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anonymous
 5 years ago
Subtract 5/(x^2xy)  5/(y^2xy). Could somebody explain to me how to do this?
anonymous
 5 years ago
Subtract 5/(x^2xy)  5/(y^2xy). Could somebody explain to me how to do this?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You must have the same denominators so :\[= [5(yx)  5(xy)]/x(xy)(yx)\]\[= (5y5x5x+5y)/x(xy)(yx) \]\[= (10y10x)/x(xy)(yx)\]\[= 10(yx)/x(xy)(yx)\]\[= 10/x(xy)\] ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what I really need is how to find the LCD? I'm kind of confused

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first factor, then you have to multiply both sides to get the same denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't the denominator xy(xy)(xy)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you notice , when you factor the denominator on the first side you'll get : x(xy) and when you factor the denominator of the other side you get : x(yx) we have x in common, but one has (xy) and the other has (yx) so we multiply the first one with (yx) and the second one with (xy) so both can have : x(xy)(yx) the same denominator ^_^ clearer now? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no laka, youre right...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no laka, it's not ^_^ you'll get x^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0starica, the second denominator, factored, gives you y(yx) not x(yx)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh right! ^^" sorry for the mess again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you'll have to multiply x and y with the others too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes laka, you are right dear, thank you ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0answer should be something like 5(xxyy)/xy(xy)(yx) or 5(xxyy)/xy(xy)^2 aka 5(xxyy)/(xy(xy)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll have to redo it again ._.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im curious starica, how are you on here like all day? O.o

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol I wasn't here all day ^_^ I come in mornings and nights only

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0outcast girl the answer is :\[= 5y(yx)  5x(xy)/xy(yx)(xy)\]\[= 5y^2 5xy 5x^2 +5xy/xy(xy)(yx)\]\[= 5(y^2x^2)/xy(xy)(yx)\]\[= 5(yx)(y+x)/xy(xy)(yx)\]\[= 5(y+x)/xy(xy)\] for the mess up, I'm sure this time ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you get the pretty notation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0click on equation and write it down ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhhhhhh, makes sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0something called equation on youer reply box :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks! I think I got it now :)

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Just a suggestion mind you. the second term that is factored as:\[5/y(yx)\] can be simplified by changing it to the following:\[+(5/(xy)\] This can be done by multiply by 1 twice, once changing the  to a plux making it and addition and then again changeing the y(yx) to y(x+y) I think thats legal. Don't know if that will help or not

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Oooops Change the last line on the post 2nd above to read y(xy)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, it's true, it can be, but I'm not sure though, wouldn't that change the function completely?

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I wondering too, but multiplying by 1 one twice is the same as multiplying by a +1 and theoretically shouldn't change it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have just answered the question in a much simpler way! thank you ^_^!
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