Solve the inequality $\displaystyle \frac{2}{x-5}<\frac{1}{x+1}$
Hence, solve $\displaystyle \frac{2}{|x|-5}<\frac{1}{|x|+1}$
For part(a), you should know you can't multiply either of the denominators like how an algebraic equation involving rational function is solved. First step, move all function to one side and simplify.
$\displaystyle \frac{2}{x-5}-\frac{1}{x+1}<0$
$\displaystyle \frac{x+7}{(x-5)(x+1)}<0
$
The solution set is {x: x<-7 , -1<x<5}
As for part (b), let |x|=a
then the solutions are a<-7 and -1<a<5
back substituting, |x|<-7 and -1<|x|<5
Obviously, |x|<-7 and |x|>-1 are not valid since $\displaystyle |x|\geq 0$
so the only solution would be |x|<5 or -5<x<5
You cannot cross multiply because x can take any value, which includes values less for example less than 1. If that happened, then you'll be multiplying everything by a negative value, which will change the sign of the inequality. Since you don't know the value of x, you mustn't multiply.
Perhaps i was a bit unclearunclear $\displaystyle \neq $ wrong -.-it's not that |x| cannot be greater than -1. In fact, it's always greater than -1 regardless of the value of x so it will not affect the inequality.
I tot you were unclear so I keep thinking |x|>-1 is not valid since because $\displaystyle x|\geq 0$ which doesn't make sense.
But in the end I still benefit from your help.
another qn:
if |x|>-0.05, there is infinite solutions, hence no solution, right?
if |x|<-0.05, there is also no solution, right?
confirming.....
This was my original reply to you. You did nothing with it.
Take case i) $\displaystyle x<-7$ so the intersection is empty.
For case ii) $\displaystyle x>-7$ so the intersection is $\displaystyle -1<x<5$.
For case iii) $\displaystyle x<-7$ so the intersection is $\displaystyle x<-7$.
Unite those set and you get: $\displaystyle \left( { - \infty , - 7} \right) \cup \left( { - 1,5} \right)$