Answer
$\{(2,1,-1)\}$.
Work Step by Step
We will use the Elimination Method. let's note the equations of the given system:
$\left\{\begin{matrix}
2x& -y &+3z&=&0& ...... (1) \\
& 2y & +z&=&1& ...... (2)\\
x& +2y &-z &=&5& ...... (3)
\end{matrix}\right.$
Step 1:- Reduce the system to two equations in two variables.
Multiply the equation (3) by $-2$.
$\Rightarrow -2x-4y +2z=-10 $ ...... (4)
Add equation (1) and (4).
$\Rightarrow 2x-y+3z-2x-4y +2z=0-10 $
Simplify.
$\Rightarrow -5y+5z =-10 $ ...... (5)
Divide equation (5) by $-5$:
$\Rightarrow y-z =2 $ ...... (6)
Step 2:- Solve the two equations in two variables.
Add equation (2) and (6).
$\Rightarrow 2y+z+y-z=1+2$
Add like terms.
$\rightarrow 3y=3$
Divide both sides by $3$:
$\Rightarrow y=1$
Step 3:- Use back-substitution in step 2 to find second variable.
Substitute the value of $y$ into equation (6).
$\Rightarrow 1-z =2 $
Add $-1$ to both sides.
$\Rightarrow 1-z-1 =2-1 $
Simplify:
$\Rightarrow -z=1$
Multiply both sides by $-1$:
$\Rightarrow z =-1 $
Step 4:- Use back-substitution into original equation to find third variable.
Substitute the value of $y$ and $z$ into equation (3).
$\Rightarrow x +2(1)-(-1) =5$
Simplify.
$\Rightarrow x +2+1 =5$
$\Rightarrow x +3 =5$
Subtract $3$ from both sides.
$\Rightarrow x +3 -3=5-3$
Simplify.
$\Rightarrow x =2$
The solution set is $\{(x,y,z)\}=\{(2,1,-1)\}$.
