Answer
$\{(-1,2,-3)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x+2y-z=6\\
2x-y+3z=-13\\
3x-2y+3z=-16
\end{cases}$
Use the elimination method. Add the first equation to the third equation to eliminate $y$. Then multiply the second equation by 2 and add it to the first to eliminate $y$:
$\begin{cases}
x+2y-z+2(2x-y+3z)=6+2(-13)\\
3x-2y+3z+x+2y-z=-16+6
\end{cases}$
$\begin{cases}
x+2y-z+4x-2y+6z=6-26\\
3x-2y+3z+x+2y-z=-16+6
\end{cases}$
$\begin{cases}
5x+5z=-20\\
4x+2z=-10
\end{cases}$
$\begin{cases}
x+z=-4\\
2x+z=-5
\end{cases}$
Multiply the first equation by -1 and add it to the second to eliminate $z$ and find $x$:
$-(x+z)+2x+z=-(-4)-5$
$-x-z+2x+z=4-5$
$x=-1$
Determine $z$:
$x+z=-4$
$-1+z=-4$
$z=-3$
Determine $y$:
$2x-y+3z=-13$
$2(-1)-y+3(-3)=-13$
$-11-y=-13$
$y=2$
The solution set is:
$\{(-1,2,-3)\}$
