If $3x+7=2$ (mod $16$), then $2x+11$ is congruent (mod $16$) to what integer between $0$ and $15$, inclusive?
We have $3x+7 \equiv 2 \pmod{16}$, so $3x \equiv -5 \equiv 11 \pmod{16}$. This solves to $x \equiv 9 \pmod{16}$.
Thus, $2x+11 \equiv 2\cdot 9+11\equiv 13 \pmod{16}$.
