Answer
No solution or $\varnothing$.
Work Step by Step
The given system of equations is
$-3x+4y=12$
$6x-8y=16$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
-3 & 4 & 12\\
6 & -8 & 16
\end{array}\right]$
Perform $R_1\rightarrow \frac{R_1}{(-3)}$.
$\Rightarrow \left[\begin{array}{cc|c}
-3/(-3) & 4/(-3) & 12/(-3)\\
6 & -8 & 16
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -4/3 & -4\\
6 & -8 & 16
\end{array}\right]$
Perform $R_2\rightarrow R_2-6\times R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -4/3 & -4\\
6-6\times 1 & -8-6\times (-4/3) & 16-6\times (-4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -4/3 & -4\\
0 & 0 & 40
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow (1)x+(−4/3)y=-4$
and
$\Rightarrow (0)x+(0)y=40$
In the second equation no values of $x$ and $y$ satisfy.
Hence, the system is inconsistent and has no solution.
The solution set is $\varnothing$.
