Question #9f8ce

Aug 8, 2017

The solution is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}5 \\ 2\end{matrix}\right)$

Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

$M = \left(\begin{matrix}3 & 4 & : & 23 \\ 5 & 3 & : & 31\end{matrix}\right)$

We perform the row operations

$R 2 \leftarrow 3 R 2 - 5 R 1$, $\implies$, $\left(\begin{matrix}3 & 4 & : & 23 \\ 0 & - 11 & : & - 22\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 11}$, $\implies$, $\left(\begin{matrix}3 & 4 & : & 23 \\ 0 & 1 & : & 2\end{matrix}\right)$

$R 1 \leftarrow R 1 - 4 R 2$, $\implies$, $\left(\begin{matrix}3 & 0 & : & 15 \\ 0 & 1 & : & 2\end{matrix}\right)$

$R 1 \leftarrow \frac{R 1}{3}$, $\implies$, $\left(\begin{matrix}1 & 0 & : & 5 \\ 0 & 1 & : & 2\end{matrix}\right)$