Relationship between Primal and Dual problems Considering the following program:
\begin{cases}
\max & 8x_1 & + 3x_2\\
& x_1 &-6x_2&\ge2\\
& 5x_1 +&7x_2&=-4\\
&x_1&&\le 0\\
&& x_2&\ge 0
\end{cases}
Why do we have 
\begin{cases}
\min & 2w_1 & -4 w_2\\
& w_1 &+5x_2&\ge8\\
& -6w_1 &+7w_2&\ge 3\\
&w_1&&\le 0\\
&& w_2 \mbox{ unrestricted}
\end{cases}
And not
\begin{cases}
\min & 2w_1 & -4 w_2\\
& w_1 &+5x_2& +w_3&\le8\\
& -6w_1 &+7w_2&&\le 3\\
& 6w_1 &-7w_2&&\le -3\\
&w_1&&&\le 0\\
&& w_2 &&\le 0
\end{cases}
I thought having this from the table 6.1 given in Mokthar S.Bazara and John J.Jarvis, Linear Programing and Network Flows p241
 A: When you write the constraints of the dual problem, you should proceed in 3 steps:


*

*Transpose the matrix of constraints, excluding the constraints of type $x_i\ge\mbox{or}\le 0$ (which you did not do, as you have 3 constraints in your dual matrix). The right hand term are the costs of the objective function. For the moment, the signs $\ge\mbox{or}\le \mbox{or} =$ are undefined:
\begin{cases}
&\omega_1+5\omega_2 &\mbox{ ? } 8\\
&-6\omega_1+7\omega_2 &\mbox{ ? } 3\\
\end{cases}

*Now, determine the signs  $\ge\mbox{or}\le \mbox{or} =$. To do so, check how variables $x_1$ and $x_2$ are constrained:$x_1$ should be negative, which means the first constraint should have a $\le$ sign. $x_2$ is positive, so the second constraint should have a $\ge$ sign:
\begin{cases}
&\omega_1+5\omega_2 &\le 8\\
&-6\omega_1+7\omega_2 &\ge 3\\
\end{cases}

*The last step is to determine how variables $\omega_1$ and $\omega_2$ are constrained. Since the first primal constraint has a $\ge$ sign, $\omega_1$ should be negative, and since the second one has a $=$ sign, $\omega_2$ is unconstrained:
\begin{cases}
&\omega_1+5\omega_2 &\le 8\\
&-6\omega_1+7\omega_2 &\ge 3\\
&\omega_1 \le 0\\
&\omega_2 \in \mathbb{R}
\end{cases}

