Use the Chinese Remainder Theorem to determine the value of $x$. I'm trying to solve the following modular arithmetic question using the Chinese Remainder Theorem, using this link. (We learned a different method in our class, but I found this easier to grasp).
$$x \equiv 1 (\text{mod} \ 5)$$
$$x \equiv 2 (\text{mod} \ 7)$$
$$x \equiv 3 (\text{mod} \ 9)$$
$$x \equiv 4 (\text{mod} \ 11)$$
I then represented $x$ as a sum of $4$ boxes, such that the first term is "related" to $\text{mod} \ 5$ (i.e. the $1^{st}$ term will not be made $0$ due to the $\text{mod} \ 5$), the second term is related to $\text{mod} \ 7$ and so on. Here's what I mean by "related":

If we only consider $\text{mod} \ 5$, the value of box $1$ is $693$, the value of box $2$ is $495$, then $693 \  \text{mod} \ 5 = 3$ but $495 \  \text{mod} \ 5 = 0$. Likewise, if we only consider $\text{mod} \ 7$, then the value of box $1$ is $693 \  \text{mod} \ 7 = 0$ but $495 \  \text{mod} \ 7=5$. And so on...

After doing all that, I have $$x = (7 \times 9 \times 11) + (5 \times 9 \times 11) + (5 \times 7 \times 11) + (5 \times 7 \times 9)$$
The next step is applying the $\text{mod} \ 5$ to $x$:
$$\begin{align}
x \ \text{mod} \ 3 &\equiv 691 \ \text{mod} \ 5 + 495 \ \text{mod} \ 5 + 385 \ \text{mod} \ 5 + 315 \  \text{mod} \ 5 \\ &\equiv 693 \ \text{mod} \ 5 + 0 + 0 + 0 \\ &\equiv 693 \ \text{mod} \ 5 \\ &\equiv 3 \ (\text{mod} \ 5)
\end{align}$$
This is where I get stuck. In the video, and the video doesn't explain how to deal with such a scenario. 
PS - If there is a more "intuitive" or more efficient version of the Chinese Remainder Theorem, I'd be grateful if you could share it.
PPS - Sorry if the question is a bit awkwardly formulated. As you can guess this is my first doing this.

I think the best way for me to solve a CRT problem is like this:
$$x \equiv1 \pmod{5} \implies x \in \{1,6,11,16,21,26\dots\}$$
$$x \equiv4 \pmod{11} \implies x \in \{4,15,26,\dots\}$$
Now one can immediately see the intersection at $x=26$, and indeed $x \equiv 26 \pmod{55}$ satisfies both $x \equiv1 \pmod{5}$ and $4 \pmod{11}$.
Similarly,
$$x \equiv 2 \pmod{7} \implies x \in \{2,9,16,23,30,\dots\}$$
$$x \equiv 3 \pmod{9} \implies x \in \{3,12,21,30,\dots\}$$
So $x \equiv 30 \pmod{63}$
Now, from there I can solve it with the casual method:
$$x \equiv26 \pmod{55} \implies x=55k+26$$
$$\implies55k+26 \equiv30 \pmod{63} \implies 55k \equiv4 \equiv 130 \pmod{63}$$
$$\implies 11k \equiv 26 \equiv 341 \pmod{63} \implies k \equiv 31 \pmod{63} \implies k=63j+31$$
$$\implies x=55(63j+31)+26=3465j+1731 \implies x \equiv 1731 \pmod{3465}$$
Noting, of course, that $3465=5\cdot7\cdot9\cdot11$
