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