## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( 3,1,5 \right)$.
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.