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Math Help - system of 3 equations

  1. #1
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    system of 3 equations

    solve for x,y, and z:

    x+y = 3z
    x^2 + y^2 = 5z
    x^3 + y^3 = 9z
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  2. #2
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    Quote Originally Posted by jmedsy View Post

    x+y = 3z

    z = \frac{x}{3}+\frac{y}{3}

    so

    x^2 + y^2 = 5\left(\frac{x}{3}+\frac{y}{3}\right)

    (You might look to complete the sqaure here.)

    and

    x^3 + y^3 = 9\left(\frac{x}{3}+\frac{y}{3}\right)
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  3. #3
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    i still don't really see how its solved without some very lengthy, tedious algebra.

    what's the next step
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  4. #4
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    Quote Originally Posted by jmedsy View Post
    without some very lengthy, tedious algebra.
    that's what I also see.
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  5. #5
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    Hello, jmedsy!

    I think I've solved it . . . I'll get you started.


    Solve for x,y,z\!:\;\;\begin{array}{cccc}x+y &=& 3z & [1] \\ x^2 + y^2 &=& 5z & [2] \\ x^3 + y^3 &=& 9z & [3] \end{array}

    Note that x=y=z=0 is a solution.
    In the following work, all the variables are nonzero.


    Square [1]: . (x+y) \:=\:(3z)^2 \quad\Rightarrow\quad x^2 + 2xy + y^2 \:=\:9z^2

    . . . \underbrace{x^2+y^2}_{\text{This is }5z} + 2xy \:=\:9z^2 \quad\Rightarrow\quad 5z + 2xy \:=\:9z^2 \quad\Rightarrow\quad xy \:=\:\frac{9z^2-5z}{2} .[1]


    Cube [1]: . (x+y)^3 \:=\:(3z)^3 \quad\Rightarrow\quad x^3 + 3x^2y + 2xy^2 + y^3 \:=\:27z^3

    . . . \underbrace{x^3+y^3}_{\text{This is }9z} + 3xy\underbrace{(x+y)}_{\text{This is }3x} \:=\:27z^2 \quad\Rightarrow\quad 9z + 9xyz \:=\:27z^3 \quad\Rightarrow\quad xy \:=\:3z^2-1 .[2]


    Equate [1] and [2]: . \frac{9z^2-5z}{2} \:=\:3z^2-1 \quad\Rightarrow\quad 3z^2 - 5z + 2 \:=\:0

    Factor: . (z-1)(3z-2) \:=\:0 \quad\Rightarrow\quad z \:=\:1,\:\tfrac{2}{3}


    Your turn . . .

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  6. #6
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    edited, thanks for the help
    Last edited by jmedsy; January 27th 2010 at 07:20 PM.
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