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

Oct 21, 2016

This set of equations can not be solved.

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} x - y - z = 2$
[2]$\textcolor{w h i t e}{\text{XXX}} 2 x + y + z = 8$
[3]$\textcolor{w h i t e}{\text{XXX}} x + y + z = 6$

Note that adding equations [1] and [2] together, we get
$\textcolor{w h i t e}{\text{XXX}} 3 x = 10 \rightarrow x = \frac{10}{3}$

But adding equations [1] and [3] together, gives us
$\textcolor{w h i t e}{\text{XXX}} 2 x = 8 \rightarrow x = 4$

Since $\frac{10}{3} \ne 4$ 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:
$\left(\begin{matrix}1 & - 1 & - 1 & 2 \\ 2 & 1 & 1 & 8 \\ 1 & 1 & 1 & 6\end{matrix}\right)$

$\Rightarrow \left(\begin{matrix}1 & - 1 & - 1 & 2 \\ 0 & 3 & 3 & 2 \\ 0 & 2 & 2 & 4\end{matrix}\right)$

$\Rightarrow \left(\begin{matrix}1 & - 1 & - 1 & 2 \\ 0 & 1 & 1 & \frac{2}{3} \\ 0 & 0 & 0 & \frac{8}{3}\end{matrix}\right)$

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 $1$