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

Lesson 3: Solving absolute value inequalities

# Solving absolute value inequalities 1

Sal solves the inequality |h|-19.5 < -12. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• How is absolute value used in the world?
• absolute value is used in real world to find magnitude or length of particular things (scalar value).

Examples:

How far San Fransisco from New York City?
Answer : x km (It will never be negative value)

Answer : 10 cm , not -10 cm
• what exactly is a SOLUTION SET? does it include the number line graphing, or is it the >>_ , or the (_,_) , or all of them together?
• The solution set of a given equation is the set of all the elements that satisfy the equation. This means that all elements in the set make the equation a true proposition.

For example, consider the equation x+3=4
The solution set is the set {1} because when you substitute x in the equation you get 1+3=4 which is a true proposition.
• In equations, we can shift the numbers/variables from one side to another But, in inequalities can we shift the numbers/variables from one side to another?
• Of course, as long as you divide and multiply by positive numbers it is fine. But as soon as you multiply or divide by negative numbers, you have to swap the inequality. The addition and subtraction can be handled with easily, and the inequality can retain its direction.
• What if we have | x | = - 2 ? ... Is this even solvable?
• Absolute values are always positive, because the distance from zero to any number is never going to be negative. Since the absolute value of no number is -2, there is no solution.
• Are there different ways to do absolute values??
cause my teacher taught me a different way and I don't exactly understand the reasoning behind what she does??
• There is technically only one way to solve absolute values, which is to make the non variable side both negative and positive, but if you are talking about simplification, there are many ways to simplify an absolute value expression to solve it.
• Okay, I answered this question on my test the other day, and I know what the answer is, but I can't figure out how to get there. I get half of the inequality right, but I can't get the other half. Here's the question:
``|(1/2)x-1|+(3/2)x<10``

I got the absolute value isolated:
``|(1/2)x-1|<10-(3/2)x``

And rewrote the equation to get rid of the absolute value signs:
``-10+(3/2)x<(1/2)x-1<10-(3/2)x``

``-9+(3/2)x<(1/2)x<11-(3/2)x``

Multiply both sides by 2:
``-18+(6/2)x<x<22-(6/2)x``

Split the inequality into two parts:
`-18+3x<x` and `x<22-3x`
`-18<-2x` and `4x<22`
Divide by -2 and 4:
`9>x` and `x<11/2`
So I get the `x<11/2` right, but the other half is supposed to be `0<x` not `9>x`. Where'd I go wrong!?
• I understand this example for the -7 1/2 is less than h, which is less than 7 1/2. But the answer to a problem such as: n-3 (is less than) 6 would be -3(is less than) n (is less than) 9. How does that work if it's not just a negative version of the answer you first got? (Sorry if this question is unclear)
• how do you do the problems with...
{x|x|<5