Graphing Solutions to Linear Inequalities

First, we’re going to graph y > 1/3x + 2. To graph a linear inequality, we’re going to graph the line as if this were an equal sign, and then, if it’s greater than or less than, it’s a dashed line.

If it’s greater than or equal to, less than or equal to, we use a solid line. So, since this is just greater than, we need to use a dashed line.


The following transcript is provided for your convenience.
This is in slope intercept form, y = mx+b, so our slope is 1/3, and our y-intercept is 2. So, we’re going to start with our y-intercept at 2, and then use our slope to find our next point, rise 1, run 1, 2, 3. And then we’ll draw our dashed line, connecting our two points.

The final thing we need to do to graph an inequality is to shade, but where do we shade? One way to figure this out is to pick a point on either side of the line and see if satisfies your inequality. If it does, then that’s the side you shade. If it doesn’t, then you shade the opposite side.

A good point to use is (0,0), because it’s pretty easy to solve for. So, y is 0. Is 0 greater than, and then again, we plug in 0 for x, 1/3 times 0 is 0, 0+2 is 2. So, is 0 > 2? No, it isn’t. Which means since (0,0) does not work, it is not a solution, then this is not the side we shade, so we shade the other side. And really, everything above this would be shaded, but I’m going to color the whole board.

So, this will be your solution to this inequality. Every point in this shaded region would make this inequality true. However, points on the line would not make the inequality true, since its y is greater than, not greater than or equal to, and that’s why this line is dashed to symbolize that the points that are on this line, the points that lie on this line do not satisfy this inequality.

Now, we’re going to graph a system of inequalities. The same rules apply, but where you shade is a little bit different. So, first, we’re going to graph y >= 2x+3. Since it’s greater than or equal to, we use a solid line. And again, it’s in slope intercept form, so our slope here is 2, or 2/1, and our y-intercept is 3.

So, we start with 3, up 1, 2, 3. And then we use our slope, rise 2, run 1. Now, we can draw our solid line, points because again, it’s a greater than or equal to. And there are a couple of different ways you can do the shading. One thing you could do is start by figuring out where you would shade for this line, so, like, again, you could use that point (0,0). 2 times 0 is 0, 0+3 is 3, is 0 >= 3? No, it isn’t, so for this line, we’d be shading on this side of the line, and we can put it like a little dot here just to remind ourselves this is the side we’d be shading for this line.

And then you can graph your other line and see where you shade for that one, where they overlap is going to be the solution, and that’s one way, and I’ll show you another way when we’re done graphing the other line. So, this next line is y < -1/2x-2. Less than, since it’s not less than or equal to, that means we’re going to be using a dashed line again. And again, it’s in slope intercept form, so our slope is -1/2, and our y-intercept is -2.

So, start with the y-intercept, -2, and then use your slope to find your next point, rise 1, or actually, we’re going down 1 since it’s a negative 1, and then over 2. And again, this is a dashed line because it’s just less than, not less than or equal to. Then we connect our points with our dashed line, and put our arrows on the end to signify tha line goes on.

So, now that we have our two lines crossing, we’ve actually created four regions. We have this region, this region, this one, and this one. For our first line, this region was what we’ve shaded. So, now, we need to see where we’re shading for our next line. Is it above the line or below this dashed line? Again, I’m just going to use (0,0) in order to check.

So, -1/2 times 0 is 0, 0-2 is -2. So, is 0 < -2? It is not, so this is not where we’re shading for this one, we’re shading down here in this region, and so, where the two shaded areas overlap would be right over here. This is your solution since for this line, we were shading on this side, and for this line, we were shading on this side, so the two shaded areas that overlap is just this region right here.

And another way you could figure this out is just to pick a point in all four of the regions, and if the point satisfies both equations, then that’s where you shade. And if your point doesn’t satisfy both equations, then we don’t shade in that region.


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