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Mathematics 14 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

or am i wrong?

OpenStudy (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? :)

OpenStudy (anonymous):

no laka, youre right...

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah i realised

OpenStudy (anonymous):

you sure?

OpenStudy (anonymous):

oh right! ^^" sorry for the mess again

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

I'll have to redo it again ._.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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 ^_^

OpenStudy (anonymous):

sorry for the mess*

OpenStudy (anonymous):

that's correct :)

OpenStudy (anonymous):

how do you get the pretty notation?

OpenStudy (anonymous):

yay , lol ^_^

OpenStudy (anonymous):

click on equation and write it down ^_^

OpenStudy (anonymous):

ohhhhhhhh, makes sense

OpenStudy (anonymous):

something called equation on youer reply box :)

OpenStudy (anonymous):

thanks! I think I got it now :)

OpenStudy (anonymous):

your*

OpenStudy (anonymous):

most welcome ^_^!

OpenStudy (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

OpenStudy (radar):

Is that legal?

OpenStudy (radar):

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

OpenStudy (anonymous):

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

OpenStudy (radar):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes thank you!

OpenStudy (anonymous):

:)

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