Giải thích các bước giải:

 ĐKXĐ:  \(x;y \ne 0\)

Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
 - x + y = 4\\
\frac{{36}}{y} + \frac{{42}}{x} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = x + 4\\
\frac{{36}}{{x + 4}} + \frac{{42}}{x} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = x + 4\\
\frac{{36x + 42\left( {x + 4} \right)}}{{x\left( {x + 4} \right)}} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = x + 4\\
\frac{{78x + 168}}{{{x^2} + 4x}} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = x + 4\\
78x + 168 = {x^2} + 4x
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = x + 4\\
{x^2} - 74x - 168 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \approx 76,2 \Rightarrow y \approx 80,2\\
x \approx  - 2,2 \Rightarrow y \approx 2,2
\end{array} \right.
\end{array}\)