problem related to duality theorem in linear programming 

Theorem
    $$\max\{c^Tx:x\ge0;Ax\le b\} = \min\{y^Tb:y\ge0;y^TA\ge c^T\}$$
    (assuming both sets are nonempty)

Use the above theorem to prove the following variant of the duality theorem:
  $$\max\{c^Tx:Ax=b;x\le0\} = \min\{y^Tb:y^TA\le c^T\}$$
  (assuming both sets are nonempty)

Please help me to prove the above variant of the duality theorem.
I am a masters student and linear programming is new to me. This question is a part of my assignment. I was not able to prove it.
 A: Rewrite the given problem into canonical form.
\begin{equation*}
  \begin{array}{lr@{}l}
    \max & c^T x & \\
    \text{s.t.} & Ax &{} = b \\
    & x &{} \le 0
  \end{array}
\end{equation*}
Replace $-x$ by $x$.
\begin{equation*}
  \begin{array}{lr@{}l}
    \max & -c^T x & \\
    \text{s.t.} & Ax &{} = -b \\
    & x &{} \ge 0
  \end{array}
\end{equation*}
Replace the equality with two inequalities.
\begin{equation*}
  \begin{array}{lr@{}l}
    \max & -c^T x & \\
    \text{s.t.} & Ax &{} \le -b \\
    & -Ax &{} \le b \\
    & x &{} \ge 0
  \end{array}
\end{equation*}
Compute its dual.  According to the (Strong Duality) Theorem, the optimal solution of the primal problem and its dual is the same.
\begin{equation*}
  \begin{array}{lr@{}l}
    \min & \begin{bmatrix} u \\ v \end{bmatrix}^T
    \begin{bmatrix} -b \\ b \end{bmatrix} & \\
    \text{s.t.} & \begin{bmatrix} u \\ v \end{bmatrix}^T
    \begin{bmatrix} -A \\ A \end{bmatrix} &{} \ge -c^T \\
    & u, v &{} \ge 0
  \end{array}
\end{equation*}
Simplify the results.
\begin{equation*}
  \begin{array}{lr@{}l}
    \min & -u^T b + v^T b & \\
    \text{s.t.} & u^T A - v^T A &{} \ge -c^T \\
    & u, v &{} \ge 0
  \end{array}
\end{equation*}
Factorise the terms.
\begin{equation*}
  \begin{array}{lr@{}l}
    \min & (v - u)^T b & \\
    \text{s.t.} & -(v - u)^T A &{} \ge -c^T \\
    & u, v &{} \ge 0
  \end{array}
\end{equation*}
Note that $y := v - u$ is free.  (i.e. $y$ can be any vector in $\mathbb{R}^m$).
\begin{equation*}
  \begin{array}{lr@{}l}
    \min & y^T b & \\
    \text{s.t.} & -y^T A &{} \ge -c^T
  \end{array}
\end{equation*}
We finally get the problem on the RHS.
\begin{equation*}
  \begin{array}{lr@{}l}
    \min & y^T b & \\
    \text{s.t.} & y^T A &{} \le c^T
  \end{array}
\end{equation*}
