# Solve 1/x + 1/(x+4) = 1/3 ?

## These are all fractions and I know I have to create a common denominator.

Nov 17, 2017

$x = 1 \pm \sqrt{13}$

#### Explanation:

Given:

$\frac{1}{x} + \frac{1}{x + 4} = \frac{1}{3}$

Multiply both sides of the equation by $3 x \left(x + 4\right)$ to get:

$3 \left(x + 4\right) + 3 x = x \left(x + 4\right)$

which multiplies out to:

$3 x + 12 + 3 x = {x}^{2} + 4 x$

That is:

$6 x + 12 = {x}^{2} + 4 x$

Subtract $6 x + 12$ from both sides to get:

$0 = {x}^{2} - 2 x - 12$

$\textcolor{w h i t e}{0} = {x}^{2} - 2 x + 1 - 13$

$\textcolor{w h i t e}{0} = {\left(x - 1\right)}^{2} - {\left(\sqrt{13}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x - 1\right) - \sqrt{13}\right) \left(\left(x - 1\right) + \sqrt{13}\right)$

$\textcolor{w h i t e}{0} = \left(x - 1 - \sqrt{13}\right) \left(x - 1 + \sqrt{13}\right)$

So:

$x = 1 \pm \sqrt{13}$

These are both solutions of the original equation since they do not cause any of the denominators to becomes zero.