jmc

If $3x+7\equiv 2(\mathrm{mod}16)$, then $$6\cdot (3x+7)\equiv 6\cdot 2(\mathrm{mod}16).$$Expanding out the right side, we have $$18x+42\equiv 12(\mathrm{mod}16).$$Reducing coefficients modulo $16$, we have $$2x+10\equiv 12(\mathrm{mod}16).$$Finally, adding $1$ to both sides, we get $$2x+11\equiv \boxed{13}(\mathrm{mod}16).$$(It's good to notice a couple of things about this solution. For one thing, why did we multiply by $6$ at the beginning? The idea is to get a $2x$ term on the left, since our goal is to compute the residue of $2x+11$. Another thing to notice is that one step in this process is not reversible. If the goal in this problem had been to solve for $x$, then it would appear from our final result that $x=1$ is a solution, yet $x=1$ doesn't actually satisfy $3x+7\equiv 2(\mathrm{mod}16)$. Why not? At what step did we introduce this bogus solution?)