How do you solve this system of equations: #x- y = 6 , 2y + 5z = 1, and 3x - 4z = 8#?

2 Answers
Shwetank Mauria
Nov 3, 2017

#x=4#, #y=-2# and #z=1#

Explanation:

We first eliminate #z# from the latter two equations. This can be done by multiplying second equation by #4# and third equation by #5# and adding the two together.

#4xx(2y+5z)+5x(3x-4z)=4xx1+5xx8#

or #8y+20z+15x-20z=4+40#

or #15x+8y=44# ..........................(A)

Multiplying #x-y=6# by #8# we get

#8x-8y=48# ..........................(B)

Adding A and B, we get

#23x=92# i..e. #x=4#

and putting this in first equation #4-y=6# i.e. #y=-2#

and putting value of #y# in #2y+5z=1#, we get

#-4+5z=1# or #5z=5# i.e. #z=1#

Manasvini Nittala
Nov 3, 2017

The solutions are #x=4,y=(-2),z=1#

Explanation:

the given equations are
#x-y=6->(1)#
#2y+5z=1->(2)#
#3x-4z=8->(3)#
from #(1)#
#rArry=x-6#
put #y=x-6# in #(2)# we get
#2x+5z=13->(4)#
multiply #(4)# by #3# and #(3)# by #2#
we get #6x+15z=39->(5)# and #6x-8z=16->(6)#
solving #(5) and (6)# we get #z=1# put #z=1# in #(4)# we get #x=4#
put #x=4# in #(1)# we get #y=-2#

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