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December 20th, 2007, 02:46 PM  #1  
Newbie Joined: Dec 2007 Posts: 5 Thanks: 0  Simplying an Algebraic function
Hi there, It would be great if someone could simply this algebraic function for me, I tried few times, but I'm not getting the same answer as the textbook. Thanks in advance. Quote:
Quote:
 
December 20th, 2007, 03:45 PM  #2 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
((3x^2y^3)  (3x^6/y)) / (y^6) = ((3x^2y^3)  (3x^6/y))(y) / (y^6)(y) = ((3x^2y^4)  (3x^6)) / (y^7) = (3x^2(y^4+x^4)) / (y^7) 
December 20th, 2007, 05:37 PM  #3  
Newbie Joined: Dec 2007 Posts: 5 Thanks: 0  Quote:
 
December 20th, 2007, 05:56 PM  #4 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
Multiplying 3x^6/y by y is important. In order to do this, we have to multiply ((3x^2y^3)  (3x^6/y)) by y. I'll generalize this below. In algebraic operations, we can say that (a  b)(c) = ac  bc. So, at ((3x^2y^3)  (3x^6/y)) / (y^6), if we want to get rid of the denominator y of 3x^6/y, we have to multiply by y, and to make the algebra easier, multiply the whole parenthesis part by y, and you can get rid of denominator y, and that's pretty much it. Let me know if you have question. 
December 20th, 2007, 06:41 PM  #5 
Newbie Joined: Dec 2007 Posts: 5 Thanks: 0 
it makes sense, johnny. Thank you ! 

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