Question 1169129
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<pre>

From the given inequality, you correctly deduced this final inequality


    y > {{{(2/3)x - 3}}}.    (1)


Geometrically, you draw the straight line y = {{{(2/3)x - 3}}},


It divides the coordinate plane in two parts: one part is above this line; the other part is below this line.


The set of solutions to the inequality (1) is the set of all points of the coordinate plane, that are strictly

above the line (excluding points on the line).


So, if you draw the line on the graph paper, which has coordinate grid,

you can easily see which points do belong to the solution set and which points do not.



    {{{drawing(500, 500, -10, 10, -10, 10,
               grid(1,1),

       graph( 500, 500, -10, 10, -10, 10,
              (2/3)x-3),
       locate(3,   8,   the),
       locate(2.5, 6, solution),
       locate(3,   4,   set)
)}}}


                           Line y = {{{(2/3)x-3}}}



If the problem asks whether the assigned points do belong to the solution set, you always can substitute their coordinates 
into the given inequality and check immediately for these points.
</pre>

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On solving inequalities in two variables graphically, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Graphs/Solving-systems-of-inequalities-in-two-unknown-graphically-in-a-coordinate-plane.lesson>Solving systems of inequalities in two unknown graphically in a coordinate plane</A> 

in this site.


Learn the method from there.



Happy learning (!)