Using the Chinese Remainder theorem to solve the systems of congruence's. We got introduced to the Chinese Reminder Theorem, but I still haven't quite grasped it and I have a problem that asks: 

Find $x \bmod 77$ if $x \equiv 2 \pmod 7$ and $x \equiv 4 \pmod {11}$.

My attempt was like so: 
$$B = 7 * 77 * 11 = 5929$$
$$B_{1} = \frac{5929}{77} = 77$$
Using $B_{1}*x_{1} \equiv 1 \pmod {77}$: 
$$77x_{1} \equiv 1 \pmod {77} $$
$$x_{1} \equiv 78 \pmod {77}$$
So then would the $x$ be $78$?      

I don't know if this is what you have on your mind, but I would do it like this:
$x=7a+2$ and $x= 11b+4$ for some integers $a,b$ so $7a= 11b+2$ and thus $7|11b+2$.
Multiply this with 2 and we get $7|22b+4$ so $7|b-3$. Now we can write $b=7t+3$ and thus $x=77t+37$. So $x \equiv 37 \pmod {77}$.
