Given
\[\left[ \begin{align}
& 3x+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-y \\
& 2y-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 \\
\end{align} \right]=\left[ \begin{align}
& 1\,\,\,\,\,2 \\
& -5\,\,\,3 \\
\end{align} \right]\]
Comparing the corresponding terms of given matrix we get
\[-y\text{ }=\text{ }2\]
Therefore \[y=-2\]
Again we have
\[\begin{array}{*{35}{l}}
3x\text{ }+\text{ }y\text{ }=\text{ }1 \\
3x\text{ }=\text{ }1\text{ }\text{ }y \\
\end{array}\]
Substituting the value of y we get
\[\begin{array}{*{35}{l}}
3x\text{ }=\text{ }1\text{ }\text{ }\left( -2 \right) \\
3x\text{ }=\text{ }1\text{ }+\text{ }2 \\
3x\text{ }=\text{ }3 \\
x\text{ }=\text{ }3/3 \\
x\text{ }=\text{ }1 \\
\end{array}\]
Hence \[x\text{ }=\text{ }1\text{ }and\text{ }y\text{ }=\text{ }-2\]