are **fully** factorised.

The bracket of the numerator is the difference of two squares and can be factorised into double brackets.

\cfrac{2(x^{2}-9)}{(x+1)(x+3)}=\cfrac{2(x-3)(x+3)}{(x+1)(x+3)}

**Simplify the fraction if possible.**

\cfrac{2(x+3)(x-3)}{(x+1)(x+3)}

We can now cancel a common binomial factor of (x+3).

\cfrac{2\cancel{(x+3)}(x-3)}{(x+1)\cancel{(x+3)}}=\cfrac{2(x-3)}{x+1}

The final answer is \cfrac{2(x-3)}{x+1}.

**Step-by-step guide:** Difference of two squares

**Cross multiplying vs multiplying fractions**

When multiplying fractions, we multiply the numerators together, and multiply the denominators together. Some students can confuse this with a method referred to as ‘cross multiplying’ which some people teach as a method for dividing by a fraction.

For example,

Incorrectly choosing to cross multiply,

The correct solution is

**Common denominators are not needed**

When we add or subtract fractions we must ensure there is a common denominator. A mistake often made by students is to believe this rule also applies to the multiplication of fractions. When we are multiplying fractions, we can choose to write the fractions with a common denominator but this simply makes the process lengthier and it is not necessary.

For example,

Without writing the fractions with a common denominator,

Writing the fractions with a common denominator first,

1) Write as a single fraction in the simplest form, \cfrac{3}{2e}\times\cfrac{8}{f}.

\cfrac{3f}{16e}

\cfrac{16e}{3f}

\cfrac{24}{2ef}

\cfrac{12}{ef}

\cfrac{3}{2e}\times\cfrac{8}{f}=\cfrac{3\times{8}}{2e\times{f}}=\cfrac{24}{2ef}=\cfrac{12}{ef}

2) Write as a single fraction in the simplest form, \cfrac{2x+3}{2}\times\cfrac{2(2x+1)}{3}.

\cfrac{6x+9}{4(2x+1)}

\cfrac{4x^{2}+8x+3}{3}

\cfrac{14x+13}{6}

\cfrac{4x^{2}+8x+3}{6}

\cfrac{2x+3}{2}\times\cfrac{2(2x+1)}{3}=\cfrac{(2x+3)\times{(2(2x+1))}}{6}=\cfrac{8x^{2}+16x+6}{6}=\cfrac{4x^{2}+8x+3}{3}

3) Write as a single fraction in the simplest form, \frac{4x+4}{7}\times\frac{3x}{2x+2}.

\cfrac{8(x+1)^{2}}{21x}

\cfrac{12x^{2}+12}{14x+4}

\cfrac{6x}{7}

\cfrac{12x^{2}}{14}

\cfrac{4x+4}{7}\times\cfrac{3x}{2x+2}=\cfrac{(4x+4)\times{3x}}{7\times(2x+2)}=\cfrac{3x\times{4(x+1)}}{7\times2(x+1)}=\cfrac{12x(x+1)}{14(x+1)}=\cfrac{6x}{7}

4) Write as a single fraction in the simplest form, \cfrac{15}{4x^{2}}\div\cfrac{3x}{2}.

\cfrac{5}{2x^{3}}

\cfrac{13}{x}

\cfrac{5}{2x}

\cfrac{5}{2x^{2}}

\cfrac{15}{4x^{2}}\div\cfrac{3x}{2}=\cfrac{15}{4x^{2}}\times\cfrac{2}{3x}=\cfrac{30}{12x^{3}}=\cfrac{5}{2x^{3}}

5) Write as a single fraction in the simplest form, \cfrac{5+x}{12}\div\cfrac{x^{2}}{4}.

\cfrac{1+x}{12-x^{2}}

\cfrac{20+x}{12x^{2}}

\cfrac{5+x}{3x^{2}}

\cfrac{5}{3x}

\cfrac{5+x}{12}\div\cfrac{x^{2}}{4}=\cfrac{5+x}{12}\times\cfrac{4}{x^{2}}=\cfrac{4(5+x)}{12x^{2}}=\cfrac{5+x}{3x^{2}}

6) Write as a single fraction in the simplest form, \cfrac{x^{2}-36}{4x-8}\div\cfrac{x-6}{2}.

\cfrac{x^{2}-72}{4(x-2)(x-6)}

\cfrac{x-18}{2}

\cfrac{x+6}{2(x-2)}

\cfrac{6-x}{2x}

\cfrac{x^{2}-36}{4x-8}\div\cfrac{x-6}{2}=\cfrac{x^{2}-36}{4x-8}\times\cfrac{2}{x-6}=\cfrac{(x^{2}-36)\times{2}}{(4x-8)\times(x-6)}=\cfrac{2(x+6)(x-6)}{4(x-2)(x-6)}

=\cfrac{2(x+6)}{4(x-2)}=\cfrac{x+6}{2(x-2)}

1. Write as a single fraction, \cfrac{2}{x}\times\cfrac{3}{y}\div\cfrac{5}{z}.

**(3 marks)**

Show answer

\cfrac{6}{xy}\div\cfrac{5}{z}

**(1)**

\cfrac{6}{xy}\times\cfrac{z}{5}

**(1)**

\cfrac{6z}{5xy}

**(1)**

2. Triangle ABC is a right angled triangle where BC is perpendicular to AC and x > 0.

(a) Calculate the area of the triangle.

(b) As the value of x increases, what happens to the area of the triangle?

Explain your answer.

**(4 marks)**

Show answer

(a) (\cfrac{1}{x}\times\cfrac{x}{2})\div{2}

**(1)**

\cfrac{x}{2x}(=\cfrac{1}{2})\div{2}

\cfrac{1}{2}\times\cfrac{1}{2}

**(1)**

\cfrac{1}{4}

**(1)**

(b) Stays the same.

The area is independent of the value of x.

**(1)**

3. Use factorisation to fully simplify \cfrac{4x^{2}-16}{x+1}\div\cfrac{x-2}{3x+3} into the form a(x+b) where a and b are integers.

**(5 marks)**

Show answer

4x^{2}-16=4(x^{2}-4)

**(1)**

4(x^{2}-4)=4(x+2)(x-2)

**(1)**

3x+3=3(x+1)

**(1)**

\cfrac{4(x+2)(x-2)}{x+1}\div\cfrac{x-2}{3(x+1)}=\cfrac{4(x+2)(x-2)}{x+1}\times\cfrac{3(x+1)}{x-2}

**(1)**

\cfrac{4(x+2)(x-2)\times{3}(x+1)}{(x+1)(x-2)}=12(x+2)

**(1)**

You have now learned how to:

- Multiply and divide with algebraic fractions

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