Answer
$(6-5t, \frac{7-3t}{2}, t)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix}-1&4&1&8\\ 2&-6&1&-9\\ 1&-6&-4&-15 \end{bmatrix}
\begin{array} \\ \\R_2+2R_1\to R_2\\R_3+R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix}-1&4&1&8\\ 0&2&3&7\\ 0&-2&-3&-7 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \end{array}$
Step 3. As the last two rows are the same, let $z=t$, the second row gives $2y+3t=7$ or $y=\frac{7-3t}{2}$, row-1 gives $-x+2(7-3t)+t=8$ or $x=6-5t$
Step 4. Conclusion: the solutions are $(6-5t, \frac{7-3t}{2}, t)$
