call : x - 4y + z = 6 (e1) == > z = 6 - x + 4y 2x + 5y - z = 7 (e2) 2x - y - z = 1 (e3)
sub z from e1 in e2 you have : 2x + 5y - 6 + x - 4y = 7 ==> 3x + y = 13 ==> 3x = 13 - y (*)sub z from e1 in e3 you have: 2x - y - 6 + x - 4y = 1 ==> 3x - 5y = 7 (**)
sub 3x from (*) in (**), you have 13 - y - 5y = 7 ==> 6 = 6y ==> y = 1sub y = 1 into (*), you have: 3x = 13 - 1 = 12 ==> x = 4
sub both y = 1 and x = 4 into (e1), you have: 4 - 4(1) + z = 6 ==> z = 6
so answer : x = 4, y = 1, z =6 (you can use these values to check other equations to see if they come out all right)