\begin{align} \min_{x} c^Tx \\ s.t.~Ax=b \\ Unrestricted \end{align}
Take $x=x_1-x_2$ \begin{align} \min c^T(x_1-x_2) \\ s.t.~A(x_1-x_2)=b \\ ~ x_1, x_2 \ge 0 \end{align}
This is can be written as \begin{align} \min {\begin {pmatrix}c \\ -c \\ \end {pmatrix} }^T \begin {pmatrix} x_1 \\ x_2 \\ \end {pmatrix} \\ s.t.~\begin {pmatrix} A, & -A \end {pmatrix} \begin {pmatrix} x_1 \\ x_2 \\ \end {pmatrix}=b \\ ~ x_1, x_2 \ge 0 \end{align}
Now, this is in the standard form of the linear program. Therefore, the dual can be written as
\begin{align} \max b^T y \\ s.t.~ \begin {pmatrix}c \\ -c \\ \end {pmatrix} - \begin {pmatrix} A^T\\ -A^T \\ \end {pmatrix} y \ge 0\\ ~ y \ge 0 \end{align}
This can be simplify as
\begin{align} \max b^T y \\ s.t.~ c-A^T y \ge 0\\ ~ -c+A^T y \ge 0\\ ~ y \ge 0 \end{align}
\begin{align} \max b^T y \\ s.t.~ A^T y \le c\\ ~ A^T y \ge c\\ ~ y \ge 0 \end{align}
This is equivalent to
\begin{align} \max b^T y \\ s.t.~ A^T y = c\\ ~ y \ge 0 \end{align}