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Course: Algebra 1>Unit 2

Lesson 6: Compound inequalities

Double inequalities

Sal solves the double inequality -16≤3x+5≤20, which is the same as the compound inequality -16≤3x+5 AND 3x+5≤20. Created by Sal Khan and Monterey Institute for Technology and Education.

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• How do you know a statement is and/or?
• The statement will have an "or" in it if it is an or inequality. For "and" inequalities it will either be in the way it was in the video(-16 ≤ 3x + 5 ≤ 20) or say: -16 ≤ 3x + 5 and 3x + 5 ≤ 20.
• So I know you can separate the inequalities, but is there a way to answer −13 > −5x + 2 > −28 by leaving it together? What I mean is that when you divide by negative 5 three times, do you like switch both signs? Or is that not a thing, and you're supposed to just separate them?
• You would be dividing by a negative number, so you should switch both signs.
• I have this problem on my summer packet and I'm confused.
7-3x≥-5 and -2≤5-7x
I solved and I got:
X≤4 and x≤1
However, on my answer key only x≤1 is listed and graphed on a number line. What makes you use only one answer and how do you know when to check for this?
• I guess the "and" makes a difference. If x<=1 it is <= 4 as well, so the latter is not needed.

It's like saying "The car is black and it is not red". If you have stated that it is black, there is no need to say it is not red.
• so can you do word problems
• How do you know whether a double inequality will be AND or OR?
• Double inequalities are always AND. Sal explains this very early in the video (@) when he splits the double inequality into -16≤3x+5 AND 3x+5≤20. He tells you that "both" inequalities must be true. The is the basic definition of an AND compound inequalities.
• Does every problem have to be figured out with a number line. Doesn't that kinda complicate things in a way??
• it depends on what type of problem it is and what its asking
• how do I solve -2<x<or equal to 5
• It has already been solved. The solution set is all real numbers larger than -2 up to and including 5.
• what about a number outside of the absolute value in the front like 5|4-3x| (greater than or equal to) 30
• Hi Sabrina,
If you have a number outside of the absolute value sticks, get it to the other side of the inequality first then proceed. So in your example:
5|4 - 3x| >= 30
divide each side by 5 to get
|4 - 3x| >= 6
then finish the problem as you normally would.
|4 - 3x| + 5 >= 30
subtract the 5 from each side to get
|4 - 3x| >= 25
As I said above, the goal is to get the absolute value by itself on one side of the inequality and then proceed.
Hope that helps :-)
• Hi! I just have one small question! How would we solve a double inequality with an x variable on two or all of the sides? Is that possible or can x only be on one side (or the one in the middle) when it comes to double inequalities? I would really appreciate your help, because I'm curious and I can't seem to find an answer when I google the question. If you have by any chance found any explanation on the internet, I would also appreciate if you link it! Thank you so much! Please keep safe in these troubling times!