Bradford T. answered 01/30/21
Retired Engineer / Upper level math instructor
Looks like z will be the free variable since there are only the 2 equations.
[1 1 1 6]
[1 6 7 34]
r2 = r2-r1
[1 1 1 6]
[0 5 6 28]
r2 = r2/5
[1 1 1 6]
[0 1 6/5 28/5]
From the last row
y = (28-6z)/5
let z = 3
y = (28-18)/5 = 2
From the first equation
x + 2 + 3 = 6 --> x = 1
x=1, y=2, z=3