Hi, and welcome. My name is Anthony Varela, and today, we are going to solve a system of linear inequalities by graphing them on the coordinate plane. So first, we're going to graph individual inequalities that make up our system, but we're going to be shading half planes according to the inequality symbol defined in that system. And then we're going to be looking at all of these solution regions to identify then the solution region of the system.
So let's quickly reviewed how to graph linear inequalities, so here, we have y is greater than 1/3 x plus two. Now, the first step is to figure out what type of line you're going to use, so look at the inequality symbol. And if it is a strict inequality symbol, so it does not include equal to. So that would just be greater than or less than. You're going to use a dashed line.
If you're inequality symbol is non-strict, so it includes equal to, you're going to be using a solid line. So let's go ahead and plot then on this graph y equals 1/3 x plus two using either a dashed line or a solid line. So I'm going to be using a dashed line, and here is y equals 1/3 x plus two.
I see my y-intercept here occurring at y equals two, and then I have a slope of 1/3, a rise of one and a run of three. So here is the line that represents y equals 1/3 x plus two. And now we have to shade in a half plane, either this area above the line or this area below the line. And once again, we're going to be looking at our inequality symbol.
Now because this has y on one side and everything else on the other side, we can just take a look then at this inequality symbol. And if it tells us greater than, we're going to be then shading above. And if it's less than, we're going to be shading below. So this tells us that we're going to shade above.
Now we're going to introduce another inequality here, so this now represents a system here. And we have x is less than negative three, so here, I've drawn the line x equals negative three. And I used a dashed line according to my inequality symbol, but now what half plane am I going to shade here?
This is x is less than negative three. And once, again, because I have x on one side by itself and everything else on the other, I'm going to look at my inequality sign and shade everything that's less than negative three. So that would then be on the left side, so I have that noted here for in terms of x. If it's less than or less than or equal to, you shade to the left. If it's greater than or greater than or equal to, you shade on the right of that vertical line.
So here then is the solution region to x is less than negative three. Now a solution then to a system of inequalities is on a graph would be where these two solution regions, yellow and blue, intersect or overlap. And that would be this green area right here, so this then represents the solution to these two inequalities as a system.
So the solution region is the intersection or overlap of all shaded regions that make up the system. So let's go ahead and practice then graphing a system of linear inequalities and identifying then that solution region. So first thing that we're going to do is get all of these inequalities plotted on our graph, so I'm going to start with x plus 2y is less than six.
Now I'm just going to rewrite this, so that I can isolate y on one side of my inequality. Because then I can just analyze that sign and decide which half plane to shade. So first thing that I'm going to do to isolate y subtract x from both sides, so I have 2y is less than negative x plus six.
Now I'm going to divide everything by two, so what I have then is y is less than negative 1/2 x plus three. So I'm going to plot this then as a line, and I'm going to be using a dotted line. And I know that I'm going to have my y-intercept at y equals three and a slope of negative 1/2. So there's my y-intercept, and here's my slope of negative 1/2. And here's my dotted line.
Now where am I going to shade, above the line or below the line? Now I'm going to be shading below the line, so there it is, the solution region for the first inequality. Now let's move on to the other one, 4x minus 3y is less than or equal to 12. Now, once again, I'm going to isolate y, so I'm going to subtract 4x from both sides of this inequality.
So now, I only have negative 3y-- don't forget the negative. --is less than or equal to negative 4x plus 12. So to isolate y, I'm going to divide everything by negative three. So now, I have y on one side of my inequality. But notice because I divided by a negative, my inequality sign has flipped. So it used to be less than or equal to. Now it's greater than or equal to, and negative four divided by negative three is positive 4/3 and positive 12 divided by negative three is a negative four.
So now, let's go ahead and put this on our graph. I'm going to be using a solid line due to my inequality symbol. And I'm going to have my y-intercept at negative four and a slope of 3/4, so a rise of four and a run of three. So here is the line y equals 4/3 x minus four.
To represent an inequality, I need to shade 1/2 of this plane. And is that going to be shading above or shading below? Well, looking at our inequality symbol here, this tells us to shade above. So now, identifying then the solution region of the system would be this green solution region right here, because it is the overlap or intersection of our two inequalities that make up our system.
So what this means then is at any point, it's in this green region satisfies our system of inequalities. It satisfies each inequality in the system, so we can go ahead and pick any point to test that. And I always like to choose the origin, because that's easy to deal with. But for the fun of it, let's pick something else just to show that this is true.
So we're going to take the point negative two, negative two. Now this exists within the solution region to our system, so we need to show that it does, indeed, satisfy both of these inequalities. So let's plug-in negative two for both x and y. And we get then that negative two minus four is less than six, and that is certainly true.
For our second inequality in the system, plugging in negative two for both x and y. So four times negative two is negative eight, and I'll say, negative three times negative two is a positive six. And negative eight plus six is certainly less than or equal to 12, so it is within the solution region.
Well, let's go through one more example here to show a system of inequalities that doesn't have a solution at all, so we're going to quickly plot on here y is less than x plus three. So here is that. We have our intercept that y equals three, a slope of one, and we're using a dashed line and shading below.
Let's go ahead and put in y is greater than two, so here is the line y equals two. We're using a dashed line, and we're shading everything above. And to put in here x is less than negative four, we're going to locate x equals negative four. That's right here.
We're using, once again, a dotted line, and we're shading everything underneath x equals negative four. So that would be everything to the left right here. Now notice that I see lots of different colors, so there's a couple of different regions of overlap. But we're looking for an intersection of all of the shaded regions in the system, and we actually don't have a region that represents an intersection of all of this individual solution regions.
So this actually has no solution, even though you're going to have a couple of inequalities, where their solution regions overlap. But not all of them, so no solution here. So let's review. We talked about when graphing linear inequalities, use a dashed line for our strict inequality symbols and a solid line for the non-strict inequality symbols.
When deciding which region to shade, if you have y or x on one side of your inequality symbol and everything else, greater than or greater than or equal to tells you to shade above or to the right. And the less than or less than or equal to symbols tell you to shade below or to the left, if it's a vertical line. The solution region of a system of linear inequalities is the intersection of all shaded regions in the system, and you could have a system with no solution, if there's no intersection of all of those shaded regions. So thanks for watching this video and solving a system of linear inequalities by graphing. I hope to see you next time.
