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System of equations problem

  1. Apr 7, 2008 #1
    [SOLVED] System of equations problem

    I'm trying to brush up on my algebra and had difficulties with a problem.

    Solve for (x,y): (x^-(2/3))(y^(2/3)) = (2x^(1/3))(y^-(1/3)) and 6 = x + 2y

    I'm mainly having problems simplifying the first equation for y but got y=0. Any help?
  2. jcsd
  3. Apr 7, 2008 #2
    [tex]x^{-a} = \frac{1}{x^a}[/tex]

    [tex]\frac{x^a}{x^b} = x^{a-b}[/tex]

    Start with that and let's see where it takes you.
  4. Apr 7, 2008 #3
    solving the second equation is easy enough - just make y the subject.

    As for the first, I find it's easier to rewrite it without using negative powers.
    [tex]x^{-2/3} = \frac{1}{x^{2/3}}[/tex]

    From there, it's just a matter of rearranging to make y the subject again.
    From there, you can find x using both equations.
  5. Apr 7, 2008 #4
    Perfect, I think it was the basic concepts that helped. Just to verify:

    I was able to simplify the first equation to y = 2x. Using that...
    y = 2(6-2y)
    y = 12 - 4y
    5y= 12

    Plug in to x = 6-2y...
    x = 6 - 2(12/5)
    x = 1.2

    So, (1.2, 2.4)

    That seems right. Thank you.
  6. Apr 7, 2008 #5
    Plug in those values back into the original question to see if the left hand side equals the right hand side to make sure you're right.
  7. Apr 7, 2008 #6

    2.4 = 2 (1.2)
    2.4 = 2.4 CHECK

    1.2 = 6 - 2(2.4)
    1.2 = 1.2 CHECK

    Everything is verified. Thanks.
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