Here's one method :
All the graphs of the lines are the same.....just the solution areas are different......so....let's just pick a point and see if it solves the original system
(0,0) is one I like to try, first......so we have
(0) + (3/2)(0) > - 3 ⇒ 0 > -3
0 - 5 ≤ - 6(0) ⇒ -5 ≤ 0
Note that both of these are true.....so ...(0,0) is in the solution region.....the only graph that has this point in the solution area is the lower right graph