## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( 1,-1,2 \right)$.
Now, it is given that: $x-y+3z=8$ ...... (I) $3x+y-2z=-2$ ...... (II) $2x+4y+z=0$ ...... (III) Now, adding equation (I) and equation (II): \begin{align} & x-y+3z=8 \\ & 3x+y-2z=-2 \end{align} $4x+0y+z=6$ ...... (IV) By multiplying equation by 4 and adding with equation (III): \begin{align} & 4x-4y+12z=32 \\ & 2x+4y+z=0 \\ \end{align} $6x+13z=32$ ...... (V) Now multiply equation (IV) by 13 and subtract it from equation (V): \begin{align} & 6x+13z=32 \\ & 52x+13z=78 \\ \end{align} Solve and we get: \begin{align} & -46x=-46 \\ & x=1 \end{align} Putting the value of $x=1$ in equation (IV): \begin{align} & 4x+z=6 \\ & 4\left( 1 \right)+z=6 \\ & z=2 \end{align} Putting the value of $x=1$ and $z=2$ in equation (I): \begin{align} & 1-y+3\left( 2 \right)=8 \\ & 1-y+6=8 \\ & -y=1 \\ & y=-1 \end{align} Thus, the order triple $\left( 1,-1,2 \right)$ satisfies the three systems.