Reduced row echelon form. How do get to this matrix? I am reading this text and I'm stuck on how they get to the reduced row echelon form of this matrix:

I've tried a ton of combinations but I still can't get there. Can someone show me the way? Here's one that I have but then I'm stuck:
$\begin{matrix} 
\\ 1 & 2 & 5 
\\ 0 & -5 & -5 
\end{matrix}$
But then how do I get rid of the 2 in the second column of the first row to 0?
 A: We have 
$$\begin{bmatrix} 
1 & 2 &5 \\
2 & -1 & 5
\end{bmatrix}$$
Denoting row $i$ as $r_i$ and assume that the left arrow $A \leftarrow B$ means that we are putting quantity $B$ in $A$. Moreover,
we know that we can do linear operations on the rows to reach the row-echelon form, i.e.
$r_2 \leftarrow r_2 - 2r_1$
$$\begin{bmatrix} 
1 & 2 &5 \\
0 & -5 & -5
\end{bmatrix}$$
$r_1 \leftarrow r_1 + \frac{2}{5}r_2$
$$\begin{bmatrix} 
1 & 0 &3 \\
0 & -5 & -5
\end{bmatrix}$$
$r_2  \leftarrow -\frac{1}{5}r_2$
$$\begin{bmatrix} 
1 & 0 &3 \\
0 & 1 & 1
\end{bmatrix}$$
A: Proceed systematically instead of trying to guess at combinations that might work. You’ve gotten a pivot in the first row and everything else in the pivot column is zero, so you move on to the second row. Find the leftmost nonzero element and divide the row by that number to make its first nonzero element a $1$. Now it should be pretty obvious what multiple of the second row to subtract from the first to zero out the element above this new pivot. If there were more to the matrix, you’d also zero out the rest of the second column, then move on to the next nonzero row. Lather, rinse, repeat.
