How do you solve the system of equations #4x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-8#?

1 Answer
Apr 17, 2017

Answer:

#x=47/11#
#y=(-51)/11#
#z=(-9)/11#

Explanation:

Given -

#4x+2y+z=7#----------------(1)
#2x+2y+4z=-4#-------------(2)
#x+3y-2z=-8#----------------(3)

Take the first two equations and eliminate #y#

#4x+2y+z=7#----------------(1)
#2x+2y+4z=-4#-------------(2) -------- #(1)-(2)#
#2x-3z=11# -------------(4)

Take equations (1) and (3)
We have to elimiate #y# term

#4x+2y+z=7#----------------(1) -------#xx3#
#x+3y-2z=-8#----------------(3) -------#xx2#

#12x+6y+3z=21# ------------(5)
#2x+6y-4z=-16#-------------(6) ----- #(5)-(6)#
#10x+7z=37# ----------------(7)

Take equations (4) and (7)

#2x-3z=11# -------------(4) --------#xx5#
#10x+7z=37# ----------------(7)

#10x-15z=55# ---------(8)
#10x+7z=37#-----------(7) ------#(8)-(7)#
#-22z=18#
#z=-18/22#

Plugin >#z=-18/22# in equation (7)

#10x+7(-18/22)=37#
#10x-126/22=37#
#10x=37+126/22=(814+126)/22=940/22=470/11#
#x=470/11xx1/10=470/110=47/11#
#x=47/11#

Plugin >#z=-18/22# and >#x=47/11# in equation (1)

#4x+2y+z=7#----------------(1)
#4(47/11)+2y-18/22=7#
#4(47/11)+2y-9/11=7#
#188/11+2y-9/11=7#
#(188-9)/11+2y=7#
#179/11+2y=7#
#2y=7-179/11=(77-179)/11=(-102)/11#
#y=(-102)/11xx1/2=(-102)/22=#
#y=(-51)/11#

#x=47/11#
#y=(-51)/11#
#z=-18/22=(-9)/11#