<s>[INST] Solve below somewhat symmetric equations Solve $x,y,z$ subject to
$$x^2+y^2 - xy = 3$$
$$(x-z)^2+(y-z)^2 - (y-z)(x-z) = 4$$
$$(x-z)^2+y^2 - y(x-z) = 1$$
$$x,y,z \in R^{+}$$
My attempts:
$(x-z)^2+(y-z)^2 - (y-z)(x-z) - ((x-z)^2+y^2 - y(x-z)) = xz - 2yz = (x-2y)z = 3$
$x^2+y^2 - xy - ((x-z)^2+y^2 - y(x-z)) = 2xz - yz - z^2 = (2x-y-z)z = 2$
divide above, we have $\frac{2x-y-z}{x-2y} = \frac{2}{3}$, or  $3z = 4x+y$
Then the rest is easy. But I start to get interested in if there are some geometric solution. Like the comment says, they are the elliptical cylinders in the positive octant
 [/INST] We multiply the thrid equation by four, subtract it from two, and then substitute with equation one solved for x.
\begin{align*}
&[ (x-z)^2+(y-z)^2 - (y-z)(x-z) ]\\  
- 4&[  (x-z)^2+y^2 - y(x-z) ]=0 \\ 
,\space &x = \dfrac{\sqrt{3}\sqrt{4 - y^2} + y}{2}\\
\\
\implies &y = \dfrac{14 - 3 \sqrt{3}}{13},
\space  z = \pm\sqrt{3}
\end{align*}
It should be easy to figure out $x$ from here.
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