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### Rational Inequalities

- To solve a rational inequality we
- Set the inequality with zero on one side
- Solve the related equation that is set equal to zero
- Find where the denominator would equal zero.
- Using the values from step 2 and 3 (critical values) we divide the x-axis into intervals
- Use a test point from inside each of the intervals to see if it satisfies the inequality
- Check the end points of each interval to see if it satisfies the inequality (Note we never include any values that make the denominator equal to zero)
- When solving the related equation, its best to combine all rational expressions into one rational expression. This will often involve adding or subtracting rational expressions.
- To better find the critical values, factoring the numerator and denominator of the rational expression will be useful.

### Rational Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:06
- Rational Inequalities 0:18
- Testing Intervals for Rational Inequalities
- Steps to Solving Rational Inequalities
- Tips to Solving Rational Inequalities
- Example 1 3:33
- Example 2 12:21

### Algebra 1 Online Course

### Transcription: Rational Inequalities

*Welcome back to www.educator.com.*0000

*In this lesson we are going to go ahead and take care of rational inequalities.*0002

*There are a few different techniques that you could use for solving rational inequalities,*0007

*but Iâ€™m just going to focus on looking at a table and keeping track of the sign for these.*0011

*Recall that when we are trying to solve an inequality that involves a polynomial, we want it in relation to 0.*0020

*The reason why we are doing that is we just have to focus on whether it is positive or negative, a lot less to handle.*0029

*This will make looking at the intervals that represent our solution a little bit easier to find.*0039

*You will see that the process for solving these rational inequalities looks a lot like*0047

*the process for solving our polynomials that we covered earlier.*0051

*We will still look at factoring it down, figuring out where each of those factors are equal to 0*0055

*and looking at tables so we can test the values around each of those 0.*0060

*The actual process for solving a rational inequality looks like this.*0067

*The very first thing that we are going to is we are going to set the inequality in relation to 0 on one side.*0072

*It means get everything shifted over, so you have that 0 sitting over there.*0078

*Then we will go ahead and solve the related rational equation.*0084

*Solve the same thing and put an equal sign in there and see when it is equal to 0.*0088

*We will also figure out where the denominator can be equal to 0.*0093

*Those are some of our restricted values that we cannot use.*0097

*The reason why it is important to solve it and find out where the bottom is 0 is at those points it could change sign.*0100

*We will call those particular points our critical values.*0108

*It is around those points that it could change in signs, we are interested in what happens.*0113

*To determine what it does around those we will use a few test points*0119

*that will help us to determine which individual interval satisfy the over all inequality.*0124

*Until we get to finding our intervals, we are not quite done yet.*0132

*We also have to pay close attention to the end points to see which ones should be included and which ones should not be included.*0135

*I will give you a few tips on that, watch for that.*0143

*Here are my tips.*0149

*Remember that when you are working with these rational inequalities and get everything over onto one side,*0150

*we might combine it into one large rational expression.*0155

*If we got 3 or 4 of them, put them all down into one.*0159

*Make sure you factor both the top and the bottom.*0164

*You need to see where each of the factors is equal to 0.*0166

*A lot of factoring.*0169

*You will never include any values that make the denominator 0.*0172

*We cannot divide by 0 even if it has an or equal in there, never include anything that makes the bottom 0.*0176

*What that said, if we are dealing with a strict inequality like greater than or less than*0184

*then we will not include any points on the end points.*0189

*If we have greater than or equal to, less than or equal to, we will include where the numerator or the top will equal 0.*0196

*Those are the only end points we have to worry about including.*0209

*That is a lot of information, let us jump in and look at some of these examples.*0215

*I want to solve when -1/x + 1 is greater than or equal to -2/x - 1.*0219

*Rather than worrying about clearing up fractions or anything like that,*0226

*let us get everything over onto one side and get it in relation to 0.*0229

*I'm going to add 2/x - 1 to both sides.*0236

*My goal here is to combine these fractions into one large fraction.*0251

*Do not attempt to clear out these fractions like you would with a rational equation.*0256

*If you clear out the fractions you will lose information about the denominators*0261

*and we want to check around points where the bottoms could be 0.*0265

*Do not clear those out.*0269

*My LCD that I will need to use will be (x + 1)( x â€“ 1)*0271

*Let us give that to each of our fractions.*0280

*-1 will give this one x -1.*0288

*We will do that on the bottom and on the top then we will give the other one x +1.*0292

*Let us combine this into a single fraction here.*0311

*We would have to do a little bit of distributing so - x + 1 + 2x + 2 / x + 1 x - 1 greater than or equal to 0.*0317

*Just a little bit more combining on the top that is a - x + 2 would be a single x and then 1 + 2 would be 3.*0337

*It is quite a bit of work but we have condensed it down into a single rational expression on the left there and we also have it in relation to 0.*0355

*That is a good thing.*0363

*I want to figure out where would this thing be equal 0?*0365

*It will equal 0 anywhere in the top would equal to 0.*0369

*The top is equal to 0 at x = -3.*0375

*We also want to check where would the denominator be 0?*0383

*The bottom equal 0 at two spots when x =- 1, 1x =1.*0389

*All three of these values are what I call our critical values,*0399

*and it is around these values that we need to check the sign of our rational expression here.*0403

*That way we can see whether it is positive or negative.*0408

*Just like before when we are dealing with those polynomial inequalities, this is where our table come into play.*0412

*First start out by drawing a number line and putting these values on a number line.*0419

*I want to start with the smallest ones so -3 and then we will work our way up from there - 1 and 1.*0425

*Along one side of this table we will go ahead and we will write down the factors that our rational inequality here.*0435

*I have x + 3 x +1 and x â€“ 1.*0442

*Now comes the part where we simply grab a test values and see what it is doing around these.*0450

*Our first test value we need to pick something that is less than -3.*0457

*Let us choose -4, that is on the correct side and we will put it into all of our factors to see what sign they have.*0461

*-4 + 3 = -4 + 1 still - and -4 -1=-5, negative.*0469

*We will select a different test value between -1 and -3.*0484

*-2 will work out just great and we will put it into all of our factors.*0490

*Let us see what that does.*0494

*-2 + 3 that would give us something positive, -2 + 1 that will be negative and -2 â€“ 1, negative.*0497

*We are doing pretty good and now we need a test value between -1 and 1.*0512

*I think 0 is a good candidate for this.*0518

*That will be a nice one to test out.*0519

*0 + 3 = 3, 0 + 1 =1, 0-1=-1.*0522

*One last test value we need something larger than 1.*0532

*I will put 2 into all of these.*0536

*2 + 3= 5, 2 + 1 =3, 2 -1 =1*0539

*My chart here is keeping track of the individual sign of all the factors.*0547

*We will look at this column by column, so we can see how they all package up for our original rational expression here.*0553

*In this first column I have a negative and it is being divided by a couple of other negative parts.*0561

*What Iâ€™m thinking of visually in my mind here is that this will look a lot like the following.*0572

*I have a negative value on top and a couple of negative values in the bottom.*0578

*When those negative values in the bottom combine they will give us another positive value.*0583

*We have a negative divided by a positive.*0591

*That means the overall result of that first interval is going to be negative.*0595

*In the first interval it is a negative.*0603

*For the next one I will have a positive Ã· negative Ã— negative and that will end up positive.*0607

*Then I will have a positive Ã· positive Ã— negative = negative.*0616

*My last I will have positive Ã· couple of other positive values, everything in there is positive.*0623

*I know what my rational expression is doing on each of the intervals.*0630

*I know when it is positive, I know when it is negative and things are looking pretty good.*0634

*Looking at our rational inequality over here I can see that I'm interested in the values that are greater than or equal to 0.*0639

*That means I want to figure out what are the positive intervals.*0647

*In this chart that we have been keeping track of all that so I can actually see where it is positive.*0652

*I have between -3 and - 1.*0659

*I have from 1, all the way up to infinity.*0662

*Both of those would be some positive values.*0665

*Let us write down those intervals.*0669

*I'm looking at between -3 and -1 and from 1 all the way up to infinity.*0670

*There is one last thing that we need to be careful of what endpoints should I include, which endpoints should I not?*0679

*Notice how we are dealing with or equals to.*0688

*I want to include places where the top of my rational expression could have been 0.*0695

*It is a good thing I highlighted it early on, it fact they are right here.*0700

*I know that the top will be equals 0 at -3.*0703

*I'm going to include that in my intervals.*0708

*We never include spots where the bottom is 0.*0712

*We have marked those out and I will use parentheses to show that those should not be included.*0716

*Let us go ahead and put our little union symbol so we can connect those two intervals.*0723

*This interval from -3 all the way up to -1 and -1 to infinity*0728

*that would be our interval that represents the solution for the rational inequality.*0734

*Let us try one that looks fairly small, but actually has a lot involved in it.*0743

*This one is x -5 / x -10 is less than or equal to 3.*0748

*Like before let us get everything over on to one side first.*0754

*That way it is in relation to 0.*0757

*x -5 / x -10 -3 is less than or equal to 0.*0760

*We need to worry about getting that common denominator.*0773

*I see I have x -10 in the denominator, let us give that x -10 to the top and bottom of our 3.*0776

*Looking much better from now, we will go ahead and combine these rational expressions right here.*0799

*I think we need to do a little bit of work, let us go ahead and distribute this -3, as long as we have it there.*0807

*x -5 - 3x + 30 / x â€“ 10 you want to know where is that less than or equal to 0.*0813

*Just a few things that we can combine we will go ahead and do so.*0827

*-2x from combining our x and let us see a 25 will be from combining 5 and 30.*0832

*We want to know where is that less than or equal to 0.*0850

*We are in good shape so far.*0855

*Now that we have crunched it down into one rational expression, we are interested in where is it equal to 0.*0857

*This will be equal to 0 wherever the top is equal to 0.*0865

*We are going to do a little bit of work with this one, but we can figure out where the top is equal to 0.*0869

*We will have to move the 25 to the other side and divide both sides by -2.*0874

*I have 25/2.*0881

*Let us make a little note is where the top is equal to 0.*0885

*Okay, that looks pretty good.*0890

*Let us figure out where the bottom equal 0, that is not so bad that will be x = 10.*0892

*It is around those two values that we will go ahead and check to see what the sign is, around 10 and 25/2.*0903

*Let us go to our table here.*0915

*The smaller of these values would actually be our 10 and 25/2.*0922

*We will put that over there.*0929

*Let us check the sign and see how these turnouts.*0933

*- 2x + 25 and the other one would be x â€“ 10.*0936

*Let us grab some sort of test value that is less than 10.*0947

*One thing I can see here is I will plug in a 0 that is less than 10.*0952

*If I plug it in the first one I will get 25.*0958

*If I plug in 0 for the other one I will get -10.*0961

*That is my sign for that one.*0965

*On to the middle interval I need something between 10 and 25/2*0969

*I think 11 will work out just fine.*0974

*If I put 11 into that top one I will get -22 + 25 that will be positive.*0979

*If I put 11 into the bottom one, 11 - 10 would be 1.*0987

*Okay, interesting.*0994

*I need to pick something larger than 25 / 2.*0997

*25/2 is about 12 Â½ let us go ahead and test out something like 20.*1002

*If I put 20 into the top one I will have -2 Ã— 20 that will be -40 + 25 that will give me a negative value.*1010

*If I put 20 into the bottom one it will be 20 â€“ 10 and that will be a 10.*1019

*These two things are being divided, to look at our over all sign we will take each of these signs and go ahead and divide them.*1026

*Positive Ã· negative = negative, positive Ã· positive = positive and negative Ã· positive = negative.*1035

*What am I interested in for this particular one I want to know where it will be less than or equal to 0.*1046

*Let us look for those negative values.*1053

*I can see one interval here everything less than 10 and I have another interval here where it is greater than 25/2.*1057

*Let us write down those intervals.*1066

*From negative infinity up to 10, from 25/2 all the way up to the other infinity.*1067

*One last thing, we need to figure out what end points we should include and which ones we should not include.*1076

*This one does have or equals 2 so I will include the places where the top is equal to 0.*1082

*We already marked those out, the top is equal to 0 at 25/2 so we will include it.*1089

*We never include spots where the bottom is equal to 0, so I will not include the 10.*1097

*We will finish off by putting a little union symbol in between.*1104

*The intervals that represent the solution are from negative infinity up to 10 and from 25/2 all the way up to infinity.*1108

*And then we can consider this inequality solved.*1118

*It is a lengthy process, but we will use that table to help you organize your information*1121

*and watch at what point you can and cannot include in the solution intervals.*1126

*Thank you for watching www.educator.com.*1132

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