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

- anonymous

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

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- chestercat

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- anonymous

You must have the same denominators so :\[= [5(y-x) - 5(x-y)]/x(x-y)(y-x)\]\[= (5y-5x-5x+5y)/x(x-y)(y-x) \]\[= (10y-10x)/x(x-y)(y-x)\]\[= 10(y-x)/x(x-y)(y-x)\]\[= 10/x(x-y)\]
^_^

- anonymous

what I really need is how to find the LCD? I'm kind of confused

- anonymous

first factor, then you have to multiply both sides to get the same denominator

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## More answers

- anonymous

isn't the denominator xy(x-y)(x-y)?

- anonymous

or am i wrong?

- anonymous

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? :)

- anonymous

no laka, youre right...

- anonymous

no laka, it's not ^_^ you'll get x^3

- anonymous

starica, the second denominator, factored, gives you y(y-x) not x(y-x)

- anonymous

yeah i realised

- anonymous

you sure?

- anonymous

oh right! ^^" sorry for the mess again

- anonymous

xy(y-x)(x-y)?

- anonymous

so you'll have to multiply x and y with the others too

- anonymous

yes laka, you are right dear, thank you ^_^

- anonymous

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

- anonymous

I'll have to redo it again ._.

- anonymous

im curious starica, how are you on here like all day? O.o

- anonymous

lol I wasn't here all day ^_^ I come in mornings and nights only

- anonymous

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 ^_^

- anonymous

sorry for the mess*

- anonymous

that's correct :)

- anonymous

how do you get the pretty notation?

- anonymous

yay , lol ^_^

- anonymous

click on equation and write it down ^_^

- anonymous

ohhhhhhhh, makes sense

- anonymous

something called equation on youer reply box :)

- anonymous

thanks! I think I got it now :)

- anonymous

your*

- anonymous

most welcome ^_^!

- radar

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

- radar

Is that legal?

- radar

Oooops Change the last line on the post 2nd above to read y(x-y)

- anonymous

hmm, it's true, it can be, but I'm not sure though, wouldn't that change the function completely?

- radar

I wondering too, but multiplying by -1 one twice is the same as multiplying by a +1 and theoretically shouldn't change it.

- anonymous

you have just answered the question in a much simpler way! thank you ^_^!

- anonymous

yes thank you!

- anonymous

:)

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