question (doubt)....

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solve to find x and y:-

  • question (doubt).... -

    try grouping

    (x+2y)(x^2 - 2xu + 4y^2) + (x+2y) = 0
    (x+2y)(x^2 - 2xy + 4y^2 + 1) = 0

    x = -2y or x^2 -2xy + 4y^2 + 1 = 0

    treat the second like a quadratic with


    x = (2y ± √(4y^2 - 4(1)(4y^2+1))/2
    = (2y ± √(-12y^2 - 4)/2
    notice the inside of the √ will always be negative so there is no real solution to that second part

    so x = -2y

    from that you can find an infinite number of solutions,
    e.g. (-2,1), (4,-2), (-5.2 , 2.6) ...

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