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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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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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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