Let a common solution be $ a $. Then we know \begin{align*} a\equiv 3 & \pmod 4\\ a\equiv 1 & \pmod 3\\ a\equiv 1 & \pmod 5 \end{align*} Since $ \gcd(3,5)=1 $, $ (2) $ and $ (3) $ together yield $ a\equiv 1\pmod {3\cdot 5} $ which is the same as $ a\equiv 1\pmod {15} $. Then there exists an integer $ n $ such that $ a=1+15n $. Substituting this into $ (1) $ gives \[1+15n\equiv 3\pmod 4\implies n\equiv 2\pmod 4\] So $ n $ has a lower bound of $ 2 $. Then $ n\ge 2\implies a=1+15n\ge 31 $. $ 31 $ satisfies the original congruences so subtracting it from both sides of each gives \begin{align*} a-31\equiv -28\equiv 0 & \pmod 4\nonumber\\ a-31\equiv -30\equiv 0 & \pmod 3\nonumber\\ a-31\equiv -30\equiv 0 & \pmod 5\nonumber \end{align*} Since $ \gcd(3,4)=\gcd(4,5)=\gcd(3,5)=1 $, we have $ a-31\equiv 0\pmod {3\cdot 4\cdot 5} $, that is, $ a\equiv 31\pmod{60} $. Note that any solution of the above congruence also satisfies the original ones. Then the two solutions are $ 31 $ and $ 60+31=91 $. Thus, $ 31+91=\boxed{122} $.