Solving these two equations

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: $$\{(2,1),(0.4,-0.2)\}$$


$\dfrac{1}{x^2+y^2}=A$ $$5x(1+A)=12\implies A=\frac{12}{5x}-1$$ $$5y(1-A)=4 \implies A=1-\frac{4}{5y}$$ 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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Linear Algebra

Polynomials

Algebra Precalculus

Systems Of Equations

Quartic Equations

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