Simple algebra!

According to the question, $x+y=15$ $. . . . . . .(i)$

$3y-x=5$ $. . . . . . .(ii)$

Adding $(ii)$ and $(i)$

$x+y+(3y-x)=15+5$

Or, $2y=10$ or, $y=\boxed{5}$

Substituting, $y=\boxed{5}$ in $(i)$ , we get,

$x+5=15$

Or, $x=\boxed{10}$

Substituting, $y=\boxed{5}$ and $x=\boxed{10}$ in, $x^{2}+2y+1$ we get,

Therefore, $x^{2}+2y+1=10^{2}+2(5)+1$

$=100+10+1=\boxed{111}$