Question 85717
16.


Start with the given system of inequalities

{{{x-3y>6}}}


{{{3x+2y>12}}}




In order to graph this system of inequalities, we need to graph each inequality one at a time.


So lets graph the first inequality


    In order to graph {{{x-3y>6}}} we need to graph the equation {{{x-3y=6}}} (just replace the inequality sign with an equal sign). So lets graph the line {{{x-3y=6}}} (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)

    {{{graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3)}}} graph of {{{x-3y=6}}}

    Now lets pick a test point, say (0,0) (any point will work, but this point is the easiest to work with), and evaluate the inequality {{{x-3y>6}}}

     {{{(0)-3(0)>6}}} Plug in x=0, y=0


     {{{0>6}}} Simplify



Since this inequality is <b>not</b> true, we shade the entire region that <b>doesn't</b> contain (0,0)



{{{drawing( 400, 300, -10, 10, -10, 10,graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-0),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-0.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-1.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-2.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-3),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-3.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-4.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-5.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-6),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-6.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-7.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-8.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-9),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-9.75))}}}

 Here is the graph of {{{x-3y>6}}} with the graph of the line({{{x-3y=6}}}) in red and the shaded region in green

(note: The red line should be a dashed line since it is <b>not</b> included in the region. Since the inequalityis a <font size=6>></font> sign, it tells us <b>not</b> to include the boundaries.)





Now lets graph the second inequality


    In order to graph {{{3x+2y>12}}} we need to graph the equation {{{3x+2y=12}}} (just replace the inequality sign with an equal sign). So lets graph the line {{{3x+2y=12}}} (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)

    {{{graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2)}}} graph of {{{3x+2y=12}}}

    Now lets pick a test point, say (0,0) (any point will work, but this point is the easiest to work with), and evaluate the inequality {{{3x+2y>12}}}

     {{{3(0)+2(0)>12}}} Plug in x=0, y=0


     {{{0>12}}} Simplify



Since this inequality is <b>not</b> true, we shade the entire region that <b>doesn't</b> contain (0,0)



{{{drawing( 400, 300, -10, 10, -10, 10,graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+0),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+0.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+1.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+2.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+3),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+3.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+4.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+5.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+6),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+6.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+7.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+8.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+9),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+9.75))}}}

 Here is the graph of {{{3x+2y>12}}} with the graph of the line({{{3x+2y=12}}}) in red and the shaded region in green

(note: The red line should be a dashed line since it is <b>not</b> included in the region. Since the inequalityis a <font size=6>></font> sign, it tells us <b>not</b> to include the boundaries.)

So we essentially have these 2 regions

{{{drawing( 400, 300, -10, 10, -10, 10,graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-0),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-0.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-1.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-2.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-3),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-3.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-4.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-5.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-6),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-6.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-7.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-8.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-9),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-9.75))}}} Region #1 which is the graph of {{{x-3y>6}}}

{{{drawing( 400, 300, -10, 10, -10, 10,graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+0),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+0.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+1.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+2.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+3),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+3.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+4.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+5.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+6),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+6.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+7.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+8.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+9),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+9.75))}}} Region #2 which is the graph of {{{3x+2y>12}}}



So these regions overlap to produce this region. It's a little hard to see, but after evenly shading each region, the intersecting region will be the most shaded in.

{{{

drawing( 400, 300, -10, 10, -10, 10, -10, 10,
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-0),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-0.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-1.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-2.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-3),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-3.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-4.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-5.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-6),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-6.75),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-7.5),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-8.25),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-9),
graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3, (6-1*x)/-3-9.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+0),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+0.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+1.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+2.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+3),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+3.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+4.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+5.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+6),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+6.75),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+7.5),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+8.25),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+9),
graph( 400, 300, -10, 10, -10, 10,(12-3*x)/2, (12-3*x)/2+9.75))}}}

Here is a cleaner look at the intersection of regions


{{{drawing( 400, 300, -10, 10, -10, 10,
  graph( 400, 300, -10, 10, -10, 10,(6-1*x)/-3,(12-3*x)/2),circle(5,-1,0.05),
 circle(5,-1,0.08),circle(6,-2,0.05),
 circle(6,-2,0.08),circle(6,-1,0.05),
 circle(6,-1,0.08),circle(7,-4,0.05),
 circle(7,-4,0.08),circle(7,-3,0.05),
 circle(7,-3,0.08),circle(7,-2,0.05),
 circle(7,-2,0.08),circle(7,-1,0.05),
 circle(7,-1,0.08),circle(7,0,0.05),
 circle(7,0,0.08),circle(8,-5,0.05),
 circle(8,-5,0.08),circle(8,-4,0.05),
 circle(8,-4,0.08),circle(8,-3,0.05),
 circle(8,-3,0.08),circle(8,-2,0.05),
 circle(8,-2,0.08),circle(8,-1,0.05),
 circle(8,-1,0.08),circle(8,0,0.05),
 circle(8,0,0.08),circle(9,-7,0.05),
 circle(9,-7,0.08),circle(9,-6,0.05),
 circle(9,-6,0.08),circle(9,-5,0.05),
 circle(9,-5,0.08),circle(9,-4,0.05),
 circle(9,-4,0.08),circle(9,-3,0.05),
 circle(9,-3,0.08),circle(9,-2,0.05),
 circle(9,-2,0.08),circle(9,-1,0.05),
 circle(9,-1,0.08),circle(9,0,0.05),
 circle(9,0,0.08))}}} Here is the intersection of the 2 regions represented by the dots