Subtract 5/(x^2-xy) - 5/(y^2-xy). Could somebody explain to me how to do this?
You must have the same denominators so :\[= [5(y-x) - 5(x-y)]/x(x-y)(y-x)\]\[= (5y-5x-5x+5y)/x(x-y)(y-x) \]\[= (10y-10x)/x(x-y)(y-x)\]\[= 10(y-x)/x(x-y)(y-x)\]\[= 10/x(x-y)\] ^_^
what I really need is how to find the LCD? I'm kind of confused
first factor, then you have to multiply both sides to get the same denominator
isn't the denominator xy(x-y)(x-y)?
or am i wrong?
if you notice , when you factor the denominator on the first side you'll get : x(x-y) and when you factor the denominator of the other side you get : x(y-x) we have x in common, but one has (x-y) and the other has (y-x) so we multiply the first one with (y-x) and the second one with (x-y) so both can have : x(x-y)(y-x) the same denominator ^_^ clearer now? :)
no laka, youre right...
no laka, it's not ^_^ you'll get x^3
starica, the second denominator, factored, gives you y(y-x) not x(y-x)
yeah i realised
you sure?
oh right! ^^" sorry for the mess again
xy(y-x)(x-y)?
so you'll have to multiply x and y with the others too
yes laka, you are right dear, thank you ^_^
answer should be something like 5(xx-yy)/xy(x-y)(y-x) or 5(xx-yy)/-xy(x-y)^2 aka -5(xx-yy)/(xy(x-y)^2
I'll have to redo it again ._.
im curious starica, how are you on here like all day? O.o
lol I wasn't here all day ^_^ I come in mornings and nights only
outcast girl the answer is :\[= 5y(y-x) - 5x(x-y)/xy(y-x)(x-y)\]\[= 5y^2 -5xy -5x^2 +5xy/xy(x-y)(y-x)\]\[= 5(y^2-x^2)/xy(x-y)(y-x)\]\[= 5(y-x)(y+x)/xy(x-y)(y-x)\]\[= 5(y+x)/xy(x-y)\] for the mess up, I'm sure this time ^_^
sorry for the mess*
that's correct :)
how do you get the pretty notation?
yay , lol ^_^
click on equation and write it down ^_^
ohhhhhhhh, makes sense
something called equation on youer reply box :)
thanks! I think I got it now :)
your*
most welcome ^_^!
Just a suggestion mind you. the second term that is factored as:\[-5/y(y-x)\] can be simplified by changing it to the following:\[+(5/(x-y)\] 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(y-x) to y(x+y) I think thats legal. Don't know if that will help or not
Is that legal?
Oooops Change the last line on the post 2nd above to read y(x-y)
hmm, it's true, it can be, but I'm not sure though, wouldn't that change the function completely?
I wondering too, but multiplying by -1 one twice is the same as multiplying by a +1 and theoretically shouldn't change it.
you have just answered the question in a much simpler way! thank you ^_^!
yes thank you!
:)
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