There are a bunch of ways in which we can solve the system of linear equations. Elimination method is illustrated here.
Let us number the equations and proceed
x-2y+z = -14 -> (1)
y - 2z = 7 -> (2)
2x+3y-z = -1 ->(3)
Because we can eliminate z from equations (1) and (3) easily by summing them we eliminate z from all the equations. You can arbitrarily choose x or y and you will get the same answer.
(1) \times 2 +(2) \implies
2x - 4y + \cancel(2z) = -28
0x + y -\cancel(2z) = 7
2x - 3y = -21 -> (4)
(1) + (3) implies
x-2y+cancel(z) = -14
2x+3y-cancel(z) = -1
3x+y = -15 -> (5)
(4)+3\times(5) implies
2x-cancel(3y) = -21
9x+cancel(3y) = -45
11x = -66
x = -6
Substitute in (5)
3x+y = -15
y = -15 - 3x
y = -15 - 3(-6)
y = -15+18
y = 3
Substitute y in (2)
y - 2z = 7
3 - 2z = 7
-2z = 4
z = -2
Substitute in any of the equation to verify the results. Equation (3) gives
2x+3y-z = -1
L.H.S implies
2(-6)+3(3)-(-2) = -12+9+2
=-12+11
=-1
Hence our solution is (x,y,z) = (-6,3,-2)
