Algebra 2 Solving Compound and Absolute Value Inequalities

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In today's lesson, we are going to be solving compound and absolute value inequalities.0002

A compound inequality consists of two or more inequalities combined by either "and" or "or."0009

To solve a compound inequality, solve each part.0016

For example, if you are given 4x - 2 < 10, and x - 1 ≥ 4, it is connected by the word and, so it is a compound inequality.0022

So, I am going to solve each part of that.0040

4x...add 2 to both sides; that would be less than 12; divide both sides by 4, so x is less than 3.0044

And if I add 1 to both sides, I am going to get that x is greater than or equal to 4.0054

We will talk, in a second, about how the solution set works for this; but the first step is to just solve both parts of the inequality.0063

The same idea applies if an inequality is a compound inequality connected by the word or.0074

For example, you might be given 3x > 6, or 4x < 4.0080

Again, solve each part: x > 2, or x < 1.0090

Breaking this down and talking about the solution sets for each: if "and" is used, the solution set is the intersection of the solution sets of the individual inequalities.0103

So, I am going to review this here--the idea of intersection--and this is also covered in detail in the Algebra I series.0113

Intersection means the common members of both solution sets.0121

Just looking at numbers: if I have 2, 4, 6, 8, and 10, and that is my first set; and then I have a second set0127

that is 6, 8, 10, 12, and 14; the intersection would be the elements common to both.0136

So, I see that I have 6 in both; 8; and 10; so the intersection would be 6, 8, and 10.0149

Apply that to inequalities connected by the word and: for example, x - 4 < 10, and 2x ≥ 4.0158

OK, solve each part as discussed; so add 4 to both sides...this means that x is less than 14;0173

and (both conditions must hold) if I divide both sides by 2, I get x ≥ 2.0179

To visualize this, look at the number line: if I have a 0 right here, and then 2, 4, 6, 8, 10, 12, 14,0187

this is telling me that x is less than 14; and that would just go on and on and on.0200

This is telling me x is greater than or equal to 2, which is going to start here and go on and on and on and on.0207

But I am looking for the intersection, or the common elements; and the common elements would be greater than or equal to 2, and less than 14.0215

So, the intersection would come out as 0, 2, 4, 6, 8, 10, 12, 14--greater than or equal to 2, and less than 14.0225

And you could write that out as an inequality, or using set notation, that x is greater than or equal to 2, and less than 14.0240

So, this is actually just a more efficient way of writing, instead of two separate sections here with the word and;0255

you can just write it out more efficiently like this; x is greater than or equal to 2 and less than 14; or showing it on the number line like this.0262

So, that is for inequalities joined by the word and.0271

If or is used in a compound inequality, the solution set is the union of the solution sets of the individual inequalities.0276

Now, reviewing what a union is, just using numbers: if you have a set, such as 4, 5, 6, 7, 8; and then you have another set,0285

-2, -1, 2, 3, 4, 5; and you are asked to find the union; well, the union is any elements that are in either one of these, or both.0298

So, if it is in one; if it is in the other; or if it is in both; that would be the union.0315

Now, I am looking, and I have 4, and that is in both; I don't need to write it twice; 5; 6 is just in this one, but it is included;0321

7, just in that one; 8; and then, I already covered 4 and 5, but I also have to include -2, -1, 2, and 3.0332

And I could rewrite this in ascending order, but it is all included here.0344

The union means that it includes the elements that are in either one of these, or both.0348

Applying this concept to inequalities: 3x + 2 > 8, or 4x - 3 < 1 is a compound inequality joined by the word or.0353

First, solve each inequality: 3x >...subtract 2 from both sides...6; divide both sides by 3 to get x > 2.0366

Or: solve this one also--if I add 3 to both sides, I am going to get 4x < 4; I am going to divide both sides by 4 to get x < 1.0380

Now, looking at this on the number line, what this is saying is that...0, 1, 2, 3, 4...x is greater than 2.0391

This is saying x is less than 1; and there is no overlap here, but that is OK, because the union0403

means that, if something is in either one of these or both, it is included.0409

This is saying that the solution set is that x is greater than 2, or x is less than 1.0414

And this can be written in set notation; so, I would write it as x is greater than 2, or x is less than 1.0420

So, when you use "and," it is the intersection of the two solution sets.0429

When "or" joins a compound inequality, the solution set is the union of the two solution sets.0432

Now, we have worked with absolute value equations; and this time, today, we are working with absolute value inequalities.0440

To solve an absolute value inequality, you need to use the definition of absolute value.0448

And recall that the definition of absolute value is the distance that a value is from 0 on the number line.0453

So, the absolute value of 3 is 3, because the distance between this and 0 on the number line is 3.0461

The absolute value of -3 is also 3, because the distance between -3 and 0 on the number line is 3.0481

So, the absolute value of 3 is 3, because it is 1, 2, 3 away from 0 on the number line.0489

The absolute value of -3 is 3, because -3 is also 3 units from 0 on the number line.0494

For more complicated problems, you may need to rewrite the inequality as a compound inequality.0505

Talking about this in a little bit more detail: you may see two different forms.0511

You may see inequalities in the form |x| < n.0516

If you see that, rewrite the inequality as a compound inequality, as follows.0523

Looking at this, I have an example: the absolute value of x is less than 4.0532

Think about what this is saying; it is saying that the absolute value of x is less than 4 units from 0.0541

OK, so anything that is less than 4...it could be 3, 2, 1, but it is less than 4 away from 0.0552

That could be satisfied by anything between 0 and 4; but it can also be satisfied by anything between 0 and -4,0562

because the distance between -4 and 0 on the number line is less than 4.0580

-3: the distance between that and 0 is less than 4.0585

So, anything in this range is going to satisfy it; so what this is really saying is that x is less than 4, and x is greater than -4.0589

So, in order for the answers to fall in this range, they can't just be less than 4 and go all the way down;0605

because then you will get numbers way over here, that have an absolute value that is much bigger than 4.0610

So, it is in this range, where x is greater than -4, but less than 4.0614

So, in general, when you see an absolute value in this form, where it is less than something, you rewrite it0621

as two related inequalities: x < 4, and x > -4.0630

We are just generalizing out: |x| < n can be rewritten as x < n, and x > -n.0637

OK, the other possibility is that you have inequalities that are in the form |x| > n.0647

For example, |x| > 3; well, what that is saying is that the absolute value of the number is more than 3 away from 0 on the number line.0660

So, anything bigger than 3 is going to have an absolute value of greater than 3.0673

4 has an absolute value greater than 3; and on up.0680

In addition, however, if I look over here on the left, any number smaller than -3 is also going to have an absolute value that is greater than 3.0685

If I took -4, the absolute value of that is 4, which is greater than 3.0695

So, this would translate to x > 3, or x < -3.0700

You need to memorize these two forms and be familiar with them.0709

And generalizing this out, this would say that |x| > n could be rewritten as x > n, or x < -n.0712

So, you need to keep these in mind: when you see |x| < n, you rewrite it as x < n, and x > -n.0722

When you see |x| > n, then x > n, or x < -n.0731

And you can apply these to more complex inequalities, such as |4x + 1| ≥ 12.0741

I would recognize that it is in this form, and I would rewrite it as 4x + 1 ≥ 12, or 4x + 1 ≤ -12.0752

So, just follow this pattern; and we will work on this in the examples.0770

OK, starting with Example 1: this is a shorthand way of a compound inequality that is joined by the word "and."0775

Instead of writing out 7 < 14x - 42, and 14x - 42 ≤ 35, they just combine the same term, put it in the middle, and left these on the outside.0785

But this is really a compound inequality that is joined by the word "and."0805

And remember that, in order to solve these, you solve each inequality; and then you find the intersection of their solution sets.0810

So, for absolute inequalities joined by the word "and," we are going to need to find the intersection of the solution sets.0818

All right, let's solve each of these; that is the first step.0824

I am going to rewrite this with the x on the left; 14x - 42 ≥ 7--more standard.0828

It is saying the same thing: 14x - 42 ≥ 7; I just reversed the sides of the inequality.0845

Next, add 42 to both sides; and that is going to give me 49; 14x ≥ 49.0854

Now, divide both sides by 14; and since that is a positive number, I don't need to reverse the inequality symbol.0869

I can simplify this, because they have a common factor of 7; so to simplify this, remove the common factor; that is going to give me 7/2.0878

I just pulled out the 7 from both the numerator and the denominator--factored it out.0890

OK, solving this inequality--my second inequality: adding 42 to both sides is going to give me 14x ≤ 77.0895

Again, dividing both sides by 14 is going to give me 77/14; again, I have a common factor of 7, so this is going to give me 11/2.0909

Now, the intersection of these, graphing this out: well, this is (2, 4, 6)...about 3.5, 3 and 1/2; and this is equal to 5 and 1/2.0922

OK, to help me graph it out, I am going to write it in decimal form: 1, 2, 3, 4, 5...a little more room....6.0945

OK, so what this is saying is x ≥ 3.5, which is right here; and it includes that point 3.5, so I am going to put a closed circle.0961

Here, the other restriction is that x ≤ 5.5; so this is greater than--it is going up this way;0975

but I have to stop when I get to here, because the intersection of the solution set has to meet both conditions.0983

It has to be greater than or equal to 3.5, and less than or equal to 5.5.0993

So, I can write it like this as an inequality; I can graph it here, or do set notation: where x is greater than or equal to 7/2 and less than or equal to 11/2.0998

OK, again, recognize that this is really a compound inequality; and it is joined by the word and, but it is just a shorthand way of writing it.1025

I wrote this out as two related inequalities, solved each, and found the intersection of their solution sets.1033

The second example is a compound inequality joined by the word or.1042

I am going to go ahead and solve both of these and find the union of the solution sets.1045

Adding 7 to both sides gives me 3y ≤ 15; divide both sides by 3: y ≤ 5.1058

On this side, I am subtracting 10 from both sides to get 2y > 16; dividing both sides by 2 gives me y > 8.1067

OK, since this is or, it is the union of the solution sets; so any element of either set,1082

or that is in both sets, would be included in the solution set for this compound inequality.1089

So here, I have y ≤ 5, and here I have y > 8.1104

So, all of this is included, and all of this is part of the solution set.1115

It could also be written as y ≤ 5, or y > 8.1122

OK, the third example involves an absolute value inequality.1132

When I look at this, I just want to think about which form this is in; and this is in the general form |x| < n.1138

And recall, for those, that you could rewrite this and get rid of the absolute value bars by saying x < n, and x > -n.1145

That is the general form; so I am going to rewrite this as 2z - 6 ≤ 8 (that is this form), and 2z - 6 ≥ -8 (which is this form).1158

It is really important to memorize these or understand them well enough that you can apply them to more complex situations.1174

Now, I am going to solve each.1181

2z - 6 ≤ 8: I am adding 6 to both sides to get 14, then dividing both sides by 2 to get z ≤ 7.1184

And I need to solve this other one.1200

I am going to add 6 to both sides to get 2z ≥ -6, z ≥ -3.1205

And since I am dividing by a positive number, I don't have to worry about reversing the inequality symbol.1218

OK, I end up with...again, I had an absolute value inequality in this form; I solved both inequalities.1226

Actually, I made a little mistake here; let me go ahead and correct that.1246

2z - 6 ≥ -8, so 2z ≥...this is actually going to be -2; so adding 6 to both sides is going to give me -2.1250

Now, if I divide both sides by 2, that is going to give me z ≥ -1.1263

OK, you can either leave this as it is, or you can go ahead and graph it out.1271

And since this is "and," the solution set needs to meet both of these conditions.1277

So, -1, 0, 1, 2, 3, 4, 5, 6, 7...a little more...let's go to 8; OK, z is less than or equal to 7, so this is going to go on and continue on;1284

but it also has to meet the condition that z is greater than or equal to -1.1301

So, anything in this range is going to be the intersection of the solution set for these inequalities.1305

Also, writing it out using set notation: what we have is that z is greater than or equal to -1, and less than or equal to 7.1313

These are three different ways of writing out the solution set.1331

OK, the next example is also an absolute value inequality.1335

And this one is in the form |x| > n, in this general form, which can be rewritten as x > n or x < -n.1340

So, rewriting this and removing the absolute value bars as two different inequalities: 3w - 8, divided by 5, is greater than 4;1354

so that is this first form; or 3w - 8, divided by 5, is less than -4.1364

OK, the first thing to do is get rid of the fraction: multiply both sides of this inequality by 5.1374

Next, add 8 to both sides; and finally, divide both sides by 3.1385

For this inequality, again, get rid of the fraction; multiply both sides by 5; add 8 to both sides; and divide by 3.1398

This is joined by the word "or"; and 28/3...I am going to rewrite that as 9 and 1/3.1415

So, w is greater than 9 and 1/3, or w is less than -4.1425

To graph this, -5, -4, -3, -2, -1, 0, 1...all the way to 9 and 1/3.1430

OK, so this is saying that w is greater than 9.3, or it is less than -4.1445

Open circle, because it is a strict inequality...1456

Set notation would be w is greater than 28/3, or w is less than -4.1460

Recognizing that this is in the general form |x| > n, I rewrote this as 3w - 8, divided by 5, is greater than 4, or the same expression is less than -4.1472

Solve each one: and the solution set is the intersection of the solution sets of those two inequalities.1487

That concludes this lesson of Educator.com; and I will see you soon!1496