[tex]\begin{cases}x+2y+z=5\\2x-y+2z=15\\3x+y-z=8\end{cases}[/tex]
First, write the system in matrix form:
[tex]\begin{bmatrix}1&2&1\\2&-1&2\\3&1&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\15\\8\end{bmatrix}[/tex]
Now take the augmented matrix,
[tex]\left[\begin{array}{ccc|c}1&2&1&5\\2&-1&2&15\\3&1&-1&8\end{array}\right][/tex]
Add -2(row 1) to row 2, and add -3(row 1) to row 3:
[tex]\left[\begin{array}{ccc|c}1&2&1&5\\0&-5&0&5\\0&-5&-4&-7\end{array}\right][/tex]
Add -1(row 2) to row 3:
[tex]\left[\begin{array}{ccc|c}1&2&1&5\\0&-5&0&5\\0&0&-4&-12\end{array}\right][/tex]
Multiply through row 2 by -1/5, and multiply through row 3 by -1/4:
[tex]\left[\begin{array}{ccc|c}1&2&1&5\\0&1&0&-1\\0&0&1&3\end{array}\right][/tex]
We can already solve for [tex]y[/tex] and [tex]z[/tex] from here, but we'll hold off on that because the matrix still isn't in RREF, since the leftmost element in row 2 (1) is not the only non-zero element in its column. We fix this by adding -2(row 2) to row 1:
[tex]\left[\begin{array}{ccc|c}1&0&1&7\\0&1&0&-1\\0&0&1&3\end{array}\right][/tex]
and adding -1(row 3) to row 1:
[tex]\left[\begin{array}{ccc|c}1&0&0&4\\0&1&0&-1\\0&0&1&3\end{array}\right][/tex]
This tells us the solution to this system is [tex](x,y,z)=\boxed{(4,-1,3)}[/tex].
