jmc

To determine the residue of $a(\mathrm{mod}10)$, we can subtract $a+b$ from $2a+b$: $$\begin{array}{rl}a& =(2a+b)-(a+b)\\ & \equiv 1-2\\ & \equiv -1\\ & \equiv 9(\mathrm{mod}10).\end{array}$$Then we know that $9+b\equiv 2(\mathrm{mod}10)$, so we can solve for $b$: $$\begin{array}{rl}b& \equiv 2-9\\ & \equiv -7\\ & \equiv 3(\mathrm{mod}10).\end{array}$$Finally, we substitute to obtain $$a-b\equiv 9-3\equiv 6(\mathrm{mod}10),$$and so the last digit of $a-b$ is $\boxed{6}$.