How do you solve the system of equations #3x-4y+2z=18#, #-2x+y-6z = -29#, and #4x+2y-4z=-10#?

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Answer 1 · 2017-07-29T11:52:31

Solution: #x= 2 , y= -1 , z = 4#

#3x-4y+2z=18 (1) , -2x+y-6z = -29 (2) , 4x+2y -4z = -10 (3)#

From equation (2) # y = 2x+6z -29 # putting #y# in equation (1)

& equation (3) we get # 3x- 4(2x+6z -29) +2z=18# or

#-5x-22z=-98 (4)# and # 4x+ 2(2x+6z -29) -4z = -10 # or

#8x+8z=48 (5)# ; #[(4)*8] ; -40x - 176z = -784 (6) # and

#[(5)*5] ; 40x +40z = 240 (7) # Adding #(6) & (7)# ewe get,

#-136z = -544 :. z = -544/136= 4 :. x = (48 - 8*4)/8 =16/8=2#

# y = 2x+6z -29 , y = 2*2 +6*4 -29= -1 #

Solution: #x= 2 , y= -1 , z = 4# [Ans]