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!

For congruences involving small coefficients, a bit of familiarity with numbers, a bit of trial and error, and knowledge of basic congruence theorems can save a lot of messing around with inverses.  Here is how I would solve the problem.  First rewrite it as
$$2x\equiv1-3y\pmod7\ .$$
Now the congruence $2x\equiv a\pmod7$ has a unique solution for each $a$, because $2$ and $7$ are coprime.  So the above congruence will be satisfied by one $x$ value for each $y$ value.  To find a solution I would say, wouldn't it be nice if the RHS were even? - then I could just cancel $2$.  So I would write
$$2x\equiv 8+4y\pmod7\quad\Leftrightarrow\quad x\equiv4+2y\pmod7\ ,$$
and this is the solution.
Exercise.  Repeat the working but starting "the other way round", that is,
$$3y\equiv1-2x\pmod7\ .$$
You will probably find that you obtain a solution which looks different, then it would be a good idea to check that it is really the same.
