jmc

Note that $14$, $46$, and $100$ all have a common factor of $2$, so we can divide it out: the solutions to $$14u\equiv 46(\mathrm{mod}100)$$ are identical to the solutions to $$7u\equiv 23(\mathrm{mod}50).$$ Make sure you see why this is the case.

Now we can multiply both sides of the congruence by $7$ to obtain $$49u\equiv 161(\mathrm{mod}50),$$ which also has the same solutions as the previous congruence, since we could reverse the step above by multiplying both sides by $7^{-1}$. (We know that $7^{-1}$ exists modulo $50$ because $7$ and $50$ are relatively prime.)

Replacing each side of $49u\equiv 161$ by a $(\mathrm{mod}50)$ equivalent, we have $$-u\equiv 11(\mathrm{mod}50),$$ and thus $$u\equiv -11(\mathrm{mod}50).$$ This is the set of solutions to our original congruence. The two smallest positive solutions are $-11+50=39$ and $-11+2\cdot 50=89$. Their average is $\boxed{64}$.