What phenomenon is this? $(2\Bbb{Z} + 1)\cup 3\Bbb{Z} = 2\Bbb{Z} \cup 3\Bbb{Z} + 3$ $(2\Bbb{Z} + 1)\cup 3\Bbb{Z} = 2\Bbb{Z} \cup 3\Bbb{Z} + 3$
Proof:
$$
\begin{align*}
2\Bbb{Z} &= \bullet \circ \bullet \circ \bullet \circ \bullet \circ \dots \\
3\Bbb{Z} &= \bullet \circ \circ \bullet \circ \circ \bullet \circ \dots \\
2\Bbb{Z}+1 &= \circ \bullet \circ \bullet \circ \bullet \circ \bullet \dots \\
2\Bbb{Z} \cup 3\Bbb{Z} &= \bullet \circ \bullet \bullet \bullet \circ \bullet \circ \dots \\
(2\Bbb{Z} + 1)\cup 3\Bbb{Z} &= \bullet \bullet \circ \bullet \circ\bullet\bullet\bullet \dots
\end{align*}
$$
Shift the last one by $\pm 3$. Conclude.  That's a visual proof.  But what is the underlying algebra?
 A: Notice that the sets $2\mathbb Z$ and $3\mathbb Z$ (as well as translations thereof) are both periodic and share $6$ as a period - that is, you can reduce the problem mod $6$ and note that, since your sets are equal mod $6$ they are equal in general. This essentially tells us that your visual proof suffices as a rigorous one.
To give a more algebraic proof of your identity though, one might note that we can start with the expression
$$2\mathbb Z \cup 3\mathbb Z + 3$$
and then "distributing" the $3$ gives the equivalent
$$(2\mathbb Z + 3) \cup (3\mathbb Z + 3)$$
and noting that $a\mathbb Z + b = a\mathbb Z + (b+na)$ for any $n$, since those sets are "periodic". Reducing both sets in parenthesis yields the desired
$$(2\mathbb Z + 1) \cup 3\mathbb Z.$$
A: It's a compact form of set-builder notation, relying on metaphor to overload the multiplication and addition operators.
$\Bbb Z$ is the set of integers.   And we can define, $a\Bbb Z{+b}$ , to mean, $\{a x + b\mid x\in \Bbb Z\}$ in set-builder notation (or, more formally, $\{y \mid \exists x\in \Bbb Z: y=ax+b\}$ ), where $a,b$ are constants (and usually integers).
Thus this is the set obtained by multiplying each integer by $a$ then adding $b$.   Otherwise known as a scale and shift operation, respectively.
Using this notation we can derive rules, such as distribution: $(a\Bbb Z\cup b\Bbb Z)+c=(a\Bbb Z+c)\cup(b\Bbb Z+c)$ and periodicity: $a\Bbb Z = a\Bbb Z+a$, and so forth, enabling us to manipulate the symbols, thusly:
$$\begin{align}(2\Bbb Z+1)\cup 3\Bbb Z & = (2\Bbb Z+2+1)\cup(3\Bbb Z + 3) 
\\ & = (2\Bbb Z\cup 3 \Bbb Z)+3 \\ \Box\end{align}$$
A: What you have is a set of linear equations. You want to find the numbers so that:
$x\equiv 1\bmod 2$ or $x\equiv 0 \bmod 3$. I suggest you look at the integers $\bmod 6$. And manually figure out which are the exact residue classes that work $\bmod 6$.
In the more general case you may be given some linear congruences
$x\equiv a_1\bmod n_2$, $x\equiv a_2\bmod n_2\dots x\equiv a_r\bmod n_r$, and asked to find the set of integers that satisfy at least one of these.
You can always look at the problem $\bmod$ $lcm(n_1,n_2\dots n_r)$ and find the solutions manually inside this larger modulo, which contains all of the residue information for each of the $n_i$.
In your case It is clear if a number is $1\bmod 2$ or a multiple of $3$ then it is $0,1,3$ or $5\bmod 6$. On the other hand the numbers that are multiples of $3$ and $2$ are $0,2,3,4$, adding $3$ gives $3,5,0,1$. This gives us $(2\mathbb Z+1)\cup3\mathbb Z = (2\mathbb Z \cup 3\mathbb Z)+3$
