Given that $x,y \in \mathbb{R}$. Solve the system of equation.



My attempt, I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.

I don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.

  • 2
    $\begingroup$ Hint: $\;\frac{12}{5x}+\frac{4}{5y}=\cdots$ $\endgroup$ – dxiv Mar 26 '17 at 3:01
  • 1
    $\begingroup$ Can't you equate through $x^2 + y^2$ term and then isolate x (or y) and then sub back in and solve? $\endgroup$ – 123 Mar 26 '17 at 3:02

As dxiv has given hint:

$$\frac{12}{5x} + \frac{4}{5y} = 2 \implies x = \frac{6y}{5y-2}$$

Now put this in any one of the relation to make the relation full in terms of $y$.


Multiply equation (1) by y and equation (2) by x,



Now add both equations,

$$10xy = 12y + 4x$$

You can find y in terms of x or x in terms of y.

  • $\begingroup$ You can try substitution. May be that is better option. $\endgroup$ – Kanwaljit Singh Mar 26 '17 at 3:55

Just to provide another way.

Use polar coordinates $$x=\rho \cos(\theta) \qquad y=\rho \sin(\theta)$$ to make the equations to be $$\frac{5 \left(\rho ^2+1\right) \cos (\theta )}{\rho }=12$$ $$\frac{5 \left(\rho ^2-1\right) \sin (\theta )}{\rho }=4$$ So,$$\cos(\theta)=\frac{12 \rho }{5 \left(\rho ^2+1\right)}\qquad \sin(\theta)=\frac{4 \rho }{5 \left(\rho ^2-1\right)}$$ Now, using $$\sin^2(\theta)+\cos^2(\theta)=1 \implies \frac{16 \rho ^2}{25 \left(\rho ^2-1\right)^2}+\frac{144 \rho ^2}{25 \left(\rho ^2+1\right)^2}=1$$ Now, let $t=\rho ^2$ to have $$\frac{16 t}{25 \left(t -1\right)^2}+\frac{144 t}{25 \left(t+1\right)^2}=1$$ Reduce to same denominator to get $$25 t^4-160 t^3+206 t^2-160 t+25=0$$ where you can notice the symmetry of coefficients.

Factoring gives $$25 t^4-160 t^3+206 t^2-160 t+25=(5-t) (5 t-1) \left(5 t^2-6 t+5\right)=0$$ The quadratic does not show real roots; this makes the solutions to be $$t_1=\frac 15\qquad t_2=5$$ then $$\rho_1=\frac 1 {\sqrt 5}\qquad \rho_2=\sqrt 5$$ and now compute the corresponding $\theta$'s from $$\tan(\theta)=\frac{\sin(\theta) }{\cos(\theta) }=\frac{t+1}{3 (t-1)}=-\frac 12$$.

  • $\begingroup$ Up for this answer $\endgroup$ – Mathxx Mar 26 '17 at 6:09

$$5x(1+\frac{1}{x^2+y^2})=12\cdots \cdots (1)$$

$$5y(1-\frac{1}{x^2+y^2})=4\cdots \cdots \cdots \cdots (2) \times i$$

$$5(x+iy)+5\cdot \frac{x-iy}{(x+iy)(x+iy)}=12+4i$$

Now adding these two equations

put $z=x+iy$ and $\bar{z} = x-iy$ and $|z|^2= z \cdot \bar{z} = x^2+y^2$

So $$5z+\frac{5\bar{z}}{z \cdot \bar{z}}=12+4i$$

So $$5z+\frac{5}{z} = 12+4i$$

So $$5z^2-(12+4i)z+5=0$$ after solving $$z=\frac{12+4i\pm \sqrt{(12+4i)^2-100}}{10} = \frac{12+4i\pm (8+6i)}{10} = 2+i\;,\frac{2-i}{5}$$

So $$z= x+iy = 2+i\;,\frac{2-i}{5}$$

So $$(x,y) = \left\{(2,1)\;\;, \left(\frac{2}{5},-\frac{1}{5}\right)\right\}$$


First of all get rid of $x^2+y^2$

$$ \frac{12}{4x} +\frac{4}{5y} = 2,\quad 2x+6y = 5xy $$

It is a rectangular hyperbola passing through origin with asymptotes parallel to axes reminding one of $ x= t, y= 1/t $ when axes are asymptotes. So solve for $x,y$

$$ x= \frac{6y}{5y-2},\, y= \frac{2x}{5x-6}$$

The asymptotes are

$$ x= \frac65 ,\, y= \frac{2}{5} $$

We take hyperbola in the form

$$ ( x-\frac65 ) (y- \frac25) = k $$

Since it goes through the origin, $k$ must be the product $k=\dfrac65 \cdot \dfrac25 = \dfrac{12}{25} $ so as not to leave behind any constant term.

We can now find $x,y$ upto a parameter $t$

$$ ( x-\frac65 ) (y- \frac25) = \frac{12}{25}$$

$$\rightarrow x= \frac65 + t \sqrt{12}/5,\, y= \frac25 + \sqrt{12}/{5t},\, $$


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