 Welcome back to our lecture series Math 1050, College Algebra for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Mistletoe. This is our first video for lecture 33, and this is going to be based upon our section 5.3 Rational Equations and Inequalities. We've been following a fairly standard pattern since chapter two basically of this series, where we introduce a new function family. We then talk about things about that function like domain, range, the graph of the function, how do you solve equations? How do you solve inequalities? How do you work with story problems? We don't necessarily always go into that order, and we also will explore certain exploratory things that also might be related to that. That's where we are right now. We've talked about rational functions. We've spent a good amount of time discussing the graphs of rational functions. In this lecture, I want to talk about how we solve rational equations and rational inequalities. We saw a little bit of that while we were graphing them. Sometimes we had to decide whether a graph crossed its horizontal asymptote or something like that. But the basic idea one has when you solve a rational equation, well, it is the following. It's called clearing the denominator. Well, because you have these fractions that show up inside of your function, and so these are going to be fractions which the denominators themselves might have variables in them. That's what a rational function is all about. The best way to get rid of, I should say to solve this equation, is to get rid of the denominators as much as possible. Because in order to add things together and combine like terms, we'd have to have a common denominator in the first place. But it turns out if we're going to go through the effort of finding a common denominator, we actually are in a position where we can just get rid of the denominators entirely, and so we can turn a rational, this rational equation into a polynomial equation. To do that, we have to identify the denominators in the equation right here. So we look at this, we have an x minus two, we have an x minus one, and we have an x minus one times x minus two. So the first thing to do here is we have to identify the least common denominator. That is, what is the least common multiple of all of the denominators present? And I like to think of finding the LCD much like as ordering the pizza for a bunch of college students and their roommates in the dorm. Because college students are notorious for not having lots of money, right? We eat mac and cheese and ramen, top ramen, all these cheap foods all the time. So once in a while, when you wanna go luxurious, we're gonna get a pizza. But amongst the three roommates here, we really only have enough money for one large pizza, so we gotta put all the toppings on it that we want. So our first roommate here, he's like, hey, I want x minus two on my pizza. That is, I want pepperoni on my pizza. And so that's the first request. We're gonna put x minus two on this pizza. Now, the second roommate, he's like, hey, you know, I want olives on my pizza. And which case, then we're really, it's like, oh, I want x minus one on my pizza. So the second fraction is gonna request x minus one on the pizza. Now, when the third roommate hears this, this roommate's like, hey, you know, I want, that sounds like a great idea. I want olives and pepperoni on my pizza. I want both of those things. So in terms of the pizza request, we're not gonna order double pepperoni or double olives because this fraction right here likes the order as it is. It doesn't need any more. And so if a factor shows up twice in these denominators, like you have this x minus two in the front and this x minus two in the back, we're not gonna order it twice. We're just gonna get a pepperoni, olive pizza right here. And so this is gonna be our least common denominator, x minus two times x minus one. There is no benefit for us whatsoever to multiply that out, keep it factored. What we're gonna do instead is we are gonna multiply the left-hand side and the right-hand side by this least common denominator. So we multiply the left by x minus two and x minus one and we do it on the right as well. What's good for the goose is good for the gander. So we have to do the same thing to both sides of the equation. Now, on the left-hand side, we're gonna end up with a three times x minus two times x minus one all over x minus two. On the right-hand side, we're gonna distribute this because we do have a plus after all. If you have a sum of two things and you multiply and you can distribute that multiplication through the sum. So we're gonna distribute onto both of the fractions there. We're gonna get x minus two times an x minus one. This sits above x minus one. And then we have a seven x minus two times x minus one sitting above an x minus one times x minus two. All right, so we chose our least common denominator so that all of the denominators should cancel out. If you look at the left-hand side, we have an x minus two on top and we have an x minus two on the bottom that cancels out. In which case, these common divisors will cancel across the fraction bar and we end up with three times x minus one. Then moving to the right-hand side, you look at the first fraction. There's an x minus one on top, a factor of x minus one. That cancels with the factor of x minus one on the denominator. And so that fraction will simplify just to be x minus two. And then finally, if you look at the last fraction, there's a factor of x minus two that cancels with the top and there's a x minus one that cancels with the bottom. In which case you leave just a seven. And so after all this multiplication, we end up with just a polynomial equation. That's honestly just a linear polynomial. If we distribute the three, we end up with three x minus three equals x minus two plus seven. You could add the seven and the negative two together you get x plus five. Combining like terms, let's subtract x from both sides. Let's add three to both sides. So we end up with a two x on the left, three x minus one. And then on the right, we have five plus three, which is an eight. And so then my next inclination is divided by two on the left and right-hand side. And this would then suggest that our solution should be x equals four. That's thus solving this equation right here. Now one should always be cautious when you're working with a rational function because unlike some function families we've seen in the past, rational functions do have restricted domains. And what is possible is that when you go about solving this equation x equals four, you might discover, and then there could be multiple solutions. In this case, there's just one. It could be that the number you found x minus four could be outside the domain of the original problem. Thinking about the rational functions in play, you look at the first rational function, three over x minus two. This thing is undefined when x equals two. So two cannot be an answer no matter what we want because that would require division by zero. We can't do that. If you look at the second fraction, it would be undefined when x equals one. So that's the domain of that rational expression. And then the third one is actually combined. It's neither defined at one nor at two. Now the good news is x equals four is neither one nor two. It's inside the domain. So that does turn out to be the solution to our rational equation in the situation. Let's look at another example. Let's take this time via situation three x over x minus one plus two equals three over x minus one. So I can already see right here as I look at the domains of these things, of these rational functions, that whatever happens, x cannot equal one. That that's not an acceptable solution no matter what happens. Now, likely chance I'll get different numbers like we saw in the last example, but just so you're aware the solution cannot be x equals one. Now we want to clear the denominators here just like the other one. When you have a rational function, a rational equation, you always want to clear the denominators. Now in this example, the only denominator present is an x minus one. It is repeated. And so remember a little pizza story right here. The first fraction's like, oh, I want to have x minus one on my pizza. The second fraction, the two, you can think of as two over one, but I'll actually write without a fraction bar there. Two is kind of like, eh, I'm not hungry. I'm not going to eat anything. And so then the last one's like, oh yeah, yeah, I want x minus one on my pizza too. So the fact that they're both asking for x minus one doesn't mean we get double pepperoni. It just means we'll get a pepperoni pizza. So we're going to times the left-hand side and the right-hand side by x minus one. What's good for the goose is good for the gander. Make sure you do the same thing to both sides of the equation so equality's preserved. We do have to distribute on the left-hand side because we have the plus. And what's going to happen is when you take the x minus one times the first fraction and it'll cancel the x minus one in the denominator, it'll give us just a three x. On the second one, the two, you'll just get two times x minus one. We'll have to distribute that out in a moment. And then for the fraction on the right-hand side, the x minus ones will cancel leaving just a three, like so. So again, we have a linear equation. Start combining like terms. I'm going to first distribute the two there. So we get three x plus two x minus two equals three. We can combine together the x is the three x plus two x is a five x, negative two plus three. So I want to move the two to the other side. So I'm going to add two to both sides. Do the inverse operation there. We're going to end up with five x is equal to five. And so then the final thing to do would be divide both sides by five and we end up with x equals one. All right, this feels like my spider-sense is tingling right now. x equals one, can that be the solution? Wait a second. If we go back and check in the original equation, which it's all messy right now, let me get rid of some of this stuff, right? Remember the original equation, the equation that has not been defiled by the filthy hands of man, right? x equals one was the only number forbidden to be a solution. One is outside the domain of this problem. So as much as we feel happy we solve this equation, turns out x equals one cannot solve this equation. But this is what we see here. The only number that could solve the equation was one, but one is also the only number not allowed to solve it. So we get this contradiction. We have to throw out x equals one as a solution. And as there's no other solutions left, we actually would conclude on this equation, there is no solution. We have this inconsistent equation, no solution, the solution side is empty. So that's a possibility. You might not get a solution to your problem. You do have to make sure you check the domain. So let's look at a third example of solving a rational equation. So we wanna solve the equation nine plus three over x minus two over x squared equals zero. And I've already commented here that x cannot equal zero because that would make the denominators in play go to zero. Well, so since we have this rational function, the numerators are denominators just times just x and x squared. Let's look for the LCD here. The LCD, well, if you ask all of your roommates, the first roommate's like, eh, I don't want a pizza. The second roommate's like, oh, I want a pepperoni pizza. And then the third roommate is like, I wanna double pepperoni pizza. Notice how you have an x squared there. This is a repeated factor. And so it's like, I really like pepperoni. Let's get double pepperoni. So the first one's like, yeah, sure, let's get double pepperoni and I'm good with that. So the least common denominator is gonna be x squared. It's not gonna be x squared times x, it would just be x squared. Because you'll notice that x squared does divide x squared because it's one times x squared. And x also divides x squared because it's x times x. So the least common denominator is gonna be x squared and we're gonna multiply both sides of the equation by x squared. Now, if you did sort of erroneously think the least common denominator was an x cubed, you couldn't multiply both sides of the equation by x cubed, that would be acceptable. And you would clear the denominators doing that. But the problem is when you multiply by x cubed, you're gonna end up with a cubic polynomial, which is one degree higher than we want. We actually can get away with a quadratic polynomial. And so the thing is when you clear the denominators for a rational equation, it turns into a polynomial equation and we don't want the degree being any bigger than it has to be. The bigger the degree, the harder it is to solve. And therefore, we're best using the least common denominator. So we get an x squared here because x squared was a repeated. It's a repeated root in the denominator there. So multiply both sides of the equation by x squared. So we do that on the left or do it on the right. If you just trip the x squared, you're gonna get nine x squared. You're gonna get a, I'll actually show the details of this one, three x squared over x. And then you're gonna negative two x squared over x squared. On the right-hand side, anything times zero which is zero, so that's easy enough. Returning to the left-hand side, you have x squares that cancel there, so you just get the constant negative two. And then for the middle term, x will go into x squared just once, leaving you just an x. So as we clear the fractions, we're gonna end with a nine x squared plus three x minus two equals zero. So this is what I was saying earlier. We ended up with a rational, not a rational, we clear the denominators of a rational equation. We end up with this quadratic polynomial here in which case, we need to try to factor it. So we could think of, let's see, factors of nine times negative two, that would be 18, negative 18. We need factors of negative 18 that we could add up to be three, positive three. I could take six and negative three. Six times negative three is still negative 18, but their sum would be three. So with this, I'm gonna break it out into factoring by groups. So we're gonna get nine x squared plus six x. That'll be my first group. And then we're gonna get negative three x minus two equals zero. From the first group, we can take out a three x. That leaves behind three x plus two. From the second group, I can just take out a negative one. That leaves behind three x plus two, which is great. You'll notice the term, the binomial in the parentheses is identical. So I like to think of like the twins from Harry Potter were identical. You factor them out. That leaves behind a three x minus one and a three x plus two equals zero. And so setting each of those factors equal to zero and solving them, we see that x should be one third or negative two thirds. So we do get these fractional solutions here. Remember, the only forbidden value was x equals zero. X could not equal zero as a solution. We found numbers other than that. So we can keep both of our solutions of one third and negative two thirds. And so these few examples here, hopefully can show us how one can solve a rational equation. The strategy I think is fairly simple whenever you have rational equations, find the LCD and multiply both sides by this least common denominator to clear out the denominators. Then it turns into a polynomial equation and we'll solve it like we did in the previous unit.