Common Core Algebra II.Unit 10.Lesson 13.Solving Rational Inequalities

All right, so ultimately, I want to find all the values of X that make that a true statement. Let array asks us, and what value of X is the ratio equal to zero is that part of the solution set. What I'd like you to do is pause the video now and try to answer both of those questions in letter a. All right. Well, let's go through it. So in order for any fractional expression or rational expression to be equal to zero, the numerator must be equal to zero. Of course I can solve that in my head as you can as well, and we get X equals 5. Now the question is, is it part of the solution set? Well, if you can tell that without doing any more work, that's great. But if not, there's always one check, which is that we put it into the inequality. And we see if it makes it true. All right? Now what that gives us is zero divided by 8 is greater than or equal to zero. Of course, zero divided by a to zero, so we get zero is greater than or equal to zero. And that is, of course, a true statement. Because zero is equal to zero, right? It's all about that equality sign. And therefore, the answer is yes. It is part of the inequality. Okay? All about that little equals being included. Now letter B asks a very similar question at what X value is that ratio undefined. Is this value part of the solution set? Again, pause the video and try to answer both of those two questions. 

All right. Well, in order for any fractional expression to be undefined rational expression to be undefined, the denominator, what you're dividing by has to be equal to zero. In which case, we get the value X equals negative three. Again, if you know off the top of your head, whether that is or isn't part of the solution set, awesome. If not, you would simply substitute it into the inequality and test it. Is that greater than or equal to zero? Well, if we do the arithmetic, we get negative 8 divided by zero is greater than or equal to zero. Now here's where you have to be a little concerned, because negative 8 divided by zero doesn't have a value. It's undefined. So can a number that's undefined be greater than or equal to zero. And the answer is no, of course not. I mean, it's not even a number, right? So that's a false inequality and the answer must be no. All right? So obviously there's something important about both 5 and negative three. And we're going to investigate that now by using our graphing calculators. 

Let's bring out the TI 84 plus. All right. Now, what I'm going to do is I'm going to use it not to graph this expression, although it's going to start to feel that way at the beginning. But to just simply see for any given X value, is that ratio, X -5 divided by X plus three is it positive? Is it negative? Is it zero? Is it undefined? That's what we're going to be looking at. So let's go into Y equals. If you have any equations and Y one or Y two, et cetera, clear them out. All right, now in Y one, what I'm going to do is I'm going to put that ratio in. Now, I don't have any fancy fraction like mode turned on. So I'm just going to put parentheses. X -5, parentheses. Divided by. Parentheses, X plus three. Parentheses. That's very, very important that you have those parentheses in there. Unless you have sort of fraction mode turned on where you can clearly see what's in the numerator and clearly see what's in the denominator. So make sure that they're in there. Take a moment. All right, let's pop into the table now. Okay? Actually, let's pop into table setup table setup. So go into table setup. And what I want to do is investigate for a variety of different values of X, whether or not that ratio is positive or negative. So I think I'll start my table at negative ten, then I'll make it go by ones. 

That's fine. So start at negative ten, going by ones. And now let's actually go into the table. Now, what you see, sorry, I have to cheat a little bit and look down at my calculator over here. Is it negative ten, right? When I have this table, make sure we can identify X and Y one. What I have an X value of negative ten, and I have that ugly Y value of 2.1429. I don't care so much about the actual value. What I care about is the fact that it's positive. All right? And that means negative ten must be part of the solution set because it makes it true. So actually, let me go down on the table a little bit, right? When I go to negative 9, the output, the Y column is still positive. When I go to negative 8, the output, the Y column is still positive. In fact, as I go down in that table, it takes a little while on this calculator. As I go down in that table, what I see is I continue to see I continue to see positive outputs until I hit X equals negative three. And at X equals negative three, and this is very predictable from letter B, I see an error, because it makes that fraction undefined. 

But what's really cool is when I go to X equals negative two, what do I have there? I've got an output of negative 7. When I go to X equals negative two, and I have an output of negative 7, that's negative. So that means by inequality is false there. Okay, it's false. So negative two is not going to be part of the solution set. And when I go to negative one, I see that the output is negative. And when I go to X equals zero, I see the output is negative. And X equals one, the output is negative. X equals two, the output is negative. In fact, I have to keep going down in that table until I hit X equals 5. And then the output is zero, which I knew from letter a the cool thing is now if I look at X equals 6, now the output has turned positive again. And at X equals 7, the output is positive. And at X equals 8, the output is positive. In fact, I can keep going down in that table as far as I want and the output will remain positive. So I'm going to draw a little number line to really summarize everything that I just saw in my graphing calculator's table. I'm going to mark that negative three, and I'm going to mark that 5. I'm not going to put anything else on this number line. 

Please don't. I'm going to put an open circle at three because the expression is undefined there. And I'm going to put a closed circle at 5, because I know it's part of the solution set. What I saw was that all the outputs were positive for numbers that were less than negative three. The output was all negative when the numbers were between negative three and 5. And the outputs were all positive when the X values were greater than 5. Now what I wanted was I wanted all values of X that gave me outputs that were greater than or equal to zero. In other words, positive. So those would be those outputs. And therefore, my solution set is all values of X, such that X is less than negative three, or X is greater than 5. And I just made a little mistake there. It should be X is greater than or equal to 5, but now you can probably barely see that. So let me just erase this so it's very clear. Greater than or equal to 5. All right. So clearly there was something very important about the zeros of this expression. What made it equal to zero? And what made it undefined? Now let's talk a little bit before we go on to the next exercise why that's happening. Now, don't forget that fractions ultimately are division. And whether or not division turns out to be positive or negative depends on whether the numerator is negative or the denominator is negative or the numerator is positive, denominator is positive. We all know that a negative divided by a negative is a positive. 

We all know that a negative divided by a positive is a negative, et cetera, right? So let's say you picked a number like X equals negative four. And you substituted it into this expression. You'd get negative four -5. Divided by negative four plus three, and that would get me negative 9 divided by negative one. The point really isn't negative 9 divided by negative one. It's negative divided by a negative and that gives you a positive. But then if you pick a number between negative three and 5, I gotta find some space here. Let's say we pick X equals zero. That's a good choice. Then we would have zero -5. Divided by zero plus three. Again, it's not really important. What we get, what's important is that the numerator is negative, the denominator is positive, and that results in an overall negative. Finally, if we pick a number in this area greater than 5, let's say X equals 6, that's convenient. And we substituted it in. We'd have 6 -5, divided by 6 plus three. Again, not really important what the final answer is, but the numerator is positive. The denominator is positive, and we all know that a positive divided by a positive so positive. All right? And therefore we get that solution set. Let me be very clear. The final answer to this problem. Is this and to a certain extent this number line. 

This is the graphical answer. And this is the algebraic answer. All right. Pause the video now and really take all of that in before we introduce a new piece of terminology. All right, I'm going to clear out the text. And let's learn about what's known as a critical value. Critical values are values of the input, so they're X values. That are either zeros of the expression. Or make the expression undefined. Zeros of the expression or make the expression undefined. So that last expression, X -5 divided by X plus three. X equals 5 is a zero. All right, when we already saw why that was, because when we substituted it in, we got zero divided by 8, which was zero. All right? On the other hand, X equals negative three made the expression undefined. Because when we substituted it in, we got negative 8 divided by zero. Which is a problem. You can't divide by zero. You'd be crazy to try to divide by zero. So the point here, though, is that 5 and negative three are both critical values of this rational expression. And critical values are going to pun intended are going to play a very critical role in solving rational inequalities. So let's take a look at how these things work. All right, exercise number two. Solve the rational inequality X squared plus three X minus four, divided by X squared -6 X plus 9 is less than zero. 

For all values of X show your solution on a number line and state its solution in either interval or set builder notation. All right, great. We want to find all the values of X that make this ratio negative. All right. And the first thing I'm going to do apparently in red is I'm going to find the zeros. All right, I'm going to take that numerator. And I'm going to set it equal to zero. All right. I'm going to use the zero product law that'll work out well. And I'm going to get X equals negative four. And X equals one. So these are critical values. They're going to show up on my number line. Then I'm going to find the ones that make it undefined. Because there are also critical values. That's a little bit more difficult, right? Then in the last problem, it's not more difficult than finding the zeros. All right, and this one factor is kind of interestingly, right? Because it factors as X minus three and X minus three. So we set X minus three equal to zero, and we just get X equals three. So there are my critical values. Negative four positive one and positive three. Now what I'm going to do is I'm going to draw out a number line. And I'm going to label all my critical values. I'm going to put negative four, positive one, and pause the three. For me, the next thing that I do is determine if any of those critical values lie in the solution set. Now, one thing is for certain wherever there is an expression that's undefined, it can not lie in the solution set. So I'm going to put a little open circle there. 

The question is, do the zeros fall in the solution set. And in this case, they don't because there is no equality when we're comparing this to zero. So I'm also going to put open circles here. And here. Now, what I'm going to now do is pick X values in each one of these intervals. And test them in my inequality to find out if it's positive or negative. And I'm going to actually test it in the factored form of the inequality. Here's the factored form of the expression. X plus four times X minus one. All divided by X minus three times X minus three. So let's take a look at how this works. Let's say in this area, I pick X equals negative 5, okay? If I put it into the first parentheses, negative 5 plus four, I get negative one. Now I'm not going to write down the negative one. I'm just going to write down a little negative. On the other hand, if I put it in this parentheses, I get negative 5 minus one, which is negative 6. I also get a negative. If I put it in this parentheses, negative 5 minus three is negative. And obviously also a negative. Now remember, this is multiplication up here and division overall. A negative times a negative is a positive. And also a positive. And then a positive divided by a positive so positive. So any number I pick in here is going to give me a positive. All right. Let's try X equals zero. That lies between negative four and one. Let me put it into the first parentheses. And I get a positive. Then in the second parentheses, and I get a negative. Then in the denominator, I get a negative. And I also get a negative. 

Again, remember that's multiplication and division, a positive times a negative is a negative. A negative times a negative is a positive. And a negative divided by a positive is a negative. So any number I pick in here is going to result in a negative. Let me go with the next value in here. Let's go with X equals two. I put it in the first parentheses. I get a positive. Second parentheses I get a positive. Then I get a negative and a negative. Again, multiplication and division gives me a positive in the numerator, and a positive in the denominator, which overall gives me a positive. Now be careful. So far, it looks like the signs just alternate. But watch what happens when I pick X equals four. When I put it into the first parentheses, I get a positive. The second one I get a positive. And the denominator I get a pause at a positive and a positive times a positive is a positive. And then the denominator I also get a positive, which gives me a plus. So they're the sign did alternate. Now that has to do with the fact that that X equals three was a double root. A double root. Signs don't tend to change at double roots. They would at a single root or triple root, but not at a double root. 

Now, after all was said and done, what in the world in my coloring in? I want to color in all values of X that resulted in this being less than zero in other words that resulted in it being negative. So I'm going to do that in red, here's my solution set. Everything from negative four to positive one. So there's my graph of it in interval notation. It would look like this negative four to one. And in set builder notation, it would look like this. Whichever you like better. So let me put an OR there because you don't need them both. All right, that's actually how we're going to do almost every one of these problems. There's going to be a little wrinkle in a moment that's going to be a bit irritating, but you'll roll with it. It'll be fine. All right. Pause the video now and write down anything you need to. Okay, I'm going to clear out the text. All right, exercise number three. Solve the rational inequality X divided by X -8 is less than or equal to 5. Represent your answer using a number line and using set builder notation. All right? So what we're going to do in this case is because we're comparing this to 5. What I really want to do is I want to compare it to zero. 

And that means I've got a little bit of work to do before I even start with anything else. So what I'm always going to do with rational inequalities is I'm always going to attempt to compare them to zero. So I'm going to move that 5 to the other side, and then I want this all as a fraction. So what that means is I've got to do a little bit of common denominator work. The common denominator is X -8. So I'm going to just do some common denominator work on the back side of my sheet. All right. And that's going to give me X -5 X plus 40. All divided by X -8, less than or equal to zero. And that's negative four X plus 40. Divided by X -8 is less than or equal to zero. All right. So no problem. All I did was took this inequality and wrote it equivalently as this inequality. All right, let me check my work. Good it all looks good. One thing that's particularly tricky is the fact that you've got this double negative here, that worked out okay. So now let's take it right from here. All right. The first thing I'm going to do is find my zero. Okay? And my zero, I'm going to find by setting that numerator equal to zero. And that's pretty easy. We get negative four X is equal to negative 40. And X is equal to ten. 

We also want to figure out where the thing is undefined. And it's going to be undefined whenever X -8 is equal to zero. Or X is equal to 8. So those two are pretty close together, but that's okay. It's not a problem on my number line. Far away, close together. I don't really care. I've got 8. I've got ten. Remember wherever it's undefined it can not be part of the solution set. It just can't. So I'm going to have an open circle there. How about that ten? Well, the ten will be part of the solution set because the equality actually is contained in the solution. All right? So now I want to start testing numbers. Now, again, I can certainly test this in the factored form of this inequality. The factored form would kind of look like this. If I factored a negative four out, it would look like this. There's nothing wrong with using this. In fact, watch, let's do it. Here's my factored form. Let me pick X equals 7. If I did that, I can't forget the negative four. That's that negative right there. But then if I put the 7 in parentheses, I'd also get a negative. And then if I put it in the denominator, I'd also get a negative. So that would be a positive divided by a negative, and that would be a negative. All right. Then let's say I picked X equals 9. Remember, I don't want to forget that negative four, but then if I put it into that parenthesis, I'd also get a negative. And then if I put it in the denominator, I'd get a positive. A negative times a negative is a positive. And a positive divided by a positive is a positive. So there we see the sides alternating. 

All right. Finally, let's say I took X equals 11. If I substituted it into the numerator, I'd get a negative from that negative four. Then I'd get a positive when I subtract ten. And then I'd get a positive in the denominator, which would give me a negative divided by a positive, which would give me a negative. All right. So what am I looking for? I'm looking for all values of X that make this inequality less than or equal to zero. That means it would be out here. And it would be out here. All right, so I'm going to use set builder notation. I'm going to say all values of X, such that X is less than 8. Or X is greater than or equal to ten. All right. And that's it. Now the irritating part about this one is having to convert it so that we can pair it to zero. But if we go through that process, right? Then it's all just about finding the critical values and then finding numbers and substituting them in to test the inequality. So pause the video now and look at all the work that we did. Okay, I'm going to clear out the text. And let's do one last problem. All right. Now, you just have to sort of keep your wits about you when you do problems like this. And you have to just work it. So that you get it compared to zero. All right? Now, in order to do that, I want to compare this inequality to zero. 

The first thing that I would do is I would move my three halves to the other side because then I have an inequality compared to zero. Now I want it to be all one fraction, so I have to get a common denominator. And that common denominator has got to have a factor of two. It's got to have a factor of X plus three, and it's got to have a factor of X so the common denominator I'm going to be looking to get is this. That's going to be my at least common denominator. So let's take that first fraction. The first fraction in order to have that denominator, I've got a multiply, the numerator, by two times X plus three. And the denominator obviously by the same, but once I do that, EI multiplied by X plus three and two, this X was already there. Then it's got that common denominator. The second fraction, I'm going to have to multiply top and bottom by two X because it's already got the X plus three. Oops. My mistake that should be an X plus two. And the last one. I'm going to have to multiply the numerator and denominator by X and X plus three. Because it already has the two. And now we have a common denominator. Oh, there's going to be some work to be done here, though. Wow. Oh my goodness. Let's see what we have. Okay, well, we all have a common denominator now, two X times X plus three. So I can put down a single denominator, but now I have to multiply stuff out and combine like terms and oh my goodness. Okay, but here we go. 

This is going to be the most challenging of all of them because I have three way multiplication. X minus one times two times X plus three. All right? So I think I'll leave that two out until last and I'll multiply the X minus one in the X plus three. That's going to be X squared plus two X minus three. Minus one multiplied this out. That's going to give me two X squared plus four X and watch out. We have a subtraction here. So that's going to give me negative three X squared. -9 X is less than zero. Keep going. All right, just have to keep work in this problem down. Here, we'll have two X squared. Plus four X -6. Plus two X squared plus four X minus three X squared -9 X is less than zero. Wow. We're not going to have any room to just do the inequality here. So let's combine like terms in the numerator. We have two X squared four X squared minus X squared sorry minus three X squared gives us one X squared. All right, so that takes care of this. This and this, let's take a look at our X's. Four X and four X is 8 X -9 X gives me negative X that takes care of this. This and this. And then we have that negative 6. All divided by two X times X plus three. Now again, you might be wondering why am I not multiplying out the denominator? Well, eventually I want to know the zeros of the denominator. So first, let's find the zeros of the numerator, which are the zeros of the overall expression X squared minus X -6 equals zero. Then I'll just factor this and I'll get X minus three, and X plus two. And I'll get X equals three, and X equals negative two.

 All right, we also want to figure out where it's undefined. And that's going to be wherever two X times X plus three is equal to zero. So that's going to give me two X equals zero and X plus three is equal to zero. X equals zero and X equals negative three. Wow. So look at all the critical values we have. One, two, three, four. And a lot of algebra to get those critical values. Wow, wow, wow. Anyway, let's put them on our number line. We've got negative three there. We've got negative two there. We've got zero there. And we've got three there. Let's do open circle closed circle kind of thing. All right. They're actually all open circles. And they're all open circles because at X equals zero and negative three, the expression is undefined. And at X equals three and negative two, it's equal to zero, but the equality is not in there. Now, I'm kind of running out of space and it's gotten to be sort of a long video, but we would continue to check all those values of X and we'd find that we had a positive there. A negative there, a positive there, a negative there. In fact, all of these do end up alternating in a positive there. So if we take values of X in between all of these and substitute them into this inequality, that's what I would work with. This inequality, because that's the one that's compared to zero. That's the sign analysis that you'd get. And what we're looking for is we're looking for values of X that make that expression less than zero. So let's actually color that in in red. It's going to be less than zero here. And it's going to be less than zero here. 

So at the end of the day, I'm going to write this in interval notation. Our solution set goes from negative three to negative two union from zero to three. So that is our final solution set. So what we can do in just a moment after we've kind of pause the video and you've looked at everything on this page, because there's a lot going on on this page, is what I can then clear out all the text and we can kind of wrap this up. So pause the video now and think about what we did on this page. And then we'll clear it out and finish it up. Okay. Let me clear it out. All right, so rational inequalities undoubtedly the hardest inequalities that you're going to work with in this course. It's all about really two things. Comparing the inequality to zero, so you might have to rearrange and do a little bit of algebra in order to do that. And also finding the critical values. Those values of X that make the numerator equal to zero, and the denominator equal to zero. Once you plot those on a number line, then it's all about testing values of X that are in between those critical values to see if they make the inequality true or false. All right. Well, thank you for joining me for another common core algebra to us and by E math instruction. My name is Kirk Weiler, and until next time, keep thinking and keep solving problems.