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Math Help - Advanced Math Refresher

  1. #1
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    Advanced Math Refresher

    What type of equation is this ? I have taken studied this type of subject years ago and now I need to refresh.

    Problem:
    Solve for x and y in

    xy + 8 + j(x^2y + y) = 4x + 4 + j(xy^2 + x)


    Thanks !
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  2. #2
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    Equating real and imaginary parts gives

    \displaystyle x\,y + 8 = 4x + 4 and \displaystyle x^2y+y=x\,y^2 + x.

    Rearranging equation 1 gives \displaystyle y = \frac{4x-4}{x}.

    Substituting into equation 2 gives

    \displaystyle x^2\left(\frac{4x-4}{x}\right) + \frac{4x-4}{x} = x\left(\frac{4x-4}{x}\right)^2 + x

    \displaystyle \frac{x^2(4x-4)}{x} + \frac{4x-4}{x} = \frac{x(4x-4)^2}{x^2} + \frac{x^3}{x^2}

    \displaystyle \frac{4x^3 - 4x^2 + 4x - 4}{x} = \frac{x(4x-4)^2 + x^3}{x^2}

    \displaystyle \frac{4x^4 - 4x^3 + 4x^2 - 4x}{x^2} = \frac{x(4x-4)^2 + x^3}{x^2}

    \displaystyle 4x^4 - 4x^3 + 4x^2 - 4x = x(4x-4)^2 + x^3

    \displaystyle 4x^4 - 4x^3 + 4x^2 - 4x = x(16x^2 - 32x + 16) + x^3

    \displaystyle 4x^4 - 4x^3 + 4x^2 - 4x = 16x^3 - 32x^2 + 16x + x^3

    \displaystyle 4x^4 - 21x^3 + 36x^2 - 20x = 0

    \displaystyle x(x-2)^2(4x-5)=0

    \displaystyle x = 0 or \displaystyle (x - 2)^2 =0 or \displaystyle 4x - 5 = 0

    It should be clear that you can not let \displaystyle x = 0, so that means \displaystyle x = 2 or \displaystyle x = \frac{5}{4}.


    Now substituting back into \displaystyle y = \frac{4x-4}{x} gives

    \displaystyle y = \frac{4(2) - 4}{2} or \displaystyle y = \frac{4\left(\frac{5}{4}\right)-4}{\frac{5}{4}}

    \displaystyle y = 2 or \displaystyle y = \frac{4}{5}.


    So the solutions are \displaystyle (x, y) = \left(2, 2\right) and \displaystyle (x, y) = \left(\frac{5}{4}, \frac{4}{5}\right).
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