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