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How would I solve the following inequality problem.

$s+1<2s+1<4$

My book answer says $s\in (0, \frac32)$ as the final answer but I cannot seem to get that answer.

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  • $\begingroup$ Can you share your reasoning? $\endgroup$
    – Alex R.
    Commented Jan 12, 2013 at 3:47
  • $\begingroup$ Well what I did was get s<2s<3 but then I would divide by s and I do not get get how it would lead to zero. $\endgroup$ Commented Jan 12, 2013 at 3:49
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    $\begingroup$ When you have $a < x < b$, that means $a < x$ and $x < b$. Do you see how to solve the problem now? $\endgroup$ Commented Jan 12, 2013 at 3:53
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    $\begingroup$ Fernando: When people take time to answer your questions, and the answers are helpful to you, it is good to upvote those that are helpful. That's one way to say "thank you" so to speak. Also, for any question you ask, you can "accept" an answer. To upvote, click on the greyed-out up-arrow to the left of the answer. To accept an answer, click on the "greyed-out" check-mark next to the answer you want to accept. That indicates that the answer was particularly helpful/ and/or that it has fully answered your question. $\endgroup$
    – amWhy
    Commented Jan 16, 2013 at 0:28
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    $\begingroup$ @amWhy, I hope you don't mind: I presented your comment above as an exemplar in this answer. $\endgroup$
    – JRN
    Commented Oct 27, 2017 at 13:32

2 Answers 2

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We have $$s+1<2s+1<4.$$ This means $2s+1<4$, and in particular, $2s<3$. Dividing by the $2$ gives $s<3/2$. Now, observing on the other hand that we have $s+1<2s+1$, we subtract $s+1$ from both sides and have $0<s$. This gives us a bound on both sides of $s$, i.e., $$0<s<\frac{3}{2}$$ as desired.

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  • $\begingroup$ Oh thanks I did not know that. $\endgroup$ Commented Jan 12, 2013 at 3:57
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You have 2 inequality in 1

1)$s+1<2s+1$ and

2)$2s+1<4$

Now,you solve first the inequality 1)

$s+1<2s+1$

$0<s$

Then, solve the inequality 2)

$2s+1<4$

$2s<3$

$s<3/2$

Then, you have both, $0<s$ and $s<3/2$, namely $0<s<3/2$ and that's the answer in your book

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