Question 884687
Split it into two parts.


{{{3k<2k}}} allows you to divide both sides by k and you have a false statement:
{{{3<2}}}; so you must expect that {{{k<0}}}.  k can be any value as long as it is NEGATIVE.


The other part is {{{2k<k-4}}},
{{{2k-k<k-4-k}}}
{{{k<k-k-4}}}
{{{k<-4}}}.


The compound inequality must have the parts of the solutions which are true for both parts.  The parts of their solutions in common is  {{{highlight(k<-4)}}}.