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

Nov 17, 2015

$x = - \frac{3}{4} + \frac{\sqrt{129}}{4} \textcolor{w h i t e}{\times}$ and $\textcolor{w h i t e}{\times} x = - \frac{3}{4} - \frac{\sqrt{129}}{4}$

#### Explanation:

First of all, please be aware that $x \ne 0$ and $x \ne - 3$ must hold so that the equation is defined.

Now, with these restrictions in mind, let's start solving. :)

First of all, we need to get rid of the denominators. To do so, multiply both sides with $x \cdot \left(x + 3\right)$:

$\textcolor{w h i t e}{\times x s s} \frac{5}{x} - 2 = \frac{2}{x + 3}$

$\iff \textcolor{w h i t e}{\times} 5 \left(x + 3\right) - 2 x \left(x + 3\right) = 2 x$
$\iff \textcolor{w h i t e}{\times} 5 x + 15 - 2 {x}^{2} - 6 x = 2 x$

... add $- 2 x$ on both sides and simplify ...

$\iff \textcolor{w h i t e}{\times} 15 - 3 x - 2 {x}^{2} = 0$

This is a quadratic equation which can be solved for example with the quadratic formula which states that the solution of $a {x}^{2} + b x + c = 0$ is

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In our case, $a = - 2$, $b = - 3$ and $c = 15$.

$x = \frac{3 \pm \sqrt{9 - 4 \cdot \left(- 2\right) \cdot 15}}{-} 4 = \frac{3 \pm \sqrt{129}}{-} 4$
$x = - \frac{3}{4} + \frac{\sqrt{129}}{4} \textcolor{w h i t e}{\times}$ and $\textcolor{w h i t e}{\times} x = - \frac{3}{4} - \frac{\sqrt{129}}{4}$