jmc

To minimize our work, we may begin by rewriting $3^{-1}+5^{-1}$ in the following way: $$\begin{array}{rl}{3}^{-1}+5^{-1}& \equiv 5\cdot 5^{-1}\cdot 3^{-1}+3\cdot 3^{-1}\cdot 5^{-1}\\ & \equiv 5\cdot 15^{-1}+3\cdot 15^{-1}\\ & \equiv (5+3)\cdot 15^{-1}\\ & \equiv 8\cdot 15^{-1},\end{array}$$where all congruence is modulo $31$. Notice that this process is just like finding a common denominator!

Now we wish to find the inverse of $8\cdot 15^{-1}$. This inverse must be $15\cdot 8^{-1}$, since $$8\cdot 15^{-1}\cdot 15\cdot 8^{-1}\equiv 8\cdot 1\cdot 8^{-1}\equiv 1(\mathrm{mod}31).$$Finally, we note that $8^{-1}\equiv 4(\mathrm{mod}31)$, since $8\cdot 4=32\equiv 1(\mathrm{mod}31)$. Therefore, we have $$\begin{array}{rl}(3^{-1}+5^{-1}{)}^{-1}& \equiv 15\cdot 8^{-1}\\ & \equiv 15\cdot 4\\ & \equiv 60\\ & \equiv \boxed{29}(\mathrm{mod}31).\end{array}$$