#2**+1 **

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

CPhill
Dec 19, 2017