Answer
$\{(2,1,-1)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +2y &-z&=&5& ...... (1) \\
2x& -y & +3z&=&0& ...... (2)\\
& 2y &+z &=&1& ...... (3)
\end{matrix}\right.$
Addition method:-
Step 1:- Reduce the system to two equations in two variables.
Multiply the equation (1) by $-2$.
$\Rightarrow -2x-4y +2z=-10 $ ...... (4)
Add equation (2) and (4).
$\Rightarrow 2x-y+3z-2x-4y +2z=0-10 $
Simplify.
$\Rightarrow -5y+5z =-10 $
Divide both sides by $-5$.
$\Rightarrow y-z =2 $ ...... (5)
Step 2:- Solve the equations from the step 1.
Add equation (3) and (5).
$\Rightarrow 2y+z+y-z =1+2 $
Add like terms.
$\Rightarrow 3y =3 $
Divide both sides $3$.
$\Rightarrow y =1 $
Step 3:- Use back-substitution to find the remaining two variables.
Substitute the value of $y$ into equation (3).
$\Rightarrow 2(1)+z=1 $
Subtract $2$ from both sides.
$\Rightarrow 2+z-2=1-2 $
Add like terms.
$\Rightarrow z=-1 $
Substitute the value of $y$ and $z$ into equation (1).
$\Rightarrow x +2(1) -(-1)=5$
Simplify.
$\Rightarrow x+3=5$
Subtract $3$ from both sides.
$\Rightarrow x+3-3=5-3$
Add like terms.
$\Rightarrow x=2$
The solution set is $\{(x,y,z)\}=\{(2,1,-1)\}$.
