# How do you solve (x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12) and check for extraneous solutions?

Oct 11, 2016

$x = \frac{7 \pm \sqrt{1401}}{4}$

#### Explanation:

As $\left(x - 4\right) \left(x - 3\right) = {x}^{2} - 7 x + 12$ - note that as they are in denominator $x \ne 3$ and $x \ne 4$

$\frac{x - 7}{x - 4} + \frac{x + 8}{x - 3} = \frac{x + 158}{{x}^{2} - 7 x + 12}$

$\Leftrightarrow \frac{\left(x - 7\right) \left(x - 3\right) + \left(x + 8\right) \left(x - 4\right)}{{x}^{2} - 7 x + 12} = \frac{x + 158}{{x}^{2} - 7 x + 12}$

or $\left(\left(x - 7\right) \left(x - 3\right) + \left(x + 8\right) \left(x - 4\right)\right) = \left(x + 158\right)$

or ${x}^{2} - 10 x + 21 + {x}^{2} + 4 x - 32 = x + 158$

or $2 {x}^{2} - 6 x - 11 - x - 158 = 0$

or $2 {x}^{2} - 7 x - 169 = 0$

and using quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ as olution for quadratic equation $a {x}^{2} + b x + c = 0$

$x = \frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \cdot {2}^{\cdot} \left(- 169\right)}}{4}$

= $\frac{7 \pm \sqrt{49 + 1352}}{4} = \frac{7 \pm \sqrt{1401}}{4}$