This doubt has been bothering me for ages. I would be truly grateful for any help.
Problem 1: $\dfrac{2}{|x-4|}>1$ Express the solutions using intervals
Solution: $x\in(2,4)\cup(4,6)$
Problem 2: $\dfrac{|x+3|+x}{x+2} >1$ Express the solutions using intervals
Solution: $x\in (-1,\infty)\cup (-5,-3)\cup (-3,-2)\Rightarrow x\in (-1,\infty)\cup (-5,-2)$
In general, let the solutions of any modulus inequality be $x\in (a,b)\cup(b,c)\cup(c,d)$ Then how do we decide when to merge two or more intervals together (as was done for the second problem) When do we $not$ merge two intervals together?
Any help and guidance would be really really appreiated.
$$\dfrac{|x+3|+x}{x+2} >1$$
Assume $x\le -3$ $$\dfrac{|x+3|+x}{x+2} >1\iff\dfrac{-(x+3)+x}{x+2} >1\iff -(x+3)+x<x+2\iff-5<x$$ So the first interval is indeed $(-5,-3]$
Now let $-3\le x < -2$ $$\dfrac{|x+3|+x}{x+2} >1\iff\dfrac{+(x+3)+x}{x+2} >1\iff +(x+3)+x<x+2\iff x<-1$$ So the second interval is $[-3,-2)$
Now $x>-2$ $$\dfrac{|x+3|+x}{x+2} >1\iff\dfrac{+(x+3)+x}{x+2} >1\iff +(x+3)+x>x+2\iff x>-1$$
So $(-1,+\infty)$ is the third interval.
Putting the three together leads to: $$S=(-5,-3]\cup [-3,-2) \cup (-1,+\infty)= (-5,-2) \cup (-1,+\infty)$$