# How do you solve (3x-2)/(x^2-17x+72)-(x-4)/(x^2-20x+96)=(x-6)/(x^2-21x+108) and check for extraneous solutions?

Jan 12, 2018

${x}_{1} = - 4$ and ${x}_{2} = 15$

#### Explanation:

$\frac{3 x - 2}{{x}^{2} - 17 x + 72} - \frac{x - 4}{{x}^{2} - 20 x + 96} = \frac{x - 6}{{x}^{2} - 21 x + 108}$

$\frac{3 x - 2}{\left(x - 8\right) \left(x - 9\right)} - \frac{x - 4}{\left(x - 8\right) \left(x - 12\right)} = \frac{x - 6}{\left(x - 9\right) \left(x - 12\right)}$

$\frac{\left(3 x - 2\right) \left(x - 12\right) - \left(x - 4\right) \left(x - 9\right)}{\left(x - 8\right) \left(x - 9\right) \left(x - 12\right)} = \frac{x - 6}{\left(x - 9\right) \left(x - 12\right)}$

$\frac{\left(3 {x}^{2} - 38 x + 24\right) - \left({x}^{2} - 13 x + 36\right)}{\left(x - 8\right) \left(x - 9\right) \left(x - 12\right)} = \frac{x - 6}{\left(x - 9\right) \left(x - 12\right)}$

$2 {x}^{2} - 25 x - 12 = \left(x - 6\right) \cdot \left(x \cdot 8\right)$

$2 {x}^{2} - 25 x - 12 = {x}^{2} - 14 x + 48$

${x}^{2} - 11 x - 60 = 0$

$\left(x + 4\right) \cdot \left(x - 15\right) = 0$

Hence ${x}_{1} = - 4$ and ${x}_{2} = 15$