7.1. Real $x, y, z$ are such that $x y + x z + y z + x + y + z = -3, x^{2} + y^{2} + z^{2} = 5$. What is $x + y + z$?
Answer: -1

Solution. Add twice the first equation to the second, we get $(x+y+z)^{2}+2(x+y+z)=-1$. Therefore, $(x+y+z+1)^{2}=0, x+y+z=-1$.