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?
 A: 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 $x=-4$, $y=-6$, $z=-7$.
A: Your system is equivalent to
\begin{align}
x(y+z+x)&=68\\
y(z+x+y)&=102\\
z(x+y+z)&=119.
\end{align}
Therefore, in effect, you have
\begin{align}
xa&=68\\
ya&=102\\
za&=119\\
a&=x+y+z.
\end{align}
This means that
$$a=\frac{68}{a}+\frac{102}{a}+\frac{119}{a}$$
or $$a^2=68+102+119\implies a=\pm 17,$$
which results in either $x= 4, y=6, z=7$ or $x=-4, y=-6, z=-7.$ (Thanks to paw88789 for pointing this out).
A: 
Solve for $x,y,z$:


$$x(y+z-x)=68-2x^2$$
$$y(z+x-y)=102-2y^2$$
$$z(x+y-z)=119-2z^2$$

$$x(x + y + z) = 68 - 2x^2 + 2x^2 = 68. \tag1$$
Similarly, $$y(x + y + z) = 102. \tag2 $$
$$z(x + y + z) = 119. \tag3 $$
Therefore, $$x::y::z = 68::102::119.\tag4 $$
You can use (4) above to eliminate (for example) the $x,y$ variables, expressing both of them in terms of $z$.  Then, if the system is solvable, routine methods should conquer the original equations.
