Answer
$\{(3,2,4)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +3y &-z&=&5& ...... (1) \\
-x& +2y & +3z&=&13& ...... (2)\\
2x& -5y &-z &=&-8& ...... (3)
\end{matrix}\right.$
Addition method:-
Step 1:- Reduce the system to two equations in two variables.
Multiply the equation (1) by $-2$.
$\Rightarrow -2x-6y +2z=-10 $ ...... (4)
Add equation (1) and (2).
$\Rightarrow x+3y -z-x+2y+3z=5+13 $
$\Rightarrow 5y +2z=18 $ ...... (5)
Add equation (3) and (4).
$\Rightarrow 2x-5y-z-2x-6y +2z=-8-10 $
Simplify.
$\Rightarrow -11y+z=-18 $ ...... (6)
Step 2:- Solve the two equations from the step 1.
Multiply the equation (6) by $-2$.
$\Rightarrow 22y-2z=36 $ ...... (7)
Add equation (5) and (7).
$\Rightarrow 5y +2z+22y-2z=18+36 $
$\Rightarrow 27y=54 $
Divide both sides by $27$.
$\Rightarrow \frac{27y}{27}=\frac{54}{27} $
Simplify.
$\Rightarrow y=2 $
Step 3:- Use back-substitution to find the remaining two variables.
Substitute the value of $y$ into equation (5).
$\Rightarrow 5(2) +2z=18 $
Simplify.
$\Rightarrow 10 +2z=18 $
Subtract $10$ from both sides.
$\Rightarrow 10 +2z-10=18-10 $
Add like terms.
$\Rightarrow 2z=8 $
Divide both sides by $2$.
$\Rightarrow z=4$
Substitute the values of $y$ and $z$ into equation (1).
$\Rightarrow x +3(2) -(4)=5$
Simplify.
$\Rightarrow x +6 -4=5$
$\Rightarrow x +2=5$
Subtract $2$ from both sides.
$\Rightarrow x +2-2=5-2$
Simplify.
$\Rightarrow x =3$
The solution set is $\{(x,y,z)\}=\{(3,2,4)\}$.
