If $A=\left[\begin{array}{c}12x\\ 3-12\end{array}\right]$  and $B=\left[\begin{array}{c}y\\ x\\ 1\end{array}\right]$ be such that $AB=\left[\begin{array}{c}6\\ 8\end{array}\right]$ , then

$AB=\left[\begin{array}{c}6\\ 8\end{array}\right]$

$AB=\left[\begin{array}{c}y+2x+x\\ 3y-x+2\end{array}\right]=\left[\begin{array}{c}6\\ 8\end{array}\right]$

$3x+y=6\to \left(1\right)$

$3y-x=6\to \left(2\right)$

Solve (1) and (2)

$y=\frac{24}{10},x=\frac{12}{10}$

$y=2x$