jmc

Let the desired number be $a$. Then $$\begin{array}{r}a\equiv 4(\mathrm{mod}5),\\ a\equiv 6(\mathrm{mod}7).\end{array}$$ The first congruence means that there exists a non-negative integer $n$ such that $a=4+5n$. Substituting this into the second congruence yields $$4+5n\equiv 6(\mathrm{mod}7)⟹n\equiv 6(\mathrm{mod}7)$$ So $n$ has a lower bound of $6$. Then $n\ge 6⟹a=4+5n\ge 34$. $\boxed{34}$ is the smallest solution since it is a lower bound of $a$ and satisfies both original congruences.