Answer
$\{(1,3,-2)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x+y-z=6\\
3x-2y+z=-5\\
x+3y-2z=14
\end{cases}$
Use the elimination method. Add the second equation to the first equation; then multiply the second equation by 2 and add it to the third, to eliminate $z$:
$\begin{cases}
x+y-z+3x-2y+z=6-5\\
x+3y-2z+2(3x-2y+z)=14+2(-5)
\end{cases}$
$\begin{cases}
4x-y=1\\
7x-y=4
\end{cases}$
Multiply the first equation by -1 and add it to the second to eliminate $y$ and determine $x$:
$\begin{cases}
-4x+y=-1\\
7x-y=4
\end{cases}$
$-4x+y+7x-y=-1+4$
$3x=3$
$x=1$
Determine $y$:
$4x-y=1$
$4(1)-y=1$
$y=3$
Determine $z$:
$x+y-z=6$
$1+3-z=6$
$z=-2$
The solution set of the system is:
$\{(1,3,-2)\}$
