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 $78$?      

Let us use the Euclidean algorithm to express $\gcd(7,11)$ as a linear combination:
$$4 = 11 - 7\\
3 = 7 - 4 = 2 \cdot 7 - 11 \\
1 = 4 - 3 = 2 \cdot 11 - 3 \cdot 7.$$
Now, $x = (2 \cdot 11 - 3 \cdot 7) x = 2 (11x) - 3 (7x)$.  By the given congruences, $11x \equiv 11 \cdot 2 = 22 \pmod{77}$ and $7x \equiv 7 \cdot 4 = 28 \pmod{77}$.  Therefore, $x \equiv 2 \cdot 22 - 3 \cdot 28 = -40 \equiv 37 \pmod{77}$.
(This is hopefully closer to what you were hinting at in your comments as the method of solving CRT problems that your source is using.)
