We have, $\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=0$ $\Rightarrow\ \begin{bmatrix}2\text{x}\\\text{x}\end{bmatrix}+\begin{bmatrix}3\text{y}\\5\text{y}\end{bmatrix}=\begin{bmatrix}-8\\-11\end{bmatrix}=0$
$\therefore\ \begin{bmatrix}2\text{x}+3\text{y}-8\\\text{x}+5\text{y}-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow\ 2\text{x}+3\text{y}-8=0\ ...(\text{i})$
and $\text{x}+5\text{y}-11=0\ ....(\text{ii})$
Solving equation (i) and (ii), we get
$\text{x}=1$ and $\text{y}=2$