(x-8)/(x^2-x-42)= 1+8/(x-7)

solve for x.

i got 7.75 but it is saying its not right any help?

2. Originally Posted by huh1
(x-8)/(x^2-x-42)= 1+8/(x-7)

solve for x.

i got 7.75 but it is saying its not right any help?
$\displaystyle \frac{x-8}{(x-7)(x+6)} = \frac{x-7}{x-7} + \frac{8}{x-7}$

$\displaystyle \frac{x-8}{(x-7)(x+6)} = \frac{x+1}{x-7}$

$\displaystyle \frac{x-8}{(x-7)(x+6)} = \frac{(x+1)(x+6)}{(x-7)(x+6)}$

numerators form the equation ...

$\displaystyle x-8 = (x+1)(x+6)$

$\displaystyle x-8 = x^2 + 7x + 6$

$\displaystyle 0 = x^2 + 6x + 14$

this quadratic has no real solution

3. Thank you so much! I figured it was no real solution, but I couldn't actually prove it. haha. Thanks again.