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Math Help - Simultaneous equations

  1. #1
    Senior Member I-Think's Avatar
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    Simultaneous equations

    Buenas noches. I have a question here that I have answered, so I hope that the forum could rate my solution, detect errors and provide alternatives. Thanks.

    Question: If x and y are distinct real numbers such that  x^3=1-y and y^3=1-x, find all possible values of xy .

    Solution:  x^3=1-y
    y^3=1-x,

    x^3-y^3=x-y
    (Dividing by (x-y) ): x^2+xy+y^2=1
    (x+y)^2-xy=1 \Rightarrow(x+y)^2-1^2=xy<br />
    (x+y+1)(x+y-1)=xy<br />
    Case 1: x+y+1=x, x+y-1=y,
    y=-1,x=1
    Case 2: x+y+1=y, x+y-1=x,
    x=-1,y=1


    Only possible value of xy is -1

    First post with latex.
    Last edited by I-Think; May 13th 2009 at 08:03 PM.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by I-Think View Post
    Buenas noches. I have a question here that I have answered, so I hope that the forum could rate my solution, detect errors and provide alternatives. Thanks.

    Question: If x and y are distinct real numbers such that  x^3=1-y and y^3=1-x, find all possible values of xy .

    Solution:  x^3=1-y
    y^3=1-x,

    x^3-y^3=x-y
    (Dividing by (x-y) ): x^2+xy+y^2=1
    (x+y)^2-xy=1 \Rightarrow(x+y)^2-1^2=xy<br />
    (x+y+1)(x+y-1)=xy<br />
    Isomorphism: Whatever follows is wrong.
    Case 1: x+y+1=x, x+y-1=y,
    y=-1,x=1
    Case 2: x+y+1=y, x+y-1=x,
    x=-1,y=1


    Only possible value of xy is -1

    First post with latex.
    One possible line of attack:

     x^2 + xy + y^2 = 1

    See it as a quadratic in x, then x = \dfrac{-y \pm \sqrt{4 - 3y^2}}{2}

    Now for a real solution to exist, 4 \ge 4 - 3y^2 \ge 0
    Thus y^2 \le \frac{4}{3}

    And then xy = \dfrac{-y^2 \pm \sqrt{4y^2 - 3y^4}}{2}.....
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  3. #3
    Senior Member I-Think's Avatar
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    Question

    I was wondering if that method was correct...
    May I ask what makes this method wrong in this case, I've seen it used in other questions. (Is it that two variables are being related to each other and not to a real number?)

    Also, does this mean there is an infinite number if values of xy when both x and y are real numbers?
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by I-Think View Post
    I was wondering if that method was correct...
    May I ask what makes this method wrong in this case, I've seen it used in other questions. (Is it that two variables are being related to each other and not to a real number?)
    Here's a counterexample: x = 0, y = 1,then (x+y+1)(x+y-1)=xy. However neither of your cases are true!!

    Also, does this mean there is an infinite number if values of xy when both x and y are real numbers?
    Yes, I think they asked for a range of values...
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