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} $