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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

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For more information, please see full course syllabus of Algebra 1
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Lecture Comments (6)

1 answer

Last reply by: Professor Eric Smith
Sun Jul 6, 2014 2:35 PM

Post by patrick guerin on July 6, 2014

I sometimes experience buffering and the lecture refreshes. So sometimes when I experience that problem I try to go to the lecture slide that I was at, but when I tried it wouldn't let me move it would just stay where it was. How do I fix that?

1 answer

Last reply by: Mrigyen Sawant
Thu May 15, 2014 11:11 PM

Post by Mrigyen Sawant on May 15, 2014

Hi Mr. Eric,
At 9:25, you say that x<1, but 3x<3 has been divided by 3 on both sides, so shouldn't the '<' symbol be flipped?

1 answer

Last reply by: Professor Eric Smith
Tue Apr 22, 2014 8:21 PM

Post by Tommy Lunceford on April 8, 2014

Isn't the first example wrong? the absolute value of -3 and 3 is not less than 3. Shouldn't the numbers be 2 and -2? Or did I miss something?

Inequalities with Absolute Values

  • When mixing together inequalities and absolute values, we must follow the techniques of both. This means we want to isolate the absolute value, and split it into two problems.
  • Depending on the direction of the inequality, we connect the two problems with AND or OR.
    • || < a use AND for this case
    • || > a use OR for this case
  • Remember to flip the inequality symbol if you multiply or divide by a negative number.
  • Some inequalities will have no solution. For example we cannot have an absolute value less than a negative number.

Inequalities with Absolute Values

Solve | b − 6 | ≥ 7
  • b − 6 ≥ 7 b − 6 ≤ − 7
  • b − 6 ≥ 7
  • b ≥ 13
  • b − 6 ≤ − 7
  • b ≤ − 1
b ≥ 13 and b ≤ − 1
| 5a − 2 | < 18
  • 5a − 2 < 185a − 2 > − 18
  • 5a − 2 < 18
  • 5a < 20
  • a < 4
  • 5a − 2 > − 18
  • 5a > − 16
  • a > − [16/5]
  • a > - 3[1/5]
a < 4 and a > − 3[1/5]
− 3[1/5] < a < 4
| w − 3 | ≤ 18
  • w − 3 ≤ 18
    w − 3 ≥ − 18
  • w − 3 ≤ 18
  • w ≤ 21
  • w − 3 ≥ − 18
  • w ≥ − 15
w ≤ 21 and w ≥ − 15
− 15 ≤ w ≤ 21
| f − 8 | < 17
  • f − 8 < 17f − 8 > − 17
  • f − 8 < 17
  • f < 25
  • f − 8 > − 17
  • f > − 9
f < 25 and f > − 9
− 9 < f < 25
| m + 7 | > 1
  • m + 7 > 1m + 7 < − 1
  • m + 7 > 1
  • m > − 6
  • m + 7 < − 1
  • m < − 8
m > − 6 and m < − 8
| r + 10 | ≤ 3
  • r + 10 ≤ 3r + 10 ≥ − 3
  • r + 10 ≤ 3
  • r ≤ − 7
  • r + 10 ≥ − 3
  • r ≥ − 13
r ≤ − 7 and r ≥ − 13
− 13 ≤ r ≤ − 7
| y + 16 | < 49
  • y + 16 < 49y + 16 > − 49
  • y + 16 < 49
  • y < 33
  • y + 16 > − 49
  • y > − 65
y < 33 and y > − 65
− 65 < y < 33
| 2d − 6 | > 10
  • 2d − 6 > 102d − 6 < − 10
  • 2d − 6 > 10
  • 2d > 16
  • d > 8
  • 2d − 6 < − 10
  • 2d < − 4
  • d < − 2
d > 8 and d < − 2
| 4x − 3 | > 13
  • 4x − 3 > 134x − 3 < − 13
  • 4x − 3 > 13
  • 4x > 16
  • x > 4
  • 4x − 3 < − 13
  • 4x < 10
  • x < [10/4]
  • x < 2[2/5]
x > 4 and x < 2[2/5]
| 5m − 10 | > 35
  • 5m − 10 > 355m − 10 < − 35
  • 5m − 10 > 35
  • 5m > 45
  • m > 9
  • 5m − 10 < − 35
  • 5m < 25
  • m < 5
m > 9 and m < 5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Inequalities with Absolute Values

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 0:07
  • Inequalities with Absolute Values 0:23
    • Recall…
  • Example 1 3:39
  • Example 2 6:06
  • Example 3 8:14
  • Example 4 10:29
  • Example 5 13:29

Transcription: Inequalities with Absolute Values

Welcome back to

In this lesson we are going to take a look at inequalities with absolute values.0002

In our previous lessons, we did take care of inequality separately and we did take care of absolute value separately.0009

Our goal for this one is to mix the two together and see if we can solve some new types of problems.0015

When you have some number A greater than 0 and you have some algebraic expression x as packaged together in the following way.0026

The absolute value of x < A, or the absolute value of x > A then you can actually split these into two different problems. 0035

Let us look at the first one, suppose that you have the absolute value of x < A, the way that we can put this is that the algebraic expression x > -A.0043

The algebraic expression of x could be less than A.0054

It is interesting to note that not only do we split it into two problems 0059

and we will see that we have picked up our connector and it actually connected the two.0063

On the other side, when we are dealing with the absolute value of x > A.0068

This will still split into two problems but now we have the x < -A or x > A.0073

This is a little bit different and that now we have a different connector between the problems.0084

We will definitely use our tools on how to deal with these connectors and solving both problems separately0091

and figure out how their solutions should be connected in the end.0096

One thing that you may be looking at and be a little bit worried is keeping track of all of your signs.0100

It might start with less than and then these guys one is greater than and one is less than.0107

These guys over here, I have greater than and then I have less than or greater than.0113

Watch for my tips on how to deal with dealing with the absolute value and the absolute value separately so we can keep them both straight.0118

To understand why it splits into all of these different cases and why the signs look so funny, you have to recall two things that we covered earlier.0131

One has to do with those absolute values.0140

When we are dealing with absolute values this will end up being split into two different problems.0145

One that handles the positive possibilities, and one that handles the negative possibilities.0157

That is why we have two different things.0162

Also, you have to remember the rule that when you multiply or divide by negative that you will end up flipping your inequality sign.0165

That is why the direction of the inequality ends up changing in both of our cases.0172

Since one is that negative possibility, the sign ends up getting flipped.0177

To help you keep these both straight this is what I recommend.0183

When you are looking at your connectors, if you have the absolute value being less than a number then connect using and.0187

If you have the absolute value greater than a number, then use or to connect the two.0198

You will see that if you use each of these rules bit by bit, rather than trying to jumble them altogether, 0205

it is not too bad being able to pick apart one of these absolute values and inequality problems. 0211

Let us start out with some very small examples and work our way up.0221

That way we can keep track of everything we need to do.0225

This one simply says that the absolute value of x < 3.0228

Here is what I'm going to do first.0232

I’m going to use my techniques for our absolute values to simply just split this into two problems.0234

Remember before that we have our positive possibility that x < 3 or we have our negative possibility -x < 3.0246

I have not done anything with that inequality symbol, I have not even touched it yet I simply split it into two problems.0258

Because we have the absolute value less than a number, I will connect these using the word and.0266

It will become especially important when we get tour final solution.0274

Now I just have to solve these separately.0278

x < 3 is already done, -x < 3 I have to multiply both sides by -1.0281

Since I’m multiplying by a negative, we need to flip our sign.0289

Here I have two things that x must be a number that is less than 3 and x must be a number greater than 3.0294

Since we are dealing with and, we are looking for all numbers that satisfied both of the possibilities.0302

Let us look at our number line first to see if we can hunt down everything that does that.0309

All numbers less than 3 would be on this side of 3 and we will put in a big open circle.0327

They must be greater than -3 and I will put in another big open circle and -3.0336

We will see that it shades in everything greater than that value.0341

Our overall solution is all numbers between -3 and up to 3.0346

Use your techniques for splitting up into two problems.0354

Be careful on how you connect them and remember that you should flip the sign when you multiply or divide by negative.0357

Let us try another small one, and again watch for this process to happen.0363

This one says the absolute value of z = 5.0369

The first thing I'm going to do is just split this into two problems. 0373

I’m doing this because I'm taking care of that absolute value.0380

z > or = 5, -z > or = 5 I have not touched my inequality symbol I just took care of my two cases.0384

Since I'm dealing with the absolute value being greater than or equal to a number over here and I will connect these using or.0397

This will become especially important when we get to our solution and we just solve each of them separately.0408

Z > or = 5 is already done and for the other one, I will multiply both sides by -1.0413

Now if I multiply it by a negative, this will flip its sign.0423

I have z > or = 5 or z < or = 5.0428

I think we are ready to start packaging things up here.0434

Since I’m dealing with the connection or, as long as it satisfies one of these possibilities I will include it in my final solution.0442

All the numbers greater than or equal to 5 that would be all of these numbers here on the number line.0450

All the ones less than or equal to 5, that would be these ones down here.0460

Now that I know more about the number line, let us go ahead and write this using our interval notation.0465

Negative infinity up to 5 brackets, because we are including from 5 to infinity, our little union symbol since it could be in either one of those intervals.0470

Now that we have the process down a little bit better for being able to split this up, let us look at ones that are just a little bit more complicated.0487

We want to solve the inequality involving our absolute value.0498

I want make sure that the absolute value is isolated on one side of my inequality sign and fortunately this one is.0503

Let us go ahead and split it into our two problems.0510

On this side we will have 3x + 2 < 5.0519

On the other side I have -3x + 2 < 5.0525

I will take note as to the direction of my inequality symbol, our inequality less than number.0533

Because of its direction I will connect these using and.0541

A little bit more work, let us go ahead and solve each of these.0547

For this one, I will subtract 2 from both sides and then divide both sides by 3 so x < 1.0551

On the other side, let us go ahead and distribute our negative sign and let us add 2 to both sides.0563

We can see that on this case I only did divide by -3 so because of that we will end up flipping our sign -7/3.0576

I need all numbers that are greater than -7/3 and they are also less than a 1.0589

On a number line I could see where -7/3 is, there is 0 and I can see where 1 is.0600

I’m just looking for all numbers in between. 0609

Since those would be the only ones that are greater than -7/3 and also less than 1.0614

We have our interval -7/3 up to 1 and that would be our solution.0620

Let us get into another one. Solve the inequality using an absolute value.0631

In this one, we do a little bit of work to isolate that absolute value first.0637

Let us go ahead and add 1 to both sides, 5 - 2x > or = 1.0642

Now that my absolute value is isolated, we will split it into two problems.0653

This will handle our positive possibility, so no changes, 0660

And the other one will take care of what happens when the inside part could have been negative.0666

Since I'm dealing with these absolute values and inequalities, let us look at the direction of everything.0674

I have the absolute value less than or equal to number.0680

This is that situation where we connect those using or.0684

I have two problems that are connected using or, let us continue solving.0689

On the left side, I will continue by subtracting 5 from both sides, then I can divide both sides by -2.0696

Since I'm dividing by negative, let us flip our sign, x < or = 2.0710

Onto the other side I'm going to distribute my negative sign in there first -5 + 2x > or = 1.0719

Let us get that x all alone and we are just a little bit closer being all alone by adding 5 to both sides.0733

And I will finally divide both sides by 2, x > or = 3.0743

I have what I’m looking for all numbers that are less or equal to 2 and all numbers that are greater than 3.0753

Let us see what that looks like on a number line.0760

All numbers less than or equal to 2 would include the 2 and everything less than that, so we will shade that in.0767

All numbers equal to 3 or greater than would be over here and so we have a big gap between 2 and 3 but that is okay. 0775

Let us graph this using our interval notation.0784

From negative infinity up to 2 included and from 3 up to infinity and we are looking at the union of both of those two.0787

That is not too bad as long as you isolate that absolute value first before splitting it into two problems, you should be fine.0799

Let us look at one last one, and this highlights why you have to be careful on how you package up your solution using those connections and and or.0807

In this one, we already have our absolute value isolated to one side 3 - 12x all in absolute value.0818

I'm going to split this immediately into two problems.0828

3 - 12x < or = -3 and 3 - 12x < or = -3 because of the direction of our inequality symbol.0838

Let us go ahead and connect these using and.0858

Solving the one on the left 3 - 12x < or = 3.0864

Let us subtract 3 from both sides then we will divide both sides by -12.0870

I’m flipping my signs because I'm dividing by negative.0883

We can reduce this fraction as normal, just ½ so x > or = ½.0889

Working with the other one, I will distribute through with my negative sign and I will add 3 to both sides. 0897

It looks like one last step and I can go ahead and divide both sides by 12, 0 ÷ 12 is simply 0.0918

Let us see exactly what this would include.0933

On a number line, I want to shade in all numbers that are greater than 1/2 and they must be less than 0.0936

Of course, that poses a huge problem.0945

There are lots of numbers greater than ½ and there is lots of numbers less than 0 but unfortunately I can not find numbers that satisfied both of them.0951

There are no numbers that satisfied both conditions.0962

In fact this one, I can say has no solution. 0966

Now, you might have caught that earlier and if you did that is okay, you can definitely save yourself a lot of work.0972

If we look all the way back here at the original problem, we have a bit of an issue.0980

Remember that the absolute value makes everything positive and sitting on the other side of our inequality is a negative number.0985

We simply can not have something positive being less than something negative, that is not going to work. 0999

That is another good reason why this one has no solution.1005

Be careful when going through these problems and make sure you split them up because of your absolute value.1009

Watch the direction of that absolute value so you know how to connect them using, say, either, and or or .1013

Remember to solve each of them separately and connect very carefully when you get to your solutions.1020

Thanks for watching