elementary number theory - How do I change to a different modulo?

I'm trying to solve the following problem:

What remainder does integer $n$ have when divided by $142$ if we know that $20n + 4$ and $72n - 12$ have the same remainder when divided by $142$?

I 'translated' the problem to maths:

$$20n + 4 \equiv 72n - 12 \pmod {142} \\ n \equiv x \pmod {142}$$

And we are essentially looking for $x$. Solving the linear congruence on top:

$$20 + 4 \equiv 72n - 12 \pmod {142} \\ 52n \equiv 16 \pmod {142} \\ 13n \equiv 4 \pmod {71} \\ 13n \equiv 4 + 8 \times 71 \pmod {71} \\ 13n \equiv 572 \pmod {71} \\ n \equiv 44 \pmod {71} \\ $$

So $n = 71k + 44$ for any $k \in \mathbb{Z}$.

The problem is that I can't answer the original question as there, the modulo is $142$ and now it is $71$. How could I somehow 'change' the modulo and get the right answer? Or what would be a way to think about this problem?

PS: I realise that I'll get more solutions. For example, if I had $2x \equiv 4 \pmod 8$, I would divide by $2$ and get $x \equiv 2 \pmod 4$. Then $x = 4k + 2$. For these numbers are for example $6, 9, 11, \ldots$. I can see that if we were dividing these by $8$, not $4$, the remainders would be $2$s and $6$s.