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Math Help - Solving for x and y

  1. #1
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    Solving for x and y

    solve [(x^2)/25 + (y^2)/4 -25=0, (x^2)/16 + (y^2)/9 -29=0, x, y] - Wolfram|Alpha

    I am not certain what the method is. I have made and equation y^2=... from the first equation and put it into the other but I did not get +or-20.
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  2. #2
    Super Member Quacky's Avatar
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    Re: Solving for x and y

    \frac{x^2}{25}+\frac{y^2}{4}=25

    Multiply through by 4:

    \frac{4x^2}{25}+y^2=100

    y^2=100-\frac{4x^2}{25}

    What did you get when you substituted this in and tried to solve?
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  3. #3
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    Re: Solving for x and y

    X^2=1299600/2601 X=+/- 22.35 x is +/-20.
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  4. #4
    Super Member Quacky's Avatar
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    Re: Solving for x and y

    I'm fairly sure that might be overcomplicating things.

    y^2=100-\frac{4x^2}{25}

    Substituting this into \frac{x^2}{16}+\frac{y^2}{9}-29=0 gives:

    \frac{x^2}{16}+\frac{100-\frac{4x^2}{25}}{9}-29=0

    Multiplying by 9:

    \frac{9}{16}x^2+100-\frac{4}{25}x^2-261=0

    Which leads to:

    \frac{161}{400}x^2=161

    Does that clarify?
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  5. #5
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    Re: Solving for x and y

    I have typed the hyperbola correctly by accident. The book wrote 21 not 29 and I found that error. Also it had a negative sign instead of positive.
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  6. #6
    Super Member Quacky's Avatar
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    Re: Solving for x and y

    The joy of serendipity.
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  7. #7
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    Re: Solving for x and y

    Another way:

    x^2/25 + y^2/4 = 25 [1]
    x^2/16 + y^2/9 = 29 [2]

    [1] *100: 4x^2 + 25y^2 = 2500 [1]
    [2] * 144: 9x^2 + 16y^2 = 4176 [1]

    36x^2 + 225y^2 = 22500 : [1]*9
    -36x^2 - 64y^2 = -16704 : [2]*-9

    Add above: 161y^2 = 5796
    Now solve for y, then substitute to solve for x
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