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