2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

\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 $(x,y)=(\frac{2}{5},-\frac{1}{5})$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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 $(1,-2)$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

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.