# How do you solve (3x-3)/(x^2-13x+40)-(x-6)/(x^2-22x+112)=(x-9)/(x^2-19x+70) and check for extraneous solutions?

Apr 19, 2017

$x \in \left\{20 , - 3\right\}$

#### Explanation:

Factor the denominators.

${x}^{2} - 13 x + 40 = \left(x - 8\right) \left(x - 5\right)$
${x}^{2} - 22 x + 112 = \left(x - 8\right) \left(x - 14\right)$
${x}^{2} - 19 x + 70 = \left(x - 14\right) \left(x - 5\right)$

Now, put each fraction on a common denominator.

$\frac{\left(3 x - 3\right) \left(x - 14\right)}{\left(x - 8\right) \left(x - 5\right) \left(x - 14\right)} - \frac{\left(x - 6\right) \left(x - 5\right)}{\left(x - 8\right) \left(x - 5\right) \left(x - 14\right)} = \frac{\left(x - 9\right) \left(x - 8\right)}{\left(x - 8\right) \left(x - 5\right) \left(x - 14\right)}$

$3 {x}^{2} - 3 x - 42 x + 42 - \left({x}^{2} - 11 x + 30\right) = {x}^{2} - 17 x + 72$

$3 {x}^{2} - 45 x + 42 - {x}^{2} + 11 x - 30 = {x}^{2} - 17 x + 72$

${x}^{2} - 17 x - 60 = 0$

$\left(x - 20\right) \left(x + 3\right) = 0$

$x = 20 \mathmr{and} - 3$

Our initial restrictions were $x \ne 5 , 8 , 14$, so all of our solutions are acceptable.

Hopefully this helps!