# How do you solve  1/(x-2) + 3/ (x+3) = 4/(x^2 +x-6)?

Mar 14, 2018

$x = \frac{7}{4}$

#### Explanation:

Note that:

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

So multiplying both sides of the given equation by ${x}^{2} + x - 6$ we get:

$\left(x + 3\right) + 3 \left(x - 2\right) = 4$

which simplifies to:

$4 x - 3 = 4$

Add $3$ to both sides to get:

$4 x = 7$

Divide both sides by $4$ to get:

$x = \frac{7}{4}$

Since this value makes none of the denominators $0$, it is a valid solution of the original equation.