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### Course: Algebra (all content)>Unit 8

Lesson 3: Solving absolute value inequalities

# Solving absolute value inequalities 2

Sal solves the inequality |p-12|+4 < 14. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Am I correct in understanding that whenever you have an absolute number in an equation or inequality, the nature of the equation or inequalty is changed and limited entirely, because then we are directly dealing only with the number relative to zero as opposed to when there are no absolute values in the query, it is possible that the query is directly about any type of quantity, for example, quantity of apples, oranges, dollars, atoms (or also positively or negatively away from zero)?
• A inequality does not yield a concrete number upon its solution, and gives a set of numbers. And no, the solution of the inequality is not limited entirely. If you say |h| > 22, this means that h has to be smaller than -22 and larger than 22. This means it stretches to infinity in both directions except for a gap of 44 in the middle on the number line. Hope this clears your doubt.
• Something Sal said got me thinking. So, If what I know is correct, the less than symbol can be anything up to and including 9.999999 away from ten. Sal said this in this video as well. But, it can't be ten, or else it would be a less than or equal to symbol. And other mathematical concepts say that .999999 equals 1. that would seem to override the less than symbol. So which is correct? Or does the .999999=1 rule not apply to linear equations? Or am I missing something simple?
• 9.999999 is less than 10. So is 9.9999999999999 for example. But 9.9999... (where the "..." means the 9's go on forever) is equal to 10.
Likewise 0.999999 is less than 1 but 0.99... = 1. (It doesn't matter that I only wrote two 9's, as long as it's understood that the "..." means that the 9's actually continue infinitely.)
• I don't understand the following question
|3+r| /7 ≤ 5

• 1) Multiply both sides by 7 to eliminate the fraction: |3 + r| ≤ 35
2) Split the inequality into a compound inequality: 3 + r ≤ 35 AND 3 + r ≥ -35
3) Solve the compound inequality. You will get: -38 ≤ r ≤ 32
Hope this helps.
• So how would you end up writing the answer on a test? Would the solution simply be 2<p<22?
• Yes, that is one way to write it. There are a few. Here are some examples:

2 < p < 22
(2, 22)
p > 2 and p < 22

Unless the teacher has taught or specified a certain way, they should all be valid. :)
• [Around ]
Okay, so I understand using -10<p-12<10 would help in this situation, but my teacher taught us to use that as a "answer holder" for AND inequalities, so why use them to find the algebraic answer when you can just do something like
p-12<10
and
p-12>-10 ?

{Example base equations:
A < X < B
A < X
B > X
X + A > B }
• It can be solved in either way. So, pick the method that you are comfortable with.
• If -2|x+2| <= -10.. how do we solve this.. Is it equal to :
|x+2| <= 5

|x+2| >= -5
• In short, the second equation that you wrote is correct: lx+2l<=-5 because you flip the inequality sign when dividing by a negative. Check out the whole solution below:

Q: Solve -2lx+2l<=-10.

1. Isolate the absolute value by dividing by -2. When you divide by a negative, the inequality sign flips, so you will end up with lx+2l>=5. Again, we changed from <= to >= because we divided by a negative.

2. We can split the inequality by saying that -5>=x+2 or x+2>=5. We do this because when following an absolute value by a > or >= symbol, it becomes an "or" inequality.

3. We want to isolate x, so we arrive at -7>=x and x>=3 by subtracting 2 from both sides of both equations.

4. Checking our work.

Let's try some values for x to see if they work: if -7>=x, then x<=-7. We can plug in any value less than or equal to -7, and it should work. Here's what happens when we plug in -10 to the original equation: -2l(-10)+2l<=-10 -> -2l-8l<=-10 -> -2(8)<=-10 -> -16<=-10, which works, so we know that -7>=x is a valid solution.

For x>=3, we can plug in any number greater than or equal to 3 and it should work. Here's what happens when we plug in 5: -2l(5)+2l<=-10 -> -2l7l<=-10 -> -2(7)<=-10 -> -14<=-10, which works, so we know that x>=3 is also a valid solution.

Let's plug in something not in our answer set to check. Here's what happens when we plug in 0: -2l(0)+2l<=-10 -> -2l2l<=-10 -> -2(2)<=-10 -> -4<=-10, which does not work. Since this number is not in our answer set, and we have checked numbers in our answer set, we know that our solutions were correct.

5. The solution is x<=-7 and x>=3. We can also write this in interval notation as (-infinity, -7]U[3, infinity).

Hope this helps!
• How can you tell if a question is an or, and problem?
• The inequality symbol tells you what to use.
-- If the problem is: |x| < a number, then you use AND. This also applies to "<=".
-- If the problem is: |x| > a number, then you use OR. This also applies to ">=".

Hope this helps.
• Would the first inequality be graphed the same if it was x<|10| instead of |x|<10? Why or why not?
(1 vote)
• Brennan,
|10| is just 10
So x<|10| is just x<10
This would include all negative numbers
including for example -20 because -20 < |10|

|x|<10 is different.
It means all values of x that the absolute distance from 0 is less than 10
If you try -20 for x then |-20| = 20 and 20 is not less than 10.
So -20 would not be part of the solution.

|x| <10 means all real numbers between -10 and +10 or
-10 < x < 10

I hope that helps make it click for you.
• The problem is :
-x|x| > 4
The way I am approaching this is for two cases. First case when x > 0:
-x*x>4
=>-x^2>4
=>x^2<-4

Case 2 when x < 0:
-x*-x > 4
=>x^2 > 4
=>x > 2 or x > -2

But when x is negative and I take a value x > -2, say -1, it does not satisfy the equation:
-(-1).1 = 1. This is not > than 4

Instead taking the value -3 which is outside the solution set satisfies the equation.

-(-3).3 = 9 which is greater than 4. Where am I going wrong?
• how do you solve |3n+7| <=23 is it 3n-7<=23 &3n+7<=23 or 3n+7<=23 &3n+7<=-23 which one is it?
• It is neither one.
It is: 3n+7<= 23 AND -(3n+7)<=23
The 2nd inequality could be written as: 3n+7>=-23 if you divide both sides by -1.

Hope this helps.

## Video transcript

Solve for p. We have the absolute value of p minus 12 plus 4 is less than 14. So, let's just do this one step at a time. The first thing we wanna do is we really just wanna isolate the part that has the absolute value. So, to do that we can just get rid of this positive 4 here. We can do that by subtracting 4 from both sides of the inequality. And so the left hand side the positive 4 and the negative 4 cancel out we're just left with the absolute value of p minus 12. And on the right hand side it, we get 14 minus 4 is 10 and we still have the less than sign. So we have the absolute value of p minus 12 is less than 10. So let's just think about it a little bit. If I were to tell you, if I were to tell you that the absolute value of x is less than 10 what does that mean? That means that the distance from x to 0 has to be less than 10. So if I were to draw a number line, if I were to draw a number line and put 0 here we can only go up 10 away and even that's too far. It has to be less than 10. So if this is positive 10 it would have to be less than positive 10 cuz 10 is exactly 10 away. We have to be less than 10 away from 0. And then we could go all the way to the left until negative 10 and even that we wouldn't be able to include because its absolute value is 10. It's not less than 10 but negative 9 negative 9.999 we could include all of those things. The absolute value of any of those things is gonna be less than 10. So another way to write this, this absolute value inequality is that x could be, x could be greater than negative 10, x could be greater than negative 10 and x needs to be less than 10 or we can write this as x is between. So negative 10 is kinda the bottom boundary and we're not going to include it. X is going to be greater than that and is less than 10. So this is another way to write the absolute value of x is less than 10. Is essentially saying that x has to be between negative 10 and 10 and it can't be either negative 10 or 10. We're not, there's no equals sign here. So, the same exact logic. Instead of an x we have a p minus 12 here. So, we can write the absolute value of p minus 12 is less than 10 is saying that negative 10 is less than p minus 12 which is less than 10. And we can just solve this compound inequality all at once by isolating the p in the middle. The best way to isolate the p in the middle we wanna get rid of this negative 12. So let's add 12 to all three, all three sections of this compound inequality. And so we get negative 10 plus 12 is positive 2. And then is less than p minus 12 plus 12 is just p. And then is less than 10 plus 12 is 22. So p is greater than 2 and less than 22. So if we were to plot it on a number line our solution set looks like this. So this might be 2 over here. This might be 22 over here maybe 0 sitting right over here. P is greater than 2. P is greater than 2. It's not greater than or equal to so we have to not fill this in. It has to be an open circle it's only greater than and it's less than 22. It's not less than or equal to so we're not gonna fill that circle in. And it's everything, it's everything in between. And we can verify it for ourselves. Let's try a value that might work out well. Well, 12 is in between these two numbers. It's in our magenta region right here. So let's try p is equal to 12. So if you have 12 minus 12. So it's the absolute value of 12 minus 12 plus 4 which should be less than 14. So this is 0 plus 4 which needs to be less than 14 and 4 is definitely less than 14. So 12 worked 0 shouldn't work. Lets try 0. Zero minus 12 so it's the absolute value of 0. I wanna do this in a different color. It is the absolute value of 0 minus 12 plus 4. This should not be less than 14. This should not work. So we get the absolute value of negative 12. Plus 4 should not be less than 14. Then we should end up with a contradiction here. So we have 12 plus 4 less than 14 we end up with 16 is less than 14 which is not true. So 0 does not work so at least we're feeling pretty good. We took something outside of our solution set didn't work something inside of our solution set it did work.