# How do you solve (x+2)/(x^2-6x-27)+(x-17)/(x^2-2x-63)=(x+1)/(x^2+10x+21) and check for extraneous solutions?

Oct 10, 2016

Multiply everything by the least command denominator to remove the fractions The solve for x

#### Explanation:

$\left({x}^{2} - 6 x - 27\right)$ can be factored into ( x -9) x ( x +3)

$\left({x}^{2} - 2 x - 62\right)$ can be factored into ( x -9) x ( x+7)

$\left({x}^{2} + 10 x + 21\right)$ can be factored into ( x + Mu7) ( x +3)

The least common denominator for all three is
( x-9) x ( x+7) x (x +3)

multiply all the terms by the least common denominator

$\left(x - 9\right) \left(x + 7\right) \left(x + 3\right) \times \left\{\frac{x + 2}{{x}^{2} - 6 x - 27} + \frac{x - 17}{{x}^{2} - 2 x - 63} = \frac{x + 1}{{x}^{2}} + 10 x + 21\right\}$ This gives

(x + 7) ( x+2) + (x+3)(x -7) = (x + 1) Multiplying this out gives

${x}^{2} + 9 x + 14 + {x}^{2} - 4 x - 7 = x + 1$ Adding the common terms gives

$2 {x}^{2} + 5 x + 7 = x + 1$ subtract 1 x from each side gives

$2 {x}^{2} + 5 x - 1 x + 7 = x - x + 1$ resulting in

$2 {x}^{2} + 4 x + 7 = 1$ subtract one from each side gives

$2 {x}^{2} + 4 x + 7 - 1 = 1 - 1$ resulting in

$2 {x}^{2} + 4 x + 6 = 0$

Use the quadratic equation to solve for the roots