# How do you solve the rational equation 4/(x+2)-3/(x-5)=15/(x^2-3x-10)?

Jan 27, 2016

x = 41

#### Explanation:

Firstly: factor the denominator on the right side.

${x}^{2} - 3 x - 10 = \left(x - 5\right) \left(x + 2\right)$

Equation now becomes :

$\frac{4}{x + 2} - \frac{3}{x - 5} = \frac{15}{\left(x - 5\right) \left(x + 2\right)}$

Multiply each term on both sides by (x-5)(x+2)

$\frac{4 \left(x - 5\right) \cancel{x + 2}}{\cancel{x + 2}} - \frac{3 \cancel{x - 5} \left(x + 2\right)}{\cancel{x - 5}}$= $\frac{15 \cancel{x - 5} \cancel{x + 2}}{\cancel{x - 5} \cancel{x + 2}}$

which simplified becomes : 4(x-5) -3(x+2) = 15

hence 4x - 20 - 3x - 6 = 15

therefore x = 15+6+20 = 41