$5x\equiv3\pmod3$ The answer from class is $x = 3 + 3t$ , $t$ belongs to $\mathbb Z$
I see that:
0 1 2 0 1 2 0 1 2 0
0 1 2 3 4 5 6 7 8 9
Am I understand this right? What is the proper way to find this answer? 
 A: If
$$5x \equiv 3 \mod 3,$$
then
$$5x = 3 + 3t_{1}$$
for some $t_{1}.$
Since
$3 \nmid 5,$
so
$x \equiv 0 \mod 3.$
Thus
$$x = 3t_{2}$$
for some $t_{2}.$
But if
$x = 3t_{2},$
then
$$5x = 15t_{2} \equiv 3 \mod 3,$$
we thus find all $x$ such that $5x \equiv 3 \mod 3.$
A: Since we are calculating modulo $3$, any number can be changed by a multiple of $3$. In particular, we can replace $3$ by $3-3=0$, and $5$ by $5-2\times3 = -1$. Thus an equivalent equation is
$$-x\equiv 0 \pmod 3$$
In other words, $-x$ is a multiple of $3$, which is the same as $x$ being a multiple of $3$.
Therefore the solution is $x = 3t, t\in\mathbb Z$.
A: Since we are solving for $x$ and the equation has a "$5x$", the instinctive thing we want to do is multiply by the inverse of $5$. 
The inverse of $5$ in mod $3$ is $2$ (indeed notice that $5 \times 2 = 10 \equiv 1 $ (mod $3$)). Multiplying both sides of the equation by $2$ we have that $10x \equiv x $ (mod $3$) $\equiv 6 $ (mod $3$) $\equiv 3 $ (mod $3$). So $x = 3 + 3t$ for some $t \in \mathbb{Z}$.
This is the answer given in your class, but observe that $3 \equiv 0 $ (mod $3$) so $x = 3t$ is a simpler answer, as pointed out by another answerer. 
Is this a more "systematic" approach which you are looking for?
A: Since the $\gcd(5,3) = 1$ there there is a solution $\pmod{3}$
Now $\pmod{3}$ only has $3$ elements $\{0,1,2\}$ and so the solution must one be of those $3$ elements
You can try all of them , you will see that $5(1) \not\equiv 3 \pmod{3}$  and also $5(2) \not \equiv 3 \pmod{3}$
The solution is $5(0) \equiv 3 \pmod{3}$.
Now since $0$ is a solution then the equivalence class of $[0]$ is the whole solution and $$[0] = \{.....,-12,-9,-6,-3-0,3,6,9,12,....... \}$$
And so $x = 3k$ where $k$ is any integer
