Find all solutions of the following linear congruence $2x+3y≡1\pmod 7$ I've never done one with two variables, I understand that $(2, 3, 7) = 1$ and $1|1$ so there are $1*7$ solutions, and $x=0,1,2,3,4,5,6$. I have no idea how to proceed from here, and any help would be greatly appreciated!

\begin{align}
2x+3y\equiv 1\bmod 7 &\iff 3y\equiv 1-2x\equiv 1+5x\bmod 7\\
&\iff y\equiv5(1+5x)\equiv5+4x \bmod 7
\end{align}
The last step is because $5$ is the inverse of $3\bmod 7$.
So this looks like a function. For every value of $x$ there is exactly one value of $y$ that makes the congruence hold.
