Suppose $x, y \in \mathbb{R}$, and $\mathcal{S_1}$ is a system of inequalities: \begin{align*} \mathcal{S_1} &= \begin{Bmatrix} x - y \geq 1\\ -x + 2y \geq 1\\ 3x - 5y \geq 2 \end{Bmatrix}\\ &= \begin{Bmatrix} x \geq 1 + y\\ x \leq 2y-1\\ x \geq \frac{2+5y}{3} \end{Bmatrix} \end{align*}

I eliminate $x$ from $S_1$ to obtain $S_2$:

\begin{align*} \mathcal{S_2} &= \begin{Bmatrix} 1+y \leq 2y -1 \\ \frac{2+5y}{3} \leq 2y-1 \end{Bmatrix}\\ &= \begin{Bmatrix} 2 \leq y \\ 2+5y \leq 6y-3 \end{Bmatrix}\\ &= \begin{Bmatrix} 2 \leq y \\ 5 \leq y \end{Bmatrix} \end{align*}

Therefore, I know that $5 \leq y < \infty$. Let $S_3$ denote the following system of inequalities

\begin{align*} \mathcal{S_3} &= \begin{Bmatrix} x - y \geq 1\\ -x + 2y \geq 1\\ 3x - 5y \geq 2\\ 5 \leq y < \infty \end{Bmatrix} \end{align*}

My question is, is $S_3 = S_1$?

  • $\begingroup$ Look at the $x-y \geq 1$ inequality. It should become $x\geq 1+y$. This changes to give $y\geq 2$ instead. Yes, the sets are the same as all operations don't change the amount of info. $\endgroup$ Commented May 7, 2021 at 1:58
  • $\begingroup$ This is the question you asked here. $\endgroup$
    – Nexball
    Commented May 7, 2021 at 2:00
  • $\begingroup$ @MSE, you're right, the two questions are similar, but technically, $S_3$ is not the end result of what one would get from doing FME. I wanted to double-check that $S_1 = S_3$ holds. $\endgroup$
    – Adrian
    Commented May 7, 2021 at 2:10
  • $\begingroup$ @AdamKarlson Sorry I'm not following. What do you mean by this changes to give $y \geq 2$ instead? $\endgroup$
    – Adrian
    Commented May 7, 2021 at 2:13
  • $\begingroup$ @Adrian Oh, I am saying that the there is a mistake in the line I pointed out. Try doing it again $\endgroup$ Commented May 7, 2021 at 2:16

1 Answer 1


Yes, $S_3=S_1$. We need only to show that


Therefore, we have,

$$3(-x+2y)+3x-5y≥3\times 1+2 $$

$$\iff y≥5.$$

But, this notation is problematic.


The correct one can be written as follows:





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