(1) x+y=7+z,
(2) 2x-3y=4+2z,
(3) x-y=23+5z, 2(3) is 2x-2y=46+10z
(4)=(1)+(3)=2x=30+6z, x=15+3z;
(5)=2(3)-(2)=-2y+3y=46-4+10z-2z, y=42+8z.
So we now have x and y in terms of z in equations (4) and (5).
This means we can take any of the equations and substitute for x and y to find z.
Take (1), for example:
15+3z+42+8z=7+z,
(3+8-1)z=7-15-42=-50, so z=-50/10=-5.
Now we have z we can find x=15+3z=15-15=0; y=42+8z=42-40=2.
SOLUTION: x=0, y=2, z=-5. All equations are satisfied.