Answer
$(6-2t,4-t,t)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&1&2\\ 1&1&3&6\\ 3&-1&5&10 \end{bmatrix}
\begin{array} \\ \\R_2-R_1\to R_2\\R_3-3R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&1&2\\ 0&2&2&4\\ 0&2&2&4 \end{bmatrix}
\begin{array} \\ \ \\ \\ \end{array}$
Step 3. The last two rows are the same. Let $z=t$, we have $y=4-t$ and $x=6-2t$ (using back-substitution)
Step 4. Conclusion: the solutions to the system are $(6-2t,4-t,t)$
