Question 642887
you are given:

{{{3x-2y >=0}}}}

that inequality simplifies to:

{{{3x >= 2y}}}....=>...{{{2y<=3x }}}=>...{{{y<=(3/2)x }}}


The {{{boundary}}} of the inequality is {{{y = (3/2)x}}}

The point ({{{0}}},{{{0}}}) is {{{on}}} the boundary line.  It {{{satisfies}}} the
equality but {{{cannot}}} be used to test the inequality {{{because}}} it is {{{on }}}the {{{boundary}}}{{{ line}}}.  

You have to use a test point that is {{{not}}} on the {{{boundary}}}{{{ line}}}.

Using ({{{-4}}}, {{{2}}}) you get {{{2<=(3/2)(-4) }}}...=>...{{{2<=(3*(-4)/2) }}}..=>...{{{2<=(-12)/2) }}}
=>...{{{2<=-6 }}}
That is {{{not}}} {{{true}}} so the half-plane does containing ({{{-4}}}, {{{2}}}) is {{{not}}} the solution set for the inequality.

*[invoke plot_any_inequality "y<=(3/2)x", -10, 10, -10, 10, 600, 600]

as you can see (when you plot it) this point ({{{-4}}}, {{{2}}}) is  {{{not}}} the solution set for the inequality