Solve the following equations
Question:

If $2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right]$, find $x-y$

Solution:

$2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}6+1 & 8+y \\ 10+0 & 2 x+1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}7 & 8+y \\ 10 & 2 x+1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right]$

$\Rightarrow 8+y=0 \quad$ and $\quad 2 x+1=5$

$\Rightarrow y=-8 \quad$ and $\quad 2 x=4$

$\Rightarrow y=-8 \quad$ and $\quad x=2$

Hence, $x-y=2-(-8)=10$.