WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we are going to be solving compound and absolute value inequalities.
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A compound inequality consists of two or more inequalities combined by either "and" or "or."
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To solve a compound inequality, solve each part.
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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.
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So, I am going to solve each part of that.
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4x...add 2 to both sides; that would be less than 12; divide both sides by 4, so x is less than 3.
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And if I add 1 to both sides, I am going to get that x is greater than or equal to 4.
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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.
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The same idea applies if an inequality is a compound inequality connected by the word or.
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For example, you might be given 3x > 6, or 4x < 4.
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Again, solve each part: x > 2, or x < 1.
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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.
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So, I am going to review this here--the idea of intersection--and this is also covered in detail in the Algebra I series.
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**Intersection** means the common members of both solution sets.
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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 set
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that is 6, 8, 10, 12, and 14; the intersection would be the elements common to both.
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So, I see that I have 6 in both; 8; and 10; so the intersection would be 6, 8, and 10.
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Apply that to inequalities connected by the word and: for example, x - 4 < 10, and 2x ≥ 4.
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OK, solve each part as discussed; so add 4 to both sides...this means that x is less than 14;
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and (both conditions must hold) if I divide both sides by 2, I get x ≥ 2.
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To visualize this, look at the number line: if I have a 0 right here, and then 2, 4, 6, 8, 10, 12, 14,
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this is telling me that x is less than 14; and that would just go on and on and on.
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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.
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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.
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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.
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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.
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So, this is actually just a more efficient way of writing, instead of two separate sections here with the word and;
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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.
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So, that is for inequalities joined by the word and.
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If or is used in a compound inequality, the solution set is the union of the solution sets of the individual inequalities.
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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,
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-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.
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So, if it is in one; if it is in the other; or if it is in both; that would be the union.
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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;
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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.
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And I could rewrite this in ascending order, but it is all included here.
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The union means that it includes the elements that are in either one of these, or both.
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Applying this concept to inequalities: 3x + 2 > 8, or 4x - 3 < 1 is a compound inequality joined by the word or.
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First, solve each inequality: 3x > ...subtract 2 from both sides...6; divide both sides by 3 to get x > 2.
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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.
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Now, looking at this on the number line, what this is saying is that...0, 1, 2, 3, 4...x is greater than 2.
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This is saying x is less than 1; and there is no overlap here, but that is OK, because the union
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means that, if something is in either one of these or both, it is included.
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This is saying that the solution set is that x is greater than 2, or x is less than 1.
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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.
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So, when you use "and," it is the intersection of the two solution sets.
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When "or" joins a compound inequality, the solution set is the union of the two solution sets.
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Now, we have worked with absolute value equations; and this time, today, we are working with absolute value inequalities.
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To solve an absolute value inequality, you need to use the definition of absolute value.
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And recall that the definition of absolute value is the distance that a value is from 0 on the number line.
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So, the absolute value of 3 is 3, because the distance between this and 0 on the number line is 3.
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The absolute value of -3 is also 3, because the distance between -3 and 0 on the number line is 3.
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So, the absolute value of 3 is 3, because it is 1, 2, 3 away from 0 on the number line.
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The absolute value of -3 is 3, because -3 is also 3 units from 0 on the number line.
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For more complicated problems, you may need to rewrite the inequality as a compound inequality.
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Talking about this in a little bit more detail: you may see two different forms.
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You may see inequalities in the form |x| < n.
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If you see that, rewrite the inequality as a compound inequality, as follows.
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Looking at this, I have an example: the absolute value of x is less than 4.
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Think about what this is saying; it is saying that the absolute value of x is less than 4 units from 0.
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OK, so anything that is less than 4...it could be 3, 2, 1, but it is less than 4 away from 0.
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That could be satisfied by anything between 0 and 4; but it can also be satisfied by anything between 0 and -4,
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because the distance between -4 and 0 on the number line is less than 4.
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-3: the distance between that and 0 is less than 4.
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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.
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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;
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because then you will get numbers way over here, that have an absolute value that is much bigger than 4.
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So, it is in this range, where x is greater than -4, but less than 4.
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So, in general, when you see an absolute value in this form, where it is less than something, you rewrite it
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as two related inequalities: x < 4, and x > -4.
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We are just generalizing out: |x| < n can be rewritten as x < n, and x > -n.
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OK, the other possibility is that you have inequalities that are in the form |x| > n.
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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.
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So, anything bigger than 3 is going to have an absolute value of greater than 3.
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4 has an absolute value greater than 3; and on up.
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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.
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If I took -4, the absolute value of that is 4, which is greater than 3.
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So, this would translate to x > 3, or x < -3.
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You need to memorize these two forms and be familiar with them.
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And generalizing this out, this would say that |x| > n could be rewritten as x > n, or x < -n.
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So, you need to keep these in mind: when you see |x| < n, you rewrite it as x < n, and x > -n.
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When you see |x| > n, then x > n, or x < -n.
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And you can apply these to more complex inequalities, such as |4x + 1| ≥ 12.
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I would recognize that it is in this form, and I would rewrite it as 4x + 1 ≥ 12, or 4x + 1 ≤ -12.
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So, just follow this pattern; and we will work on this in the examples.
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OK, starting with Example 1: this is a shorthand way of a compound inequality that is joined by the word "and."
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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.
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But this is really a compound inequality that is joined by the word "and."
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And remember that, in order to solve these, you solve each inequality; and then you find the intersection of their solution sets.
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So, for absolute inequalities joined by the word "and," we are going to need to find the intersection of the solution sets.
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All right, let's solve each of these; that is the first step.
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I am going to rewrite this with the x on the left; 14x - 42 ≥ 7--more standard.
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It is saying the same thing: 14x - 42 ≥ 7; I just reversed the sides of the inequality.
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Next, add 42 to both sides; and that is going to give me 49; 14x ≥ 49.
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Now, divide both sides by 14; and since that is a positive number, I don't need to reverse the inequality symbol.
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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.
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I just pulled out the 7 from both the numerator and the denominator--factored it out.
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OK, solving this inequality--my second inequality: adding 42 to both sides is going to give me 14x ≤ 77.
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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.
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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.
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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.
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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.
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Here, the other restriction is that x ≤ 5.5; so this is greater than--it is going up this way;
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but I have to stop when I get to here, because the intersection of the solution set has to meet both conditions.
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It has to be greater than or equal to 3.5, and less than or equal to 5.5.
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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.
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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.
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I wrote this out as two related inequalities, solved each, and found the intersection of their solution sets.
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The second example is a compound inequality joined by the word or.
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I am going to go ahead and solve both of these and find the union of the solution sets.
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Adding 7 to both sides gives me 3y ≤ 15; divide both sides by 3: y ≤ 5.
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On this side, I am subtracting 10 from both sides to get 2y > 16; dividing both sides by 2 gives me y > 8.
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OK, since this is or, it is the union of the solution sets; so any element of either set,
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or that is in both sets, would be included in the solution set for this compound inequality.
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So here, I have y ≤ 5, and here I have y > 8.
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So, all of this is included, and all of this is part of the solution set.
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It could also be written as y ≤ 5, or y > 8.
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OK, the third example involves an absolute value inequality.
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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.
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And recall, for those, that you could rewrite this and get rid of the absolute value bars by saying x < n, and x > -n.
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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).
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It is really important to memorize these or understand them well enough that you can apply them to more complex situations.
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Now, I am going to solve each.
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2z - 6 ≤ 8: I am adding 6 to both sides to get 14, then dividing both sides by 2 to get z ≤ 7.
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And I need to solve this other one.
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I am going to add 6 to both sides to get 2z ≥ -6, z ≥ -3.
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And since I am dividing by a positive number, I don't have to worry about reversing the inequality symbol.
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OK, I end up with...again, I had an absolute value inequality in this form; I solved both inequalities.
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Actually, I made a little mistake here; let me go ahead and correct that.
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2z - 6 ≥ -8, so 2z ≥...this is actually going to be -2; so adding 6 to both sides is going to give me -2.
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Now, if I divide both sides by 2, that is going to give me z ≥ -1.
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OK, you can either leave this as it is, or you can go ahead and graph it out.
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And since this is "and," the solution set needs to meet both of these conditions.
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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;
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but it also has to meet the condition that z is greater than or equal to -1.
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So, anything in this range is going to be the intersection of the solution set for these inequalities.
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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.
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These are three different ways of writing out the solution set.
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OK, the next example is also an absolute value inequality.
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And this one is in the form |x| > n, in this general form, which can be rewritten as x > n or x < -n.
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So, rewriting this and removing the absolute value bars as two different inequalities: 3w - 8, divided by 5, is greater than 4;
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so that is this first form; or 3w - 8, divided by 5, is less than -4.
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OK, the first thing to do is get rid of the fraction: multiply both sides of this inequality by 5.
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Next, add 8 to both sides; and finally, divide both sides by 3.
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For this inequality, again, get rid of the fraction; multiply both sides by 5; add 8 to both sides; and divide by 3.
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This is joined by the word "or"; and 28/3...I am going to rewrite that as 9 and 1/3.
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So, w is greater than 9 and 1/3, or w is less than -4.
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To graph this, -5, -4, -3, -2, -1, 0, 1...all the way to 9 and 1/3.
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OK, so this is saying that w is greater than 9.3, or it is less than -4.
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Open circle, because it is a strict inequality...
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Set notation would be w is greater than 28/3, or w is less than -4.
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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.
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Solve each one: and the solution set is the intersection of the solution sets of those two inequalities.
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That concludes this lesson of Educator.com; and I will see you soon!