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Solve for $x$ and $y$ $$5x(1+\frac{1}{x^2+y^2})=12$$ $$5y(1-\frac{1}{x^2+y^2})=4$$

Combining these two equations we get $$5x^3-15x^2y+5xy^2-15y^3+5x+15y=0$$ Which could be factored if it had $-15y$ instead of $+15y$ for the last term.
Generally I do not post questions involving solving simple equations, but I find it really hard!

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  • $\begingroup$ Complexify: $$12 + 4i = 5(x+iy) + \frac{5(x-iy)}{x^2+y^2} = 5\biggl( z + \frac{1}{z}\biggr).$$ You get a quadratic equation in $z$. $\endgroup$ – Daniel Fischer Oct 21 '17 at 15:55
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We have $$1+\frac{1}{x^2+y^2}=\frac{12}{5x}$$ and $$1-\frac{1}{x^2+y^2}=\frac{4}{5y},$$ which gives $$\frac{6}{x}+\frac{2}{y}=5$$ or $$y=\frac{2x}{5x-6},$$ which after substitution to the first equation gives

$$25x^4-120x^3+209x^2-156x+36=0$$ or $$(5x^2-12x+6.5)^2-2.5^2=0$$ or $$(x-2)(5x-2)(5x^2-12x+9)=0.$$ Id est, we got the answer: $$\{(2,1),(0.4,-0.2)\}$$

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$\dfrac{1}{x^2+y^2}=A$ $$5x(1+A)=12\implies A=\frac{12}{5x}-1$$ $$5y(1-A)=4 \implies A=1-\frac{4}{5y}$$ we now know that bu $A=A$ $$\frac{12}{5x}-1=1-\frac{4}{5y} (*)$$ when we write $y$ in terms of $x$ or the other way around in either of these equations (above of course) we are done, a hint for the remaining multiply the system $(*)$ with $5xy$...

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