Find a formula for all integers $x$ such that $5x-1$ is divisible by $13$ and $19x-12$ is divisible by $23$ Find a formula for all integers $x$ such that $5x-1$ is divisible by $13$ and $19x-12$ is divisible by $23$.
Hello. I am working on a review sheet for my test tomorrow and I am stuck on this question. 
I found earlier that all integers $x$ such that $19x-12$ is divisible by $23$ is 
$x\equiv 20 \pmod{23}$, but I am not sure if this helps.

Yes, $19x\equiv 12\pmod{23}$ is equivalent to $x\equiv 20\pmod{23}$. Similarly, $5x\equiv 1\pmod{13}$ is equivalent to $x\equiv 8\pmod{13}$.
So we want to solve the system of congruences $x\equiv 8\pmod{13}$, $x\equiv 20\pmod{23}$. In general to solve systems of linear congruences we use the machinery of the Chinese Remainder Theorem. But for a "small" problem like ours, one can find the answer by playing around. We want $20+23k$ to have remainder $8$ on division by $13$. We can try $k=0$ to $12$, or use more machinery.
We want $7+10k\equiv 8\pmod{13}$ or equivalently $10k\equiv 1\pmod{13}$. The inverse of $10$ modulo $13$ is $4$, so we want $k=4$. Thus the general solution is $20+4\cdot 23\pmod{(13)(23)}$.
