Really good idea you have there, @aaronhma !
One thing though (soo close!)-- since you're kinda looking at the problem from the perspective of the filling pipe, we actually have to flip the sign of the \(\frac{1}{111}\). Here's why:
Let's look at the first equation:
$$\frac{1}{185} + \frac{1}{74} - \frac{1}{x} = \frac{1}{111}$$
What this equation literally means is that, in \(1\) minute, the large pipe drains \(\frac{1}{185}\) of the pool, the small pipe drains \(\frac{1}{74}\) of the pool, and the filling pipe fills or "un-drains" \(\frac{1}{x}\) of the pool, which is why we subtract the \(\frac{1}{x}\). This altogether is equal to draining \(\frac{1}{111}\) of the pool.
But looking at the second equation, which is from the perspective of the filling pipe, we have:
$$\frac{1}{x} - \frac{1}{74} - \frac{1}{185} = \frac{1}{111}$$
This equation says that, in \(1\) minute, the filling pipe fills \(\frac{1}{x}\) of the pool, the small pipe drains or "un-fills" \(\frac{1}{74}\) of the pool, which is why we subtract \(\frac{1}{74}\), the large pipe drains or "un-fills" \(\frac{1}{185}\) of the pool, which is why we subtract\(\frac{1}{185}\).
But here's the catch: this altogether is equal to draining \(\frac{1}{111}\) of the pool. We have "un-filled" the pool by that much. This means we have less water than when we started.
So, the correct equation would be actually equal to \(-\frac{1}{111}\):
$$\frac{1}{x} - \frac{1}{74} - \frac{1}{185} = -\frac{1}{111}$$
Notice how this is exactly the same as Prof. Loh's equation but multiplied all the way on both sides by \(-1\).
Really great thinking you did there, and it really goes to show how many different ways there are to solve the same problem! Well done 🙂
