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

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

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!