System of equations in x and y

Question

Solve for $x, y \in \mathbb{R} $

$$ 5x \left(1+\frac{1}{x^2+y^2}\right) =12$$ $$ 5y \left(1-\frac{1}{x^2+y^2}\right) =4$$

I need a Different Approach apart from what i posted..Thank You

Answer 1

I Called... $ x^2+y^2=t$, So

$$ 5x= \frac{12t}{t+1} $$ $$ 5y= \frac{4t}{t-1} $$

Squaring and Adding:

$$ 25t= \left(\frac{12t}{t+1}\right)^2+\left(\frac{4t}{t-1}\right)^2 $$ $\implies$

$$ 25= \left(\frac{144t}{t^2+2t+1}\right)+\left(\frac{16t}{t^2-2t+1}\right)$$ $\implies$

$$ 25= \left(\frac{144}{t+\frac{1}{t}+2}\right)+\left(\frac{16}{t+\frac{1}{t}-2}\right)$$ Let $$t+\frac{1}{t} =z$$, Then

$$ 25=\left(\frac{144}{z+2}\right)+\left(\frac{16}{z-2}\right)$$ $\implies$

$$ 25z^2-160z+156=0$$ $\implies$

$$ z=\frac{6}{5}, \frac{26}{5} \implies t=5, \frac{1}{5} \implies x=2,y=1 $$

Answer 2

with $x,y>0$,we have the equivalevent system $$\begin{cases} \dfrac{1}{x^2+y^2}=\dfrac{6}{5x}-\dfrac{2}{5y}\\ 1=\dfrac{6}{5x}+\dfrac{2}{5y} \end{cases}$$

so $$\dfrac{1}{x^2+y^2}=\dfrac{36}{25x^2}-\dfrac{4}{25y^2}$$ $$\Longrightarrow x=2y$$