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Solving A Non Linear System With a Cube

  1. Apr 24, 2012 #1

    B18

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    1. The problem statement, all variables and given/known data
    Find all solutions for the following system:
    X^2-xy+2y^2=8
    X^3-xy^2=0


    2. Questions
    What steps could I take to solve this? I've tried just about everything I can think of. I don't
    know if I should solve the cubic equation for a variable and plug it in or what. I am petty stumped here guys.

    3. The attempt at a solution
    I took an x out of the x^3-xy^2 making it x(x^2-y^2)=0 and that is basically the only logical thing I have thought of. Thanks for any of your help!
     
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 24, 2012 #2
    There are a couple of ways to go about this. You're on the right track with the second equation, just factor x^2 - y^2 and you'll be able to find three solutions to try in the first equation. Plug them in to the first equation and solve for y. All in all you should get 4 solutions for y and 5 for x, making the total number of (x,y) pairs = 5*4 = 20
     
  4. Apr 24, 2012 #3

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    Ok so when I factor x^3-xy^2=0 I will get x=0 and x^2-y^2=0 which then results in x=±y. How else could I get answers for that. Does x=±y count as two answers? So my three answers from that second equation would be→x=0, x=y, x=-y.
     
  5. Apr 24, 2012 #4

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    I came up with (0,2), (0,-2), (-y,2), (-y,2), (y,2), (y,-2). And when I put y= 2 in I end up getting the same first two answers.
     
  6. Apr 24, 2012 #5

    Ray Vickson

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    Yes, of course y = ± x is two solutions; after all, they say two different things.

    RGV
     
  7. Apr 24, 2012 #6

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    I agree, thanks for your help guys. I am still struggling a bit but I will try some more.
     
  8. Apr 24, 2012 #7

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    Does anyone else have another way about this I am not getting as many answers as I expect.
     
  9. Apr 25, 2012 #8
    Well you've already found three solutions from the second equation(x = 0, x=-y and x=y), what happens when you plug these into the first equation and solve?
     
  10. Apr 25, 2012 #9

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    I end up with (0,2) (0,-2) and for the x= y I get some unfactorable stuff. I got (-y, root 2) and (-y, - root 2) I don't think that is correct tho but that is what I get hen substituting in -y for x in the first equation.
     
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