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# Simplifying rational expressions: two variables

CCSS.Math:

## Video transcript

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 here I mean sometimes you have to but for it looks like every term here is divisible by five so let's factor out a 5 first so the numerator I can rewrite as 5 times 5 times you factor out a 5 here you get x squared factor out of 5 here you get plus 4 actually I'm going to rewrite it as 4y X and you'll see in a second and why I'm doing that actually I'll tell you why 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 it seems to hit your the pattern of how we're used to seeing quadratics so let's see so you have x squared plus 4y X you can view the 4y is the coefficient on the first degree in X term on the X term right over there plus 4y squared 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 to two numbers or I guess so you say do we know two expressions that would add that when you multiply them you get negative 6y squared and then when you add them you get negative x y so or sorry when you add them you get negative y that's actually why I liked writing it like this so actually let me rewrite this this is the same thing as negative Y X and so you can view the coefficient here as negative Y and so we need to think of two numbers so or two expressions a times B that is equal to negative 6y squared and when I add them a plus B I get that is equal to negative Y and so you can imagine both of them are going to be expressions a and B are going to be expressions that involve Y and so let's see if you if this was just a negative one and if the just a negative six what we would do we would do negative three and positive two and let's see if we did negative 3y and positive 2y that indeed is going to be equal to negative 6y squared and negative 3y plus 2i does indeed equal negative Y so that's our a and B right over there and if it seems a little mysterious how did Sal just all of a sudden get negative 2y or negative 3y and positive 2y let me write a analogous quadratic here that only has one variable if I were to write x squared minus X minus 6 and I were to ask you to factor that out you say okay well this is going I have negative 2 I have negative 3 times negative 2 is negative 6 and if I add them well that's going to be negative 1 so you would say well that's going to be that's going to be X minus 3 times X plus 2 and so the only difference between this and that is instead of having just a negative 1 here you have a negative 1y instead of having just a negative 6 here you have a negative 6y squared and so you could just think of this instead of just negative 3 and positive 2 negative 3y and positive 2y hopefully that makes sense and if it doesn't I encourage you to to kind of play around with this multiply these out a little bit a little bit of more more familiar with this but once we now that we know that can be factored like this let's rewrite this this is going to be X minus 3y times X plus 2y and 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 a very similar exercise to what we did just now what two expressions if I multiply them I get 4y squared and if I add them I get 4y looks like 2y would do the trick so it seems like we can rewrite the numerator this is going to be so let me draw a little line here to make it clear that this is this is going to be equal to five times X plus 2y times I could say just x plus 2y squared or I can just say X plus 2y times X plus 2y once again 2y times 2 is 4 y squared - y plus 2y is 4y and so and that's all going to be over that is all going to be over X minus 3y x minus 3y times X plus 2i and so now I have a common factor X plus 2y in both the numerator and the denominator so I can Candle x plus 2y divided by X plus 2i well that's just going to be 1 if we assume that X plus 2y does not equal 0 and 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 plus 2i cannot be equal to 0 or another way you could say it is that X cannot be equal to cannot be equal to negative 2y I just subtracted 2y from both sides there and so what you're left with and we can redistribute this five if we want to write it out in an expanded form we could rewrite it as the numerator would be 5x let me write it over here 5x + 10 y and the denominator is X minus 3y but once again if we want them to be algebraically equivalent we would have to say X cannot be equal to X cannot be equal to negative 2y and now this is algebraically equivalent to what we had up here and you can argue that it's a little bit simpler