Simplifying rational expressions: two variables | High School Math | Khan Academy

Let's see if we can simplify this expression, and like always, pause the video and have a go at it. Now, this one is interesting because it involves two variables, but it's really the same ideas that we've done when we factored things with one variable.

So, for example, up here in the numerator, well, I never like having a non-one coefficient on the second degree term. I mean, sometimes you have to, but it looks like every term here is divisible by five. So let's factor out a five first. The numerator I can rewrite as five times... five times, you factor out a five here, you get x; factor out a five here, you get plus 4. Actually, I'm going to rewrite it as 4YX, and you'll see in a second why I'm doing that.

Actually, I'll tell you why I'm doing that right now. Why I'm writing the Y there is that this way it seems to hit the pattern of how we're used to seeing quadratics. So, let's see. You have X plus 4YX; you can view the 4Y as a coefficient on the first-degree X term right over there, plus 4Y^2. And it's going to be over... over, now the denominator here; can we factor this out?

Well, let's just think about it. Do we know two numbers, or I guess you could say do we know two expressions, that when you multiply them you get -6Y^2 and then when you add them, you get -Y? That's actually why I liked writing it like this. So, actually, let me rewrite this. This is the same thing as negative YX, and so you can view the coefficient here as -1Y.

Now, we need to think of two numbers, or two expressions, A and B, that are equal to -6Y^2, and when I add them A + B, I get -Y. So, you can imagine both of them are going to be expressions that involve Y. Let's see if this was just a -1 and if this is just a -6. Well, we would do -3 and positive 2.

Let's see if we did -3Y and positive 2Y. That indeed is going to be equal to -6Y^2, and -3Y + 2Y does indeed equal -Y. So that's our A and B right over there. If it seems a little mysterious how I just all of a sudden got -2Y or -3Y and positive 2Y, let me write an analogous quadratic here that only has one variable.

If I were to write X^2 - X - 6 and I were to ask you to factor that out, you'd say, "Oh, okay, well, this is going to be -2; I have -3 * 2, which is -6, and if I add them, well, that's going to be -1." So you would say, "Well, that's going to be X - 3 and X + 2." The only difference between this and that is instead of having just a negative one here, you have a -1Y. Instead of having just a 6 here, you have a -6Y^2.

So you could just think of this instead of just -3 and positive 2 as -3Y and positive 2Y. Hopefully, that makes sense, and if it doesn't, I encourage you to kind of play around with this, multiply these out a little bit, and get a little bit more familiar with this.

But now that we know that it can be factored like this, let's rewrite this. This is going to be X - 3Y times X + 2Y. Nothing seems to simplify out just yet, but it looks like what we have in magenta here could be simplified further, and we're going to do a very similar exercise to what we did just now.

What two expressions, if I multiply them, I get 4Y^2, and if I add them, I get 4Y? It looks like 2Y would do the trick. So, it seems like we can rewrite the numerator. This is going to be, let me draw a little line here to make it clear that this is going to be equal to 5 times (X + 2Y) times (X + 2Y).

Once again, 2Y times 2Y is 4Y^2, and 2Y + 2Y is 4Y. That's all going to be over... that is all going to be over (X - 3Y)(X + 2Y). So now, I have a common factor (X + 2Y) in both the numerator and the denominator.

So I can cancel (X + 2Y) / (X + 2Y). Well, that's just going to be one if we assume that (X + 2Y) does not equal zero. That's actually an important constraint because once we cancel this out, you lose that information.

If you want this to be algebraically equivalent, we could say that (X + 2Y) cannot be equal to zero. Alternatively, you could say that X cannot be equal to -2Y. I just subtracted 2Y from both sides there. So what you're left with, and we can redistribute this five if we want to write it out in expanded form, we could rewrite it as the numerator would be 5X + 10Y, and the denominator is X - 3Y.

But once again, if we want it to be algebraically equivalent, we would have to say X cannot be equal to -2Y. Now this is algebraically equivalent to what we had up here, and you can argue that it's a little bit simpler.