# How do you set up and solve the following system using augmented matrices 3x-2y=8, 6x-4y=1?

Mar 8, 2017

No solution

#### Explanation:

The typesetting here does not support row augmented matrices.

But, if we are going with: $3 x - 2 y = 8 , 6 x - 4 y = 1$ ...then we start with:

$\left(\begin{matrix}3 & - 2 \\ 6 & - 4\end{matrix}\right) \left(\begin{matrix}8 \\ 1\end{matrix}\right)$

However, there is an immediate problem, we have linearly dependent rows, this is a singular matrix :(

$\left(\begin{matrix}3 & - 2 \\ 6 & - 4\end{matrix}\right) = \left(\begin{matrix}\frac{1}{2} \left(6\right) & \frac{1}{2} \left(- 4\right) \\ 6 & - 4\end{matrix}\right)$

So this either has no solutions or infinitely many:

$\left(\begin{matrix}3 & - 2 \\ 6 & - 4\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}8 \\ 1\end{matrix}\right)$

Then go back and look at the equations and you will see that they represent parallel lines:

$3 x - 2 y = \textcolor{red}{8}$

$2 \left(3 x\right) - 2 \left(2 y\right) = \textcolor{b l u e}{2 \left(\frac{1}{2}\right)}$