# 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.

• Can you share your reasoning? Commented Jan 12, 2013 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. Commented Jan 12, 2013 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? Commented Jan 12, 2013 at 3:53
• @amWhy, I hope you don't mind: I presented your comment above as an exemplar in this answer.
– JRN
Commented Oct 27, 2017 at 13:32

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.

• Oh thanks I did not know that. Commented Jan 12, 2013 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