simultaneous equation help

x^2-2xy=3...1
3x^2-y^2=26...2

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asked by oscar
  1. From #1, y = (x^2-3)/(2x)
    so, in #2,

    3x^2 - ((x^2-3)/(2x))^2 = 26

    Or, assuming x≠0,

    (11x^4-98x^2-9)/(4x^2) = 0
    (x-3)(x+3)(11x^2+1) = 0
    x = -3 or 3 or complex values.

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    posted by Steve

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