I am given the system of linear congruences:
$x\equiv 2$ mod $7$, $x\equiv 3$ mod $11$, and $x\equiv 4$ mod $13$.
I start to solve by substituting the first equation into the second, yielding: $x=2+7(-3+11m)$ for some $m\in\mathbb{Z}$. With some algebra, we find $x=-19+77m$.
I then plug that result for $x$ into the 3rd equation to find: $77m-13n=23$ for some $m,n\in\mathbb{Z}$. Then, I use Euclid's Algorithm to solve that equation, where I find $m=-23$ and $n=-138$.
What do I do with these solutions? How can I continue with the original problem? Edit: Please keep answers within the realms of this method.
I dont know if this is the most efficient but it works. You have three primes $p$,$q$,$r$. Find $a,b,c$ such that $$aqr \equiv 1 \mod p$$ $$bpr \equiv 1 \mod q$$ $$cpq \equiv 1 \mod r$$
Then to solve $$x\equiv \alpha \mod p$$ $$x\equiv \beta \mod q$$ $$x\equiv \gamma \mod r$$
you have $aqr\alpha+bpr\beta+cpq\gamma$.
In this case we have $$(-2)11\cdot 13 \equiv 1 \mod 7$$ $$(4)7\cdot 13 \equiv 1 \mod 11$$ $$(-1)7\cdot 11 \equiv 1 \mod 13$$
So your solution is $$2(-2)11\cdot 13 +3(4)7\cdot 13 +4(-1)7\cdot 11 =212$$
Which you can check is indeed a solution.