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 🙂