There are 4 equations but only 3 unknowns. This means one equation is unnecessary or there is inconsistency.
Call the equations in order A, B, C and D.
B and C are the same so we can remove one. Let's remove C.
3A+B: 3x+3y+3z+2x-3y+4z=27+13; 5x+7z=40. So 5x=40-7z. We have x in terms of z.
D-3A: y+2z=40-27=13. So y=13-2z. We have y in terms of z.
We can substitute for x and y in one equation (choose A) to leave z as the only variable:
(40-7z)/5+13-2z+z=9. Multiply through by 5: 40-7z+65-10z+5z=45; -12z+60=0, so 12z=60 and z=5.
We can now find x and y: 5x=40-7z=40-35=5, making x=1. y=13-2z=13-10=3.
So the solution is x=1, y=3 and z=5.
Substitute these values in the original equations to check them out.
Gauss-Jordan method:
Write the equations in matrix format:
[ 1 1 1 | 9 ]
[ 2 -3 4 | 13 ]
[ 3 4 5 | 40 ]
R2→R2-2R1
[ 1 1 1 | 9 ]
[ 0 -5 2 | -5]
[ 3 4 5 | 40 ]
R3→R3-3R1:
[ 1 1 1 | 9 ]
[ 0 -5 2 | -5 ]
[ 0 1 2 | 13 ]
R3→5R3+R2:
[ 1 1 1 | 9 ]
[ 0 -5 2 | -5 ]
[ 0 0 12 | 60 ]
R2→-R2+R3 then R2→(1/5)R2 and R3→(1/12)R3:
[ 1 1 1 | 9 ]
[ 0 1 2 | 13 ]
[ 0 0 1 | 5 ]
R2→R2-R3:
[ 1 1 1 | 9 ]
[ 0 1 1 | 8 ]
[ 0 0 1 | 5 ]
R1→R1-R2:
[ 1 0 0 | 1 ]
[ 0 1 1 | 8 ]
[ 0 0 1 | 5 ]
R2→R2-R3:
[ 1 0 0 | 1 ]
[ 0 1 0 | 3 ]
[ 0 0 1 | 5 ]
From this identity matrix x=1, y=3 and z=5.