# How do you solve the system x+y+z=-3, 3x+y-z=13, and 3x+y-2z=18?

Apr 25, 2017

The augmented matrix looks something like this:

$\left(\begin{matrix}1 & 1 & 1 \\ 3 & 1 & - 1 \\ 3 & 1 & - 2\end{matrix}\right) \left(\begin{matrix}- 3 \\ 13 \\ 18\end{matrix}\right)$

Then: $R 2 \to R 2 - 3 R 1 , R 3 \to R 3 - 3 R 1$

$\left(\begin{matrix}1 & 1 & 1 \\ 0 & - 2 & - 4 \\ 0 & - 2 & - 5\end{matrix}\right) \left(\begin{matrix}- 3 \\ 22 \\ 27\end{matrix}\right)$

Then: $R 3 \to R 3 - R 2$

$\left(\begin{matrix}1 & 1 & 1 \\ 0 & - 2 & - 4 \\ 0 & 0 & - 1\end{matrix}\right) \left(\begin{matrix}- 3 \\ 22 \\ 5\end{matrix}\right)$

So now we back substitute:

$- z = 5 \implies z = - 5$

$- 2 y - 4 z = 22 \implies y = - 1$

$x + y + z = - 3 \implies x = 3$