\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}