# How do you solve (x+7)/(x^2-3x-18)+(x-17)/(x^2-x-30)=(x+1)/(x^2+8x+15) and check for extraneous solutions?

Sep 11, 2016

$x = 2$ is the Soln., while, $x = - 5$ is extraneous.

#### Explanation:

We first factorise all the $D r s .$ of the Eqn. Hence,

$\frac{x + 7}{\left(x - 6\right) \left(x + 3\right)} + \frac{x - 17}{\left(x - 6\right) \left(x + 5\right)} = \frac{x + 1}{\left(x + 5\right) \left(x + 3\right)}$.

Now, we multiply the eqn. by $\left(x - 6\right) \left(x + 3\right) \left(x + 5\right)$ & get

$\left(x + 7\right) \left(x + 5\right) + \left(x - 17\right) \left(x + 3\right) = \left(x + 1\right) \left(x - 6\right)$

$\therefore {x}^{2} + 12 x + 35 + {x}^{2} - 14 x - 51 = {x}^{2} - 5 x - 6$

$\therefore 2 {x}^{2} - 2 x - 16 - \left({x}^{2} - 5 x - 6\right) = 0$

$\therefore {x}^{2} + 3 x - 10 = 0$

$\therefore \left(x + 5\right) \left(x - 2\right) = 0$

$\therefore x = - 5 , x = 2$

Among these, $x = - 5$ is extraneous soln., since it makes the second and third terms of the eqn. meaningless. $\therefore x \ne - 5$.

Further, $x = 2$ satisfies the eqn.

Hence, $x = 2$ is the Soln., while, $x = - 5$ is extraneous.

Enjoy Maths.!