# How do you solve -3 + \frac{1}{x+1}=\frac{2}{x} by finding the least common multiple?

#### Explanation:

Given that

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

$- 3 + \frac{1}{x + 1} - \frac{2}{x} = 0$

$\frac{- 3 x \left(x + 1\right) + x - 2 \left(x + 1\right)}{x \left(x + 1\right)} = 0$

$\frac{- 3 {x}^{2} - 3 x + x - 2 x - 2}{x \left(x + 1\right)} = 0$

$\frac{- 3 {x}^{2} - 4 x - 2}{x \left(x + 1\right)} = 0$

$- 3 {x}^{2} - 4 x - 2 = 0 \setminus \quad \left(\setminus \forall \setminus x \setminus \ne 0 , x \setminus \ne - 1\right)$

$3 {x}^{2} + 4 x + 2 = 0$

${B}^{2} - 4 A C = {4}^{2} - 4 \left(3\right) \left(2\right) = - 8 < 0$

The given equation has no real root .

The complex roots are given by quadratic formula as follows

$x = \setminus \frac{- 4 \setminus \pm \setminus \sqrt{{\left(4\right)}^{2} - 4 \left(3\right) \left(2\right)}}{2 \left(3\right)}$

$= \setminus \frac{- 4 \setminus \pm 2 i \setminus \sqrt{2}}{6}$

$= \setminus \frac{- 2 \setminus \pm i \setminus \sqrt{2}}{3}$