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About this Lesson
 Type: Video Tutorial
 Length: 11:05
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 120 MB
 Posted: 08/18/2009
This lesson is part of the following series:
College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Systems of Equations (33 lessons, $44.55)
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/collegealgebra. The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
 Thinkwell
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Systems of Equations
Systems of Inequalities
Graphing the Solution Set of a System of Inequalities
So, now let’s take a look at this graphing and take that a step further. Let’s start seeing the graphs of solutions to systems of inequalities. So, let’s try on right off the bat and see how we do. Let’s solve 1 x < 2 and at the same time 0< y 3. I want to see what the solutions set looks like for this thing. So, the first step is to draw some axes. Let’s draw some axes. So, what I’m going to do now is I’m going to graph both of these things separately. So, let’s take a look at this. Now the first thing I say is, “Okay, now if x = 1 where is that.” So, x = 1 is a vertical line, all the places where x = 1. That would be right here. So, I put in a vertical line and since I see an equal sign I’m allowed to actually include it. So, let me actually included, make it solid, x = 1.
Now the other possibility is x = 2. Now x = 2, of course, is way over here. But notice that I have strict inequality. So, I’m going to put a dotted line right there. A dotted vertical line to represent that right hand side. Now, where do I fill the stuff in? Well, notice now I actually have three regions and I have to think about where I’m filling in. Well, you can think about it and then see where x is between 1 and 2. So, which x’s are between 1 and 2? Well, these x’s right in here. So, these must be the x’s that satisfy this. So, I’m going to shade that in. So, I shade in those x’s and thus any y that has an x between 1 and 2.
What this means, by the way, is that I can include this line, I can touch that line, but I can’t touch this line. I can just get as close as I want, but I can’t touch it. It has to be strictly less than 2. Now, what about the y thing? Well, y > 0, but 3. So, what would that look like if I were to graph that? Well, if I were to graph that y = 0 is just the xaxis and I have to put a dotted line there. So, I guess I would put this sort of right on top it even though technically it’s right on it, but just to show you it’s dotted. I’m going to put this like this, just to remind you of the fact it’s not included. Then y 3, so I have to go up to 3 and I’m allowed to equal 3. So, I put a solid horizontal line.
Now, where do I fill in? I want all the values where the y’s are between these two. So, that would be in this band right here. So, in fact, I would color this in. Color this band right in there. So, what’s the solution set of this? Well, you can see it’s where they overlap. So, it would be this little rectangular region here, where I can include this edge and I can include this edge, but I don’t include this edge and I don’t include this edge. So I include these pieces, but not these pieces. These are dotted. So this is the solution. Sometimes you can actually write solution and sort of point right into this region right here.
Let’s try another one. It’s sort of fun actually. How about this one, 3x + 2y 1 and x + 2y < 1. Well, the first thing I would do here is actually just write these equations. You could write them in point slope form or you could just use them in standard form as they are. Imagine there’s equal signs there. Let me actually show you a little slick way of graphing these things if you just want to graph them as is and not solve for anything. So, the auxiliary thing I’d look at for the first one is 3x + 2y = 1. Now what would the graph of that look like? Well, all I’m going to do is I’m going to put in my axes and you know what I’m going to do? I’m just going to find the intercepts.
So, how do I find the y intercept? That’s where x = 0. So, just let x = 0 and solve for y. Look, I immediately see y = ½. So, I know this passes through this point at ½. So, I have ½ right here. Let’s suppose this is one, so now I’m at ½. Let’s see what the x value, the value where we cross the xaxis. That’s where y = 0. So, I put zero in there and I see immediately x = 1/3. So, you see it’s actually pretty easy to graph these things. I just connect them. I see that I have greater than or equal to so I’m allowed to put a solid line right there. So, there’s this graph right there.
Now, the question is do I shade above or below? Well, let’s think about it. Let’s put in (0,0) and see if it satisfies that? If I put in (0,0) I’d see 0 1, that’s false. So, therefore, that must not be the region. The region I want must be up here, so I’d shade all that in or you could use sort of hitech stuff and since I have it, I’ll use it. So, in fact, that will be the region. I would just shade that in. What about the next graph? Well, the next graph is also a line. Let’s see what its intercepts are. Use a different color now.
So, let’s see, if x were zero what would I see y would equal? Y would have to equal, make believe there’s an equal sign there, y would have to equal ½. So, what I would see is I’d go down to ½ and what’s the x intercept? Well, if y is zero, I look at this and I see that x = 1. So, I see x = 1. That point is there. So, now I’m going to connect them. Now, let’s be careful. There’s strict inequality here. Since there’s strict inequality, that means I have a dotted line. So, I’ve got a dotted line going on here because I’m not allowed to include that boundary. It’s a strict inequality. No matter where the boundary is we know we’re not going to include it. Let’s make a little bit more here. More down there. Cool.
Now, where are we in terms of this line? Well, just pick the origin and see. So, for example, if I put in (0,0) does that satisfy this? Is 0 < 1? No it’s not. So, therefore, I must want this region right here. So, I pick the region below and where’s the intersection? Well, you can see the intersection. It’s right at this wedge right here. So, it’s sort of this triangular region. That’s the point of intersection. That’s the solution. Those are all the points that satisfy both of these at the same time. I’m not allowed to include this boundary, but I am allowed to include that boundary because that’s a solid line here. So, that’s how you graph these things.
Let’s try one last one just to show you what it would look like if you were to do this with nonlinear functions. How about this, x² + y² 9 and x  y + 1 > 0. Let’s graph those things and see what we’ve got. Well, I recognize the top thing as actually a circle, because I see x² and y ² like that. It’s centered at the origin and its radius is the which is 3. So, I see this is going to be one, two, three. So, I’ve got to now draw a circle and of course you can imagine how nervous I am. Because drawing circles live without any instrument at all is really hard. So, you have to use your imagination here. And it’s solid. I had to make it solid, but it is solid because I have an equal sign here. So, I can make it solid.
Now, do I fill the inside of the circle or the outside? Well, let’s pick the origin and plug it in. The (0,0), if I put in 0 here I get 0. Is that less than or equal to 9? Sure is. So, I shade in the inside o the circle. Now, what does the second thing look like? Well, that second thing is just a line and you could write it out if you wanted to. How could you write it? You could write it as the line itself would be x + 1 = y. I just brought that over to this side and put an equal sign there. So, I see y = x + 1. So, the intercept is one and the slope is one. So, this just looks like a good, old fashion line. It passes through here and has slope one. However, notice there’s strict inequality. So, I’m going to put a dotted line.
Now, where do I go? Do I fill above or below? Well, here’s what I do. I just take the origin and put it in and see if it’s true. Put a (0,0) in here. I see zero, zero plus 1. That’s 1. Is 1 > 0? It sure is. So, I’m going to want to include that point. So, in fact I’m going to shade this way. You can see where the shading intersects. You know, when I was in school and learned this stuff, my teacher called these hash marks. I just thought that was great. So, you see the hash marks intersect sort of right in this little part circle, part line region. Where we don’t include the line, but we include the circle. That would be the solution set for these two inequalities at once. Any point in there will satisfy both these at the same time.
Okay, so by graphing you can actually visualize what the solution set is to a whole bunch of inequalities and find out where they’re equal. Why is this a value? You’ll see up next in some interesting applications that sort of revolve around this notion of linear programming. Not a big deal. Don’t let the name scare you. I’ll see you soon.
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