Changes in Primal and Dual Solutions When Scaled

Question

Suppose that we have an LP in the standard form
$$\text{(P) min }c^Tx: Ax=b, x\le 0$$
and it's dual would be
$$\text{(D) max }b^Tx: Ax\le c$$
Let $\overline{x}$ and $\overline{y}$ be the optimal solutions to (P) and (D) respectively.

How do these solutions change if we multiply the constraint $Ax=b$ by a scalar? I think we'd have the following
$$\text{(P$\alpha$) min }c^Tx: \alpha Ax=b, x\le 0$$
and it's dual would be
$$\text{(D$\alpha$) max }b^Tx: \alpha Ax\le c$$
Where $\alpha \in \mathbb{R}-\{0\}$

My initial thoughts are that
1. The optimal value's of (P), (D), (P$\alpha$), and (D$\alpha$) won't change since the feasible region wont' change.
2. The optimal solutions will just be $\overline{x}/\alpha$ and $\overline{y}/\alpha$ for (P$\alpha$) and (D$\alpha$) respectively.


Yet, this seems wrong.
I'm rather new to Linear Programming and duality and would love an explanation of why my intuition is/isn't correct.

Answer 1

We would have

$$\text{(P$\alpha$) min }c^Tx: \alpha Ax=\color{blue}\alpha b, x\le 0$$
and it's dual would be
$$\text{(D$\alpha$) max }\color{blue}{\alpha}b^Tx: \alpha Ax\le c$$
Where $\alpha \in \mathbb{R}\setminus \{0\}$

The primal solution would remain the same since the feasible set remains the same. The corresponding dual solution would be $\frac1{\alpha}\bar{y}$.