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Beg Algebra: Introduction to Word Problems


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About this Lesson

  • Type: Video Tutorial
  • Length: 11:26
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 122 MB
  • Posted: 06/26/2009

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: 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 at 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 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.

About this Author

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Founded in 1997, Thinkwell has succeeded in creating "next-generation" 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 technology-based textbooks. For more information about Thinkwell, please visit or visit Thinkwell's Video Lesson Store at

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You know, even though you may think that algebra is a completely useless thing, and these "story questions" that we're asked to consider; trains leaving at midnight from Chicago and heading toward Detroit or whatever seem absolutely useless, the reality is that algebra is a very powerful tool for solving real world issues. That's just true, and I want us to begin to think in that template. That, in fact, we can take a real world issue and translate it into an algebraic expression. The power of that is really clear. If we could actually take a real world complicated circumstance that we're trying to understand or trying to resolve and we can convert it into some sort of algebraic expression that captures the essence of the question, then we can use all the power of algebra, all the tools, all the techniques, everything that we've been practicing up to this point, in order to resolve it. And that's the power of algebra.
What I would like to do is just to point out that quite often, if we have a real world issue, we can translate it into an algebraic template. Now, what I'd like to do is talk about how one does that translation and then also how do we resolve issues. So this raises the question; suppose you're faced with a problem. How do you solve it? Well, I'm going to be talking in the context of the stuff we're talking about together, namely algebra, but I think actually this method transcends to any subject you're thinking about and even spills into your every day life.
Here are the three basic steps required to solve any problem that may arise. Number one is to understand the question. What is it that we're trying to do? What are we trying to find? What are we trying to figure out, and bring great clarity to understand what the question is. That's the first step. That's pretty hard actually.
The second step is to figure out what you know, which means what do we have at our disposal, what facts are we given, what facts can we use as part of our arsenal, and then collect all that information. And then if you think about it, it's sort of interesting. We are located and we have a world of things that we know, and our desire is to move to solve an issue, a question that we don't know. So what do we need? We need some bridge which is a connection between the things we know and the things that we're trying to figure out. And in the context of the stuff that you and I will do together, it's actually going to be an algebraic bridge, but in the rest of your life, it might not be algebra, but it might be some other thing. But you need a connection that links the known stuff to the unknown stuff. And if you find that connection, then together, in this course, we can then solve the algebra stuff, and that's the way we're going to proceed.
So let me, first of all, just give you some generic sense of how to set things up, and then I want to actually look at an example together. So here is a sample of the kind of things we can use to convert from one, sort of the real world phrases to sort of the algebra analogy. So if somewhere we see somewhere the product of two numbers, if that appears in some sort of word question, that just means we have two numbers, (a) and (b), and we're taking their product. Three times the number, if we see that somewhere, just means 3 times some number. Twice the number, correspondingly would be twice some number. These are just random variables that are unknown. There's nothing special about (a) or (b) here. The distance multiplied by 10 just means that you need to multiply the number by 10, whatever that number is. 10 percent of a number, well, we know this, right? 10% means 10 over 100, of means times, so this is 10/100 which is 1/10 times the number, which notice is .1 o (a). So here you can see how to translate the English words into algebraic lexicon, and there's lots of others here. I don't want to spend tons of time, but here is some for describing addition. The sum of two numbers is just the sum. Three more than the number means you take the number and you add 3. The number plus five, that's a typo, of course. It should be (a) + 5. Just seeing if you're on your toes, and apparently, the staff is seeing if I'm on my toes. The distance increased by 10. That means we take whatever the unknown is and we add 10 to it. 2 added to the number just means you add 2. Distance 1 and distance 2, "and" of course always means you add; you take the first distance and add the second distance, so not a big deal, and then you can describe division in real world. The quotient of a number and 3. That means you take the number and divide it by 3. 10 divided by the number is 10 divided by the number. So here's the math. 1/3 of the number means the number divided by 3.
So, again, this is just showing you how the words can be informed by the mathematics. Now describing subtraction, we have the difference of the numbers, and that would be (a) - (b), three less than the number is this, the number minus 5, at least we're consistent, the distance decreased by 10. Even I'm bored. Okay. You get the idea.
All right. Let's take a look at an actual real world issue and see how we can use this thinking process together with algebra to set up a resolution. Okay. Here we go. Take a look on the side here. So, you want to buy a truck, and so in fact, here's the truck you might want to buy. Isn't it pretty? You're going to buy that truck. It's very, very nice. So you want to buy this truck, great, and you see that it costs $1,000, which is pretty bad when you consider the truck is only like 3 inches long. It seems like a rip off to me, but suppose you want to buy the truck for $1,000, and it turns out you have, at your disposal, $100. Nice. The smell of money--nothing sweeter. So you have $100, but the truck costs $1,000. It turns out that the job you have, you make $10.00 an hour. The question is, how long do you have to work to be able to buy the car. You already have $100, the car costs $1,000, you make $10 an hour. Well, let's try to use this process of resolving issues to see how we might actually proceed. So what do we know? So we know that the cost of the car, so we know that the cost of the car, or the truck rather, is $1,000. We know also that we have $100, and finally we know that we make $10 per hour. Now, what do we want to find out? What do we want? We want to know how many hours must we work in order to buy the car. So that's what we know, and that's what we want, and now all we need is a connection. In this case, it's going to be an algebraic connection that links these things. Well, how many hours--we don't know what that number of hours is, so now we're going to introduce a variable, and we can call it anything we want. We can use any letter. We can call it (x) if you want, we can call it anything at all. I'll call it (t) just to represent time. So now let's let the unknown answer be the number of hours, and we'll call that (t). So you see, I just call it (t). I don't know what it is. It's unknown. It's a variable, but I'm going to call it (t). Now, what do I know for sure? Putting these things together, I see that since the car costs $1,000 and I have $100, I need 1,000 - 100, or $900. I also know that if I work (t) number of hours, how much will I make? Well, I make $10 for every hour I work, so if I work (t) hours, (t) hours gives 10(t) dollars. Let's think about that. If you work (t) hours, you work, you get 10(t) dollars. Now, what do I want? Well, I want to earn $900, and here's the connection. So here's the algebraic bridge between what I want to find out and what I know. I need $900, for every hour I work I make $10, and so the question is 10 times what variable, what unknown will produce $900. So this is the connection. So now we can actually go off and solve this using algebra. In fact, we'll see how to do this, but you can just see, the main thing is to figure out what this expression is. Now that we've got the expression, we can use the power of algebra to solve it. In fact, if you think about it, if I just divide both sides by 10, we'll talk about this later, but just so you can see the resolution to the question, we see that I get (t) alone, 10 ÷ 10 = 1. So (t) = $900 ÷ 10 which you will notice is 90, and these units are hours. So if I work 90 hours, let's check it, 90 o 10, notice, is $900. So if I get $10 an hour, and I work 90 hours, I get $900. So, basically, that's about what? So 80 hours--so that's about what, two weeks or so, or maybe one week. I guess it depends on if you're working part time or not, but it's 90 hours and you can buy the car.
The important thing is, we figured out exactly what we know, we figured out exactly what we wanted, we introduced an unknown, and then we found a connection. And that algebraic bridge between the known and the unknown is the key to resolving the issue, and now we can go off and use algebra, as we'll see, in the upcoming lectures to actually solve this. So now the point is we need to figure out how to solve algebraic equations. Neat stuff. And by the way, I hope you enjoy your car. See you at the next lecture.
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