Answer
Infinitely many solutions: dependent equations.
Work Step by Step
The given system of equations is
$3x-6y=1$
$2x-4y=2/3$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
3 & -6 & 1\\
2 & -4 & 2/3
\end{array}\right]$
Perform $R_1\rightarrow \frac{R_1}{3}$.
$\Rightarrow \left[\begin{array}{cc|c}
3/3 & -6/3 & 1/3\\
2 & -4 & 2/3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -2 & 1/3\\
2 & -4 & 2/3
\end{array}\right]$
Perform $R_2\rightarrow R_2-2\times R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -2 & 1/3\\
2-2\times 1 &- 4-2\times (-2) & 2/3-2\times (1/3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -2 & 1/3\\
0 & 0 & 0
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow (1)x+(-2)y=1/3$
and
$\Rightarrow (0)x+(0)y=0$
From the second equation we can say that the system has infinitely many solutions: dependent equations.
