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 $y$ ?
$$\begin{array}{l}-2x+7y-8z=30\\ 3x-3y+4z=-18\\ -6x+3y-8z=6\end{array}$$

Step 1 ：First, we write the system of equations as an augmented matrix: $$\left(\begin{array}{cccc}-2& 7& -8& 30\\ 3& -3& 4& -18\\ -6& 3& -8& 6\end{array}\right)$$

Step 2 ：Next, we swap the first and second rows to get a leading 1 in the first row: $$\left(\begin{array}{cccc}3& -3& 4& -18\\ -2& 7& -8& 30\\ -6& 3& -8& 6\end{array}\right)$$

Step 3 ：Then, we multiply the first row by 1/3 to get a leading 1: $$\left(\begin{array}{cccc}1& -1& 4/3& -6\\ -2& 7& -8& 30\\ -6& 3& -8& 6\end{array}\right)$$

Step 4 ：We add 2 times the first row to the second row and 6 times the first row to the third row: $$\left(\begin{array}{cccc}1& -1& 4/3& -6\\ 0& 5& -4& 18\\ 0& -3& 4& -24\end{array}\right)$$

Step 5 ：We multiply the second row by 1/5 and the third row by -1/3 to get leading 1s: $$\left(\begin{array}{cccc}1& -1& 4/3& -6\\ 0& 1& -4/5& 18/5\\ 0& 1& -4/3& 8\end{array}\right)$$

Step 6 ：We subtract the second row from the third row: $$\left(\begin{array}{cccc}1& -1& 4/3& -6\\ 0& 1& -4/5& 18/5\\ 0& 0& 2/15& 2/5\end{array}\right)$$

Step 7 ：We multiply the third row by 15/2 to get a leading 1: $$\left(\begin{array}{cccc}1& -1& 4/3& -6\\ 0& 1& -4/5& 18/5\\ 0& 0& 1& 3\end{array}\right)$$

Step 8 ：We add the second row to the first row and subtract 4/3 times the third row from the first row: $$\left(\begin{array}{cccc}1& 0& 0& -4\\ 0& 1& -4/5& 18/5\\ 0& 0& 1& 3\end{array}\right)$$

Step 9 ：We add 4/5 times the third row to the second row: $$\left(\begin{array}{cccc}1& 0& 0& -4\\ 0& 1& 0& 2\\ 0& 0& 1& 3\end{array}\right)$$

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