Answer
The solution of the system is $\left( 3,1,5 \right)$.
Work Step by Step
Let us consider the given equation:
$2x-4y+3z=17$ …… (I)
$x+2y-z=0$ …… (II)
$4x-y-z=6$ …… (III)
Multiply equation (II) by 3 and add equation (I):
$\begin{align}
& 2x-4y+3z=17 \\
& \underline{3x+6y-3z=0} \\
& 5x+2y=17 \\
\end{align}$
$5x+2y=17$ ….. (IV)
Now, subtract equation (III) from equation (II):
$\begin{align}
& x+2y-z=0 \\
& 4x-y-z=6 \\
& \underline{-\text{ + + -}} \\
& -3x+3y=-6 \\
\end{align}$
$-3x+3y=-6$ …… (V)
Now, multiply equation (IV) by 3 and multiply equation (V) by 5 and add both equations,
$\begin{align}
& \text{ }15x+6y=51 \\
& \underline{-15x+15y=-30} \\
& \text{ }21y=21 \\
& \text{ }y=1 \\
\end{align}$
Putting the value of $y=1$ in equation (V):
$\begin{align}
& -3x+3(1)=-6 \\
& -3x=-9 \\
& x=3
\end{align}$
Now, putting the value of $x=3$ and $y=1$ in equation (I):
$\begin{align}
& 2\left( 3 \right)-4\left( 1 \right)+3z=17 \\
& 6-4+3z=17 \\
& z=5
\end{align}$
Thus, the ordered triple $\left( 3,1,5 \right)$ satisfies the given system.
