At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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
oh right! ^^" sorry for the mess again
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 :)
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!