# How do you solve using gaussian elimination or gauss-jordan elimination, 2x - 3y = 5, 3x + 4y = -1?

Nov 27, 2016

$x = 1 , y = - 1$

#### Explanation:

We can write the system of linear equation in matrix vector form

$A \underline{x} = \underline{b}$

$\left(\begin{matrix}2 & - 3 \\ 3 & 4\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}5 \\ - 1\end{matrix}\right)$

So the Augmented Matrix is:

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

We can now perform elementary row operations:

$\left(\begin{matrix}2 & - 3 & | & 5 \\ 3 & 4 & | & - 1\end{matrix}\right) \stackrel{\frac{1}{2} {R}_{1} \rightarrow {R}_{1}}{\rightarrow} \left(\begin{matrix}1 & - \frac{3}{2} & | & \frac{5}{2} \\ 3 & 4 & | & - 1\end{matrix}\right)$

$\left(\begin{matrix}1 & - \frac{3}{2} & | & \frac{5}{2} \\ 3 & 4 & | & - 1\end{matrix}\right) \stackrel{{R}_{2} - 3 {R}_{1} \rightarrow {R}_{2}}{\rightarrow} \left(\begin{matrix}1 & - \frac{3}{2} & | & \frac{5}{2} \\ 0 & \frac{17}{2} & | & - \frac{17}{2}\end{matrix}\right)$

And so:
$\frac{17}{2} y = - \frac{17}{2} \implies y = - 1$

Back substituting into R1:
$x - 3 \frac{y}{2} = \frac{5}{2} \implies x + \frac{3}{2} = \frac{5}{2} \implies x = 1$