How do you graph #3/2x+2/3y>1# on the coordinate plane?

We can simplify the given inequality: #3/2x+2/3y > 1# by multiplying all terms, on both sides, by #6#, to get rid of the fractions:

#9x+4y > 6#

Next we need to determine the plot line for #9x+4y=6#
and then determine which side of the line to shade in as #9x+4y > 6#

By testing a series of arbitrary integer values for #x# we can eventually find some corresponding integer values for #y#.
(I find it easier to plot with integer coordinates).

#color(white)("X")ul(x)color(white)("xxxxx")ul(y)#
#-2color(white)("xxxx")+6#
#+2color(white)("xxxx")-3#
#+6color(white)("xxxx")-12#

Plotting these coordinate pairs and drawing a line through them should give a graph like:

...BUT we only want the points for which #3/2x+2/3y>1#
That is we only want the points on one side of this line.

We notice that #(x,y)=(0,0)# does not satisfy the given inequality,
so the origin, #(0,0)# is not on the side we want.

We need to shade the side that does not include the origin (and hollow out the circles on the equality line to show that it is not included):