# How to solve 36/[x^2-9]= (2x)/[x + 3]- 1/[1]?

Mar 24, 2017

$x = 9$

The extraneous solution $x = - 3$ is rejected.

#### Explanation:

$\frac{36}{{x}^{2} - 9} = \frac{2 x}{x + 3} - \frac{1}{1}$

$\frac{1}{1}$ can be anything, make the right side denominators equal.

$\frac{36}{\textcolor{red}{{x}^{2} - 9}} = \textcolor{b l u e}{\frac{2 x}{x + 3} - \frac{x + 3}{x + 3}} \text{ } \leftarrow$ factorise and simplfy

$\frac{36}{\textcolor{red}{\left(x + 3\right) \left(x - 3\right)}} = \textcolor{b l u e}{\frac{2 x - \left(x + 3\right)}{x + 3}}$

$\frac{36}{\left(x + 3\right) \left(x - 3\right)} = \frac{2 x - x - 3}{x + 3}$

$\frac{36}{\left(x + 3\right) \left(x - 3\right)} = \frac{x - 3}{x + 3} \text{ } \textcolor{f \mathmr{and} e s t g r e e n}{\times \left(x + 3\right)}$

$\frac{36 \textcolor{f \mathmr{and} e s t g r e e n}{\times \cancel{\left(x + 3\right)}}}{\cancel{\left(x + 3\right)} \left(x - 3\right)} = \frac{\left(x - 3\right) \textcolor{f \mathmr{and} e s t g r e e n}{\times \cancel{\left(x + 3\right)}}}{\cancel{\left(x + 3\right)}} \text{ } \leftarrow$ cancel

$\frac{36}{\left(x - 3\right)} = \left(x - 3\right) \text{ } \leftarrow$ cross multiply

${\left(x - 3\right)}^{2} = 36 \text{ } \leftarrow$ square root both sides

$x - 3 = \pm 6$

If $x - 3 = + 6 \text{ } \rightarrow x = 9$

If $x - 3 = - 6 \text{ } \rightarrow x = - 3$

However, $x = - 3$ will make the denominators 0, so this is an extraneous solution which we will reject.