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.

There should be $x = (7 \times 9 \times 11)\cdot(7 \times 9 \times 11)^{-1}_5\cdot 1 $ ${}+ (5 \times 9 \times 11)\cdot(5 \times 9 \times 11)^{-1}_7\cdot 2 $ ${}+ (5 \times 7 \times 11)\cdot(5 \times 7 \times 11)^{-1}_9\cdot 3 $ ${}+ (5 \times 7 \times 9)\cdot (5 \times 7 \times 9)^{-1}_{11}\cdot 4$ for this approach.
