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About this Lesson
 Type: Video Tutorial
 Length: 11:10
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 119 MB
 Posted: 06/26/2009
This lesson is part of the following series:
College Algebra: Linear Equation Word Problems 1 (10 lessons, $13.86)
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Beginning Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/beginningalgebra. The full course covers linear equations, inequalities, polynomials, rational expressions, relations and functions, roots and radicals, quadratic equations and systems of equations.
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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11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
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ALGEBRAIC EXPRESSIONS
Well, now we're about to begin the real heart of algebra. This is where we're going to start to put all the basic ideas that we've been thinking about together into play, and it really is a move toward an abstraction, and so I want us to think about this together. In particular, what I want to talk about are what are things called algebraic expressions, and this really now is at the heart of what we're going to be thinking about.
What is an algebraic expression? It's an expression that involves lots of numbers or placeholders for numbers that are combined in an algebraic way. Let me give you an actual example, and then let's talk about it.
Here is an example of an algebraic expression: 3x + 1. Now, what's going on here? What is the (x), first of all? Well, I want you to think of the (x) as a variable. Okay? So a variable is just some unknown number. So the idea here is the following; this is some number, but we don't know what it is. So instead of writing a question markin fact, I'll tell you, if it were me, if I were rewriting all of mathematics, I would do it this way. So this would be the Professor Burger rewrite of all of mathematics. I would do things like this; now, I have to admit, it'd be harder to write. You gotta use two different colors, and you gotta also have the box and the question mark, but that's exactly what's going on here. What this means is that we have some unknown number, and remember, when there's no operation in between two quantities, it means we're multiplying. So it's saying we have some unknown number, some variable, and we're going to multiply it by 3 and we're going to add 1. So this, in some sense, is a rule. It's saying "you know what, whenever I tell you what (x) is, you can figure out what this quantity is" because all you do is insert that question mark value and compute away. Do the algebra. That's why this is called an algebraic expression. It's an expression that involvesit's being hung together with algebra stuff. In this case, addition and multiplication, but there is this unknown thing here. Now I don't want you to be scared by (x). (x) is just some mysterious thing. In fact, you are better than (x), and I'll tell you why. See, (x) is so afraid, it doesn't want to be known. It doesn't even want to tell you who it is. Where you, at least, will admit who you are, right? Of course. So, in fact, you are better than (x). So don't be intimidated by this. Just understand it's some number yet to be determined.
For example, if I tell you that (x) = 5, well, then you know exactly what this number is. What would the number be? The number would be 3 o 5 + 1. Well, 3 o 5 would be 15 and 1 is 16. So then, this thing would represent 16. If instead, I said oh, no, no, no. You know what, actually (x) = 0. Then I have 3 o 0 = 0 + 1 = 1. Then this expression becomes 1. Notice that this value will change, will vary as I change the unknown. So that's why it's called a variable, because as I vary the unknown, this quantity will vary correspondingly.
So that's all that an algebraic expression is. Now, all these things have names by the way. Let me point them out. If you've got some constant number that's multiplying the mysterious variable, that constant number is referred to as a coefficient. If you just have a number dangling in the wind like this 1 is, that sometimes, this 1 is called a constant, because notice, that is not a variable. 1 is 1no matter how you slice it, you get 1. This is a variable. Who knows? It's a mystery, but 1 is 1.
So the 3 here is called a coefficient, the 1 is called a constant. So there is some language that is sometimes used in talking about these things, but I don't really care about the language, and I don't want you to really care about it. I want you to understand what this really represents. It represents a varying quantity, and as (x) changes, this expression will change accordingly, and if I give you any number (x) at all, you can figure out what this is. For example, if I say (x) is 2/3, well, then you'll put in 2/3 here, and the 3's cancel. I'm left with just 2 + 1 = 3. That is what an algebraic expression is.
Let's take a look at some more algebraic expressions. So they can get pretty nasty in fact. Look at this one. 3(x + y)  3x(y  2) + y(x + 4)this looks really awful. In fact, you know what, maybe I shouldn't even write this down. I feel sort of embarrassed. It seems almost obscene. Look at that. It fills up the entire screen. But that is an algebraic expression. Why? Because all we have are things that are connected with algebraic operations. In this case we've got subtraction, we have some multiplication. Remember, 3 times something, I mean, with the no symbol there means multiplication, and we've got addition and we've got everything here except division.
You will also notice that we have two different letters appearing. Do you see that? We've got the (x), so that's the same (x) everywhere. That represents the same unknown number, whereas the (y) represents a potentially different number. Now all of the (y)'s are the same, but they might be different from the (x). So here we have two mysterious quantities. So if I give you the (x) value and I give you the (y) value, you can actually compute, figure out this algebraic expression.
For example, let's just try one ourselves. Suppose that (x) were to be 0 and (y) were to be 1. What would happen? Well, wherever I see an (x), I put in a 0, and wherever I see a (y), I put in a 1, and we can compute this. So here I see 3 o 1, so that's just a 3. So this would be 3 o (0 + 1)  3 o 0 o (1  2) + 1 o (0 + 4). Notice, all I've done is rewritten the same exact expression, although a little bit smaller, but I put a 0 wherever I see an (x) and I put a 1 wherever I see a (y). Now, you might be saying, "where did you come up with 0 and 1?" I just made them up, because the (x) and (y) are variables. They can vary and be anything, but in this case, once I tell you that (x) = 0 and (y) = 1, you can figure out what number that represents. In this case, it represents 3 o 1 which is 3, and notice that is 3 o 0 o stuff, that's just 0, and here I have just +4, and so I see this yields 7. So this very complicated algebraic expression is just the number 7 when (x) = 0 and (y) = 1. If I change the (y) or I change the (x), or I change them both, then this number will, again, change correspondingly.
Now this actually can be simplified, and let me quickly show you how you can use the laws of algebra to simply this. First, I'll distribute, use the distributive law and write this as 3(x) + 3(y). I'll use the distributive law here. So I have this 3(x) which I'm going to distribute to here and to here, and so I see here 3(x) o (y) and then a minus times a minus we've seen is a plus, 2 o 3 = 6(x), and then I'm going to distribute here, and I see +(y)(x) + (y)4.
Now, one of the fundamental principles of the stuff that you and I are thinking about together is that we can combine, we can add, like terms. So if you look at this, you will see there are some like terms here. Here I've got three (x)'s and here I have six more (x)'s, so I have a total of, in fact, nine (x)'s. So this actually equals 9(x). Here, notice I have three (y)'s, and remember (y) o 4 is the same as 4(y) because multiplication is commutative. So here I've got four (y)'s and here I have three (y)'s, so how many (y)'s do I have together? I've got seven. So I have plus 7(y) and here I have 3(x)(y) and I add an (x)(y). So if I have 3(x)(y) and I add 1(x)(y), now I have only 2(x)(y), and so this algebraic expression, which seems very long, it fills up the whole screen, actually can be reduced down to here.
Notice that if I take this and let (x) = 0 and (y) = 1, what happens? Well 0 o 9 = 0 and this thing, 0 times this stuff is 0, and 7 times 1 is again 7. So we see the same answer we got when we plugged in, on this long expression, but this simplified version is a lot easier to deal with. So simplifying algebraic expressions using the basic rules and properties of algebra, allows us to produce an equivalent algebraic expression, but again, we have these mysterious quantities that we have to fill in.
Well, let me try one last thing maybe just for fun. What if I let (x) take on the value of 1 and (y) take on the value of 1. Well, then I plug in a 1 for (x) and a 1 for (y), and then this thing would turn out to be 9 o 1, which is 9 + 7  2. Well, what's that? Well 9 + 7 = 16  2, and so we get an answer of 14. So with this choice of the variable or the unknown (x) and this choice of the variable (y), we see this quantity becomes 14. Notice how it varies, it changes, and that is what we mean by an algebraic expression. In this particular representation, we see that 9 is a coefficient, 7 is a coefficient and 2 is another coefficient.
Anyway, these are the objects that we are going to be manipulating and trying to solve in order to really understand not only algebra math stuff, but to try to model our world through mathematics. Great, we're making the first step towards our understanding of everything.
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