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Given something like $\frac{2n+1}{n+2} < 8$ how do you solve that step by step to get all the possible intervals?

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5 Answers

up vote 2 down vote accepted

it presents two cases, when $(n+2)$ > 0 and when $(n+2)<0$

$(2n+1)< 8(n+2)$ $for$ $n\not=2$ and $n>-2 \ $ it follows that $n> -\frac{5}{2}$ hence $ n \in(-2,\infty)$

$(2n+1)> 8(n+2)$ $for$ $n\not=2$ and $n<-2 \ $ then $n< -\frac{5}{2}$ and $n<-2$ hence $ n \in(-\infty,-\frac{5}{2})$

The answer is the union of both sets $(-\infty,-\frac{5}{2}) \cup (-2,\infty) $

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The answer arrived at is incorrect. –  André Nicolas Oct 20 '13 at 18:07
    
why? can you explain it? –  JulianP Oct 21 '13 at 5:16
    
The $5/3$ is a minor arithmetical slip. Your answer says basically the inequality holds everywhere except at a couple of points. But the inequality fails at $-2.2,-2.1,-2.05$, lots of places. –  André Nicolas Oct 21 '13 at 5:22
    
I see it, i will edit the answer –  JulianP Oct 21 '13 at 5:24
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There are two cases to consider: (i) $n+2\gt 0$, or equivalently $n\gt -2$; (ii) $n+2\lt 0$. In each case we will multiply both sides of the inequality by $n+2$. We use the fact that multiplying by a positive number preserves an inequality, while multiplying by a negative number reverses the inequality.

Case 1: The inequality is equivalent to $2n+1\lt 8(n+1)$, or equivalently $6n\gt -15$, or equivalently $n\gt -\frac{5}{2}$. But recall that Case (i) assumes that $n\gt -2$. So Case (i) has the solutions $n\gt -2$.

Case (ii): The inequality is equivalent to $2n+1\gt 8(n+1)$, which simplifies to $n\lt -\frac{5}{2}$. The condition $n\lt -2$ is compatible with this, so we get the solutions $n\lt -\frac{5}{2}$.

Thus the solutions to the original inequality are all numbers in the interval $(2,\infty)$, together with all numbers in the interval $(-\infty, -\frac{5}{2})$.

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Original Inequation: $$\frac{2n+1}{n+2} < 8$$

Subtracting both sides by 8 and taking the LCM:

$$\frac{2n + 1 - 8(n + 2)}{n+2} <0$$ This simplifies to: $$\frac{6n +15}{n+2} >0$$

Now take the cases when both the numerator and denominator are positive, when one of them is positive and the other is negative, and when both are negative. Which case gives you what you want (i.e. a positive value for the fraction?)

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Observe that $\displaystyle \lim_{n+2\to0^-}\frac{2n+1}{n+2}\to+\infty$ and $\displaystyle \lim_{n+2\to0^+}\frac{2n+1}{n+2}\to-\infty$

If $n+2\ne0,(n+2)^2>0$ for real $n$

$$\frac{2n+1}{n+2}<8\iff \frac{2n+1}{n+2}-8<0$$

Now, $$\frac{2n+1}{n+2}-8=-\frac{6n+15}{n+2}=-3 \frac{(2n+5)}{n+2}$$

$$\implies -3 \frac{(2n+5)}{n+2}<0\iff\frac{(2n+5)}{n+2}>0$$

Method $1:$

If $2n+5>0\iff n>-\frac52, n+2>0\iff n>-2\implies n>-2$

or if $2n+5<0\iff n<-\frac52, n+2<0\iff n<-2\implies n<-\frac52$

Method $2:$

Multiplying either sides by $(n+2)^2>0$ assuming $n+2\ne0$

$$ (2n+5)(n+2)>0 \iff\left(n+\frac52\right)(n+2)>0$$

We know, if $(x-a)(x-b)>0$ where $a,b$ we shall have $x<a$ or $x>b$

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How did you go from $\frac{2n+5}{n+2} > 0$ to $(2n+5)(n+2) > 0$? –  matthras Oct 20 '13 at 5:46
    
@matthras, multiplying either sides by $(n+2)^2$ which is $>0$ unless $n+2=0$ –  lab bhattacharjee Oct 20 '13 at 5:47
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$$\frac{2n+1}{n+2} < 8\tag{1}$$ In this inequality we have a fraction, where the unknown term ($n$ in this case) is present in both the numerator and the denominator. So the first thing will have to do is to find someway to eradicate that $n$ from the denominator and to see what happens next. To do that, will have to use the following properties:

Property $\bf 1:$ Let $a$,$b$ and $c$ be real numbers, where $a\gt b$. We have that $ac>bc$ if and only if $c\gt0.$

Property $\bf 2:$ If $a$,$b$ and $c$ are real numbers, where $a\gt b$ and $c$ is strictly negative, we have that $ac\lt bc$.

Note: For every non-zero real number $x$, we have that: $\dfrac1x\cdot x=1$

So to remove that $n+2$ in the denominator, will have to multiply it by its inverse, which is of course $n+2$. But wait a second! From the properties, we should know whether we're multiplying by a positive or negative number since the inequality sign changes with the latter. But here, we simply don't know whether $n+2$ is positive or negative since it can have any value! So here we should do two cases, one where $n+2$ is positive and the other where $n+2$ is negative.

Case $\bf 1:$ We have: $n+2\gt0$

$$\begin{align} \color{red}{(n+2)}\cdot\frac{2n+1}{n+2} &\overset{\bf Pr.1}{<} 8\cdot\color{red}{(n+2)}\tag{$\underline{\underline{\text{btw} \,\,n+2\neq0}}$} \\&\Downarrow\\ 2n+1 &< 8n+16 \\ \color{blue}{-2n}+2n+1 &< 8n+16 \color{blue}{-2n}\\ 1+\color{blue}{-16} & < 6n +16 \color{blue}{-16} \\ -15 &<6n \\&\Downarrow\\ -15 \cdot\color{red}{\dfrac16}&\overset{\bf Pr.1}{<}6n\cdot\color{red}{\dfrac16}\\\,\\ -\dfrac{15}6&\lt n \,\\\,\\ -\dfrac{5}2&<n\iff n\in(-\infty,-5/2)=\mathcal A\\ \end{align}$$

Case $\bf 2:$ We have: $n+2<0$.

I think now that you have the necessary tools to solve it. After that you'll have a set $\mathcal B$ and so the solutions to the inequality $\text{(1)}$ will just be $\mathcal A\cup\mathcal B$ without forgetting to exclude the element $-2$ for which $\text{(1)}$ is undefined. (since $n+2=0\iff n=-2$.)

I hope this helps and have fun!

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