How do you solve the system #x-y-z=2#, #2x+y+z=8#, and #x+y+z=6#?

1 Answer
Oct 21, 2016

Answer:

This set of equations can not be solved.

Explanation:

Given
[1]#color(white)("XXX")x-y-z=2#
[2]#color(white)("XXX")2x+y+z=8#
[3]#color(white)("XXX")x+y+z=6#

Note that adding equations [1] and [2] together, we get
#color(white)("XXX")3x=10 rarr x=10/3#

But adding equations [1] and [3] together, gives us
#color(white)("XXX")2x=8 rarr x=4#

Since #10/3 != 4# these equations are inconsistent.

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Since this was asked under "Solving a System of Equations Using a Matrix"
if you need the matrix version of this, here it is (using Gaussian elimination)

Augmented matrix:
#((1,-1,-1,2),(2,1,1,8),(1,1,1,6))#

#rArr((1,-1,-1,2),(0,3,3,2),(0,2,2,4))#

#rArr ((1,-1,-1,2),(0,1,1,2/3),(0,0,0,8/3))#

The value in the pivot position for #z#: row 3, column 3
has become #0# and can not be converted into the required value of #1#