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Homework Help: A bit of algebra

  1. Mar 13, 2005 #1
    I'm not sure how to do this:

    [3x-1/3 + x5/3y] / x2/3

    this is what i did:

    [3 + x5/3y ]/ x1/3x2/3

    [3+ x5/3y]/x

    the answer to the problem is the same as what I have except instead of x^5/3, its x^2.

    help.
     
  2. jcsd
  3. Mar 13, 2005 #2

    dextercioby

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    Of course it is.Put it like that:

    [tex] \frac{\frac{3}{x^{\frac{1}{3}}}+x^{\frac{5}{3}}y}{x^{\frac{2}{3}}} [/tex]

    Do you see where that x^{2} is coming from...?

    Daniel.
     
  4. Mar 13, 2005 #3

    arildno

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    Your 1.step, that is, extracting [tex]x^{-\frac{1}{3}}[/tex] from the parenthesis has been done incorrectly.
    Try to figure out why.
     
  5. Mar 13, 2005 #4
    u can take x^2/3 in factor for the numerator, u will have:

    [x^2/3(3x^-1 +xy)]/x^2/3

    then u simplify and obtain

    3/x + xy

    which leads to (3 + yx^2)/x
     
  6. Mar 13, 2005 #5
    no.
    I don't see it.

    how are you factoring a x^2/3 when the smallest power is -1/3 ? if you factor that out, x^5/3 should be x^1/3.


    help.
     
  7. Mar 13, 2005 #6
    [tex]\frac{\frac{3}{x^{\frac{1}{3}}} + x^{\frac{5}{3}}y}{x^{\frac{2}{3}}} = { \left( \frac{3}{x^{\frac{1}{3}}} + x^{\frac{5}{3}}y \right) \over x^{\frac{2}{3}}}\cdot \frac{x^{\frac{1}{3}}}{x^{\frac{1}{3}}} = \frac{3 + x^2y}{x}[/tex]

    I only posted this because I think you're getting more confused by the replies so far. Try to take a lesson from the method I used to simplify (that is, multiplying by 1 is a good idea sometimes!).
     
    Last edited: Mar 13, 2005
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