Answer
$(2,1,-1)$
Work Step by Step
We are given the system:
$\begin{cases}
x+2y-z=5\\
2x-y+3z=0\\
2y+z=1
\end{cases}$
We will use the addition method. Multiply Equation 1 by -2 and add it to Equation 2 to eliminate $x$:
$\begin{cases}
-2(x+2y-z)==-2(5)\\
2x-y+3z=0\\
2y+z=1
\end{cases}$
$\begin{cases}
-2x-4y+2z=-10\\
2x-y+3z=0\\
2y+z=1
\end{cases}$
$\begin{cases}
-2x-4y+2z+2x-y+3z=-10+0\\
2y+z=1
\end{cases}$
$\begin{cases}
-5y+5z=-10\\
2y+z=1
\end{cases}$
$\begin{cases}
y-z=2\\
2y+z=1
\end{cases}$
$y-z+2y+z=2+1$
$3y=3$
$y=1$
Substitute the value of $y$ in the Equation $y-z=2$ to determine $z$:
$y-z=2$
$1-z=2$
$z=-1$
Substitute the values of $y, z$ is Equation 1 of the given system to find $x$:
$x+2y-z=5$
$x+2(1)-(-1)=5$
$x+3=5$
$x=2$
The system's solution is:
$(2,1,-1)$
