Determine the solution to the given system of linear equations, by creating an augmented matrix and using Gauss-Jordan elimination to convert it to reduced row echelon form. What is the solution of $y$ ?
\[
\begin{array}{l}
-2 x+7 y-8 z=30 \\
3 x-3 y+4 z=-18 \\
-6 x+3 y-8 z=6
\end{array}
\]

Step 1 ：First, we write the system of equations as an augmented matrix: \[\begin{pmatrix} -2 & 7 & -8 & 30 \\ 3 & -3 & 4 & -18 \\ -6 & 3 & -8 & 6 \end{pmatrix}\]

Step 2 ：Next, we swap the first and second rows to get a leading 1 in the first row: \[\begin{pmatrix} 3 & -3 & 4 & -18 \\ -2 & 7 & -8 & 30 \\ -6 & 3 & -8 & 6 \end{pmatrix}\]

Step 3 ：Then, we multiply the first row by 1/3 to get a leading 1: \[\begin{pmatrix} 1 & -1 & 4/3 & -6 \\ -2 & 7 & -8 & 30 \\ -6 & 3 & -8 & 6 \end{pmatrix}\]

Step 4 ：We add 2 times the first row to the second row and 6 times the first row to the third row: \[\begin{pmatrix} 1 & -1 & 4/3 & -6 \\ 0 & 5 & -4 & 18 \\ 0 & -3 & 4 & -24 \end{pmatrix}\]

Step 5 ：We multiply the second row by 1/5 and the third row by -1/3 to get leading 1s: \[\begin{pmatrix} 1 & -1 & 4/3 & -6 \\ 0 & 1 & -4/5 & 18/5 \\ 0 & 1 & -4/3 & 8 \end{pmatrix}\]

Step 6 ：We subtract the second row from the third row: \[\begin{pmatrix} 1 & -1 & 4/3 & -6 \\ 0 & 1 & -4/5 & 18/5 \\ 0 & 0 & 2/15 & 2/5 \end{pmatrix}\]

Step 7 ：We multiply the third row by 15/2 to get a leading 1: \[\begin{pmatrix} 1 & -1 & 4/3 & -6 \\ 0 & 1 & -4/5 & 18/5 \\ 0 & 0 & 1 & 3 \end{pmatrix}\]

Step 8 ：We add the second row to the first row and subtract 4/3 times the third row from the first row: \[\begin{pmatrix} 1 & 0 & 0 & -4 \\ 0 & 1 & -4/5 & 18/5 \\ 0 & 0 & 1 & 3 \end{pmatrix}\]

Step 9 ：We add 4/5 times the third row to the second row: \[\begin{pmatrix} 1 & 0 & 0 & -4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{pmatrix}\]

Step 10 ：Finally, we can read off the solutions from the last column of the matrix. The solution for $y$ is \(\boxed{2}\)