Find all solutions, if any, to the equation $[28][x]-[22] = [33]$ in $\mathbb{Z}_{51}$. Find all solutions, if any, to the equation 
$[28][x]-[22] = [33]$ in $\mathbb{Z}_{51}$.
I know this simplifies to 
$[28][x] = [55]$,
which can be rewritten as 
$28x \equiv 55 \bmod{51}$.
From here do I use SMT and split it, then proceed to use CRT?
Any suggestions are much appreciated, thanks.

You can certainly solve the congruence $\ 28x \equiv 55 \bmod{51}\ $ (which you can rewrite as $\ 7x \equiv 1 \bmod{51}\ $) by using the same procedure used in the proof of the Chinese remainder theorem.  That is, if $\ x_1\ $ satisfies the congruence $\ x_1 \equiv 1 \bmod{3}\ $, and $\ x_2\ $ the congruence $\ 7x_2 \equiv 1 \bmod{17}\ $, then the unique $\ x\in \left\{0,1,\dots, 50\right\}\ $ simultaneously satisfying both of the congruences $\ x\equiv x_1 \bmod{3}\ $ and $\ x\equiv x_2 \bmod{17}\ $ will be a solution of your original congruence, $\ 7x \equiv 1 \bmod{51}\ $.
Alternatively, since $\ \gcd\left(7,51\right)= 1\ $, you can also use the extended Euclidean algorithm to find integers $\ a\ $ and $\ b\ $ satisfying the equation $\ 7a + 51b = 1\ $. Then $\ x=a\ $ will be a solution of the congruence $\ 7x \equiv 1 \bmod{51}\ $.
