On Chinese remainder theorem Suppose we know $x\equiv 3 \bmod 11$ and $x\equiv 7\bmod 13$ and $0<x<143$ holds then CRT gives that  $x=3\times 13[13^{-1}\bmod 11] + 7\times 11[11^{-1}\bmod 13]=39\times 6 + 77\times 6=696$ gives a solution but it is not within $0$ and $143$. So we take $696\bmod 143$ and choose $124$ as solution.
My query is say instead of $3$ we choose $3+11k$ and instead of $7$ we choose $7+13\ell$ for some $k,\ell\in\Bbb Z$ can we still recover $124$? 
Instead of $696$ is there a direct way to get $124$?
We have $696=124+4(143)$. In general what is the quantity that goes in instead of $4$?

Some people do CRT theorem problems like this.  If $x\equiv 3\pmod{11},$ then $x = 3+11k$.  Plug that into $x \equiv 7 \pmod{13}$ to get $3+11k \equiv 7 \pmod{13}$ which reduces to 
$$11k \equiv 4 \pmod{13}$$
$$-2k \equiv 4 \pmod{13}$$
$$k\equiv -2 \equiv 11 \pmod{13}$$
So $x = 3+11\cdot 11 =124.$
