algebra 2

Find the points where these quadratic relations intersect. 16x^2-5y^2=64, 16x^2+25y^2-96x=256

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  1. just substitute:

    16x^2 - 96x + 5(16x^2-64) = 256
    16x^2 - 96x + 80x^2 - 320 = 256
    96x^2 - 96x - 576 = 0
    x^2 - x - 6 = 0
    (x+2)(x-3) = 0

    Now you have x, get y.

    see the solution at

    http://www.wolframalpha.com/input/?i=solve+16x^2-5y^2%3D64%2C+16x^2%2B25y^2-96x%3D256

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    posted by Steve
  2. subtract them :
    30y^2 - 96x = 192
    5y^2 = 32 + 16x

    back into the first:
    16x^2 - (32+16x) = 64
    16x^2 -16x - 96 = 0
    x^2 - x - 6 = 0
    (x-3)(x+2) = 0
    x = 3 or x = -2

    if x = 3,
    5y^2 = 32+48 = 80
    y^2 = 16
    y = ± 4

    if x = -2
    5y^2 = 32-32 = 0
    y = 0

    so we have 3 points:
    (-2,0) , (3,4) and (3,-4)

    verification:
    http://www.wolframalpha.com/input/?i=solve++16x%5E2-5y%5E2%3D64%2C+16x%5E2%2B25y%5E2-96x%3D256

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    posted by Reiny

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