How do you solve 2/(x+1) + 5/(x-2)=-2?

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

Explanation:

$\setminus \frac{2}{x + 1} + \setminus \frac{5}{x - 2} = - 2$

$\setminus \frac{2}{x + 1} + \setminus \frac{5}{x - 2} + 2 = 0$

$\setminus \frac{2 \left(x - 2\right) + 5 \left(x + 1\right) + 2 \left(x - 2\right) \left(x + 1\right)}{\left(x - 2\right) \left(x + 1\right)} = 0$

$\setminus \frac{2 x - 4 + 5 x + 5 + 2 {x}^{2} - 2 x - 4}{\left(x - 2\right) \left(x + 1\right)} = 0$

$\setminus \frac{2 {x}^{2} + 5 x - 3}{\left(x - 2\right) \left(x + 1\right)} = 0$

$\setminus \frac{2 {x}^{2} + 6 x - x - 3}{\left(x - 2\right) \left(x + 1\right)} = 0$

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

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

$\left(x + 3\right) \left(2 x - 1\right) = 0 \setminus \quad \left(\setminus \forall \setminus \setminus x \setminus \ne - 1 , x \setminus \ne 2\right)$

$x + 3 = 0 , 2 x - 1 = 0$

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