Question 549347
{{{3x + 8 < 2}}}     (1)
{{{x + 12 > 2 - x}}} (2)

Looking at (1) we can see that:
{{{3x < 2-8}}}  
{{{3x <-6}}} 
{{{3x/3 <-6/3}}} 
{{{x <-2}}}


Looking at (2) we can see that     
{{{x + 12+x > 2 - x+x}}}
{{{2x + 12 >2}}}
{{{2x >2-12}}}
{{{2x >-10}}}
{{{x >-10/2}}}
{{{x >-5}}}



So to satisfy both (1) and (2) we must have {{{-5<x<-2}}}