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Algebraic Manipulation

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    As part of a calculus question, the solutions manual takes [tex]\frac{2y^{2}}{4+y^{2}}[/tex]
    And somehow turns it into [tex]\left(2-\frac{8}{4+y^{2}}\right)[/tex]

    Ive scribbled all the things i can thinkof on paper and still cant seem to get from one to the other, its driving me nuts!

    Any help would be much appreciated :D.

    Cheers.
     
  2. jcsd
  3. Apr 11, 2009 #2

    cristo

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    They appear to add 8 and subtract 8 from the numerator. That is,

    [tex]\frac{2y^2}{4+y^2}=\frac{2y^2+8-8}{4+y^2}=\frac{2(4+y^2)-8}{4+y^2}=2-\frac{8}{4+y^2}[/tex]
     
  4. Apr 11, 2009 #3
    Ahhh, i feel like such a fool for not seeing that.

    Thanks ;)
     
  5. Apr 11, 2009 #4

    HallsofIvy

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    Another way to see that is simply "long division". [itex]y^2+ 0y+ 4[/itex] divides into [itex]2y^2+ 0y+ 0[/itex] 2 times with a remainder of [itex]2y^2- 2(y^2+ 4)= 2y^2-2y^2- 8= -8[/tex] so
    [tex]\frac{2y^2}{y^2+ 4}= 2- \frac{8}{y^2+ 4}[/tex]
     
  6. Apr 12, 2009 #5
    Thanks ;). Very clear to me now.
     
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