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Thread: 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 $\displaystyle x$ and $\displaystyle y$ are distinct real numbers such that$\displaystyle x^3=1-y$ and $\displaystyle y^3=1-x,$ find all possible values of $\displaystyle xy$ .

    Solution: $\displaystyle x^3=1-y$
    $\displaystyle y^3=1-x,$

    $\displaystyle x^3-y^3=x-y$
    (Dividing by $\displaystyle (x-y)$ ): $\displaystyle x^2+xy+y^2=1$
    $\displaystyle (x+y)^2-xy=1 \Rightarrow(x+y)^2-1^2=xy
    $
    $\displaystyle (x+y+1)(x+y-1)=xy
    $
    Case 1: $\displaystyle x+y+1=x$, $\displaystyle x+y-1=y$,
    $\displaystyle y=-1,x=1$
    Case 2: $\displaystyle x+y+1=y$, $\displaystyle x+y-1=x$,
    $\displaystyle x=-1,y=1$


    Only possible value of $\displaystyle xy$ is $\displaystyle -1$

    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 $\displaystyle x$ and $\displaystyle y$ are distinct real numbers such that$\displaystyle x^3=1-y$ and $\displaystyle y^3=1-x,$ find all possible values of $\displaystyle xy$ .

    Solution: $\displaystyle x^3=1-y$
    $\displaystyle y^3=1-x,$

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


    Only possible value of $\displaystyle xy$ is $\displaystyle -1$

    First post with latex.
    One possible line of attack:

    $\displaystyle x^2 + xy + y^2 = 1$

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

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

    And then $\displaystyle xy = \dfrac{-y^2 \pm \sqrt{4y^2 - 3y^4}}{2}$.....
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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 $\displaystyle xy$ when both $\displaystyle x $and $\displaystyle 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: $\displaystyle x = 0, y = 1$,then $\displaystyle (x+y+1)(x+y-1)=xy$. However neither of your cases are true!!

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