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 $16 \pmod{47}$. 
 A: you can write $$x\equiv \frac{3}{9}=\frac{1}{3}\equiv \frac{48}{3}\equiv 16\mod 47$$
A: You are overcomplicating this. First of all$$9x\equiv3\pmod{47}\iff47\mid3.(3x-1)\iff47\mid3x-1.$$Besides, the inverse of $3$ $\pmod{47}$ is$16$. So$$47\mid3x-1\iff47\mid16.(3x-1)\iff47|x-16.$$So, the solutions are thos integers $x$ such that $x\equiv16\pmod{47}$.
A: 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*}
A: By Bezout's identity since $\gcd(9,47)=1$ we know that exist $a,b$ integers such that
$$9a+47b=1$$
then
$$\iff 9a=1-47 b \iff 9a\equiv 1 \pmod{47}$$
The integer "$a$" is defined the multiplicative inverse of $9 \mod 47$ and can be easily found by Euclidean Algorithm. In this case we found that 


*

*$21\cdot9-4\cdot 47=1\implies a=21$


Once we have the inverse we can find the solution that is


*

*$9x \equiv 3 \pmod{47}\iff 21\cdot9x \equiv 21\cdot3 \pmod{47}\iff x\equiv 63\equiv 16 \pmod{47}$

A: Using the Extended Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&5&4&2\\\hline
1&0&1&-4&9\\
0&1&-5&21&-47\\
47&9&2&1&0\\
\end{array}
$$
shows that $21\cdot9-4\cdot47=1$ so we want $63\cdot9-12\cdot47=3$ which can be reduced, by subtracting $47$ from $63$ and $9$ from $12$, to
$$
16\cdot9-3\cdot47=3
$$
