Solve the following systems of equations:

The given equations are:

$3 x-\frac{y+7}{11}+2=10$

$\Rightarrow 3 x-\frac{(y+7)}{11}=8$

$\Rightarrow \frac{33 x-y-7}{11}=8$

$\Rightarrow 33 x-y-7=88$

$\Rightarrow 33 x-y-7=88$

$\Rightarrow 33 x-y=95 \quad \ldots \ldots \ldots(1)$

$2 y+\frac{x+11}{7}=10$

$\Rightarrow \frac{14 y+x+11}{7}=10$

$\Rightarrow 14 y+x+11=70$

$\Rightarrow 14 y+x=59$

$\Rightarrow x+14 y=59 \quad \ldots \ldots \ldots(2)$

Multiply equation (1) by 14 , we get

$462 x-14 y=1330 \quad \ldots \ldots(3)$

adding $(2)$ and $(3)$, we get

$(x+14 y)+(462 x-14 y)=59+1330$

$\Rightarrow 463 x=1389$

$\Rightarrow x=3$

Substituting the value of $x$ in $(2)$, we get

$3+14 y=59$

$\Rightarrow 14 y=59-3$

$\Rightarrow 14 y=56$

$\Rightarrow y=4$

Hence the value of $x$ and $y$ are $x=3$ and $y=4$