# What is the solution set for 30/(x^2-9) - 5/(x-3) = 9/(x+3)?

Aug 11, 2015

I found no real solution!

#### Explanation:

You can write it as:
$\frac{30}{\left(x + 3\right) \left(x - 3\right)} - \frac{5}{x - 3} = \frac{9}{x + 3}$
the common denominator can be: $\left(x + 3\right) \left(x - 3\right)$; so you get:
$\frac{30 - 5 \left(x + 3\right)}{\left(x + 3\right) \left(x - 3\right)} = \frac{9 \left(x - 3\right)}{\left(x + 3\right) \left(x - 3\right)}$
$\frac{30 - 5 \left(x + 3\right)}{\cancel{\left(\left(x + 3\right) \left(x - 3\right)\right)}} = \frac{9 \left(x - 3\right)}{\cancel{\left(\left(x + 3\right) \left(x - 3\right)\right)}}$
$30 - 5 x - 15 = 9 x - 27$
collect $x$ on the left:
$- 14 x = - 42$
$x = \frac{42}{14} = 3$
BUT substituting $x = 3$ into the original equation you get a division by zero!!! We have no real solutions.