SOLUTION: graph the linear equalities 3x + 4y <12 x + 3y < 6 x > 0 y > 0

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Question 107952: graph the linear equalities

3x + 4y <12
x + 3y < 6
x > 0
y > 0

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Start with the given system of inequalities
3x%2B4y%3C12
x%2B3y%3C6
x%3E0
y%3E0

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


First lets graph the first inequality 3x%2B4y%3C12
In order to graph 3x%2B4y%3C12, we need to graph the equation 3x%2B4y=12 (just replace the inequality sign with an equal sign).
So lets graph the line 3x%2B4y=12 (note: if you need help with graphing, check out this solver)
+graph%28+500%2C+500%2C+-20%2C+20%2C+-20%2C+20%2C+-%283%2F4%29x%2B3%29+ graph of 3x%2B4y=12
Now lets pick a test point, say (0,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality 3x%2B4y%3C12 with the test point

Substitute (0,0) into the inequality
3%280%29%2B4%280%29%3C12 Plug in x=0 and y=0
0%3C12 Simplify



(note: for some reason, some of the following images do not display correctly in Internet Explorer. So I would recommend the use of
Firefox to see these images.)


Since this inequality is true, we simply shade the entire region that contains (0,0)
Graph of 3x%2B4y%3C12 with the boundary (which is the line 3x%2B4y=12 in red) and the shaded region (in green)
(note: since the inequality contains a less-than sign, this means the boundary is excluded. This means the solid red line is really a dashed line)
---------------------------------------------------------------


Now lets graph the second inequality x%2B3y%3C6
In order to graph x%2B3y%3C6, we need to graph the equation x%2B3y=6 (just replace the inequality sign with an equal sign).
So lets graph the line x%2B3y=6 (note: if you need help with graphing, check out this solver)
+graph%28+500%2C+500%2C+-20%2C+20%2C+-20%2C+20%2C+-%281%2F3%29x%2B2%29+ graph of x%2B3y=6
Now lets pick a test point, say (0,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality x%2B3y%3C6 with the test point

Substitute (0,0) into the inequality
%280%29%2B3%280%29%3C6 Plug in x=0 and y=0
0%3C6 Simplify



Since this inequality is true, we simply shade the entire region that contains (0,0)
Graph of x%2B3y%3C6 with the boundary (which is the line x%2B3y=6 in red) and the shaded region (in green)
(note: since the inequality contains a less-than sign, this means the boundary is excluded. This means the solid red line is really a dashed line)
---------------------------------------------------------------


Now lets graph the third inequality x%3E0
In order to graph x%3E0, we need to graph the equation x=0 (just replace the inequality sign with an equal sign).
So lets graph the line x=0 (simply draw a vertical line through x=0)
+graph%28+500%2C+500%2C+-20%2C+20%2C+-20%2C+20%2C+1000%28x-0%29%29+ graph of x=0 (note:the graph is the line that is overlapping the y-axis. So it may be hard to see)
Now lets pick a test point, say (1,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality x%3E0 with the test point

Substitute (1,0) into the inequality
%281%29%3E0 Plug in x=1 and y=0
1%3E0 Simplify



Since this inequality is true, we simply shade the entire region that contains (1,0)
Graph of x%3E0 with the boundary (which is the line x=0 in red) and the shaded region (in green)
(note: since the inequality contains a greater-than sign, this means the boundary is excluded. This means the solid red line is really a dashed line)
---------------------------------------------------------------


Now lets graph the fourth inequality y%3E0
In order to graph y%3E0, we need to graph the equation y=0 (just replace the inequality sign with an equal sign).
So lets graph the line y=0 (simply draw a horizontal line through y=0)
+graph%28+500%2C+500%2C+-20%2C+20%2C+-20%2C+20%2C+0%29+ graph of y=0 (note:the graph is the line that is overlapping the x-axis. So it may be hard to see)
Now lets pick a test point, say (0,1). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality y%3E0 with the test point

Substitute (0,1) into the inequality
%281%29%3E0 Plug in x=0 and y=1
1%3E0 Simplify



Since this inequality is true, we simply shade the entire region that contains (0,1)
Graph of y%3E0 with the boundary (which is the line y=0 in red) and the shaded region (in green)
(note: since the inequality contains a greater-than sign, this means the boundary is excluded. This means the solid red line is really a dashed line)
---------------------------------------------------------------


So we essentially have these 4 regions:

Region #1
Graph of 3x%2B4y%3C12


Region #2
Graph of x%2B3y%3C6


Region #3
Graph of x%3E0


Region #4
Graph of y%3E0




When these inequalities are graphed on the same coordinate system, the 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.







Here is a cleaner look at the intersection of regions




Here is the intersection of the 4 regions represented by the series of dots