smc

If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2-10x+y^2+6y=-34$. Adding $34$ to both sides, we have $x^2-10x+y^2+6y+34=0$. We can split the $34$ into $25$ and $9$ to get $(x-5{)}^2+(y+3{)}^2=0$. Notice this is a circle with radius $0$, which only contains one point. So, the only point is $(5,-3)$, so the sum is $5+(-3)=2⟹\boxed{2}$. ~ asdf334