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Math Help - nonlinear system of equations problem

  1. #1
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    nonlinear system of equations problem

    I can't figure out where I'm going wrong in solving this system of equations.

    x2+2xy-y2=14
    x2-y2= -16

    Elimination

    x2+2xy-y2=14
    -x2-0xy+y2= 16

    2xy=30

    Substitution

    x2+30-y2=14

    x2-y2+30=14

    -16+30=14
    14=14

    I keep getting results like this any way I try to substitute or rearrange the equations.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: nonlinear system of equations problem

    Whenever you get an expression like 14=14 that means yuo have essentially made the same substitution twice, so you no longer have two independent equations.

    Try this: note that x^2 + 2xy + y^2 = (x+y)^2, and that x^2-y^2 is the difference of two squares. So the two equations become:

     (x-y)^2 = 14
     (x+y)(x-y) = -16

    Can you take it from here?
    Thanks from Shakarri
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  3. #3
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    Re: nonlinear system of equations problem

    The first equation was x^2+2xy-y^2=14 (not x^2 + 2xy + y^2 = 16)

    I don't think x^2+2xy-y^2 can be factored any further. Can it?
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: nonlinear system of equations problem

    Quote Originally Posted by cdbowman42 View Post
    The first equation was x^2+2xy-y^2=14 (not x^2 + 2xy + y^2 = 16)

    I don't think x^2+2xy-y^2 can be factored any further. Can it?
    You're right - my mistake. I'm afraid I misread it.

    OK, try this then: sub the second equation into the first:

     x^2 + 2xy - y^2 = 14 \to (x^2-y^2) + 2xy = 14 \to -16 + 2xy = 14 \to xy = 15


    So we have xy=15 and y = \frac {15} x. Sub these values back into the first equation:

     x^2 + 30 - \frac {15^2}{x^2} = 14

     x^4 + 30 x^2 - 225 = 14 x^2

     x^4 + 16x^2 - 225 = 0

    Let w = x^2, to give a quadratic equation that you can factor:

     w^2 + 16w -225 = (w-9)(w+25) = 0

    Hence w = 9 or -25, and x = +/- 3 and +/- 5i. The corresponding values for y are +/- 5 and -/+ 3i.
    Last edited by ebaines; May 15th 2013 at 11:03 AM.
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  5. #5
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    Re: nonlinear system of equations problem

    Hello, cdbowman42!

    \begin{array}{ccc}x^2 + 2xy - y^2 \:=\:14 & [1] \\ x^2-y^2 \:=\:\text{-}16 & [2] \end{array}

    Subtract [1] - [2]: . 2xy \:=\:30 \quad\Rightarrow\quad y \:=\:\tfrac{15}{x}\;\;[3]

    Substitute into [2]: . x^2 - \left(\tfrac{15}{x}\right)^2 \:=\:\text{-}16 \quad\Rightarrow\quad x^2 - \tfrac{225}{x^2} \:=\:\text{-}16

    Multiply by x^2\!:\;\;x^4 - 225 \:=\:\text{-}16x^2 \quad\Rightarrow\quad x^4 + 16x^2 - 225 \:=\:0

    Factor: . (x^2+25)(x^2-9) \:=\:0


    We have: . \begin{Bmatrix}x^2 + 25 \:=\:0 & \Rightarrow& x^2 \:=\:\text{-}25 &\Rightarrow& \text{no real roots} \\ x^2 - 9 \:=\:0 & \Rightarrow & x^2 \:=\:9 & \Rightarrow& x \:=\:\pm3 \end{Bmatrix}

    Substitute into [3]: . y \:=\:\tfrac{15}{\pm3} \:=\:\pm5


    Therefore: . (x,y) \;=\;(3,5),\;(\text{-}3,\text{-}5)
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