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Math Help - two equations

  1. #1
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    two equations

    hi
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  2. #2
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    Re: two equations

    You have a circle and an ellipse there, and need to find their points of intersection. Fundamentally it is no different than other systems of two equations with two unknowns.
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  3. #3
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    Re: two equations

    x^2 + y^2 = 12 - x - y \quad So, \quad 12 - x - y - xy = 9 \Rightarrow x + y + xy = 3 \Rightarrow x + y = 3 - xy \Rightarrow (x + y)^2 = (3 - xy)^2 \Rightarrow x^2 + 2xy + y^2 = 9 - 6xy + (xy)^2 \Rightarrow x^2 + y^2 + 8xy = 9 + (xy)^2

    You continue. At the end you will find: x = 0, y = 0, xy = 9. When x = 0 => y = +-3, when y = 0 => x = +-3, when xy = 9 => x^2 + y^2 = 18 \quad and \quad x = \frac{9}{y} ....
    Last edited by ManosG; May 5th 2013 at 02:26 PM.
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  4. #4
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    Re: two equations

    here is my trial
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  5. #5
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    Re: two equations

    Quote Originally Posted by abualabed View Post
    here is my trial
    You can't really get anywhere from there. The usual case for that would be that the RHS is zero. Then we could say that since ab = 0, we have that a = 0 or b = 0. But you don't have that.

    ManosG's method seems to be a good one.

    -Dan
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  6. #6
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    Re: two equations

    Hello
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  7. #7
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    Re: two equations

    Hello, abualabed!

    \text{If }x,y\text{ are integers, solve this system:}

    . . x^2 + y^2 + x + y \:=\:12\;\;[1]

    . . . . x^2 + y^2 - xy \:=\:9\;\;[2]

    Subtract [1] - [2]: . x + y + xy \:=\:3

    Solve for y\!:\;\;y \:=\:\frac{3-x}{1+x}

    Substitute into [2]: . x^2 + \left(\frac{3-x}{1+x}\right)^2 - x\left(\frac{3-x}{1+x}\right) \:=\:9

    . . which simplifies to: . x^4 + 3x^3 - 9x^2 - 27x \:=\:0

    . . and which factors: . x(x-3)(x+3)^2 \:=\:0

    Hence: . \begin{Bmatrix}x &=& 0,3,\text{-}3 \\ y &=& 3,0,\text{-}3 \end{Bmatrix}


    Solutions: . (x,y) \;=\;(3,0),\;(0,3),\;(\text{-}3,\text{-}3)
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  8. #8
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    Re: two equations

    I don't suffer from MHF addiction... I enjoy every minute of it!


    and

    Proof is what sets mathematics apart from every other science.



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