How do you solve the system #x-y+z=3#, #2x+y+z=8#, and #3x+y-z=1#?

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Answer 1 · 2017-01-27T03:01:25

Solution is #x=1#, #y=2# and #z=4#

We are given three equations in three variables, which are

#x-y+z=3# ..........................................(1)

#2x+y+z=8# ..........................................(2)

and #3x+y-z=1# ..........................................(3)

To solve them, what we need is to eliminate some vatiables to get the value of others.

It is apparent from (1) nd (3) that as signs of #y# and #z# are opposite, they will cancel out on adding. Hence adding them, we get

#x-y+z+3x+y-z=3+1# or #4x=4# i.e. #x=1#

Now subtracting (2) from (1), we get

#x-y+z-2x-y-z=3-8# or #-x-2y=-5#.

But as #x=1#, we have

#-1-2y=-5# i.e. #-2y=-4# or #y=(-4)/-2=2#

Putting values of #x# and #y# in (1), we get

#1-2+z=3# or #z=3-1+2=4#

Hence, solution is #x=1#, #y=2# and #z=4#