Graph the inequality.
2x + 3y > 6
First form the equation of the boundary line,
which is found by replacing the < by =
2x + 3y = 6
Draw this line by finding the intercepts (0,2) and (3,0)
Draw the line dotted because the inequality is
> and not >
Now we only have to decide whether we are to shade the
upper side of the line or the lower side of it. To do
this we pick any point which IS NOT on the line. Say we
choose as a test point the point (2,3) marked below with
an o, which is above the line.
Now we substitute
(x,y) = (2,3) into the inequality
2x + 3y > 6
If we get a true inequality, then we will shade the
same side of the line that the test point is on. If
we get a false equationj, we shade the other side:
2x + 3y > 6
2(2) + 3(3) > 6
4 + 9 > 6
13 > 6
This is true, so we shade the same side of the line
which the test point that we picked is on, namely the
upper side. I can't shade on here but you can on
your paper:
A shortcut would have been to have chosen the origin
(0,0) as the test point. You could have. Then you
could have substituted it in your head to decide
which side of the line to shade. Substituting
(x,y) = (0,0) you would have gotten simply
2x + 3y > 6
2(0) + 3(0) > 6
0 + 0 > 6
0 > 6
which is FALSE and which you could have done in your
head! Since it is FALSE, you would have known to
shade the side of the line which the origin (0,0)
is NOT on, that is the UPPER side of the line. Notice
that the origin (0,0) is on the lower side
of the line, which is OPPOSITE from the side we
determined that we should shade by using the test point
(2,3).
So why didn't I choose the origin (0,0) as a test point?
I would have normally, if I weren't teaching you. But
if I had you might have gotten the wrong idea that you can
ALWAYS choose the origin as a test point. But you cannot
choose the origin as a test point when the line passes
through the origin. However that is the only
time you cannot choose the origin (0,0) as a test point.
So feel free to use the origin as a test point except
when the line goes through the origin.
Edwin
