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)x−2y+z=−14→(1)
y - 2z = 7 -> (2)y−2z=7→(2)
2x+3y-z = -1 ->(3)2x+3y−z=−1→(3)
Because we can eliminate zz from equations (1) and (3) easily by summing them we eliminate zz from all the equations. You can arbitrarily choose xx or yy and you will get the same answer.
(1) \times 2 +(2) \implies(1)×2+(2)⇒
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)
