de vergelijking algebraisch oplossen

$\eqalign{
  & 20\left( {\frac{1}{x} + \frac{1}{{x + 9}}} \right) = 1  \cr
  & \frac{1}{x} + \frac{1}{{x + 9}} = \frac{1}{{20}}  \cr
  & \frac{{x + 9}}{{x\left( {x + 9} \right)}} + \frac{x}{{x(x + 9)}} = \frac{1}{{20}}  \cr
  & \frac{{2x + 9}}{{x(x + 9)}} = \frac{1}{{20}}  \cr
  & 40x + 180 = {x^2} + 9x  \cr
  & x{}^2 - 21x - 180 = 0  \cr
  & (x + 5)(x - 36) = 0  \cr
  & x =  - 5\,\,of\,\,x = 36 \cr} $