Rationalize the denominator and simplify.
\[\frac{ 2\sqrt{x}-3\sqrt{y} }{ \sqrt{x}+\sqrt{y} }\]
You need a "conjugate" to exploit the property: \((a+b)(a-b) = a^{2} - b^{2}\) You have \(\sqrt{x}+\sqrt{y}\). What do you suppose its "conjugate" might be?
\[\sqrt{x}-\sqrt{y}\]
That's the conjugate. So now what?
Perfect. Multiply both numerator and denominator by that. The denominator is easy. \(x - y\)
Do what? Lol
The numerator: \((2\sqrt{x}-3\sqrt{y})(\sqrt{x}-\sqrt{y})\)
\[2x+3y-2\sqrt{xy}-3\sqrt{xy}\]
Would this be it? \[(2\sqrt{x}+3\sqrt{y})(\sqrt{x}+\sqrt{y})\]
So like TK said, once you get the bottom, multiply the conjugate to the top. That \(-2\sqrt{xy}-3\sqrt{xy} \) can be simplified.
You distributed right the first time :)
Would that equation be right?
Yes, now just simplify it, those two can be combined: \(2x+3y-\color{red}{2\sqrt{xy}-3\sqrt{xy}} \)
think of the \(\sqrt{xy} \) as x's if that helps. they're the same and can be combined: \(2x+3y-\color{red}{2x-3x} \) Seem easier now??
-3x+3y ?
Oh, sorry about that, x was a terrible sub since it's already there.. err let's make it k: \(2x+3y-\color{green}{2k-3k} \) Noww try it :P
-5k+2x+3y
Good :) So now when we put our fraction back together after taking it apart we end up with this: \(\Large \frac{2x+3y-5\sqrt{xy}}{x-y} \) I doubt we can simplify that easily, so that will be what we are looking for :P
oh, yea the \(k=\sqrt{xy} \)
This is what the format of the answer has to look like
Ugh never mind. It won't load.
Sorry about that :/
So anyways, now what? Lol
Since it can't be simplified that'll be your answerr
What is?
I have leave now too, hope that made sense, and good luck :)
Thank you so much! :D
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