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Course: Algebra (all content)>Unit 8

Lesson 3: Solving absolute value inequalities

Solving absolute value inequalities: fractions

Sal solves the inequality |2r-3 1/4| < 2 1/2. Created by Sal Khan and Monterey Institute for Technology and Education.

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• I have trouble with solving combined absolute value inequalities. For example how would you solve and graph: 9< |4x-5| <23
(10 votes)
• One way to solve this problem would be to start by rewriting is as 9 < |4x - 5| and |4x - 5| < 23. Then solve each part.

9 < |4x - 5|
|4x - 5| > 9
4x - 5 > 9 or 4x - 5 < -9
4x > 14 or 4x < -4
x > 7/2 or x < -1

|4x - 5| < 23
4x - 5 < 23 and 4x - 5 > -23
4x < 28 and 4x > -18
x < 7 and x > -9/2
-9/2 < x < 7

Finally, the solution is the numbers that satisfy the criteria between -9/2 and 7 for which x > 7/2 or x < -1. So the answer is -9/2 < x < -1 or 7/2 < x < 7. On a number line, this would be represented by shading the numbers between -9/2 and -1 and the numbers between 7/2 and 7.
(17 votes)
• i have a really confusing problem related to this and i need some help on it!:
2|4x+1|-5≤ -1
(5 votes)
• Okay you want the absolute value by itself. Add 5 to both sides to get 2|4x+1| ≤ 4. Next you divide both sides by 2 to get |4x+1| ≤ 2. Ok now you put 4x+1 ≤ 2 AND 4x+1 ≥ -2. The first one (4x+1 ≤ 2) is simplified to x ≤ 1/4. The second one (4x+1 ≥ -2 ) works out to x ≥ -3/4. Thus, the answer is -3/4 ≤ x ≤ 1/4.
(10 votes)
• how so you know if its a "or" or an "and"?
(3 votes)
• Think GreaTOR and Less thAND
(17 votes)
• How is the absolute value used in the real world?
(4 votes)
• ok I think you understand David that - is like debt and + is making money
(4 votes)
• how do you know when to put "and" or "or"?
(3 votes)
• If the absolute value quantity is less than the other value, use "and". If the absolute value quantity is greater than the other value use "or".
(5 votes)
• I have done this before and for example 5647= isn't it the same number?
(4 votes)
• I don't know double check it ! ;-)
(5 votes)
• When you say, "we'll take the absolute value of it." what do you mean? Are you changing anything such as adding, subtracting, dividing, ect?
(4 votes)
• You are simply making the value of the integer positive. It is helpful when considering time and distance, because it is difficult to have negative values for distance and time. That is, unless you're a timelord.
(3 votes)
• The absolute value of an expression less than some other number is an AND problem by definition?

So |x|<12 is an AND problem and |x|>12 would be an OR problem? You can tell that by the sign?
(2 votes)
• Yes, you are correct. |x|<12 is the same as x<12 AND x>-12, while |x|>12 is the same as x>12 OR x<-12. So if a problem has a less than sign, it is an AND problem and if a problem has a greater than sign, it is an OR problem
(5 votes)
• What software does Sal use for writing down everything?
(2 votes)
• Is this an overall fraction??
(2 votes)

Video transcript

We have the absolute value of 2r minus 3 and 1/4 is less than 2 and 1/2, and we want to solve for r. So right from the get go we have to deal with this absolute value. And just as a bit of a review, if I were to say that the absolute value of x is less than, well let's just say, less than 2 and 1/2, that means that the distance from x to 0 is less than 2 and 1/2. That means that x would have to be less than 2 and 1/2, and x would have to be greater than negative 2 and 1/2. And just think about it for a second. If I were to draw it on a number line right here, that is 0, that is 2 and 1/2, and that is negative 2 and 1/2. These two numbers are exactly 2 and 1/2 away from 0, because both of their absolute values is 2 and 1/2. Now, if we want all of the numbers whose absolute value is less than 2 and 1/2, or that are less than 2 and 1/2 away from 0, it would be all of the numbers in between. And that's exactly what these two statements are saying. x has to be less than 2 and 1/2, and it has to be greater than negative 2 and 1/2. If this absolute value were the other way, that the absolute value of x has to be greater than 2 and 1/2, then it would be the numbers outside of this, and it would be an or. But we're dealing with the less than situation right there, so let's just do what we were able to figure out when it was just an x. The distance from this thing to 0 has to be less than 2 and 1/2, so we can write that 2r minus 3 and 1/4 has to be less than 2 and 1/2 and 2r minus 3 and 1/4 has to be greater than negative 2 and 1/2. Same exact reasoning here. Let me draw a line so we don't get confused. Same exact reasoning here. This quantity right here has to be between negative 2 and 1/2. It has to be greater than negative 2 and 1/2 right there. And it has to be less than 2 and 1/2, so that's all I wrote there. So let's solve each of these independently. Well, this first went over here, you've learned before that I don't like improper fractions, and I don't like fractions in general. So let's make all of these fractions. Sorry, I don't like mixed numbers. I want them to be improper fractions. So let's turn all of these into improper fractions. So if I were to rewrite it, we get 2r minus 3 and 1/4 is the same thing as 3 times 4 is 12, plus 1 is 13. 2r minus 13/4 is less than-- 2 times 2 is 4, plus 1 is five-- is less than 5/2. So that's the first equation. And then the second question-- and do the same thing here-- we have 2r minus 13 over 4 has to be a greater than negative 5/2. All right, now let's solve each of these independently. To get rid of the fractions, the easiest thing to do is to multiply both sides of this equation by 4. That'll eliminate all of the fractions, so let's do that. Let's multiply-- let me scroll to the left a little bit-- let's multiply both sides of this equation by 4. And what do we get? 4 times 2r is 8r, 4 times negative 13 over 4 is negative 13, is less than-- and I multiplied by a positive number so I didn't have to worry about swapping the inequality-- is less than 5/2 times 4 is 10, right? You get a 2 and a 1, it's 10. So you get 8r minus 13 is less than 10. Now we can add 13 to both sides of this equation so that we get rid of it on the left-hand side. Add 13 to both sides and we get 8r-- these guys cancel out-- is less than 23, and then we divide both sides by 8. And once again, we didn't have to worry about the inequality because we're dividing by a positive number. And we get r is less than 23 over 8. Or, if you want to write that as a mixed number, r is less than-- what is that-- 2 and 7/8. So that's one condition, but we still have to worry about this other condition. There was an and right here. Let's worry about it. So our other condition tells us 2r minus 13 over 4 has to be greater than negative 5/2. Let's multiply both sides of this equation by 4. So 4 times 2r is 8r. 4 times negative 13 over 4 is negative 13, is greater than negative 5/2 times 4 is negative 10. Now we add 13 to both sides of this equation. The left-hand side-- these guys cancel out, you're just left with 8r-- is greater than negative 10 plus 13 is 3. Or divide both sides of this by 8, and you're left with r has to be greater than 3/8. So our two conditions, r has to be less than 2 and 7/8 and greater than 3/8. Or we can just write it like this: r is greater than 3/8, so it's greater than-- maybe I should say 3/8 is less than r, which is less than 2 and 7/8. So if we were to plot the solution on the number line-- which I'm about to do, so that's my number line-- this is 0 right here, maybe this is 1, 2, and 3. We have 2 and 7/8. We have to be less than 2 and 7/8. Let's say that this is 2 and 7/8 right there. And we have to be greater than 3/8. Say that is 3/8, so 3/8 will be some place right around there. And everything in between is a valid solution. And we could try it out. Let's try out something that, based on what I just drew, should be a valid solution. 1 should be a valid solution. Let's try it out here. 2 times 1 minus 3 and 1/4, what is that? That's 2 minus 3 and 1/4. And so what is that? 2 minus 3 and 1/4 is-- well, 3 and 1/4 minus 2 is 1 and 1/4, so this will be negative 1 and 1/4. But we're taking the absolute value of it, so we take the absolute value of it, which is equal to 1 and 1/4, which is indeed less than 2 and 1/2. Now let's try another number. Let's try 0. Based on this, 0 should not work. So what happens if we put 0 here? You get 2 times 0, which is 0, minus 3 and 1/4. If you take the absolute value of negative 3 and 1/4, you'll get positive 3 and 1/4, which won't work. 3 and 1/4 is greater than 2 and 1/2, so that's true, that works out. And same thing for 3. 2 times 3 is 6, minus 3 and 1/4 is 2 and 3/4. Take the absolute value, it's 2 and 3/4, still bigger than 2 and 1/2, so it won't work. So at least the points that we tried out seem to validate this solution that we got.