## Precalculus (6th Edition) Blitzer

The complete solution is $\left\{ 2z-\frac{5}{4},\ \ \frac{13}{4},\ z \right\}$.
We start with: \begin{align} & x+y-2z=2 \\ & 3x-y-6z=-7 \end{align} Now converting it into an augmented matrix, we get, $\left[ \left. \begin{matrix} \begin{matrix} 1 & 1 & -2 \\ \end{matrix} \\ \begin{matrix} 3 & -1 & -6 \\ \end{matrix} \\ \end{matrix} \right|\begin{matrix} 2 \\ -7 \\ \end{matrix} \right]$ $\left[ \left. \begin{matrix} \begin{matrix} 1 & 1 & -2 \\ \end{matrix} \\ \begin{matrix} 0 & -4 & 0 \\ \end{matrix} \\ \end{matrix} \right|\begin{matrix} 2 \\ -13 \\ \end{matrix} \right]$ ${{R}_{2}}-3{{R}_{1}}\to {{R}_{2}}$ Now we will express this matrix into equations as below:. \begin{align} & x+y-2z=5 \\ & -4y=-13 \end{align} Further we will solve the equations for values of $x,y$ and $z$. \begin{align} & x=2z-\frac{5}{4} \\ & y=\frac{13}{4} \\ & z=z \end{align} Hence, the complete solution is $\left\{ 2z-\frac{5}{4},\ \ \frac{13}{4},\ z \right\}$.