How do you solve the system of equations algebraically #-x-3y+z=54, 4x+2y-3z=-32, 2y+8z=78#?

1 Answer
Aug 6, 2017

Answer:

Solution : #x =11 , y = -17 , z =14#

Explanation:

#-x-3y+z = 54 (1) ,4x +2y -3z = -32 (2) #

#2y+8z= 78(3) : y = (78-8z)/2= 39-4z# Putting # y = 39-4z#

In equation (1) and (2) we get # -x -3(39-4z)+ z =54# or

# -x +13z =171 (4) and 4x +2(39-4z) -3z = -32 or 4x-11z = -110 (5)#

#(4)*4: -4x +52z = 684 (6) # Adding (5) and (6) we get

#41 z = 574 :. z = 1 4 :. 2y + 8*14 =78 :. 2y = 78-112#

or# 2y = -34 :. y = -17 ; -x - 3* -17 + 14 =54 or x = 51+14-54=11#

Solution : #x =11 , y = -17 , z =14# [Ans]