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Thread: Nonlinear System Elimination Method

  1. #1
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    Nonlinear System Elimination Method

    Daughter preparing for test.

    $\displaystyle 4x^2$+$\displaystyle 3y^2$=48
    $\displaystyle 3x^2$+$\displaystyle 2y^2$=35

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

    $\displaystyle 4x^2$+$\displaystyle 3(2)^2$=48

    $\displaystyle 4x^2$=36
    $\displaystyle x^2$=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.

    $\displaystyle 3x^2$-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 $\displaystyle x^2$ 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; Nov 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

    $\displaystyle 9x^2+6y^2=105$

    $\displaystyle 8x^2+6y^2=96$

    subtract to elimate the common term.

    $\displaystyle x^2 = 9$

    so

    $\displaystyle x=\pm 3$

    now plug 9 into the equations to solve for y

    $\displaystyle 8(9) + 6y^2 = 96$

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

    $\displaystyle y=\pm 2$
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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.

    $\displaystyle y=-2-2x$

    $\displaystyle 3x^2-(-2-2x)=11$

    so solve the quadratic

    $\displaystyle 3x^2+2x-9=0$

    use the form $\displaystyle ax^2+bx+c=p(x+q)^2+r$ 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?

    $\displaystyle x-23=5y$

    so

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

    so solve

    $\displaystyle x^2-23x+120=0$

    dont forget to solve for the $\displaystyle y$ 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; Nov 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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