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College Algebra: Solving Two Equation Systems

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About this Lesson

  • Type: Video Tutorial
  • Length: 8:22
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 90 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra Review (30 lessons, $59.40)
College Algebra: Systems of Equations (33 lessons, $44.55)
College Algebra: Linear Systems in Three Variables (5 lessons, $7.92)

In this lesson, we start by reviewing three-equation sets that give us independent systems (meet at one point), inconsistent solutions (don't have a solution) and dependent systems (meet on a line). Next, you will move onto the instance In which you have three variables but only two equations. To solve dual-equation system problems, you first work to cancel out one of the included variables. Next, you start over to eliminate a different one of the two variables. In the end, you may come up with an answer that implies an infinite number of solutions (a line) or no solution (an instance when the two equations can never intersect - generally a situation where two planes are running parallel to each other).

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 http://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.

About this Author

Thinkwell
Thinkwell
2174 lessons
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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 www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

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So we’ve been seeing the possibilities if we have three linear equations and three variables. They could all meet at a point and we have one solution. That’s great. Those are called the independent systems. We could have a possibility where in fact they just can’t all be satisfied at the same time. They sort of don’t all meet up at some point. So that would be an inconsistent solution. Or we could have them sort of all meeting up right along a line. In fact there’s a line of solutions, infinitely many and that’s called a dependent system. Okay, well that’s if you have three linear equations and three variables.
What if I only give you two linear equations and three variables? What are the possibilities? Well, there might be no solution. In fact, if those two planes are parallel, just like that, they never touch. Or the other possibility is this. You see the two planes actually meet up along a line. So, we would expect a line of solutions. We can’t just have one answer, because if two planes are going to meet together they’re going to meet along a line. So, actually it would be very much like that dependent system that we saw earlier where the three equations actually all meet along the line.
So how would this work? Well it would work very much the same way, but happily only two equations. So life gets a little bit better. For example, how about if we have 2x - 2y = 14. We put that together with x + 4y = -2. Well actually I’m forgetting my z’s. This is easy to do. You might be saying, “What happened to the z’s?” Let’s try it again, 2x - 2y +5z, there we go, equals 14 and then x + 4y - 3z =-2. When I look at this I say, “Hey, there are two equations and three unknowns.” So there’s either two possibilities. Either they’re going to meet like this, in which case I’m going to have a whole line of solutions or they’re going to be parallel, in which case there’ll be no solutions.
So, how can I see what’s going on? Well, again it’s the divide and conquer. One thing I can do, for example, is to first try to get rid of the x’s. I can get rid of the x’s by multiplying this whole thing through by 2 and then subtracting these two equations. So, if I multiply this whole thing by 2, what does that look like? Well, this times 2 would be what? 2x + 8y - 6z = -4, multiplied everything through by 2. Now, what if I just copied this down exactly as it is? So we have 2x - 2y + 5z = 14. Well, then if I subtract these two from one another and I’ve got to subtract everything by the way. But 2x - 2x is 0. So that drops out, which is the whole point. I have a -2y - 8y, so that’s a -10y. Then here I have a 5z - (-6z), so that’s 5z + 6z, which is plus 11z. Well, that’s going to 14 -(-4), so 14 + 4, which is 18.
So, I see this fact. This is sort of one equation, right there. In fact, what I could o to get another equation in terms of x and z is to do the following. I could multiply the top equation through by 2 and then add it with the bottom. Now why am I doing this, by the way, you might be saying. Well, let me remind you that what I’m trying to do is, I know that either there’ll be no solution so I’ll get something that’s just impossible or if there will be a solution it is going to be infinitely many. So, what I’m trying to do is say, “Well let’s just pretend that in fact z will be any number at all.” Z is just some random thing. If z is some random thing, I now know what y is. I can solve this for y. I now have to figure out what x is. What I want to is get a relationship between x and z.
I can do that in the following way. I can take this equation here now. Let’s see, I just did what here? I just took the bottom one and I multiplied it by 2. So let me take now the top one and multiply it by 2 to have the cancellation of the y’s. So I multiplied this thing through by 2 and what do I see. I see 4x -4y + 10z = 28. Multiplied everything through by 2. Then I copy down the second equation as is. So, x + 4y - 3z = -2. Now, I’m not going to subtract these people, because then I’ll be in trouble. But notice if I add them right now, they all--these things drop out. So, I’m going add right now. 4x+ x is 5x, -4y + 4y is 0 and 10z - 3z is +7z and that equals 28 - 2 which is 26.
Okay, great. So that gives me a relationship between x and z. I have one between y and z, so that means the following. If I suppose now that z is just any number, anything at all--it can be anything. I’ll call it c. Then I could actually figure out what x has to equal. What would x equal? Well x would have to be what? If I put in c here, I would see 26 and then -7x, if I brought that over and then divide it by 5. Did all that in one step. What would y have to be? Well, y would be--well, if I solve this for y, I would see 11c, if I put in c here minus the 18 all divided by 10. So I just solve this for y. I brought the -10y to this side, it makes it a +10y, 18 to this side is a minus 18 and divide it through by the 10.
So I see now this does have a solution and in fact, therefore infinitely many, because it’s only two equations and three unknowns. Z could be anything at all. Once you know z that forces x and y to be these values exactly. So, an infinite family of solutions. Infinite family. Okay, one last one before we move on. How about this, 4x - 2y + 6z = 5. At the same time 2x - y + 3z = 2. Okay, what could we do here? Well, one thing I could do here, let’s just take this bottom equation and multiply it by 2 in order for me to get rid of the x’s. So, I’m going to multiply the bottom equation by2. What do I get? I get 4x - 2y + 6z = 4. I multiplied everything through by 2.
Now I copy down the first equation. So, copy that down exactly as it is and I’ll subtract. Now what happens when we subtract? Let’s see the net result. I’ve got to subtract everything now. 4x - 4x, that’s 0, -2y - (-2y), that’s 0, 6z - 6z is 0, 5 -4 is 1. So, I see that 0 = 1 and that doesn’t happen here on the web. So 0 = 1 tells me that these two equations cannot be satisfied at the same time and what really is happening here is that these two equations are actually two planes that run parallel and they never ever intercept. So, in fact, there are no solutions to this. The previous example we did, they did meet. They met along a line, so there were infinitely many and we can parameterize that in terms of any value c can take on. We can figure out the x and the y that are appropriate. Okay, so try these two equations and three variable all linear all the time.

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