How do you solve x+3y-z=13, 2x-5z=23, 4x-y-2z=14?

1 Answer
Oct 8, 2015

Answer:

#x=366/114#

#y=738/342#

#z=(-189)/57#

Explanation:

#x+3y-z=13# ------------------(1)
#2x-5z=23#------------------(2)
#4x-y-2z=14# ----------------(3)

Solve equations (1) and (2)

#x+3y-z=13# ------------------(1)
#2x-5z=23#------------------(2)

Multiply equation (1) with #2# to eliminate #x# term

#x+3y-z=13 xx 2# ------------------(1)
#2x+0y-5z=23#------------------(2)

#2x+6y-2z=26 # ------------------(1) Subtract (2) from (1)
#2x+0y-5z=23#------------------(2)
# 6y+3z=3# ----------------(4)

Solve equations (2) and (3) and eliminate #x# term

#2x+0y-5z=23#------------------(2)
#4x-y-2z=14# ----------------(3)

Multiply (2) with #2#

#2x+0y-5z=23 xx 2#------------------(2)
#4x-y-2z=14# ----------------(3)

#4x+0y-10z=46#------------------(2) Subtract (3) from (2)
#4x-y-2z=14# ----------------(3)
#y-8z=32# ----------------(5)

Take equations (4) and (5)

#6y+3z=3# ----------------(4)
#y-8z=32# ----------------(5)

Multiply equation (5) with #6#

#6y+3z=3# ----------------(4) Subtract (5) from (4)
#6y-54z=192# ----------------(5)
#57z=-189#

#z=(-189)/57#

Substitute the value of #z# in equation (2)

#2x-5z=23#------------------(2)
#2x-5((-189)/57)=23#------------------(2)
#2x+945/57=23#------------------(2) Multiply both sides by #57#
#114x+945=1311#-------- Solve it for #x#
#114x=1311-945=366#

#x=366/114#

Substitute the value of #x# and #z# in equation (1)

#x+3y-z=13# ------------------(1)
#366/114+3y-(-189)/57=13#
#366/114+3y+189/57=13# ---- Multiply both sides with #114#
#366+342y+378=1482#-------solve it for #y#

#744+342y=1482#
#342y=1482-744=738#

#y=738/342#