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College Algebra: More On Compound Inequalities

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  • Type: Video Tutorial
  • Length: 9:16
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 100 MB
  • Posted: 06/27/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Equations & Inequalities (50 lessons, $65.34)
College Algebra: Compound Inequalities (3 lessons, $4.95)

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 athttp://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 UT-Austin 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, p-adic 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.

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All right, well let me try to illustrate the notion of a compound inequality where you're having a whole bunch of inequalities that you want to be satisfied at once, with a couple more examples.
Suppose I want you to find all the x's that satisfy the following: 1 = 2x = <0, or 6 + 3x = <0. So what's the solution set? What are all the xs that will satisfy this? Remember what an or means, or means either x satisfies one or the other. The x doesn't have to satisfy both at the same time and so we're looking for here a union so when you think of or you should think of union. So how do you do this? Well, what I'm going to do, I'm going to solve this inequality for x and then I'll solve this inequality for x and I'll look on a number line and see what the solution is.
So to solve this let me first look at this, so it requires a little bit of work. What am I going to do? I want to isolate the x so I'm going to subtract the 1 on both sides and bring this -1 over and so I see -2x is < -1 and then I'll divide both sides by -2, but remember when you multiply or divide by a negative, the sign flips. So the moment I do that we get a flip in the sign and so now I see that x is in fact now greater and a negative divided by a negative is positive, greater than . So there's the solution to the first inequality, x is bigger than .
Okay, what's the solution to the second inequality? Well, let me now solve this, so I'm going to come out here and do a little work and see what happens. What I see here is if I bring the 6 over by subtracting 6 from both sides, -6, I'd see 3x <-6 and now I divide both sides by 3, do I flip the sign? Absolutely not, because 3 is positive. I only flip when I multiply or divide by a negative number so in this case I see that x is still less than and -6 Ă· 3 = -2. And remember the conjunction is or. So now I see what my answer is, it's either x is greater than or x is less than -2. So let's write that out on a number line. Again, I only put in the key players. You might want to mark all the points in effect. In fact, if you want, I can do that too. Would you like me to do that? So here's 0, here's 1, here's 2, here's -1, here's -2. I just don't like cluttering things up. What do we have though? The players here are -2-1/2 so I'm actually going to write in the half since that's so special we're going to talk about that a lot.
Okay, so what do the graphs of these things look like? x is larger than . That means that I look at all the points to the right of and I don't include , I have no equal sign there, so I put an open parentheses pointing this out like this. There's the solution to that, what's the solution to x is less than -2? Well, I don't include -2, but I include all points to the left. So where's the solution to this or that? Well, if you have this graph drawn you can read off the answer. I go from negative infinity to -2, but I don't include -2, see how I just copied that down from the picture here? The same thing here; I go from one half with an open parentheses out to infinity. And we never have a closed parentheses on infinity because infinity is not a number, so we always have open parentheses around the infinities. And since I'm one or the other I denote that by putting a union in which means as long as you're in one or the other, I'm happy. Any of those things would be a solution. For example: If you pick something like a -3, is -3 in here? Well, let's see. I'm asking, is -3 greater than or is -3 less than -2? Well, -3 is not greater than , that's wrong but notice that -3 is less than -2, so in fact, this whole thing is okay because at least one of them is satisfied. What about zero though, is zero bigger than or is zero less than -2? Well, it turns out the answer is no, zero is not bigger than the half and zero is not less than -2 so zero is not a solution to this because it doesn't satisfy either one of them. So or, you put everything together.
Let me try one last one because sometimes I can put a compound inequality together in one huge multiplex inequality. This inequality actually means something. It means that I have two inequalities going on in the background.
Let me show you what those two inequalities are. Whenever you see an equality like this, first of all what that means is zero is less than or equal to 2x + 6 and also 2x + 6 = <18 so in fact, this translates into the following: zero less than or equal to 2x + 6 and 2x + 6 = <18. So whenever you see this sort of compound inequality on one line, it really is an invisible and problem, this and at the same time as that. The really neat thing about writing it this way is that solving this actually allows you to solve both inequalities at the same time. Just treat this thing as one huge algebraic string. What every you do, though, don't think of one side, think of three sides. So whatever you do to one side, you do to the other side. So for example if I want to solve for x, the first thing I would do is subtract 6, but I subtract 6 from everything, so I subtract 6 from here, it's like a balance scale with three arms. And now if I combined, I see -6 and I have the same inequality less than or equal to 2x, because this cancels, and now less than, and then 18 - 6 = 12. Do you see how I'm performing the arithmetic but I'm doing all the steps at once very carefully? The inequalities stay the same as they were before, but I can do arithmetic to all of them at once and the balance of power stays intact.
Well, what would I do now? Well, now I'm going to want to get rid of that multiplicative factor of 2 by dividing by 2, but I divide everything, therefore, by 2 to keep everything the same and balanced. Will my signs flip? No, because 2 is positive, but if 2 were negative, then both of these signs would actually flip. But since, in fact, 2 is positive, everything stays as is, I see a -6 divided by 2 is a -3 less than or equal to x, which in turn is less than 6. So there's the solution, there's the answer, what does it look like graphically? Well, graphically it looks like this, I have -3 and then all the way out to 6 and what does this say? It says x is in between these two regions. So in fact, the solution is really everything in between, things that are bigger than -3, but smaller than 6. What about the end points? Do I include 6? No, I don't include 6 because there's no equal sign here so I put a open parentheses. Do I include the -3? Absolutely, because there's an allowable equals sign there. So in fact, I could write this answer as an interval notation as [-3](6), these two statements are identical. This statement and this statement are the same, this is an interval notation; this is an inequality notation. And notice, by the way, this really was an intersection because if I just look at this piece right there, -3 is less than or equal to x. What that means is that's this region right here, -3 is less than or equal to x, so x is all this stuff up here and x is less than 6, that's this region right here and since I put them together with that invisible and, I look for their intersection and notice the intersection is exactly the orange that I put in there.
So you can see that just by using these inequalities at once it gives you the answer immediately rather than doing one solution, the other solution and seeing where they overlap, this gives it to you automatically in the orange.
Okay, well try some of these on your own and get in the habit of working with these inequalities where you have more than one inequality in one line, multiplex inequality. Welcome to the 21^st Century.
Equations and Inequalities
Solving Inequalities
More on Compound Inequalities Page [2 of 2]

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