# Inequality problem algebra?

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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Can you share your reasoning? –  Alex R. Jan 12 '13 at 3:47
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. –  Fernando Martinez Jan 12 '13 at 3:49
When you have $a < x < b$, that means $a < x$ and $x < b$. Do you see how to solve the problem now? –  George V. Williams Jan 12 '13 at 3:53
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. –  amWhy Jan 16 '13 at 0:28

## 2 Answers

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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Oh thanks I did not know that. –  Fernando Martinez Jan 12 '13 at 3:57

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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