How do you solve the following linear system: # x-2y+z=-14, y-2z=7, 2x+3y-z=-1 #?

1 Answer
Dec 31, 2017

Answer:

#(x,y,z) = (-6,3,-2)#

Explanation:

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)#