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Can anyone help me solve this equation system ?

  1. Nov 15, 2008 #1
  2. jcsd
  3. Nov 15, 2008 #2


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    Homework Helper
    Gold Member

    You could try solving the first equation for y and substituting it int the second....that will give you an equation in terms of x only.
  4. Nov 15, 2008 #3


    Staff: Mentor

    Or you could solve for x in the 2nd equation, which is quadratic in x, by using the quadratic formula.
    [tex]yx^2 + 2y^2 x + y^3 -2 = 0[/tex]
    You will of course get two equations for x.
  5. Nov 17, 2008 #4
    You could expand (x+y)3 and then substitute the values for the two expressions that you have. That would be a lot simpler.
  6. Nov 17, 2008 #5


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    Science Advisor

    Notice that both equations only have terms of degree 3 in both variables. So one thing you could do is let r=x/y, and divide both equations by y^3. Each left-hand side will then depend only on r, and each right-hand side only on y^3. Eliminating y^3 will give you a cubic equation in r. Solve this by guessing a root, or by graphing to get a root. Once you know one root you can reduce it to a quadratic to get the other two.
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