Solution:

The given eq. are:
$\begin{array}{l}
\frac{x}{3}+\frac{y}{4}=11 \\
\Rightarrow 4 x+3 y=132 \ldots \ldots(i)
\end{array}$
and $\frac{5 x}{6}-\frac{y}{3}+7=0$
$\Rightarrow 5 x-2 y=-42 \ldots \ldots \text { (ii) }$
When multiplying equation(i) by 2 and equation(ii) by 3, we obtain:
$\begin{array}{l}
8 x+6 y=264 \ldots \ldots (iii) \\
15 x-6 y=-126 \ldots(iv)
\end{array}$
Adding equation(iii) and equation(iv), we obtain:
$\begin{array}{l}
23 x=138 \\
\Rightarrow x=6
\end{array}$
When substituting $\mathrm{x}=6$ in equation(i), we obtain:
$\begin{array}{l}
24+3 y=132 \\
\Rightarrow 3 y=(132-24)=108 \\
\Rightarrow y=36
\end{array}$
As a result, the solution is $x=6$ and $y=36$