Find all $x$ such that $11\mid 3x+7$ I found this question in Beachy and Blair: Abstract algebra book, they even have a solution to this but its not satisfactory for me. They only say "$x\equiv 5 \pmod{11}$ ". Which one can "feel" simply by trial and error. I would like to know what is the proper approach. Thank you in advance!

$11\mid 3x+7$ is the same thing as saying there exists an integer $y$ such that $3x + 7 = 11y$.
Trying $y=0, 1$ and finally $y=2$, we find that $x=5$ and $y=2$ is a solution. So we consider solutions of the form $x = 5+u$ and $y=2+v$.
\begin{align}
   3x + 7 &= 11y \\
   3(5+u) + 7 &= 11(2+v) \\
   15 + 3u + 7 &= 22 + 11v \\
   3u + 22 &= 11v + 22 \\
   3u &= 11v \\
   3 &\mid 11v \\
   3 &\mid v
\end{align}
So $v = 3t$ and $y=2+3t$ for some integer $t$. The "for some" part is a problem. It suggests that, sometimes, $y=2+3t$ may not be a solution. It is very easy to show $y=2+3t$ is always going to be a solution.
Start with $3x+7=11y$ and let $y = 2+3t$.
\begin{align}
   3x+7 &= 11y \\
   3x + 7 &= 11(2+3t) \\
   3x + 7 &= 22 + 33t \\
   3x &= 33t + 15 \\
   x &= 11t + 5
\end{align}
But, of course, saying $x = 11t+5$ for any integer $t$ means that $x \equiv 5 \pmod{11}$.
