Answer
$(-1,2,3)$
Work Step by Step
We build the system:
$\begin{cases}
3x+y+2z=5\\
x-3y+3z=2\\
2x+3y-z=1
\end{cases}$
We will use the addition method. Multiply Equation 1 by 3 and add it to Equation 2 to eliminate $y$. Also add Equation 2 to Equation 3 to eliminate $y$:
$\begin{cases}
3(3x+y+2z)+x-3y+3z=3(5)+2\\
x-3y+3z+2x+3y-z=2+1
\end{cases}$
$\begin{cases}
9x+3y+6z+x-3y+3z=15+2\\
3x+2z=3
\end{cases}$
$\begin{cases}
10x+9z=17\\
3x+2z=3
\end{cases}$
Multiply Equation 1 by 2, Equation 2 by -9 and add them to eliminate $z$ and determine $x$:
$\begin{cases}
2(10x+9z)=2(17)\\
-9(3x+2z)=-9(3)
\end{cases}$
$\begin{cases}
20x+18z=34\\
-27x-18z=-27
\end{cases}$
$20x+18z-27x-18z=34-27$
$-7x=7$
$x=-1$
Substitute the value of $x$ in the Equation $3x+2z=3$ to determine $z$:
$3(-1)+2z=3$
$-3+2z=3$
$2z=6$
$z=3$
Substitute the values of $x, z$ is Equation 1 of the given system to find $y$:
$3x+y+2z=5$
$3(-1)+y+2(3)=5$
$3+y=5$
$y=2$
The system's solution is:
$(-1,2,3)$
