0
$\begingroup$

I'm currently studying for a class where the teacher's notes were given, but there are many errors here and there so I need to make sure that everything on it is correct. I'm given the following system of inequations:

$$ \left\{\begin{matrix} 0 \leq t'_1 + t_2 \leq 10 \;\;\;\;\; (a) \\ 5\leq t_2 \leq 15 \;\;\;\;\;\;\;\;\;\;\;\; (b) \\ 12\leq t'_3+t_2\leq 22 \;\;\; (c) \end{matrix}\right. $$

Now I need to eliminate t2 form (a) and (c) so on the notes here is the resulting system:

$$ \left\{\begin{matrix}-5 \leq t'_1 \leq 5 \;\;\;(a+(-b)) \\2\leq t'_3 \leq 17\;\;\; (c+(-b)) \\5\leq t_2\leq15 \;\;\;\;\;\;\;\;\;\;\;\;\;\;(b) \end{matrix}\right. $$

Now the question is: can someone guide me through the process of eliminating t2 ? Are the resulting inequations actually correct ?

$\endgroup$

1 Answer 1

0
$\begingroup$

To eliminate $t_2$, first we multiply the second inequation by $-1$ and then we add it to the first one:
$$5 \leq t_2 \leq 15 \implies -5 \geq -t_2 \geq -15$$ We need to flip the inequation signs since we're multiplying by a negative number.
Now, we have a problem here as we can't add 2 inequations if the inequation signs are opposite; we can only subtract them. Since subtracting them is equivalent to just adding them before multiplying by $-1$, we see that either way this won't work in eliminating $t_2$. The notes are definitely wrong; they forgot to flip the inequation signs. (Don't even ask me about the havoc they wreaked on the last inequation; for starters, $(b)$ isn't even the same inequation in the top than in the bottom!)

As for how I would solve this, nothing glares at me in the face, but maybe splitting each compound inequation into two and solving for one of two variables could lead to extracting more information and solving it.

$\endgroup$
2
  • $\begingroup$ Thanks in advance, I'll check your answer later. I'm very sorry I made a mistake on inequation b. It is the same on the notes I just made an error typing, sorry about that. I edited so that now it is correct (at least (b) in both systems) $\endgroup$
    – G4bri3l
    Commented Oct 9, 2014 at 17:13
  • $\begingroup$ Don't be sorry - everyone slips up (Believe me, I should know!) I just thought it was from the notes too. I'll edit my post and add the inequation's solutions if I figure something out. It seems that it should be easy, but I haven't dealt much with systems like this before so... $\endgroup$
    – Daccache
    Commented Oct 10, 2014 at 1:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .