Giải thích các bước giải:

 ĐKXĐ: $x,y\ne 0$

Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
x + y = 90\\
\dfrac{{90}}{x} - \dfrac{{90}}{y} = \dfrac{9}{{20}}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + y = 90\\
\dfrac{{90\left( {y - x} \right)}}{{xy}} = \dfrac{9}{{20}}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + y = 90\\
200\left( {y - x} \right) = xy
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 90 - y\\
200\left( {y - \left( {90 - y} \right)} \right) = \left( {90 - y} \right)y
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 90 - y\\
200\left( {2y - 90} \right) + {y^2} - 90y = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 90 - y\\
{y^2} + 310y - 18000 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 90 - y\\
\left( {y + 360} \right)\left( {y - 50} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 90 - y\\
\left[ \begin{array}{l}
y =  - 360\\
y = 50
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
y = 50;x = 40\\
y =  - 360;x = 450
\end{array} \right.\left( {tm} \right)
\end{array}$

Vậy hệ có tập nghiệm sau $S = \left\{ {\left( {450; - 360} \right);\left( {40;50} \right)} \right\}$