<s>[INST] Solving the congruence $9x \equiv 3 \pmod{47}$ For this question $9x \equiv 3 \pmod{47}$.
I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$
when $a=47$ and $b=9$
I subbed back into the given equation:
\begin{align*}
(9)(21) & \equiv 3 \pmod{47}\\
189 & \equiv 3 \pmod{47}\\
63 & \equiv 1 \pmod{47}
\end{align*}
and I'm stuck, should I divide $63$ by $9$ to get $7$? but it does not comply to the given equation as when $x=7$, it would become $16 \pmod{47}$. 
 [/INST] We wish to solve the congruence $9x \equiv 3 \pmod{47}$.  You correctly found that $21$ is the multiplicative inverse of $9$ modulo $47$.  However, you made an error in this step:
$$9 \cdot 21 \equiv \color{red}{3} \pmod{47}$$
If we multiply one side of the congruence by $21$, we must multiply the other side of the congruence by $21$.  You should have obtained 
\begin{align*}
21 \cdot 9x & \equiv 21 \cdot 3 \pmod{47}\\
1x & \equiv 63 \pmod{47}\\
x & \equiv 16 \pmod{47}
\end{align*}
</s>