$\left[\begin{array}{cc}\mathit{x}+\mathit{y}& 3\mathit{y}\\ 3\mathit{x}& \mathit{x}+3\end{array}\right]=\left[\begin{array}{cc}9& 4\mathit{x}+\mathit{y}\\ \mathit{x}+6& \mathit{y}\end{array}\right]$
If metrices are equal, then their corresponding entries must be equal
$\mathrm{So}\mathit{x}+\mathit{y}=9\mathrm{}...\left(\mathit{i}\right)$
$3\mathit{y}=4\mathit{x}+\mathit{y}\mathrm{}...\left(\mathit{i}\mathit{i}\right)$
$3\mathit{x}=\mathit{x}+6\mathrm{}...\left(\mathit{i}\mathit{i}\mathit{i}\right)$
$\mathit{x}+3=\mathit{y}\mathrm{}...\left(\mathit{i}\mathit{v}\right)$
From (iii) $2\mathit{x}=6\Rightarrow \mathit{x}+3$
Put x = 3 in (iv)
$\Rightarrow 3+3=\mathit{y}$
$\Rightarrow \mathrm{}\mathit{y}=6$
Now (x - y) = (3 - 6) = -3