Something about this sure is misty!

Just consider the third equation: $x^2 + y^2 + xy = -\pi$

Assuming this has a solution in the real numbers yields $\begin{aligned}-xy - \pi = x^2 + y^2 \geq 0 \qquad &\Rightarrow \qquad xy \leq -\pi \\ xy - \pi = x^2 + 2xy + y^2 = (x+y)^2 \geq 0 \qquad &\Rightarrow \qquad xy \geq \pi.\end{aligned}$

Therefore $\pi \leq xy \leq -\pi,$ which is absurd, so there are no real solutions to that equation, so there are no real solutions to the system.

Let us take the above systems of equations and relate them according to:

$x^2 + y^2 + xy = 3x^2 - y^2 = -\pi;$

$2x(x^2 + y^2) = xy(x + y) = \pi^2$

Taking the second equation, we can simplify it to:

$2(x^2 + y^2) = xy + y^2 \Rightarrow xy = 2x^2 + y^2$ (i)

and substituting (i) into first equation now yields:

$x^2 + y^2 + (2x^2 + y^2) = 3x^2 - y^2 \Rightarrow 3x^2 + 2y^2 = 3x^2 - y^2 \Rightarrow y = 0$ .

If $y = 0$ , then we require $x = 0$ in (i) and the required solution is $\boxed{(x,y) = (0, 0)}$ . But this solution poses a contradiction in our original systems of equations once we substitute back in and check!

This might seem a little out of place!!

Note that a cubic polynomial (in one variable and having real co-efficients) can have either 2 non-real roots or zero non-zero real roots.

Since every cubic can by effective transformation be converted to one whose $x^{2}$ co-efficient is zero,let us find the conditions for the aforementioned two possibilities on a cubic whose sum of roots is zero.

• 2 Non-Real Roots

since the roots are going to be of the form $a+ib,a-ib,-2a$ , the cubic is $x^{3}+ x(b^{2}-3a^{2})+2a(a^{2}+b^{2})=0$

• 0 Non-Real Roots

since the roots are of the form $a,b,-(a+b)$ the cubic is $x^{3}-x(a^{2}+b^{2}+ab)+ab(a+b)=0$

Now,as earlier stated $x^{3}+\pi x+\pi^{2}=0$ can have only one of the above possibilities,the systems given in the question cannot simultaneously have solutions

Adding the first and third $(y/x)^2+(y/x)-2=0$ .So $y/x=1,-2$ .The required division with $x$ could be done as x doesn't take the value Zero.Plugging $y=x$ in the $2nd$ and $4th$ yields $x^3=π^2/2$ and $x^3=π^2/4$ that's definitely absurd.Similarly working the other $y=-2x$ (in $2nd$ and $4th$ ) yields $x^3=π^2/10$ and $x^3=π^2/2$ that's absurd...So no common solution exists to the system of equations.