Graph the absolute value inequalities:
|x + 2y| > 6
The boundary equation is found by replacing
the inequality symbol by an equal sign:
|x + 2y| = 6
which is equivalent to
x + 2y = 6 OR x + 2y = -6
So we graph those two boundary lines dotted
since the inequality is > and not >.
This says that the boundary lines are not
included in the graph.
The boundary lines divide the xy-system into
three regions:
1. the region above the upper dotted line.
2. the region between the two dotted lines.
3. the region below the lower dotted line.
Now we find out which of these regions
contain the solutions and which ones don't,
so that we can shade the regions that do,
and leave the regions that don't unshaded.
So we pick an arbitrary test point in each of
the three regions and substitute it into the
original inequality. If we get a true inequality,
then we shade that region. If we get a false
inequality, then we do not shade that region.
In the upper region, let's pick as a test point,
say, the point (6,5), and substitute x=6 and y=5
into the original inequality:
|x + 2y| > 6
|6 + 2(5)| > 6
|6 + 10| > 6
|16| > 6
16 > 6
This is true so the test point (6,5) is a solution,
so we shade the upper region.
In the middle region, that is, the region between
the two lines, let's pick as a test point, say,
the point (1,1), and substitute x=1 and y=1 into
the original inequality:
|x + 2y| > 6
|1 + 2(1)| > 6
|1 + 2| > 6
|3| > 6
3 > 6
That is false so this test point (1,1) is not a
solution so we do not shade the region between
the two lines.
In the lower region, let's pick as a test point, say,
the point (3,-7), and substitute x=3 and y=-7 into
the original inequality:
|x + 2y| > 6
|3 + 2(-7)| > 6
|3 - 14| > 6
|-11| > 6
11 > 6
This is true so, the test point (3,-7) is a solution
so we shade the lower region.
I can't shade on here, but you can on your paper:
Edwin
