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Math Help - Nonlinear System Elimination Method

  1. #1
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    Nonlinear System Elimination Method

    Daughter preparing for test.

    4x^2+ 3y^2=48
    3x^2+ 2y^2=35

    ----------------------------------------
    -12x^2+ -9y^2=-144
    12x^2+ +8y^2=140
    -----------------------------------------
    -y^2=-4
    y^2=4
    y=+-2

    4x^2+ 3(2)^2=48

    4x^2=36
    x^2=9
    x=+-3


    (I think I answered the first part of the question while typing it.)

    I know I will have a question in a moment
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  2. #2
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    Ok, now I am stuck.

    3x^2-y=11
    2x+y=-2

    Dang this has been more than 20 years ago for me.

    how do I proceed with this one. The second part does not have an  x^2 in it.
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  3. #3
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    Solution

    Solve for y in the second equation and substitute into the first equation and the solution will shake out.
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  4. #4
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    Followup thought

    I hope you remember how to solve a quadratic equation.
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  5. #5
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    2x+y=-2

    2x+2=-y

    Ok...I am dumb. I just follow the pattern in her study guide. How do I solve for Y?


    Even if I get to where I know what is going on then I have to explain it to her. She works as a nurses aid full time and wonders what she will do with this.
    Last edited by jonb; November 29th 2009 at 12:51 PM.
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  6. #6
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    quadratic equation

    I will review that...but not sure how to plug that into this.
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  7. #7
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    Hi there,

    i thought you were supposed to solve by elimination?

    you can elimate either x or y by using an appropriate factor.

    multiplying equation 2 by 3 and equation 1 by 2 gives you

    9x^2+6y^2=105

    8x^2+6y^2=96

    subtract to elimate the common term.

    x^2 = 9

    so

    x=\pm 3

    now plug 9 into the equations to solve for y

    8(9) + 6y^2 = 96

    y^2 = \frac{96-72}{6} = 4

    y=\pm 2
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  8. #8
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    Ah, same answers....different method


    This one is still giving me fits.

    -y=11
    2x+y=-2
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  9. #9
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    hi,

    you can use substituion on that one.

    y=-2-2x

    3x^2-(-2-2x)=11

    so solve the quadratic

    3x^2+2x-9=0

    use the form ax^2+bx+c=p(x+q)^2+r or the formula to solve as discriminant is > 0 but not a perfect square. ie solution has two real roots but cannot be factored to integers.
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  10. #10
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    Ok...I can see it when someone else does it. Thanks

    what if there is nothing that is squared?

    X-5y=23
    XY=-24
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  11. #11
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    hi,

    are you sure its not squared?

    x-23=5y

    so

    x\frac{(x-23)}{5}=-24

    so solve

    x^2-23x+120=0

    dont forget to solve for the y values in this and previous case. The discriminant on this one is > 0 and is a perfect square so will factor to integers.
    Last edited by sammy28; November 29th 2009 at 02:23 PM. Reason: typo + observation
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  12. #12
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    Thanks for the help.

    I think I was able to do the next 6 on my own. I think
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