How do you solve the system of equations #x+y+z=1#, #x+2y+z=2#, #x-y+z=-5# ?

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Answer 1 · 2017-03-18T00:11:32

This system has no solutions.

Given:

#{(x+y+z=1), (x+2y+z=2), (x-y+z=-5):}#

Adding the first and third equations, we get:

#2x+2z = -4#

Hence:

#x+z = -2" "...# (i)

Adding twice the third equation to the second equation, we get:

#3x+3z = -8#

Hence:

#x+z = -8/3" "...# (ii)

Equations (i) and (ii) are incompatible, since #-2 != -8/3#

So there is no solution to this system of equations.