Solve for all real values of $x$ and $y$

Question

Solve the system of equations for all real values of $x$ and $y$ $$5x(1 + {\frac {1}{x^2 +y^2}})=12$$ $$5y(1 - {\frac {1}{x^2 +y^2}})=4$$

I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5}}\le y \le {\frac 45}$}

I don't know what to do next.

Answer 1

Hint: Notice $x,y \neq 0 $

$$ \frac{12}{5x} + \frac{4}{5y} = 2 $$

Answer 2

for $$x,y\ne 0$$ we obtain $$1+\frac{1}{x^2+y^2}=\frac{12}{5x}$$ $$1-\frac{1}{x^2+y^2}=\frac{4}{5y}$$ adding both we get $$5=\frac{6}{x}+\frac{2}{y}$$ from here we obtain $$y=\frac{2x}{5x-6}$$ can you proceed? after substitution and factorizing we get this equation: $$- \left( x-2 \right) \left( 5\,x-2 \right) \left( 5\,{x}^{2}-12\,x+9 \right)=0 $$

Answer 3

Let $z=x + iy \in \mathbb{C}$, now for $|z|^2 = x^2+y^2 \not = 0$ have that:

$$\left(5x + \frac{5x}{x^2+y^2}\right) + i\left(5y - \frac{5y}{x^2+y^2}\right) = 12 + 4i$$

$$5(x+iy) + \frac{5(x-iy)}{x^2+y^2} = 12 + 4i$$

$$5(x+iy) + \frac{5}{x+iy} = 12 + 4i$$

$$5z + \frac{5}{z} = 12 + 4i$$

$$5z^2 - (12+4i)z + 5 = 0$$

Solving this complex equation we get $z_1=\frac 25 - \frac 15i$ and $z_2 =2+i$, corresponding to the solution $(x,y) = \left\{\left(\frac 25, - \frac 15 \right), (2,1)\right\}$