# How do you solve 5/(x-3)=(2x)/(x^2-9)?

Jun 11, 2018

$x = - 5$

#### Explanation:

Note that $x$ must be different from $\setminus \pm 3$. In fact, the first denominator vanished is $x = 3$, while the second vanishes if $x = \setminus \pm 3$.

Note that ${x}^{2} - 9 = \left(x + 3\right) \left(x - 3\right)$, so if we multiply and divide the left hand side by $x + 3$ (and this is legit since $x \setminus \ne - 3$) we get

$\setminus \frac{5}{x - 3} \cdot \setminus \frac{x + 3}{x + 3} = \setminus \frac{2 x}{{x}^{2} - 9}$

The left hand side can be evaluated as we noted above, so we have

$\setminus \frac{5 \left(x + 3\right)}{{x}^{2} - 9} = \setminus \frac{2 x}{{x}^{2} - 9}$

We can multiply both sides by the denominator, since it's the same for both and it can't be zero:

$5 \left(x + 3\right) = 2 x \setminus \iff 5 x + 15 = 2 x \setminus \iff 3 x = - 15 \setminus \iff x = - 5$