Consider the system of five equations in five variables, \begin{align*} x_1 - x_2 -2 x_3 + x_4 + 11 x_5 &= 13 \\ x_1 - x_2 + x_3+ x_4 + 5 x_5 &= 16 \\ 2 x_1 -2 x_2 + x_4 + 10 x_5 &= 21 \\ 2 x_1 -2 x_2 - x_3 + 3 x_4 + 20 x_5 &= 38 \\ 2 x_1 -2 x_ 2 + x_3 + x_4 + 8 x_ 5&= 22 \end{align*} whose augmented matrix row-reduces to \begin{equation*} \begin{bmatrix} \leading{1} & -1 & 0 & 0 & 3 & 6 \\ 0 & 0 & \leading{1} & 0 & -2 & 1 \\ 0 & 0 & 0 & \leading{1} & 4 & 9 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
Columns 1, 3 and 4 are pivot columns, so $D=\set{1,\,3,\,4}$. From this we know that the variables $x_1$, $x_3$ and $x_4$ will be dependent variables, and each of the $r=3$ nonzero rows of the row-reduced matrix will yield an expression for one of these three variables. The set $F$ is all the remaining column indices, $F=\set{2,\,5,\,6}$. The column index $6$ in $F$ means that the final column is not a pivot column, and thus the system is consistent (Theorem RCLS). The remaining indices in $F$ indicate free variables, so $x_2$ and $x_5$ (the remaining variables) are our free variables. The resulting three equations that describe our solution set are then, \begin{align*} (x_{d_1}=x_1)& & x_1&=6+x_2-3x_5\\ (x_{d_2}=x_3)& & x_3&=1+2x_5\\ (x_{d_3}=x_4)& & x_4&=9-4x_5 \end{align*}
Make sure you understand where these three equations came from, and notice how the location of the pivot columns determined the variables on the left-hand side of each equation. We can compactly describe the solution set as, \begin{equation*} S= \setparts{ \colvector{6+x_2-3x_5\\x_2\\1+2x_5\\9-4x_5\\x_5} }{x_2,\,x_5\in\complex{\null}} \end{equation*}
Notice how we express the freedom for $x_2$ and $x_5$: $x_2,\,x_5\in\complex{\null}$.