SOLUTION: 2. Graph x + 3y ≤ 6, indicating the solution set with crosshatching or shading. Explain how you determined where to draw the line and shade the area that represents the solut

Algebra ->  Graphs -> SOLUTION: 2. Graph x + 3y ≤ 6, indicating the solution set with crosshatching or shading. Explain how you determined where to draw the line and shade the area that represents the solut      Log On


   

Question 880266: 2. Graph x + 3y ≤ 6, indicating the solution set with crosshatching or shading. Explain how you determined where to draw the line and shade the area that represents the solution set.
Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Graph the line x+%2B+3y+=+6. This line goes through (0,2) and (6,0). Let me know if you need me to walk you through this portion.

This line is a solid line.

Now plug in the test point (0,0)

x+%2B+3y+%3C=+6
0+%2B+3%280%29+%3C=+6
0+%3C=+6

The last inequality is true. So (0,0) is in the solution set. So you shade below the solid line.

This is what the graph will look like



Shaded region in green with the solid black boundary line.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
x + 3y ≤ 6

First we draw the boundary line whose equation is

x + 3y = 6

We get the intercepts:

 x | y
 0 |
   | 0

If x = 0, x + 3y = 6
          0 + 3y = 6
               y = 2

 x | y
 0 | 2
   | 0
 
If y=0,   x + 3y = 6
        x + 3(0) = 6
           x + 0 = 6
               x = 6

 x | y
 0 | 2
 6 | 0

So the intercepts are (0,2) and (6,0)

So we plot those two points



Then we notice that the inequality is ≤ and not <.
If it were < we would draw the line dotted to show
that it was not part of the solution set, but since
it is ≤ instead, the boundary line is part of the
solution so we draw the line solid through those
intercepts:



Next we must decide which side of the line to shade.

We pick a test point on either side of the line.  The test point
must NOT be ON the line.

Any point not on the line will do as a test point.  But the easiest
point to test, when it is not on the line is (0,0)

We substitute (x,y) = (0,0) in the original inequality:

  x + 3y ≤ 6
0 + 3(0) ≤ 6    
       0 ≤ 6

That's true so since the origin is a solution and since it is
on the lower side of the line we shade the lower side, like
this:

 
 
Edwin