(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