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

1 Answer

Oct 21, 2016

This set of equations can not be solved.

Explanation:

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

Note that adding equations [1] and [2] together, we get
$XXX 3 x = 10 \to x = \frac{10}{3}$ $color(white)("XXX")3x=10 rarr x=10/3$

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

Since $\frac{10}{3} \neq 4$ $10/3 != 4$ 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 $1$

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