# How do you solve  5 /(x^2 + 4x) = 3 / x - 2/(x + 4)?

Jun 2, 2015

We have : $\frac{5}{{x}^{2} + 4 x} = \frac{3}{x} - \frac{2}{x + 4}$.

Let's write all the fractions with the same denominator :

$\frac{5}{{x}^{2} + 4 x} = \frac{3}{x} - \frac{2}{x + 4}$

$\frac{5}{{x}^{2} + 4 x} = \frac{3 \cdot \left(x + 4\right)}{x \cdot \left(x + 4\right)} - \frac{2 \cdot x}{\left(x + 4\right) \cdot x}$

$\frac{5}{{x}^{2} + 4 x} = \frac{3 \cdot \left(x + 4\right)}{{x}^{2} + 4 x} - \frac{2 \cdot x}{{x}^{2} + 4 x}$

Now, let's put all the fractions on the left :

$\frac{5}{{x}^{2} + 4 x} = \frac{3 x + 12}{{x}^{2} + 4 x} - \frac{2 x}{{x}^{2} + 4 x}$

$\frac{5}{{x}^{2} + 4 x} - \frac{3 x + 12}{{x}^{2} + 4 x} + \frac{2 x}{{x}^{2} + 4 x} = 0$

$\frac{5 - \left(3 x + 12\right) + \left(2 x\right)}{{x}^{2} + 4 x} = 0$

$\frac{5 - 3 x - 12 + 2 x}{{x}^{2} + 4 x} = 0$

$\frac{- x - 7}{{x}^{2} + 4 x} = 0$

$- \frac{x + 7}{{x}^{2} + 4 x} = 0$

We can multiply all the equation by $- \left({x}^{2} + 4 x\right)$ :

$x + 7 = 0$

That equation $= 0$ when $x + 7 = 0 \implies x = - 7$.