Answer
$\left\{\left(8,2,0\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-y=6\\
2x-3z=16\\
2y+z=4
\end{cases}$
Use the elimination method. Multiply the third equation by 3 and add it to the second equation to eliminate $z$:
$\begin{cases}
x-y=6\\
2x-3z=16\\
3(2y+z)=3(4)
\end{cases}$
$\begin{cases}
x-y=6\\
2x-3z=16\\
6y+3z=12
\end{cases}$
$\begin{cases}
x-y=6\\
2x-3z+6y+3z=16+12
\end{cases}$
$\begin{cases}
x-y=6\\
2x+6y=28
\end{cases}$
Multiply the first equation by 6 and add it to the second equation to eliminate $y$ and determine $x$:
$\begin{cases}
6x-6y=6(6)\\
2x+6y=28
\end{cases}$
$6x-6y+2x+6y=36+28$
$8x=64$
$x=8$
Determine $y$:
$x-y=6$
$8-y=6$
$y=2$
Determine $z$:
$2y+z=4$
$2(2)+z=4$
$4+z=4$
$z=0$
The solution set of the system is:
$\left\{\left(8,2,0\right)\right\}$
