Algebraic Fractions

Just like with numerical fractions, simplifying algebraic fractions is simply about finding common factors in the numerator and denominator. The only difference is how we find the factors β with algebraic expressions, we have to factorise them first.

We need to start by factorising both the numerator and denominator. In this case, we can take out a factor of 3 from both.

\begin{aligned} \frac{3p+9}{9p+15} = \frac{3(p+3)}{3(3p+5)} \end{aligned}

Itβs now easy to see that the numerator and denominator have a common factor of 3, so we can divide both the numerator and denominator by 3.

\begin{aligned} \frac{3(p+3)}{3(3p+5)} = \frac{p+3}{3p+5} \end{aligned}

Since we took out the highest common factor at the beginning, the fraction is in its simplest form.

The process is exactly the same as in example 1; we just have slightly more complex expressions to factorise. It can help to consider them separately, rather than in the fraction.

$3x^2+12x = 3x(x+4)\\x^2-x-20=(x-5)(x+4)$

Now, we can put this back into our fraction:

\begin{aligned} \frac{3x^2+12}{x^2-x-20} = \frac{3(x+4)}{(x-5)(x+4)} \end{aligned}

As you can see, (π₯ + 4) is common to the numerator and denominator so we can divide both by (π₯ + 4).

\begin{aligned} \frac{3(x+4)}{(x-5)(x+4)} = \frac{3}{(x-5)} \end{aligned}

Just as before, we need to factorise the numerator and denominator. This time, itβs easier to factorise the denominator so weβll start with that.

$y^2-49=(y+7)(y-7)$

Now, we can assume that (π¦ + 7) or (π¦ β 7) is a factor of the numerator β this makes it much easier to factorise.

$5y^2-27y-56=(5y+8)(y-7)$

This time, (π¦ β 7) is common to the numerator and denominator so we can divide both by (π¦ β 7).

\begin{aligned} \frac{5y^2-27y-56}{y^2-4} &= \frac{(5y+8)(y-7)}{(y+7)(y-7)} \\ &= \frac{5y+8}{y-7}\end{aligned}

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