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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

Transcription: Rational Inequalities

Welcome back to

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 I’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 like0047

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 00055

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 points0119

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 than0184

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 denominators0261

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 – 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 – 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 – 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 =10539

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 I’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 ÷ negative × negative and that will end up positive.0607

Then I will have a positive ÷ positive × negative = negative.0616

My last I will have positive ÷ 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 infinity0728

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 – 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 – 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 25/20969

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 ½ let us go ahead and test out something like 20.1002

If I put 20 into the top one I will have -2 × 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 – 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 ÷ negative = negative, positive ÷ positive = positive and negative ÷ 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 information1121

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

Thank you for watching