Today we’re going to graph linear inequalities such as y is greater than or equal to mx plus b. Just an example of a slope intercept form using an inequality instead of the equals sign.

This is actually a whole lot easier than it looks or than I have seen some books make it out to be. Think about how you graphed on a number line. When you graphed something like x is greater than three. How did you do that’ You drew your number line. Then you found what number was your boundary, you marked it with a dot and put your line on there. So it really is not terribly difficulty and graphing an equation of a line that is in a linear inequality format is no different. So here ́s what we’re going to do.

Let’s take an example. We are going to graph y is less than or equal to two x plus three. I’m going to treat this very similar to an equality, so we’re graphing a straight line. So for now let’s just kind of look at it that way, y equals two x plus three and notice that it’s in the slope intercept form which is my personal favorite. Mx plus b and the reason I like that is because we can identify our slope and our y intercept very easily.

So if I’m to look at ...ah, may have trouble drawing a straight line here...we want to look at our Cartesian ?? put our coordinates on it. My b equals three, means the y intercept and my slope being two, it’s the same as rise over run. So I rise two and I run one. We could go the other way. If I had a ruler I could make it perfectly straight. And there we have our line.

But I don’t just have an equal sign here. I have a less than or equal sign. That means that all of the points on one half of this graph are going to be included as possible solutions. So for example. I am going to pick random numbers here. I’m going to pick this point here and pick A. I’m picking a point on either side of the line. Point A means –negative two, five. And point B is zero, zero. So what I’m going to do now is take a look at that set of ordered pairs, I have an x and a y and I’m going to substitute it in to the equation or this actual linear inequality. So y is less than or equal to two x plus three. If I plug in five for y for the first point A and negative two for x, on the right, I get two times negative two which is negative four plus three which is negative one. And now my inequality says five is less than or equal to negative one. Which is not true, five is not smaller than or equal to a negative one. So that point cannot be part of my solution. So let’s take a look at zero, zero now. So actually when we do this...and let’s try it with zero, zero. So, zero plugged in for y , zero plugged in for x and the on the left of course is our zero and on the right two times zero plus three, zero is indeed less than or equal to three. So that means this point that I labeled B is a solution to this inequality. And what we can do is we can then shade in everything on that side of the line with those in the same area – I’m not doing a good job with my drawing here – it goes in the same area as that point. So basically with an inequality you’re splitting the entire Cartesian Plane in half. Now I realize that it goes on and on forever, so you can’t really split it in half but within your viewing area you split it in half with your line and you ́re going to shade one half or the other half. And you’re going to do this by choosing random points. I always like to go with zero, zero as one of my points because it is so easy to do calculations with it. When you substitute that into your inequality, it’s very easy to work with. And it’s that simple.

Now, the other piece you have to remember is when you do your number line examples like the kind we looked at earlier. What was that example I gave you earlier? X is greater than three. What we did was at three we put an open dot. And a line that went to the right. It was an open dot because there was no equals sign under the inequality. So this would have been the same thing. This would have been a dotted line if there were no inequality. But we had an equals sign so we made it a solid line. And all that means is by saying it is solid, it is including all of the numbers on the line. So if I had chosen this point down here, negative two, negative one, it would have satisfied this inequality, by substituting in those numbers I would have gotten an inequality that was correct. So what that means is, here’s my x, here my y, I’m going to try to substitute in negative one for y is less than or equal to two times negative two plus three. On the right I have two times negative two which is negative four plus three which is negative one. Is it true that negative one is less than or equal to negative one? Well, it certainly is not less than but it is equal to, so that point would be included to in that answer. And that is why your line is solid.

Ok, let’s try this example here. I have to put it in to the format that I love so much. Y equals mx plus b. And I realize this is an inequality all I really want is y by itself. So I am going to add two x to both sides and on the left negative three y, the exes wipe out and on the left I have two x and a positive six. So, I divide by negative three and here’s the tricky part. What you do to one you do to both sides. My negative threes wipe out here and the rule when you divide by a negative is you switch your inequality sign. So I have y is greater than negative two thirds x plus negative two. That works if you look at a real example here. Let’s see if I can make this work here. Negative one is definitely less than two. If I divide those sides by negative one. On the left, I end up with positive one and on the right I end up with a negative two and if we take a look at those numbers, positive one is bigger than negative two and that’s why we have to switch those signs. Ok. And that’s just a little side explanation.

Ok, so here we have mx plus b. So my m is negative two thirds and my b is a negative two. So this is going to make it a whole lot easier to graph because I can identify m and b and notice my inequality is an inequality without an equals sign. So I’m not going to include the line itself that line is just going to be all the points that kind of go up against it and it’s using it as a boundary so I know where to shade. So my b is negative two. So here’s negative two on the y axis. And my m is negative two thirds so I’m going to rise negative two and run three so I’m going down two and running three. Down two running three. And it is a dotted line and that dotted line is my boundary. And again I am going to choose a point on either side of that line. I’m going to choose zero, zero because it is easiest to substitute in and I am going to choose one, negative...actually, you know what I’m going to go with? I’m going to go with three, negative five. And you’ll see why in a minute. So I’m going to call point A my zero, zero. And substituting that, I get y is greater than zero plus I’m going to substitute zero in for y and zero in for x and I am going to get zero is greater than negative two. And is that true? Yes it is. So this would be the side that I would shade in. And that means that all points in that half of the image that we’re shading here are possible solutions. Every single point in that area, all of the points right up to that line are included as a possible solution as well.

You can check it by just checking the other point and I do recommend that you do that in case, you know? It’s easy to make a mistake. But in case you do mess up, you will catch it in checking the other point. So y is greater than a negative five is greater than, negative two thirds times x which is three plus a negative two. Notice why I chose three, negative five as a point, I can get rid of my fractions. You know, no one loves to deal with fractions. But I love to manipulate them so I can just get rid of them. So here my threes cancel. And I am left with negative five is greater than negative two plus negative two which is negative four. Is that true? Is negative five bigger than negative four? No, it’s not. Therefore, I chose the right side and I graphed the correct side.