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College Algebra: Solving a Linear Equation


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About this Lesson

  • Type: Video Tutorial
  • Length: 7:56
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 85 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 Equations (8 lessons, $11.88)

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 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.

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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All right! Well, you know, one of the great things that we're going to do a lot of is solving equations. We're going to want to solve everything, because you have an equation, someone gives you an equation, you want to find out what the solutions are. Okay, great!
Well, there are some sort of basic techniques that you can use to solve equations, and I thought that one of the things we should do is just take a look at some basic equations and see how you would solve them. All these equations are actually linear equations, which just means that when you see the variables, they're all just going to be to the first power. There won't be any x^3 or x^217. We're not going to see that now. That'll be in maybe the next lecture or somewhere. But, for right now, it's all going to be just stuff with just x's and other things in it. So let's take a look and line up and solve an equation, sort of like a limbo line. Can you see that? They're lining up just to solve an equation. Where else but on the Web?
Okay, let's try an example. Suppose someone gives us this equation: 2x - 5 = x + 7? What we want to do is we want to find out what value of x will satisfy this. Now, let me just say just two words here to remind you of what's going on here: x is really just a placeholder. x represents some mysterious number that someone has hidden from us. Think of it as hide and go seek. Someone's hiding and we have to find x. So how do you do that? Well, you know that this relationship holds, so what we want to do is we want to isolate x, have x all by itself, and then figure out what x is. At the end of the day after work, we want to have x = what? So what I want to do is try to get x's on one side and everything else on the other. And the way I'm going to do that is the following: to bring the x over, I'm just going to subtract the x from both sides. Since x = x, I can subtract it from both sides and nothing's going to change. So here I'm going to see 2x, but then I subtract an x. That just gives me x - 5. And now x - x, that drops out as zero and I still have that 7 there. And now, I want x all by itself, so I just add 5 to both sides, which doesn't change the value of anything, and on this side I just have x, 5 - 5 = 0. They drop out. And then here I have 7 + 5 = 12. So I see that x = 12. x has been exposed.
Anyway, you can always check your answer, by the way, by taking the 12 and placing it back in on both sides for x, and make sure this is okay. Let's just do that together in our heads. If I put in a 2 here, 2 12 = 24, 24 - 5 = 19. And if I put 12 in on this side and I take 12 and add it to 7, I get 19. So I see 19 = 19, we're okay.
So that's all there is to solving linear equations. Let's try another one, just for grins here. How about 5(y 3) + 4y - 5 = -(2y - 4)? Well, there's a lot of stuff here. Again, the theme always is to find y, but what I want to do is try to get all the y's sort of collected together here, an equal sign, and everything else on the other side.
So to that end, the first thing that I'm going to do is make all these distributive things here. I want to distribute that 5. This parenthesis means that this 5 has to be multiplied not only by the y, but also it's got to hit the 3. And similarly, this negative sign, that's actually a -1 in disguise, that thing has to not only hit the 2y, but also has to hit the -4. So I'm going to do all that distributing right now, just so we can see where we are. We'd be at 5y + 15 + 4y - 5 = -2y, and then a minus times a minus becomes a plus 4. That's what it looks like.
Now what? Well now I'm going to try to put all the y's together. So I have a 5y here and a 4y here, that's 9y. And, let's see, I can combine these numbers. I have 15 - 5, which would be 10. And that equals -2y + 4. Now I'll bring all the y's on the same side. I'll move this to this side and, by moving it, I've got to actually change the sign; or another way of thinking about it is I'm going to add 2y to both sides. How do I know I'm to add 2y? Well, because I'm subtracting it here and I want these to cancel out, so I do the opposite of whatever I'm doing to move it over. Here I'm going to have 11y + 10 equals - and these cancel out. I'm just left with the 4. Great!
Now I want to bring that 10 over, so I'll subtract 10 from both sides. And if I subtract 10 from both sides, what would I see? I would see 11y equals, if I bring this 10 over, I'd see -6, because 4 - 10 = -6. And so y = .
And again, you might want to go back and take and plug it in for each value of y here and each value of y there and make sure that this number equals that number. And you can check that and see that this answer is correct.
So that's all there is to solving these things. Let me try one last one maybe, try to stir things up a little teeny bit more. It's hard to stir things up just looking at these linear equations. There's not much stirring involved. Shaken, not stirred.
Suppose I have and I set that equal to and I want to find out what value of t will make this thing true? Well, it looks like I have a whole bunch of fractions here, so that might be sort of scary. And, to be honest with you, fractions frighten me a little bit, so I like to not have them there any more. So one way to get rid of them would be to multiply through both sides by the same number, so that the fractions would go away. Well, one thing I could do, for example, is multiply everything through, both sides, by 5. If I did that, notice what would happen. The 5 here would cancel with the 5 here. Unfortunately, I still have that 3 over here. But if I multiplied everything through by 3, then I could cancel that. So it seems like 15 is a great thing to multiply both sides by. So I'm going to multiply this side by 15 and I'm going to multiply this side by 15. If I multiply two sides of an equal side by the same number, it's not going to change the value. So if you have an equality, multiplying both sides by 15 is not going to change things.
But now, look at the cancellation. This is fantastic. The bottom goes away and it costs me a 3 on top. The 3 goes away and it costs me a 5 on top. Some people, by the way, think of this as cross-multiplying. You might think of 5 times this times 3 times this, and it's the exact same thing, if you think about it. Anyway, now I've got to distribute. Don't forget to distribute that all the way through. So that's going to be 6t + 15 - see, the 15 is because I have to distribute, and now I'm going to distribute the other way and see 5t + 10. And now I'm going to start to do a lot of manipulations just verbally. This 5t, to get it to that side, I have to subtract it from both sides. So imagine a -5t here. I have 6t - 5t, is just t alone. And if I take that 15, if I want to get it to that side, I should subtract 15 from both sides. And so that would be a zero here, and then here I have a -15. And so -15 + 10 = -5. So, in fact, t = -5.
So even when you have sort of denominators with numbers, you can clear up those denominators by multiplying through by a common multiple and, all of a sudden, you have a simple linear equation and can solve it.
Equations and Inequalities
Linear Equations
Solving a Linear Equation Page [2 of 2]

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