• Dec 18th 2010, 12:20 AM
dugongster
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

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

Thanks !
• Dec 18th 2010, 12:44 AM
Prove It
Equating real and imaginary parts gives

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

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

Substituting into equation 2 gives

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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