jmc

We are looking for pairs $(x,y)$ that satisfy both $y\equiv 5x+2$ and $y\equiv 11x+12(\mathrm{mod}16)$. Thus, the $x$-coordinates in all such pairs satisfy $$5x+2\equiv 11x+12(\mathrm{mod}16).$$Subtracting $5x+2$ from both sides of this congruence, we have $$0\equiv 6x+10(\mathrm{mod}16),$$which is equivalent to $$0\equiv 6x-6(\mathrm{mod}16)$$(since $10\equiv -6(\mathrm{mod}16)$).

Thus the solutions we seek are values $x$ in the range $0\le x<16$ such that $16$ divides $6(x-1)$. The solutions are $x=1,$ $x=9,$ so the sum of $x$-coordinates is $1+9=\boxed{10}$.

(As a check, note that the pairs $(1,7)$ and $(9,15)$ satisfy both of the original congruences, so these are the points shared by the two graphs.)