Let -4x-5y-z=18···(1), -2x-5y-2z=12···(2), and -2x+5y+2z=4···(3)
Each of 3 equations has its own 5y-term, so we eliminate y-terms first.
1. Subtract (2) from(1): -4x-(-2x)-5y-(-5y)-z-(-2z)=18-12. We have -2x+z=6···(4)
2. Add (3) to (1): -4x+(-2x)-5y+(5y)-z+(2z)=18+4, We have: -6x+z=22···(5)
3. Subtract (5) from (4) to eliminate z: -2x-(-6x)+z-(z)=6-22, that is: 4x=-16. We have: x=-4
4. substitute x=-4 for x in (4): -2(-4)+z=6, that is 8+z=6. We have: z=-2
5. Plug x=-4 and z=-2 into (1): -4(-4)-5y-(-2)=18, that is: -5y=0. We have: y=0
Here, we have: x=-4, y=0 and z=-2
CK: Plug these values into the left side of (2): -2(-4)-5(0)-2(-2)=12, so LHS=RHS CKD.
The answer is: (x,y,z)=(-4,0,-2)