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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?