jmc

We apply Simon's Favorite Factoring Trick to the left side. We first find a pair of binomials whose product produces the three terms on the left side: $(2x+3)(5y+7)=10xy+14x+15y+21$. So, we add $21$ to both sides of the original equation to get $10xy+14x+15y+21=187$. Factoring then gives $(2x+3)(5y+7)=187=11\cdot 17$. If either $(2x+3)$ or $(5y+7)$ equal 1, then $x$ or $y$ would be negative. If $5y+7$ equals 11, then $y$ would not be an integer. So, $5y+7=17$ and $2x+3=11$. Solving for $(x,y)$ gives $(4,2)$. Therefore, $x+y=\boxed{6}$.