How to solve this linear congruence equation and more general cases? Okay so I'm trying to solve $5x \equiv 7 \mod 11$ and this is the particular example that I can't do. Can someone help me learn how to solve these and more general examples $ax \equiv b \mod n$. I believe there is only one solution (well infinitely many but they are all the same $\mod n$) $a$ is coprime to $b$  however I still need help solving these and also how to solve it when it is a simultaneous linear congruence. Any help?
Thanks.

Write $5x\equiv7\mod11$ as $5x=11k+7,k\in\mathbb{N}$. Now divide both sides by $5$:
$$x=\dfrac{11k+7}{5}$$
Notice that $x$ is integer iff $k\equiv3\mod5$, so for that $k$ we have $x\equiv\text{const}\mod11$. From this you can see that
$$x\equiv\dfrac{11\cdot3+7}{5}\equiv8\mod11$$
