$$ \begin{cases} x^2 - y^2 + 12y - 21 = 0\\ 2x^2 + y^2 + 2xy + x = 0 \end{cases} $$

I've tried the change of variables: $u = x + y$, $v = x - y$

After it I've got: $$ \begin{cases} uv + 12\frac{u - v}{2} - 21 = 0\\ (\frac{u + v}{2})^2 + u^2 + \frac{u + v}{2} = 0 \end{cases} $$

After this: $$ \begin{cases} uv + 6(u - v) - 21 = 0\\ 5u^2 + 2uv + v^2 +2u + 2v = 0 \end{cases} $$ I've tried to express $v$ from the first equation and put it into the second, but I think, it is not the correct way

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    $\begingroup$ Are you sure about the equations ? These do not show real solutions. $\endgroup$ – Claude Leibovici Mar 15 '15 at 7:46
  • $\begingroup$ Indeed, WolframAlpha shows imaginary solutions for these equations. No wonder they seemed so tricky... $\endgroup$ – Newb Mar 15 '15 at 7:47
  • $\begingroup$ Yes, according to the answer, the problem has no solution. But it should be proven $\endgroup$ – Photon Mar 15 '15 at 7:53

If we graph the equations, we see that the graphs of those two simultaneous equations do not overlap, so there is no solution. In searching for a proof of this, we see that the first equation's graph is much closer to the origin that the second one's graph. This gives us the idea to set limits on the distances, or the squares of their distances, from the origin. Therefore we seek to place limits on $x^2+y^2$ for each graph. The limits do not need to be the best possible, just enough to show that the two graphs do not overlap. enter image description here

$$x^2-y^2+12y-21=0$$ $$x^2-y^2+12y-21+(2y^2-12y+21)=2y^2-12y+21$$ $$x^2+y^2=2y^2-12y+21$$ $$=2(y-3)^2+3$$

So we see that, for the first equation,

$$x^2+y^2\ge 3$$

For the second equation,

$$2x^2 + y^2 + 2xy + x = 0$$ $$(2)x^2+(2y+1)x+(y^2)=0$$

The discriminant $b^2-4ac$ of that last quadratic equation in $x$ must be non-negative for it to ever be true for real $x$ and $y$, so

$$(2y+1)^2-4\cdot 2\cdot y^2\ge 0$$ $$-4y^2+4y+1\ge 0$$

This has the solution

$$\frac{1-\sqrt 2}2\le y\le \frac{1+\sqrt 2}2$$

The upper limit has a greater absolute value, so we conclude that

$$|y|\le \frac{1+\sqrt 2}2$$

Going back to the second equation,

$$2x^2 + y^2 + 2xy + x = 0$$ $$y^2+(2x)y+(2x^2+x)=0$$

Again, the discriminant must be non-negative, so

$$(2x)^2-4\cdot 1\cdot (2x^2+x)\ge 0$$ $$-4x^2-4x\ge 0$$

This has the solution

$$-1\le x\le 0$$

Then clearly

$$|x|\le 1$$

Combining those restrictions on $|y|$ and $|x|$,

$$x^2+y^2\le (-1)^2+\left(\frac{1+\sqrt 2}2\right)^2\approx 2.457$$

We see that the restrictions (in the gray shading) for the two equations are inconsistent. Therefore, we have proved that there is no solution to those simultaneous equations.


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