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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    You must have the same denominators so :\[= [5(y-x) - 5(x-y)]/x(x-y)(y-x)\]\[= (5y-5x-5x+5y)/x(x-y)(y-x) \]\[= (10y-10x)/x(x-y)(y-x)\]\[= 10(y-x)/x(x-y)(y-x)\]\[= 10/x(x-y)\] ^_^

  2. anonymous
    • 5 years ago
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    what I really need is how to find the LCD? I'm kind of confused

  3. anonymous
    • 5 years ago
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    first factor, then you have to multiply both sides to get the same denominator

  4. anonymous
    • 5 years ago
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    isn't the denominator xy(x-y)(x-y)?

  5. anonymous
    • 5 years ago
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    or am i wrong?

  6. anonymous
    • 5 years ago
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    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? :)

  7. anonymous
    • 5 years ago
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    no laka, youre right...

  8. anonymous
    • 5 years ago
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    no laka, it's not ^_^ you'll get x^3

  9. anonymous
    • 5 years ago
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    starica, the second denominator, factored, gives you y(y-x) not x(y-x)

  10. anonymous
    • 5 years ago
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    yeah i realised

  11. anonymous
    • 5 years ago
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    you sure?

  12. anonymous
    • 5 years ago
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    oh right! ^^" sorry for the mess again

  13. anonymous
    • 5 years ago
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    xy(y-x)(x-y)?

  14. anonymous
    • 5 years ago
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    so you'll have to multiply x and y with the others too

  15. anonymous
    • 5 years ago
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    yes laka, you are right dear, thank you ^_^

  16. anonymous
    • 5 years ago
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    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

  17. anonymous
    • 5 years ago
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    I'll have to redo it again ._.

  18. anonymous
    • 5 years ago
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    im curious starica, how are you on here like all day? O.o

  19. anonymous
    • 5 years ago
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    lol I wasn't here all day ^_^ I come in mornings and nights only

  20. anonymous
    • 5 years ago
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    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 ^_^

  21. anonymous
    • 5 years ago
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    sorry for the mess*

  22. anonymous
    • 5 years ago
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    that's correct :)

  23. anonymous
    • 5 years ago
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    how do you get the pretty notation?

  24. anonymous
    • 5 years ago
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    yay , lol ^_^

  25. anonymous
    • 5 years ago
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    click on equation and write it down ^_^

  26. anonymous
    • 5 years ago
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    ohhhhhhhh, makes sense

  27. anonymous
    • 5 years ago
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    something called equation on youer reply box :)

  28. anonymous
    • 5 years ago
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    thanks! I think I got it now :)

  29. anonymous
    • 5 years ago
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    your*

  30. anonymous
    • 5 years ago
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    most welcome ^_^!

  31. radar
    • 5 years ago
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    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

  32. radar
    • 5 years ago
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    Is that legal?

  33. radar
    • 5 years ago
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    Oooops Change the last line on the post 2nd above to read y(x-y)

  34. anonymous
    • 5 years ago
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    hmm, it's true, it can be, but I'm not sure though, wouldn't that change the function completely?

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

  36. anonymous
    • 5 years ago
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    you have just answered the question in a much simpler way! thank you ^_^!

  37. anonymous
    • 5 years ago
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    yes thank you!

  38. anonymous
    • 5 years ago
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    :)

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