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

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.
