Solving each system by eliminating

x-y+z = - 1

x+y+3z = - 3

2x-y+2z = 0

There's only one system of equations here.

Let's eliminate x first,

x - y + z = - 1

x = - 1 + y - z

Substituting into x + y + 3z = - 3 gives

(-1 + y - z) + y + 3z = - 3

2y + 2z = - 2

y + z = - 1

Substituting for x into 2x-y+2z = 0 gives

2 (-1 + y - z) - y + 2z = 0

-2 + 2y - 2z - y + 2z = 0

y = 2

Got lucky on that one.

z = - 1 - y = - 1 - 2 = - 3

x = - 1 + y - z = - 1 + 2 - - 3 = 4

Answer: x=4, y=2, z = - 3

Check:

x-y+z = 4 - 2 + - 3 = - 1, good

x+y+3x = 4+2+3 (-3) = - 3, good

2x-y+2x = 2 (4) - 2+2 (-3) = 0, good