Compute $(3^{-1}+5^{-1})^{-1}\pmod{31}$. Express your answer as an integer from $0$ to $30$, inclusive.
To minimize our work, we may begin by rewriting $3^{-1}+5^{-1}$ in the following way: \begin{align*}
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{align*}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 $$8\cdot 15^{-1}\cdot 15\cdot 8^{-1} \equiv 8\cdot 1\cdot 8^{-1} \equiv 1 \pmod{31}.$$Finally, we note that $8^{-1}\equiv 4\pmod{31}$, since $8\cdot 4 = 32\equiv 1\pmod{31}$. Therefore, we have \begin{align*}
(3^{-1}+5^{-1})^{-1} &\equiv 15\cdot 8^{-1} \\
&\equiv 15\cdot 4 \\
&\equiv 60 \\
&\equiv \boxed{29} \quad\pmod{31}.
\end{align*}