# How do you solve \frac { 6} { 3x } + \frac { 1} { 2x } = - \frac { 1} { 6}?

May 10, 2018

$x = 0 \mathmr{and} - 15$

#### Explanation:

As per the given question, we have

$\frac{6}{3 x} + \frac{1}{2 x} = - \frac{1}{6}$

$\therefore \frac{12 x + 3 x}{6 {x}^{2}} = - \frac{1}{6}$

$\therefore 15 x = 6 {x}^{2} \times \left(- \frac{1}{6}\right)$

$\therefore 15 x = \cancel{6} {x}^{2} \times \left(- \frac{1}{\cancel{6}}\right)$

$\therefore 15 x = - {x}^{2}$

$\therefore {x}^{2} + 15 x = 0$

$\therefore x \left(x + 15\right) = 0$

$\therefore x = 0 \mathmr{and} x = - 15$