Question

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

Answer 1

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`