Answering this question,
I came upon the following family of equivalences
$$\left\{\begin{align} 12n+8 &\equiv 0 \bmod{(x+4)}, \\ 12n+9 &\equiv 0 \bmod{(x+3)}, \\ 12n+10 &\equiv 0 \bmod{(x+2)}, \\ 12n+12 &\equiv 0 \bmod{x} \end{align} \right \}$$
where I am seeking non negative integer pairs $(x,n)$ that solve all of the equivalences.
I tried $x=1,2,3,\dots$ until I found the solutions $$(x,n) = (12, 138+140m)$$
I'm pretty sure that there are other values of $x$ that lead to other solutions.
What I'm hoping for is a way to solve families of congruences similar to the above example without iterating over all possible values of $x$.
A few suggestions or hints would be nice.