Precalculus (6th Edition) Blitzer

$(7t-5, 3t+1, t)$
Step 1. Based on the system of equations, write an augmented matrix as $\begin{bmatrix} 1 & 1 & -10 & | & -4 \\ 1 & 0 & -7 & | & -5 \\ 3 & 5 & -36 & | & -10 \end{bmatrix} \begin{array} ..\\R1-R2\to R2\\ -3R1+R3\to R3 \end{array}$ Step 2. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & 1 & -10 & | & -4 \\ 0 & 1 & -3 & | & 1 \\ 0 & 2 & -6 & | & 2 \end{bmatrix} \begin{array} ..\\..\\ -2R2+R3\to R3 \end{array}$ Step 3. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & 1 & -10 & | & -4 \\ 0 & 1 & -3 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \begin{array} ..\\..\\ .. \end{array}$ Step 4. The last row of the matrix indicates a dependent system. Let $z=t$; from the second equation, we have $y-3t=1$ and $y=3t+1$ Step 5. From the original second equation, we have $x-7t=-5$; thus $x=7t-5$ Step 6. The solution set can be written as $(7t-5, 3t+1, t)$