How to show steps for 11x - 7 $\equiv$ 0 (mod 13)? I know the answer is $x = 3$, but I'm having trouble showing the steps for it. All I have is $11x \equiv 7 (\mod 13)$.
There are general methods here based on Euclid's division algorithm, and these always work. You can look that up.

Sometimes there are short cuts you can spot, as here. Don't forget that there are negative numbers in congruence classes. So here, modulo $13$ $$11x\equiv -2x$$ and you then only need to divide by $2$. So $7\equiv -6$ or $7\equiv 20$ get you to an even residue.

Whence the equation can be rewritten as $-2x+6\equiv 0$ with $x\equiv 3$ by simple division ($2$ and $13$ are coprime), or you could write $-2x-20\equiv 0$, from which you obtain $x\equiv -10\equiv 3$.
The problem 11x − 7 ≡ 0 (mod 13) is equivalent to 11 x0 − 7 = 13 x1 where x0, x1 ∈ . One way to solve the linear equation would be to express the original variables (x0, x1) in terms of parameter variables (t0, t1) where each of the parameter variables take on any value from . To
find such as expression you can judiciously introduce the parameter variables and corresponding equation one at a time.

After generating the additional equations (the ones with parameter variables) you can take them plus the original equation as a system on linear equations. The system of equations can be solved using back substitution. After each back substitution step if the resulting equation has a greatest common factor then the greatest common factor should be removed before continuing.

The PDF document "Showing the Steps to Solve 11x − 7 ≡ 0 (mod 13)" applies these ideas in a step-by-step fashion. The following diagram is a snap shop of the PDF page: