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# Algebraic Fractions: Multiplication

### Algebraic Fractions: Multiplication

Fractions containing algebra can be multiplied together, just like fractions containing integers.

An algebraic fraction is just like any other fraction except it contains at least one letter - normally $x$ or $y$

We call that letter a variable

Which of these fractions is an algebraic fraction?

Algebraic fractions can contain letters in the numerator or the denominator or both.

$\dfrac{2}{x}$, $\dfrac{3y}{10}$ and $\dfrac{a^2}{b^4}$ are all algebraic fractions.

When we multiply a numerical fraction, we multiply the numerators and then we multiply the denominators.

$\dfrac{3}{5}\times \dfrac{3}{4}=\dfrac{9}{20}$

We do the same with algebraic fractions. What is $\dfrac{3}{x}\times \dfrac{2}{5x}$?

Try another one. What is $\dfrac{a}{4}\times \dfrac{5a}{3}$?

As with numerical fractions, we often find there are common factors in algebraic fractions.

We need to cancel these common factors to simplify our answer.

Let's look at this one. What is $\dfrac{b}{4}\times \dfrac{2b}{3}$ without cancelling the common factor?

What is the common factor in the numerator and denominator of our answer $\dfrac{2b^2}{12}$.

We have a common factor of $2$ in $\dfrac{2b^2}{12}$. What is the simplified version of this fraction?

Try this one - what is the unsimplified answer to $\dfrac{x^2}{5}\times \dfrac{3}{x}$?

What is the common factor in the numerator and denominator in this answer? $\dfrac{3x^2}{5x}$

This time our common factor is the variable $x$. What is the simplified version of $\dfrac{3x^2}{5x}$?

Multiply $\dfrac{x^2}{y} \times \dfrac{4}{x}$ then cancel any common factors.

In our previous examples, we have simplified after multiplying out.

You can also simplify, or cancel, before multiplying.

Let's look at $\dfrac{3x}{4}\times \dfrac{8x}{5}$.

We can cancel terms *diagonally* across the fractions.

Looking at the denominator in the first fraction and the numerator in the second, what is the common factor? $\dfrac{3x}{{\color{orange}{4}}}\times \dfrac{{\color{orange}{8x}}}{5}$

Now we have spotted our common factor of $4$ we can cancel as follows:

$\dfrac{3x}{{\cancel{4}}1}\times \dfrac{2{\cancel{8}}x}{5}$

Sometimes this can get a little messy!

$\dfrac{3x}{{\cancel{4}}1}\times \dfrac{2{\cancel{8}}x}{5}$ , note we have replaced the cancelled figures with the simplified one.

What is the final answer for $\dfrac{3x}{{\cancel{4}}1}\times \dfrac{2{\cancel{8}}x}{5}$?

Our final answer is $\dfrac{6x^2}{5}$.

We cannot simplify this further! 👍

You can choose whether to cancel before or after multiplying.

But always check that your final answer can't be simplified further.

What is $\dfrac{3a}{4b} \times \dfrac{5b}{4}$? You can choose which method you prefer, either cancel first and then multiply, or multiply then simplify.

Let's try a slightly trickier one.

$\dfrac{7}{2x+10}\times \dfrac{4}{3x}$

With $\dfrac{7}{2x+10}\times \dfrac{4}{3x}$ the denominator of the first fraction is $2x+10$.

If there are any common factors, they must divide into both parts of the denominator.

To make it easier, let's see if there is a common factor for the terms in the first denominator, $2x+10$. What can both these terms be divided by?

Now we have found the common factor in the denominator, let's re-write the question.

$\dfrac{7}{2x+10}\times \dfrac{4}{3x}=\dfrac{7}{2(x+5)}\times \dfrac{4}{3x}$

What common factor is there across the fractions? $\dfrac{7}{2(x+5)}\times \dfrac{4}{3x}$

By cancelling the common factor $2$ we get $\dfrac{7}{{\cancel{2}}(x+5)}\times \dfrac{{\cancel{4}}2}{3x}$. What is the result of this?

Our final answer is $\dfrac{7}{3x^2+15x}$.

Good work - this was a tricky question! 👍🏽

To summarise, an algebraic fraction is a fraction containing a letter.

$\dfrac{4}{x}$ and $\dfrac{7}{a-1}$ are examples of algebraic fractions.

To multiply an algebraic fraction, multiply the numerator then multiply the denominators.

$\dfrac{4x}{5}\times \dfrac{2x}{3}=\dfrac{8x^2}{15}$

Look out for common factors which may be letters of numbers.

$\dfrac{4x}{5}\times \dfrac{10}{x^2}$ has common factors $5$ and $x$.

You can cancel common factors either before multiplying:

$\dfrac{4{\cancel{x}}}{{\cancel{5}}}\times \dfrac{{\cancel{10}}2}{x^{\cancel{2}}}=\dfrac{8}{x}$

Or multiply first then cancel:

$\dfrac{4x}{5}\times \dfrac{10}{x^2}=\dfrac{40x}{5x^2}=\dfrac{8}{x}$