# How do you solve (3x-3)/(x^2-13x+40)-(x-7)/(x^2-19x+70)=(x-2)/(x^2-22x+112) and check for extraneous solutions?

Aug 25, 2016

The eqn. has solns. $x = 24 , x = - 1$, and no extraneous soln.

#### Explanation:

Factorising all the Drs.

$\frac{3 \left(x - 1\right)}{\left(x - 8\right) \left(x - 5\right)} - \frac{x - 7}{\left(x - 14\right) \left(x - 5\right)} = \frac{x - 2}{\left(x - 14\right) \left(x - 8\right)}$.

Multiplying both sides by $\left(x - 5\right) \left(x - 8\right) \left(x - 14\right)$, we get,

$3 \left(x - 1\right) \left(x - 14\right) - \left(x - 7\right) \left(x - 8\right) = \left(x - 2\right) \left(x - 5\right)$.

$\therefore 3 {x}^{2} - 45 x + 42 - {x}^{2} + 15 x - 56 = {x}^{2} - 7 x + 10$.

$\therefore {x}^{2} - 23 x - 24 = 0$.

$\therefore \underline{{x}^{2} - 24 x} + \underline{x - 24} = 0$.

$\therefore x \left(x - 24\right) + 1 \left(x - 24\right) = 0$.

$\therefore \left(x - 24\right) \left(x + 1\right) = 0$.

$\therefore x = 24 , x = - 1$

These roots satisfy the given eqn.

Hence, the eqn. has solns. $x = 24 , x = - 1$, and no extraneous soln.