Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried to subtract one equation from the other,  but got nothing. No idea what to do.

There are some basic division tricks.  $7*13*11=1001$ so to have $xyz138\equiv 0 \pmod 7$ implies that $xyz138 \equiv xyz138 - 1001*xyz = 138-xyz \equiv 0$ so $xyz \equiv 138\pmod 5\pmod 7$.
Likewise $138xyz \equiv 138xyz - 1001*138\equiv xyz -138\equiv 6 \pmod {13}$ so $xyz \equiv 144\equiv 1 \pmod {13}$.
Both together means $xyz \equiv 5+7k \equiv 1+13j\pmod{7\times 13}$ for some $k,j$. So $xyz \equiv 40\pmod{91}$
We can do the same trick for $11$ to get $x1y-3z8 \equiv 5\pmod {11}$.  Or we can do the other trick that $x1y3z8\equiv 1+3+8 - (x+y+z)\equiv 5\pmod {11}$.
So we have $x+y+x \equiv 7\pmod 11$.
So $xyz = 40 + 91k$ for some $k$ where $x+y+z\equiv 7\pmod {11}$.
$40$ yield $4+0\equiv 4\pmod {11}$.
$131\implies 1+3+1\equiv 5\pmod {11}$.
$222\implies 2+2+2\equiv 6\pmod{11}$.
$313\implies 3+1+3\equiv 7\pmod {11}$.
So that'll do
$313138\equiv 0\pmod 7$
$138313 \equiv 6\pmod 13$
And $311338 \equiv 5\pmod{11}$
