What are x,y and z? x+y=-3 x+z=-2 xy+yz+xz=2

First, we'll multiply the 1st and the 2nd
equations:

(x+y)(x+z) =
(-3)*(-2)

We'll remove the
brackets:

x^2 + xz + xy + yz =
6

We notice that the sum xz + xy + yz can be replaced by
the 3rd equation:

x^2 + 2=
6

x^2 = 6-2

x^2 =
4

x1 = 2 and x2 = -2

We'll
substitute x1 in the 1st equation:

2 + y =
-3

y = -2-3

y =
-5

We'll substitute x1 in the 2nd
equation:

2  +z = -2

z =
-4

We'll substitute x2 in the 1st
equation:

-2 + y = -3

y =
-1

We'll substitute x2 in the 2nd
equation:

-2 + z = -2

z =
0

The pairs of solutions are: (2 ; -5 ; -4)
and (-2  ;-1 ; 0).