Answer
$\{(0,1,2)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
2x& -y &+z&=&1& ...... (1) \\
3x& -3y & +4z&=&5& ...... (2)\\
4x& -2y &+3z &=&4& ...... (3)
\end{matrix}\right.$
Addition method:-
Step 1:- Reduce the system to two equations in two variables.
Multiply the equation (1) by $-3$.
$\Rightarrow -6x+3y -3z=-3 $ ...... (4)
Multiply the equation (1) by $-2$.
$\Rightarrow -4x+2y -2z=-2 $ ...... (5)
Add equation (2) and (4).
$\Rightarrow 3x-3y+4z-6x+3y -3z=5-3 $
Simplify.
$\Rightarrow -3x+z =2 $ ...... (6)
Add equation (3) and (5).
$\Rightarrow 4x-2y+3z-4x+2y -2z=4-2 $
$\Rightarrow z=2 $ ...... (7)
Step 2:- Solve the two equations from step 1.
Substitute the value of $z$ into equation (6).
$\Rightarrow -3x+(2) =2 $
Subtract $2$ from both sides.
$\Rightarrow -3x+2-2 =2-2 $
Simplify and isolate $x$.
$\Rightarrow x=0$
Step 3:- Use back-substitution to find the remaining two variables.
Substitute the value of $x$ and $z$ into equation (1).
$\Rightarrow 2(0) -y +(2)=1$
Simplify.
$\Rightarrow 0 -y +2=1$
Subtract $2$ from both sides.
$\Rightarrow -y +2-2=1-2$
Add like terms.
$\Rightarrow -y =-1$
Multiply both sides by $-1$.
$\Rightarrow y =1$
The solution set is $\{(x,y,z)\}=\{(0,1,2)\}$.
