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!

  • $\begingroup$ 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 $z$. $\endgroup$ Oct 21, 2017 at 15:55

2 Answers 2


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 $5xy$...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.