 Welcome back to our lecture series Math 1050, College Algebra for Students at Southern Utah University. As usual, I will be your professor today, Dr. Andrew Misseldine. So in this video, lecture 45, we're starting the beginning of the end. We're going to be in our last chapter entitled, Chapter 7 Precalculus, for which this Chapter 7 has essentially three goals for our lecture series on College Algebra. Let me illustrate to those to you really quickly before we get into some examples. Our first goal of this Precalculus chapter is really to review topics we had seen throughout this entire lecture series. In Chapter 1, when we introduced functions, we didn't know a lot about specific function families, but we saw general ideas of domain, range, solving equations, solving inequalities. Then throughout the series, Chapter 2, Chapter 3, et cetera, et cetera, we started focusing on specific function families like linear functions, quadratic functions, polynomial functions, exponential functions, again, just to name a few. Part of this chapter as we end our lecture series is to review these important function topics, again, domain, systems of equations, inverses, graphing, difference quotients, all those important function topics. But now we want to revisit it when now that we have a better understanding of these function families. That also leads us to the second topic, which actually was an idea I got from a colleague some time ago, is that many students who take a class like College Algebra, they're taken as a prerequisites to calculus, which they'll be taking in a future semester or something like that. College Algebra is good at teaching you how to work with certain function families in isolation. It's like, this is a polynomial problem, this is a rational problem, this is a radical problem, this is a logarithm problem, and we're really good with those in isolation. But in calculus, all of these different tribes of functions co-mingle with each other. You can have an equation which involves simultaneously trigonometric functions and exponential functions. You can have an equation that involves radical functions and logarithms simultaneously. Now, in this Algebra series, we're not going to deal with any trigonometric functions. We'll leave that for the series Math 1060. But the idea of bringing together these different functions and working with them simultaneously, is a pretty important skill for calculus, which the usual way that College Algebra is pretty insufficient in that regard. As we review topics from the previous topics of the series, we're going to do that with the lens that we're going to put together things that weren't previously connected. For example, this lecture, 45 is going to be on the topic of nonlinear systems. We've talked about linear systems previously, so we've used techniques like substitution and elimination to solve systems of linear equations, as well as some matrix methods as well. But we've also dealt with nonlinear equations throughout this lecture series, quadratic equations, polynomial equations, exponential equations. What happens when you put the two ideas together? What if we want to solve a system of nonlinear equations? We have to combine the techniques of substitution and elimination we learned previously with some of these other techniques we've learned about these nonlinear equations. So we're going to marry together different topics we learned in isolation together, because that's how it's done in calculus. That's how it's done in real life. And hence, this chapter is called pre-calculus. The third reason why we have this chapter is because, well, honestly, there are some annoying section headings I have that I just need to continue the jokes we previously had, right? It turns out that Harry Potter and Star Wars kept on making movies after the original lectures were made, and therefore we have to continue the puns. I mean, what does dancing robots have to do with systems of equations? Well, if you understand the system is down, they're taking over. If you get that reference, then this one fits in there as well. But I don't want to spoil it for anyone. All right, so let's get to the heart of this first video for section 7.1. We want to study linear systems, and I claim that the methods of substitution and elimination we learned previously are applicable even in a nonlinear setting. And so matrix methods like Gaussian elimination would apply as well. To a limited degree, I'm not saying substitution elimination solves every system of linear equations, nonlinear equations, excuse me, they do solve every linear system. But we can apply those techniques to help us out in even in a nonlinear setting. So let's try this first video. We're gonna use the technique of substitution and adapt it to systems of nonlinear equations. So consider the following two by two system. That is two equations, two unknowns. So the first equation that you see here on the screen is the equation x squared plus y squared equals 100, which we've seen previously in this series. This represents a circle. It's a circle of radius 10, right? 100 of course is 10 squared. And so we're thinking of x squared plus y squared equals r squared. This is a circle centered at the origin whose radius would then be r, there would be 10. The more general circle equation we get r minus h squared plus y minus k squared equals r squared. So we're using that right here. And so we can think of the first graph is the graph of a circle, all right? That is a nonlinear graph. The graph of that is not aligned. On the other hand, we could take the equation 3x minus y equals 10. This is course just aligned in the usual sense. We could put it in slope intercept form, for example. If you add y to both sides and you subtract 10 from both sides, you would see that y equals 3x minus 10 as the slope intercept form of this equation. Before we try to solve this nonlinear system algebraically, let's consider for a moment, what is the nature of the solution set? Like what are the possible solutions you could get by solving the system? Well, like I said, the first equation gives us a circle. The second equation gives us a line. What are the possible intersections of an equation with a line? Well, as you see on the screen right here, you can see a circle and a line. The circle center at the origin, the line, I've just drawn a line. I'm not claiming that this is the graph of our circle or a line. It's just a picture right here. But what are the possible solutions? So in one situation, you see the following. You see the circle with a line. And this is an example in geometry what we call a secant line. It's possible that the circle and the line could intersect each other at two distinct locations because the line passes through like that. Another possibility is you could get something called a tangent line. That is, there could be a line that comes and just intersects the circle at just a unique location like you see on the screen right now. So we could maybe just have one solution to this system of equation. There's a possibility of that. Or it could be that the circle and the line are completely parallel with each other. That is, they have no intersection whatsoever. In this situation, you would have no solution, right? This would be an example of the system of equation being inconsistent. And so those really are the three possibilities. We either have a parallel line, a tangent line, or a secant line, which that's what I'm gonna do. I'm gonna call this last case parallel line. That is, it's parallel to the circle. They don't intersect each other whatsoever. And in this situation, of course, I'm using, as my definition of parallel, nothing to do with slopes being the same just that they're not intersecting curves, all right? So those are the three possibilities. So algebraically, how are you gonna find this? Because geometrically, we could try to graph it like we see here on the screen, which again, this is not the correct graph. This is just an illustration of what could happen. Let's try to solve this algebraically. Well, in the process of writing the line in slope intercept form, you'll notice that I actually kind of solved for y, y equals two x minus 10. And so if we try to solve this via substitution, like we said we would before, we could substitute in the, instead of the value y, we could take the expression three x minus 10. What if we were to plug that into the other equation? If we did that, we'd end up with x squared plus, instead of y, we're gonna be three x minus 10 squared equals 100. This is now, if you look at the x, which is why it's now gone, this is now a quadratic equation in terms of x. Let's FOIL out that three x minus 10. You're gonna get x squared plus, when you FOIL that out, you're gonna get three x times three x, which is a nine x squared. You're going to get, when you do the, when you do an x one, three x times negative 10, that's gonna give you a negative 30x. But then when you FOIL, you're gonna get two of those, so it'll double to give a negative 60x. And then you're gonna get negative 10 times negative 10, which is a positive 10 equals 100. So we FOIL that out, we have these terms. Now let's combine like terms. You can subtract 100 from both sides of the equation. Those will just cancel out. And then we have these x squared terms, which combined together to give us a 10x square minus 60x equals zero. So after we did the substitution, we acquired this quadratic equation. How do we try to solve the quadratic equation? We try to do it by factoring. The right-hand side's already equal to zero. So on the left-hand side, we can factor out a 10x, the common divisor there. That leaves behind x minus six equals zero. And so by the zero product property, we get that either 10x equals zero, which would imply x equals zero, or we get that x minus six equals zero, which implies x equals six. So how can there be two solutions here? Well, this comes back to the picture we had earlier. What we have in play here is a secant line, right? We see that there are two possible x-coordinates that are acceptable in this picture. And so we have, so okay, we're gonna have two solutions. We have the secant, but what are the y-coordinates? Well, with the y-coordinates, we're gonna have to come back to the equation we had before, right? So we know y equals three x minus 10. So let's insert those values of x that we found. So one of the values of x, remember, was x equals zero. If we plug that in here, we're gonna get that y equals three times zero minus 10. That is y equals negative 10. This gives us a solution. The first solution we find here is that when x is zero, y equals negative 10. So that's our first solution. The next solution would come about by plugging in into this equation, x equals six. So when x equals six, we see that y is gonna equal three times six minus 10. That is 18 minus 10, we get an eight. And so the second point would be six comma, six comma eight. And so we can check that both of these points are solutions to this system of equations. If you try zero and 10, right? In the first equation, we get zero squared plus 10 squared, which is zero plus 100, which is 100, that works out. In the second equation, we get zero minus a negative 10, which is positive 10, so that works. So the first one passes the test. For the second one, six and eight, if you put into the first equation, you'll get six squared plus eight squared, which is 36 plus 64, which is 100, that passes. And then for the second one, we just kind of saw this one as well, three times six is 18 minus eight will give you 10. So these are both real McCoy solutions to the same systems of equations. Those are the only ones we were able to do. We were able to solve this system using substitution. Let's look at another example here. So let's consider the picture. Well, let's sorry, let's consider the equations. You have x minus y squared equals zero, and you get y minus x squared equals zero as well. So when you look at these two graphs, they both look like parabolas, right? So the first one, you have an x minus y squared equals zero. You have the second one, you have a y minus x squared equals zero, these are two parabolas. Now, when you look at these parabolas, the first one, y equals x squared, if you were to solve for x, you would see that x equals y squared. Notice that the y is actually the thing that's squared. This suggests that this is gonna be a parabola that's gonna be concave to the right. Concave to the right here. As opposed to the other parabola, which if we take that one and solve for y, we're gonna get y equals x squared. This is our standard parabola we're used to. This is a concave upward parabola. And so I'm actually gonna give you the picture of these things right here. So you have y equals x squared and y and x equals y squared right there. We have these two parabolas. What are the possible ways they can intersect each other? Well, you see the picture right here. Let's just leave y equals x squared where it is. You could have y equals x squared, excuse me, x equals y squared. You can get something like this. The two pictures could intersect each other at two locations. Is it possible that they intersect each other at maybe like one location? It could be, maybe, just kind of thinking it, could be like that the parabolas are somehow tangent to each other, right? Maybe they just touch each other at one point. That's a lot easier to see if we were like concave downward. They maybe they only share the vertex or something. So that could be a possibility. If there was some rotation to the parabola, you could get that they were tangent like so. So at one point of intersections possible, this one shows us two. We also could get four points of intersection. So we could have something like the following. We have a parabola, maybe does something like this. So maybe there's like four points of intersection, right? Could you do three points of intersection? Yeah, you could have it so like the vertex touches one side and then breaks away like this. So we get three points, something like that's a possibility. Another possibility is they could just be parallel to each other, right? The other parabola is just not intersected whatsoever. And so when you talk about like the number of solutions here, the number of solutions, well, we could get zero, we could get one, we could get two, we could get three, we could get four, could you get more than four? Could you get five solutions? Now I want to get a little bit trickier if you try to start, if you try graphing these parabolas, try and try it as you might, you're not gonna find five solutions here. Then there's actually a nice algebraic argument that's gonna happen with this. We can see this from substitution in fact. So if you take these two equations again, solve for the linear factor, notice you have x and a y, solve for the x, solve for the y, you get something like this, okay? Let's substitute one into the other. For example, what if we were to substitute, say, x? You know, if we were to substitute x from this equation down into this one right here. So x is equal to y squared. This would tell us that y equals, well, instead of x, we're now gonna get a y squared. But then you're squaring that thing. So our equation becomes y equals y to the fourth. This is now a polynomial equation, a polynomial equation for which it's degree four, right? So at most you're gonna have at most four solutions to this thing, all right? And as we try to solve it, what I would do is I would set one side equal to zero. So we're gonna get y to the fourth minus y equals zero. Factor the left-hand side. The left-hand side, you take out a factor of y getting y cubed minus one. And then factoring that, we will take the difference of cubes factorization. You get y times y minus one, and then you get y squared plus y plus one, like so, equals zero. For which this quadratic right here is irreducible. The discriminant you can check for that thing is gonna be negative three. So it has no real solutions, which means in terms of intersections, there's not gonna be any, there's no, you can't have an imaginary intersection. It's gotta be a real number in that case. So we can throw out y squared plus y plus one right here, and so from the first factor, we're gonna get that y equals zero. From the second factor, we're gonna get that y equals one, like so. And so there's two possible solutions here. And so this gets back to this idea of the number of solutions. When you take a parabola and insert it into another parabola, worst case scenario, you're gonna get a degree four polynomial equation, which has the most four solutions. And so although any number less than four is possible, you can't get more than four solutions here. Now let's come back up to our original equations, right? When x equals, excuse me, we've now discovered that y equals either zero or one. What's the corresponding x coordinate of these things? Well, you can plug it into these equations, right? Notice if you plug in y right here, you're gonna see that when y equals zero, x is gonna equal zero squared, which is also zero, which gave us our first point of intersection right here, zero and zero. And then if you plug in one into this one, you're gonna get one squared, which is gonna be one. And so our second point of intersection is gonna be one, one right here.