We define group A as such:
$A=\{n\epsilon\mathbb{N}:1\le n\le 90\}$
We define relation R on A as such:
$aRb\iff23|(a+22b)$
(where "$|$" represents divisibility)
What is the equivalence class $[22]_R$ equal to?
An explanation of how you found this would be very helpful as I have been unable to find an explanation for any similar problems.
On
$aRb\iff23|(a+22b)$
$22Rb\iff23|(22+22b)\iff23|(22(1+b))$
$22$ and $23$ are coprime, so $23$ must divide $1+b$, that is
$1+b=23m\iff b=23m-1=23m'+22$ for some $m,m'\in\mathbf{Z}$.
So $22Rb\iff b=23m+22$
And therefore $$[22]_R=\left\{c\in\mathbf{Z}\left|\right.\exists k\in\mathbf{Z}\, c=23k+22\right\}$$
Hint: Note that $23 \mid (a+22b)$ iff $23 \mid (a-b)$ iff $a$ and $b$ leave the same remainder when divided by $23$.
If $a \in \mathbb N$ then $a= 23q+r$ with $0 \le r \le 22$. Then $[a]_R=[r]_R=\{ 23q+r : q \in \mathbb N \} = r +23 \mathbb N $.