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Beg Algebra: Equations


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About this Lesson

  • Type: Video Tutorial
  • Length: 8:37
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 92 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Introduction to Solving Equations (3 lessons, $4.95)

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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Now that we have this idea of an algebraic expression, what we are going to want to do is put algebraic expressions together by setting one algebraic expression equal to another, and then try to solve for the unknown quantities, trying to find the variable, and turn that variable into the answer, or in some cases maybe several different answers that will resolve that particular equation. Now a lot of times if we are looking at applications we are going to take some issue where we know some things, and there is this question, and then there is an algebraic bridge that links the known stuff to the unknown stuff. That algebraic bridge will in fact be an algebraic equation. We will have two things that we set equal to each other. The question is then what is the solution, what should that unknown quantity be in order to balance it? So when we have an algebraic equation, what we want to be able to do is we need to solve it.
So, first of all let me tell how I think about an algebraic equation. I think about an algebraic equation like a balance scale. So here is an equal sign and then there is some stuff written here, and then there is some stuff written here, and then either by the fact that this was just handed to us, this algebraic equation, or it evolved organically through some real world issue. We see that we have some relationship where this is the same as that. Now I want to solve this algebraic equation. Well how do we go about solving an algebraic equation? This is the fundamental aspect of algebra, it is solving such things. The basic strategy to keep in mind is that this delicate balance cannot be disturbed. In fact, solving equations is a delicate balance. If you think that way then you will not go wrong.
Let me show you what I mean. If I want to do something to one side, I have to do the same thing to the other side. Suppose I add 4 to this side, well that would actually bring things down. But, if I don't do anything to this side, I am in trouble. Suppose that instead I put a 5 in on this side. Well, again this still doesn't make the original thing any more equal, because this is heavier. So that is just wrong. If I put a 4 on one side, I must in fact put a 4 on the other side. So whatever I do on one side, whether it is adding, whether it is multiplying, whether it is dividing, whether it is subtracting, what I have to do is do the same thing on the other side. If we do that, then the scales of algebraic justice will remain balanced and we will produce a correct answer to the issue.
Now, how do you actually go from this very eloquent law notion of scales of algebraic justice to actually doing this stuff? There is going to be an unknown quantity, and later if you continue taking math, there may be more than one unknown quantity. But, when there is one unknown quantity, the thing to do is to try to isolate that unknown quantity. Let me show you what I mean. For example, an expression might look like this. This is an example of an algebraic equation. So what does this mean? Well, notice that's an algebraic expression: 4 times some unknown thing plus 6 is going to now equal 10. So the equal 10 part makes this entire expression now an algebraic equation. And the goal is to do some algebraic gymnastics, and then at the end of the day convert this, keeping the scales of justice the same, to an expression that looks just like this. If I can take this expression, and using all the properties that we know about algebra, to an expression that says "x" equals a number, then I have the answer "x" equals that number.
So the purpose now is to work at getting "x" equals a number. What does that mean? Well, that means, for example, that in this expression I have several things to do. First of all I have this 6, this constant term here, which I wouldn't want. So I have to do something to both sides in order to even that out. How would I get rid of that 6? Well, I would just subtract it. But, if I subtract it from this side, I have to subtract it from this side. So I just do that, and notice by writing the equal sign here, I see that the balance of justice remains the same. So here I see 10 - 6 = 4, and on this side I see 4x, and then 6 - 6 = 0. So notice I am actually closer to this kind of thing. In fact, here you could almost guess an answer couldn't you? What do I have to multiply 4 by to in order to make it equal to 4? Well, the answer is 1. So here I could just eyeball it and see that a guess would be that x = 1, and of course as always we can check our answer by plugging back 1 into the original expression, which is 4 x 1 + 6. Well 4 x 1 = 4, plus 6 is 10, and it checks. Notice that I started with this algebraic equation and I ended with an equivalent equation, but one where I can just read off the answer, x = 1. So this journey is always a journey from taking a complicated thing, and massaging it, and converting it without changing it's meaning into something where the answer becomes easy. Here the answer is not so easy and here the answer is a trivial x = 1, and of the same equivalent expression. That is the journey through algebra.
Let me show you another algebraic expression which is much more complicated, and requires a lot more work. This expression looks like this: 3 o (x + 2) = 5 o (2x + 3 - 6x), wow! Maybe this came from some word real life world issues, and we set up some sort of equation, and we got this really complicated expression. Well the first thing we want to do is to try to isolate the unknown, so we do some algebraic steps. We can distribute, for example here, and we can distribute here, and what do I see? Well, here I see 3x + 6 = 10x + 15 - 30x. I can now remember one of the fundamental rules of this course, which is that we are aloud to combine like terms. So, here these are x's, and here there are no x's, so I can't combine those at all, because these are 3 x's and this is just 6. But, notice here I have 10 x's and someone is removing 30 x's, so I have a net loss of 20 x's. I can combine the 10 and the -30 to produce -20x, and then I have the +15. So I have actually converted this complicated expression to this slightly simpler one.
Still the solution is not at all clear, and in the up and coming lectures we will see how to actually resolve this. But, the way to resolve it is to try to isolate the variables and bring them all onto one side, and the constants to the other side, keeping the balance equal. The way we will do that is we will subtract 6 from both sides, to remove that 6, and then somehow we have to move this over to that side. The way we are going to do that is by adding 20x to both sides. If we do that we actually get the answer, but I am going to hold that for a future lecture.
Anyway the important thing is that algebraic equations are simply algebraic expressions where one is set equal to the other, and our goal is to go from that cryptic expression to the very simple expression, x = . See you at the next lecture.
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