Answer
$\left\{\left(2,-1,1\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-2y+3z=7\\
2x+y+z=4\\
-3x+2y-2z=-10
\end{cases}$
Use the elimination method. Multiply the first equation by -2 and add it to the second equation to eliminate $x$. Multiply the first equation by 3 and add it to the third equation to eliminate $x$.
$\begin{cases}
2x+y+z-2(x-2y+3z)=4-2(7)\\
-3x+2y-2z+3(x-2y+3z)=-10+3(7)
\end{cases}$
$\begin{cases}
2x+y+z-2x+4y-6z=-10\\
-3x+2y-2z+3x-6y+9z=11
\end{cases}$
$\begin{cases}
5y-5z=-10\\
-4y+7z=11
\end{cases}$
$\begin{cases}
y-z=-2\\
-4y+7z=11
\end{cases}$
Multiply the first equation by 4 and add it to the second equation to eliminate $y$ and determine $z$:
$\begin{cases}
4y-4z=4(-2)\\
-4y+7z=11
\end{cases}$
$4y-4z-4y+7z=-8+11$
$3z=3$
$z=1$
Determine $y$:
$y-z=-2$
$y-1=-2$
$y=-1$
Determine $x$:
$x-2y+3z=7$
$x-2(-1)+3(1)=7$
$x+5=7$
$x=2$
The solution set of the system is:
$\left\{\left(2,-1,1\right)\right\}$
