Finding all the sets of three real numbers that satisfy specific equations I am having trouble finding a set of three numbers of real numbers $(x, y, z)$ satisfying $x + y + z = xy + xz + yz = 3$. I have tried factoring the equations around but I'm not having any luck and I don't think I'm going on the right path.

You have two equations in three unknowns, so you should expect to be able to be able to pick one variable as a parameter and solve for the other two.  Since the variables are equivalent, we will take $a$ as the parameter and solve for $b,c$.  We have $$a+b+c=3 \\ ab + bc + ac=3 \\b=3-a-c\\(a+c)(3-a-c)+ac=3\\3a+3c-a^2-2ac-c^2+ac=3\\c^2+(a-3)c+3-3a+a^2=0 \\c=\frac 12\left((3-a)\pm \sqrt{(a-3)^2-4(3-3a+a^2)}\right)\\c=\frac 12 \left((3-a)\pm \sqrt{a^2-6a+9-12+12a-4a^2}\right)\\c=\frac 12 \left((3-a)\pm \sqrt{6a-3-3a^2}\right)\\c=\frac 12 \left((3-a)\pm \sqrt{-3(a-1)^2}\right)$$  The answers are complex unless $a=1$, which shows the only real solution is $a=b=c=1$
