jmc

Note that 6 divides both $30x$ and $42$, and since 6 is relatively prime to 47, we can write $5x\equiv 7(\mathrm{mod}47)$. Note that $5\cdot 19=95=2(47)+1$, so 19 is the modular inverse of 5, modulo 47. We multiply both sides of the given congruence by 19 to obtain $95x\equiv 19(7)(\mathrm{mod}47)⟹x\equiv \boxed{39}(\mathrm{mod}47)$.