# How do you solve (x-5)/(x-8)+(x-6)/(x-7)=(x+97)/(x^2-15x+56) and check for extraneous solutions?

Sep 6, 2017

$x = - \frac{1}{2}$ or $x = 14$

#### Explanation:

In $\frac{x - 5}{x - 8} + \frac{x - 6}{x - 7} = \frac{x + 97}{{x}^{2} - 15 x + 56}$, observe that as ${x}^{2} - 15 x + 56 = \left(x - 8\right) \left(x - 7\right)$, we cannot have $x = 8$ or $x = 7$.

$\frac{x - 5}{x - 8} + \frac{x - 6}{x - 7} = \frac{x + 97}{{x}^{2} - 15 x + 56}$

$\Leftrightarrow \frac{\left(x - 5\right) \left(x - 7\right) + \left(x - 6\right) \left(x - 8\right)}{{x}^{2} - 15 x + 56} = \frac{x + 97}{{x}^{2} - 15 x + 56}$

or $\left(x - 5\right) \left(x - 7\right) + \left(x - 6\right) \left(x - 8\right) = x + 97$

or ${x}^{2} - 12 x + 35 + {x}^{2} - 14 x + 48 = x + 97$

or $2 {x}^{2} - 26 x + 83 - x - 97 = 0$

or $2 {x}^{2} - 27 x - 14 = 0$

or $2 {x}^{2} - 28 x + x - 14 = 0$

or $2 x \left(x - 14\right) + 1 \left(x - 14\right) = 0$

i.e. $\left(2 x + 1\right) \left(x - 14\right) = 0$

and $x = - \frac{1}{2}$ or $x = 14$

and there is no extraneous solution as none of these is $7$ or $8$