Solving the congruence $7x \equiv 41 \mod{13}$ I have to solve the following linear congruence: 
$$7x \equiv 41 \mod{13}$$
The question where I got this from comes in two parts. The first is that it asks to find the set of the inverses of $7 \mod 13$, which turns out to be $[2]_{13}$ but, I found that solution by solving the linear diophantine equation $$7x - 13k - 41$$ which is just the original equation in disguise i.e. $7x \equiv 41 \mod{13}$. 
Now, I'm terribly confused as to the solution of the congruence equation. Where am I going wrong and how should I proceed?
Thanks!
EDIT: From @Bill's hint
\begin{align}
7x &\equiv 41 \mod{13} \\
x &\equiv \frac{41}{7} \mod{13} \\
x &\equiv \frac{28}{7} \mod{13} \quad (?) \\
x &\equiv 4 \mod{13} \\
\end{align}
Then, this gives me the solution $[4]_{13}$.

The OP states

... but, I found that solution by solving the linear diophantine equation

From this site I found

Here the equation is $7x\equiv 41 \pmod {13}$ or
$$\tag 1 7x\equiv 2 \pmod {13}$$
Here also, multiplication $7 \times x$ is repeated addition. So you can also get the inverse of ${[7]_{13}}^{-1}$ and/or the solution of $\text{(1)}$ using the same technique (might as well cycle through it all):
$\quad 7 \times 1 \equiv 0 + 7 \equiv 7 $
$\quad 7 \times 2 \equiv 7 + 7 \equiv 1 $
$\quad 7 \times 3 \equiv 1 + 7 \equiv 8 $
$\quad 7 \times 4 \equiv 8 + 7 \equiv 2 $
$\quad 7 \times 5 \equiv 2 + 7 \equiv 9 $
$\quad 7 \times 6 \equiv 9 + 7 \equiv 3 $
$\quad 7 \times 7 \equiv 3 + 7 \equiv 10$
$\quad 7 \times 8 \equiv 10 + 7 \equiv 4$
$\quad 7 \times 9 \equiv 4 + 7 \equiv 11$
$\quad 7 \times 10 \equiv 11 + 7 \equiv 5$
$\quad 7 \times 11 \equiv 5 + 7 \equiv 12$
$\quad 7 \times 12 \equiv 12 + 7 \equiv 6$
$\quad 7 \times 13 \equiv 6 + 7 \equiv 0 $
Looking over your work you can pick out both the inverse and the solution.
You can set up a spreadsheet to handle these problems and/or check your work.
