Solve this system of equations for real $x$ and $y$:

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

I juggled with those equations and got $x-y+\dfrac{x+y}{x^2+y^2}=\dfrac{8}{5}$, from where I guessed a solution $(2,1)$.

But I don't know how to approach mathematically.

Please help me.


\begin{align} 5x\left(1+\frac{1}{x^2+y^2}\right)&=12\\ 5y\left(1-\frac{1}{x^2+y^2}\right)&=4\\ \end{align}

Obviously, $x\neq0$ and $y\neq0$.

\begin{align} \left(1+\frac{1}{x^2+y^2}\right)&=\frac{12}{5x}\\ \left(1-\frac{1}{x^2+y^2}\right)&=\frac{4}{5y}\\ 1&=\frac{6}{5x}+\frac{2}{5y}&=\frac{6y+2x}{5xy}\\ \frac{1}{x^2+y^2}&=\frac{6}{5x}-\frac{2}{5y}&=\frac{6y-2x}{5xy}\\ x^2+y^2&=\frac{6y+2x}{6y-2x}\\ 5xy&=6y+2x&=(x^2+y^2)(6y-2x)\\ \end{align}

Now, consider $x=y.\alpha$. Remember that $y\neq 0$

\begin{align} y(5\alpha) &=6+2\alpha&=y^2(1+\alpha^2)(6-2\alpha) \end{align}

$$y=\frac{6+2\alpha}{5\alpha}$$ $$25\alpha^2=(1+\alpha^2)(36-4\alpha^2)$$ $$4\alpha^4-7\alpha^2-36=0$$ $$\alpha=\pm 2$$ (There are two other imaginary roots for $\alpha$ that gives two more solutions in $\mathbb C$)

Hence two solutions : $(x,y)=(2,1)$ or $(x,y)=(\frac{2}{5},-\frac{1}{5})$

  • $\begingroup$ For that last point, don't you mean $(\frac 25, -\frac 15)$? $\endgroup$ – Rory Daulton Mar 22 '15 at 7:41
  • $\begingroup$ @RoryDaulton yes, you're right :) $\endgroup$ – Xoff Mar 22 '15 at 7:43
  • $\begingroup$ Can we prove that solutions to the system of equations above are unique? I mean, how can we be sure that we have found all the solutions and no more exists? $\endgroup$ – Enthusiastic Engineer Mar 22 '15 at 9:01
  • $\begingroup$ @EnthusiasticStudent All equations are obtained from previous one by addition or multiplication by non-zero term. Hence, it does not remove any initial solutions. At the end, by solving the last equation for $\alpha^2$, we find two real soutions for alpha and two imaginary ones. All leads to 4 solutions (2 reals and 2 complexes) that you can verify in the initial system. Then, no other solutions could exist. At some point, the only solution that we add are $(x,y)=0$ but we just have to remember that this is not a solution in the initial system. $\endgroup$ – Xoff Mar 22 '15 at 9:36
  • $\begingroup$ You considered solutions in form of $x=y.\alpha$; do not we have solutions in form of $x=y.\alpha+1$, $x=y.\alpha+2$ and etc.? $\endgroup$ – Enthusiastic Engineer Mar 22 '15 at 9:56

$$\left\{ {\begin{array}{*{20}{c}}{5x\left( {1 + \frac{1}{{{x^2} + {y^2}}}} \right) = 12}\\{5y\left( {1 - \frac{1}{{{x^2} + {y^2}}}} \right) = 4}\end{array}} \right.$$

If you sum the above equations, you can find formula of a line all the points on it are solutions to that system of equations;

$$\left\{ {\begin{array}{*{20}{c}}{1 + \frac{1}{{{x^2} + {y^2}}} = \frac{{12}}{{5x}}}\\{1 - \frac{1}{{{x^2} + {y^2}}} = \frac{4}{{5y}}}\end{array}} \right.$$

Here is the formula of the line:

$$1 = \frac{6}{{5x}} + \frac{2}{{5y}}$$

Obviously, zero value for $x$ or $y$ is not in the domain of solutions to the equations. As verification, all the results above can be found by this formula: $(x,y)=(\frac{2}{5},-\frac{1}{5})$, $(2,1)$ and $(1,-2)$.


Hint: Use the transformation $x=r\cos \theta,\ y=r\sin \theta$ and solve for $r$ first, by eliminating $\theta$ and then solve for $\theta$, once $r$ is solved for.


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