Question

the solution set of the inequation 3/(|x| + 2 <_ 1 is​

Answer 1

$\sf {\frac{3}{|x|} +2 ≤1}$

» $\sf{ \frac{3 + 2|x|}{|x|} ≤1}$

» $\sf{ 3 + 2|x| ≤ |x|}$

» $\sf { 3 ≤ -|x| }$

Multiply by -1, and reverse the inequality sign:

» $\sf { |x| ≤ -3}$

This isn't possible, since the modulus of any real number is always positive and not less than -3.