Answer
$\{(2,-1,3)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
2x& -y &-2z&=&-1& ...... (1) \\
x& -2y & -z&=&1& ...... (2)\\
x& +y &+z &=&4& ...... (3)
\end{matrix}\right.$
Addition method:-
Step 1:- Reduce the system to two equations in two variables.
Multiply the equation (3) by $2$.
$\Rightarrow 2x+2y +2z=8 $ ...... (4)
Add equation (1) and (4).
$\Rightarrow 2x-y-2z+2x+2y +2z=-1+8 $
Simplify.
$\Rightarrow 4x+y =7 $...... (5)
Add equation (2) and (3).
$\Rightarrow x-2y-z+x+y +z=1+4 $
$\Rightarrow 2x-y=5 $ ...... (6)
Step 2:- Solve the two equations from the step 1.
Add equation (5) and (6).
$\Rightarrow 4x+y+2x-y=7+5$
Simplify.
$\Rightarrow 6x=12$
Divide both sides by $6$.
$\Rightarrow \frac{6x}{6}=\frac{12}{6}$
Simplify.
$\Rightarrow x=2$
Step 3:- Use back-substitution in one of the equations from step 2.
Substitute the value of $x$ into equation (5).
$\Rightarrow 4(2)+y =7 $
Simplify.
$\Rightarrow 8+y=7$
Subtract $8$ from both sides.
$\Rightarrow 8+y -8=7-8$
Add like terms.
$\Rightarrow y =-1$
Step 4:- Back substitute both variables into the original equation to find the third variable.
Substitute the value of $x$ and $y$ into equation (3).
$\Rightarrow 2 +(-1) +z=4 $
Simplify.
$\Rightarrow 2 -1 +z=4 $
$\Rightarrow 1 +z=4 $
Subtract $1$ from both sides.
$\Rightarrow 1 +z-1=4 -1$
Simplify.
$\Rightarrow z=3$
The solution set is $\{(x,y,z)\}=\{(2,-1,3)\}$.
