2. Solve the system of equations:

$$
(x+y)^{2}-z^{2}=1, \quad(y+z)^{2}-x^{2}=5, \quad(z+x)^{2}-y^{2}=10
$$
2. From $(x+y)^{2}-z^{2}=1, \quad(y+z)^{2}-x^{2}=5, \quad(z+x)^{2}-y^{2}=10$

by factoring, it follows that

$(x+y+z) \cdot(x+y-z)=1$,

$(x+y+z) \cdot(y+z-x)=5$

$(x+y+z) \cdot(z+x-y)=10$

If we add all three equations, it holds that $(x+y+z) \cdot[(x+y-z)+(y+z-x)+(z+x-y)]=16$,
or $(x+y+z)^{2}=16$, from which it follows that $x+y+z=4$ or $x+y+z=-4$.