Equivalent representation in linear programming

Question

I have a very simple linear problem: $$\begin{split}\min_{x}&\ x^2\\\text{s.t. }&\ a_1x_1+a_2x_2=b\end{split}$$ Suppose I want to write this problem equivalently as in Find the equivalent linear program. Unlike the problem in the link, I have equality. Can I write it equivalently as: $$\begin{split}\min_{x,\alpha,\beta}&\ x^2\\\text{s.t. }&\ a_1x_1=\alpha b,\ a_2x_2=\beta b,\ \alpha+\beta=1.\end{split}$$ The converse is intuitive: Given $\{x,\alpha,\beta\}$ feasible for the second problem, adding the first and second constraints gives the constraint of the first problem. But the forward part is not clear, especially because I have never seen an equality constraint written like this. Any help would be highly appreciated.

Answer 1

You are correct. Suppose $x$ is feasible. Then $a_1x_1 + a_2x_2 = b \iff (a_1x_1 + a_2x_2 = \left(\alpha + \beta)b \text{ and } \alpha + \beta = 1\right)$. Therefore any solution satisfying the latter conditions is also feasible.

A simple example to explain what I think is the part you are unclear about is the following. Look at the following two problems:

1. \begin{split}\min_{x}&\ x\\ \text{s.t. }&\ x \geq 2 \\ & x \in \mathbb{R} \end{split}

2.
\begin{split}\min_{x}&\ x\\ \text{s.t. }&\ x \geq 2 \\ & x\geq 1 \\ & x \in \mathbb{R}\end{split}

They obviously generate the same feasible solution; the constraint added is weaker than the existing constraint. Similarly the following problem also generates the same feasible solution:

\begin{split}\min_{x,a}&\ x\\ \text{s.t. }&\ x \geq a \\ & a = 2 \\ & x \in \mathbb{R}\end{split}