# How do you solve 1/(x+3) + 1/(x-3) = 1/(x^2-9)?

Feb 28, 2016

The only configuration that yields a logical answer is:
$\frac{1}{x + 3} + \frac{1}{x - 3} = \frac{1}{{x}^{2} - 9}$

In which case$\text{ } x = \frac{1}{2}$

#### Explanation:

Considering different configuration:

$\textcolor{b l u e}{\text{Configuration 1}}$
Suppose the Left hand side was meant to be $\text{ } \frac{1}{x + 3} + \frac{1}{x - 3}$

Then the left would be:

$\frac{\left(x + 3\right) + \left(x - 3\right)}{{x}^{2} - 9}$

Comparing left to right gives

$x + 3 + x - 3 = 1$

$2 x = 1$

$x = \frac{1}{2}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Configuration 2}}$

Suppose the Left hand side was meant to be $\text{ } \frac{1}{x + 3} - \frac{1}{x - 3}$

Then the left would be:

$\frac{\left(x + 3\right) - \left(x - 3\right)}{{x}^{2} - 9} = \frac{6}{{x}^{2} - 9}$

Comparing Left to right would mean that it would have to be true for

6=1" "color(red)("Clearly this is a contradiction so it is not the case")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{The only possible scenario is for configuration 1}}$

$\textcolor{m a \ge n t a}{\text{So the answer is } x = \frac{1}{2}}$