# How do you simplify \frac { 2x } { x ^ { 2} + 8x - 9} \div ( \frac { x ^ { 2} + 4x } { x ^ { 2} - 81} \cdot \frac { x - 9} { x + 4} )?

Oct 5, 2017

$\frac{2 x}{{x}^{2} + 8 x - 9} \div \left(\frac{{x}^{2} + 4 x}{{x}^{2} - 81} \cdot \frac{x - 9}{x + 4}\right) = \frac{2}{x - 1}$

#### Explanation:

This question is best done by factorising expressions wherever you can, and then going on from there.

$\frac{2 x}{{x}^{2} + 8 x - 9} \div \left(\frac{{x}^{2} + 4 x}{{x}^{2} - 81} \cdot \frac{x - 9}{x + 4}\right)$

$= \frac{2 x}{\left(x - 1\right) \left(x + 9\right)} \div \left(\frac{x \left(x + 4\right)}{\left(x - 9\right) \left(x + 9\right)} \cdot \frac{x - 9}{x + 4}\right)$

Following BODMAS, we work with what's in the brackets first.

$= \frac{2 x}{\left(x - 1\right) \left(x + 9\right)} \div \left(\frac{x \left(\cancel{x + 4}\right)}{\left(\cancel{x - 9}\right) \left(x + 9\right)} \cdot \frac{\cancel{x - 9}}{\cancel{x + 4}}\right)$

$= \frac{2 x}{\left(x - 1\right) \left(x + 9\right)} \div \left(\frac{x}{x + 9}\right)$

We can change this division sign into a multiplication sign by flipping the fraction on the right.

$= \frac{2 \left(\cancel{x}\right)}{\left(x - 1\right) \left(\cancel{x + 9}\right)} \times \frac{\cancel{x + 9}}{\cancel{x}}$

$= \frac{2}{x - 1}$

So, $\frac{2 x}{{x}^{2} + 8 x - 9} \div \left(\frac{{x}^{2} + 4 x}{{x}^{2} - 81} \cdot \frac{x - 9}{x + 4}\right) = \frac{2}{x - 1}$