Find $x,y,z$ satisfying $x(y+z-x)=68-2x^2$, $y(z+x-y)=102-2y^2$, $z(x+y-z)=119-2z^2$ Solve for $x,y,z$:
$$x(y+z-x)=68-2x^2$$
$$y(z+x-y)=102-2y^2$$
$$z(x+y-z)=119-2z^2$$
After some manipulation, I obtain
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
After combining equations, I get
$$y=\frac{-51-x^2+z^2}{x-z}$$
This seems too tedious. Is there a simpler way?

If you add
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
you get $(x+y+z)^2=17^2$, and so $x+y+z=\pm 17$. If you substitute $x+y=\pm 17-z$ in the third of your equation, you get $\pm 17 z=119$, and so $z=7$ or $z=-7$; now it is easy to get the two solution $x=4$, $y=6$, $z=7$ and $x=-4$, $y=-6$, $z=-7$.
