Day 9 HW to Solving Rational Equations and Inequalities

All right. In this video, we are going to practice adding rational expressions such as this one. will also show you how to solve system of rational equations such as this one. And will show you how to solve rational inequality like this one. the first thing you should do is see if there's common factor that you can pull out. And in this case, there's common factor of three. So if were to rewrite this, would have + 1 over so the GCF is three. So that's going to leave 3x - 7. All right. And I'm going to space this out little bit. So I've got 2/3. Now you see left all this space over here. That's because know when I'm adding fractions, need like denominators. because pulled out the GCF, see there's already three. So what need now is 3x - 7. And I'm going to have to put that in the top and bottom. So I'm not really changing anything. Now, I'm not going to cancel these out, and certainly will not cancel out what just put in. Instead, because have like denominators, I'm going to wind up putting these together. So, it's going to be like this. I'm going to keep my like denominator of 3 * 3x - 7. And now here, I've got + 1 sitting there. Now over here, I've got little distributive property happening. So that's going to be plus Whoops. That's going to be plus 6x -4. Okay. So combining like terms, I've got and 6x. So that makes 7x. Then one and -14 makes 13. All right. And that's going to be over 3x - 7. All right. You should always look and see if there's another GCF that you can do anything with. don't see anything else. so this can be our final answer. Okay, now let's go ahead and solve this system of rational functions. So, the first thing you want to do when you're solving system is to set them equal to each other. but don't forget when you see equals this and equals that, we're going to not only find but we're going to find So when you see don't forget that you need to find as well. So if set these equal to each other, then I've got - 255 over and you know what? might as well go ahead and factor this. So x^2 So I've got * 10 is 2 * 5 -2 and positive 5 will give me the three and negative times positive is negative. So there we factored it. And so continuing on, I've got 5 overx - 2. Whoops, moved it. All right, might as well put the xus 2 in parentheses. And I'm going to leave some space because know what's coming. And I'm going to set this equal to over + 5. Okay, now we need like denominators. There there are many ways to solve these problems, but the method I'm going to do now is I'm going to make like denominators and then I'm going to cancel them all out. So, this has little bit of everything. It's got xus 2 and an + 5. So, I'm going to make all the rest of the denominators like this one. So all this fraction needs is the x+ 5. Got to do it in the top and the bottom. Now this fraction needs the xus2. So I'll do - 2 in the top and the bottom. Okay. Once the denominators are all the same, you can cancel them out. So I'm going to cancel out all the denominators. All right. So that leaves me with - 255 plus okay I'm going to do the distributive property here. So that's going to be 5x + 25 equals and right here I'm going to have x^2 - 2x. All right. So see that I've got x^2 and So I'm going to get zero on one side. Now, if combine these like terms, the - 255 and the positive 25, they cancel each other out anyway. So, I'm just going to subtract 5x from both sides. So, that's going to give me 0 is equal to x^2 - 7x. So that's going to give me 0 is equal to and so I'll just can factor this by pulling out common factor of So that' be like and - 7. so setting these equal to zero. All right, it's like I'm splitting these apart. So I've got x= 0 and over here I'm going to get x= 7. So these are my two values. All right. But need to find the values that go with them. in other words, I've got 0 comma something and I've got 7 comma something. For my final answer, need those values. So, what you do is we're going to use our TI30Xs multiv- view. take one of the two original problems and put it in your table feature of your calculator. I'm going to use this one because it's much simpler. All right, so we hit the table button and let's type in overx + 5. Okay. like to be in ask mode when do this. All right. So, now can type in any value and get the y-value that goes with it. So, have 0 and seven. So, I'm going to type in zero and I'm going to type in seven. So, 0 comma 0 and seven 7 12ths. So 0 comma 0 and seven is 7 12ths. Okay. And that is how you do that. Okay. For the last problem on this video, we're going going to solve this inequality. Now, as soon as you see it's an inequality, you'll remember that drawing number line is going to be the way to go. We're going to draw number line to help us out. But what to put on our number line? Well, first thing you should do is make this an equation and solve it. So, you know what? I'll bring back the number line in minute. might need some space. So, let's make this into an equation and solve it. And that solution is going to be one of the things that will go on my number line. So, let's see. I've got 5 over 3x 3 over 4x = 5 over 6. So, when I'm solving an equation, there are many ways to go about this, but like the technique of making all the denominators the same and then canceling them out. So, I'm looking at 3, 4, and 6, and I'm wondering to myself, what number can can these all be turned into by multiplication? Well, 3 * 4 is 12. 2 * 6 is 12. So, see lot of 12 going on here. So, I'm going to make these all into 12x because see have an So, can make this into 12x multiplying it by four. All right, you have to do the same things to the numerator and denominator. But I've got 12x. Now, if multiply this one by three, that makes this 12x. Got to go top and bottom. this would have to multiply by two and to make 12x. So I'm going to multiply by 2 and in the numerator. So now have 12x 12x 12x in the denominator. Once the denominators are the same, you can cancel them out. Okay, so now have 20 - 9 is equal to 10 All right. Wow, that's simple little equation. So 20 - 9 is going to be 11. So that's going to be 11 = 10 And to get by itself, will divide both sides by 10. All right. So, is equal to 11 over 10. So, this is one of the things that is going to go on the number line. Okay. it would help to have this as decimal just to help me place it on the number line. 10 over mean 11 over 10 is very close to 1. let me check it out. 11 / 10, that's 1.1. you know, that was silly. Okay, obviously it's 1.1. You move the decimal over. Silly me. so that's one thing that's going to go on the number line. Okay. Now, what's the other thing? Any excluded values. So, there's only one excluded value here. Looking at the denominators, see the 3x and the 4x. this tells me that cannot equal 0. mean, if you want, you can think of it as setting 3x is equal to zero and solving. All right. 0 divided by 3 is 0. So cannot equal 0. So there it is. Those are the two things. cannot equal 0. does equal 1.1. These are the two things that will put on the number line. All right. Well, now it's time for that number line. So here we go. All right. So, I've got zero and 1.1. So, here is zero. Okay. And here is 1.1. also known as 11/10ths. Okay. So, this creates three different zones that we need to test out. So in this zone we could test the value let's say -1. All right. Now looking back see this is greater than sign. now ultimately we're going to use our calculator to help us test these zones. but to put this in our calculator we need zero on one side. So what would do is would move this over by subtracting from both sides. All right. Basically, this is going to move over and and change signs. So the thing that's going to go in my calculator is this. will have 5 over 3x minus 3 over 4x and then now will have - and have is greater than zero. So this is what I'm going to wind up putting in my calculator. All right, but need to see is it greater than zero? Where is it greater than zero? That's what we want. So I'm going to be testing to see is it greater than zero here. now in here need something between 0 and 1.1. So could just use the number one. All right, because one is less than 1.1. So could see is this greater than zero. could also use decimal like 0.5 or something like that. And over here just need something that's bigger than 1.1. So something like two for example. Okay. now as far as the border points, the end points, I'm going to put some circles there and they will either be open or closed. now excluded values will always be open circles. So this will definitely be an open circle. Now the solution, all right, the 1.1, the 11 over 10 will either be open or closed depending on this symbol. If it's or equal to then it will be closed circle. but since it's just greater than then it will be an open circle. So will have an open circle right here to show that right at these points this is where it would be equal to 1.1 and but we need it to be greater than so we'll leave it an open circle. So now let's test the values in between. So like said I'm going to put this into my calculator in the table feature. So, hit table and I've got 5 over 3x. So, 5 over 3 and then I've got minus 3 over 4x. So, 3 over 4x and then minus 5 over 6. So five over six. Enter. Now like to do this in ask mode. So I'll go over to ask mode and hit enter couple of times. So now it's waiting for me to put in any value want. And it will tell me the value of the function, you know, the value that goes with it. So, like said, I'm going to test out these three values -1, 1, and 2. So, here go. -1, 1, and two. Okay. Now, want to know where is it greater than zero. in fact, let's just write this down. So, we have -74s So the question is is -74s greater than zero? no it is not. Okay. what about at one we had 112th? So the question is is 1112th greater than zero? Why yes it is. That's positive number. So it is greater than zero. At two haveg -38s. So that means is -38 greater than zero. No, that's negative number. It is not greater than zero. So the yes tells me where the solutions are. All right, that means the value of one is solution to the inequality. that means all the points in between here will be solutions to the inequality. So I'll show that by drawing this segment. If these had been yeses out here, would have drawn the arrows going outward. So this is graph of my solution set. And of course would like to write this in interval notation. So it goes from 0 to 11/10th. So will put zero and then comma and 11th. All right. Now, because this is an open circle, I'm going to use parenthesis. Same over here. If this had been closed circle, would have done square bracket. All right? So, it looks like point. It looks like an ordered pair, but it is not. This is an interval saying from 0 to 11/10, that's where we will find the solutions to this inequality. All right. So that's how you solve rational inequality.