Given: $\begin{bmatrix}a+4 & 3b\\8 & -6\end{bmatrix}=\begin{bmatrix}2a+2 & b+2\\8 & a-8b\end{bmatrix}\qquad$
By comparing the given two matrices of equal order, we can see that $a+4 = 2a+2, \:\:3b=b+2,\:\:-6=a-8b$
Solving first two equations,$\Rightarrow\:a=2,\:\:b=1$
which satisfy the equation $-6=a-8b$
$\therefore\:a-2b=2-2\times1=0$