Pls help thanks

This is kind of a guess and check answer.

The only way this could work is if x y and z equal 3, 0, 0 in some order, or 1, 2, 2. Since 3, 0, 0 isn't very good for the second equation, we'll go with 1, 2, 2. When you plug it in we get 1+4+4=9 for the first equation and 2+4+2=8 for the second equation. So we get the sum is 5

How about this:

 

\((x+y+z)^2=(x^2+y^2+z^2)+2(xy+xz+zy)\qquad \\\text{I have just expanded and simplified}\\ (x+y+z)^2=(9)+2(8)\\ (x+y+z)^2=25\\ x+y+z=\pm5\\ \text{but it has to be positive so}\\ x+y+z=5\)