Multiplying rational expressions: multiple variables

Multiply and express as a simplified rational. State the domain. We'll start with the domain. The only numbers that will make this expression undefined are the ones that would make the denominator equal to 0, and those are the situation, or that situation would occur, when either a, b, x, or y is equal to 0. If any of those are equal to 0, then we have an undefined expression. We could say the domain is all real a's, b's, x, and y's, except 0. Or we could be specific: except a, b, x, and y can't equal to 0. Or you could write this: given that a, b, x, and y does not equal 0, that none of them can be equal to 0. These are just multiple ways of stating the same thing. With that stated, let's actually multiply and simplify this rational expression. So when we multiply, you just multiply the numerator and multiply the denominator, so you have 3x squared y times 14a squared b in the numerator. Then in the denominator, we have 2ab times 18xy squared. Let's see where we can simplify this thing. We can divide the 14 by 2, and the 2 by 2, and we get 14 divided by 2 is 7, and 2 divided by 2 is 1. We could divide the 3 by 3 and get 1, and divide the 18 by 3, and get 6. Every time we divided the numerator and the denominator by 2, now the numerator and the denominator by 3, so we're not changing the expression. Then we can divide a squared divided by a, so you're just left with an a in the numerator, and a divided by a is just 1. You have a b over a b. Those guys cancel each other out. You have an x squared divided by an x, so x squared divided by x is x, and x divided by x is just a 1, so this becomes an x over 1, or just an x. Finally, you have a y over a y squared. if. You divide the numerator by y, you get 1. If you divide the denominator by y, you just get a y, and so what are we left with? We're left with in the numerator, these 1's we can ignore. That doesn't really change the number. We have a 7 times a times x. That's what we have in the numerator. In the denominator, we just have a 6y. And we have to add the constraint that a, b, x, and y cannot equal 0. When you just look at this expression right here, you're like, hey, what's wrong with x? There isn't even any b here, so it's a little bit of a bizarre statement, but you say, hey, why can't x or a be equal to 0 over here? It doesn't make it undefined. But in order for these to really be the same expressions, they have to have the same domains. Or actually, if these were functions that equal them, in order for that f of x to be equal to this f of x right over here, you'd have to constrain the domain in a similar way. This is a fundamentally different expression if you allow x's and a's. In this one, you can't allow x's and a's. For them to be really the same, you have to put the same constraints on it.