 The problem reads, pump A alone can fill an empty pool in 15 hours. Together, pump A and pump B fill the same empty pool in 10 hours. How long does it take pump B alone to fill the empty pool? So, we draw our line and we start our solution. And first of all, what is the question? Look up here and we find the question mark. And we say, it's how long does it take pump B alone to fill the empty pool? So, is that a reasonable thing to use as a variable? Yes, because we have a similar piece of data here. So, we would say x equals the time is hours. So, we always write our unit. So, that's how long pump B alone fill. This tells us exactly what we're going to get. We're going to get a value that's going to tell us the number of hours that takes pump B to fill the pool by itself. Okay. So, now, how do we do problems like this? We have to say we need the part of the work, the work done per unit time. What is our unit time here? Our unit time here is hours, right? So, we have hours. So, we're looking for a part of the work, work done, and what do we know? We know that pump A can fill the pool in 15 hours. So, what part of the pool does it fill in one hour? That's our question. And we say pump A does 1 over 15th of the work in one hour. Then we say, okay, what about pump B? We don't know how long it takes, but we've set up our variable to be that. So, we can use that instead of 15. And we can say 1 over x of the work in one hour. And what else do we know? Together, they fill the same pool in 10 hours. So, if they're working together, if they work together, work together, what is their rate of filling per hour? What is their work done per hour? So, 1 15th of the work does pump A plus pump B does 1 of x over the work. So, this is work per hour together. That's the part of the work per hour. And what do we finally have? We say 10 times that, 10 hours, 1 over 15 plus 1 over x work per hour. That gives me how much work is done. And so, when I cancel the hours, that gives me how much work is done. And I say, oh, the whole thing is spoiled, so it's a whole one work. And I say, this is our equation that we need to solve. So, then we would write, we use the distributive rule. We would say 10 over 15 plus 10 over x equals 1. And then we usually write least common, but we tell them, I tell them any multiple we'll do, but you have to write down all of the denominators. So, 15 and if x, and if there's another denominator, you have to write it down. And so, this one is easy, so 15 x. So, then I say, you have to write 15 x every time you have a plus or minus. So, 15 x, 15 x, because we have to make sure we multiply every single 1 by our 15 x. So, 10 over 15 plus 10 over x equals, and I make them write the 1 there. And so, then they can easily cancel, because it's not big ones. And they would get 10 x here. And they would cancel the x here and get 150. And there's nothing to cancel here, so this is 15 x. So, then they have learned from school, because that's the way they learned it, just to write that this equals 5 x. And so, then they would write, this is their biggest problem, 150 over 5 equals x, so x equals 30. And then they're required to write actually answered. So, it would be, pump b needs 30 hours. I don't see any way to do this problem without using fractions.