 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# Solving "Less Than" Absolute Value Inequalities - Concept

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

Share

So now we're going to talk about Absolute Value inequalities where we're actually dealing with the Absolute Value being less than a number.
Okay so a nice example we're going to look at is the Absolute Value of x is less than 4, so just using some logic let's actually think of some numbers where this works. So if we think of a positive number anything less than 4 say x=2. Absolute Value of 2 is less than 4 that would work. What if we think about negative numbers? Remember the Absolute Value is going to make this negative number positive so if we say you know x=-3 the Absolute Value of that turns to positive 3 is less than 4 this will work, so whenever we're dealing with Absolute Values less than the way I like to remember this is we actually have to make this into two equations one being x is less than 4 and the other one is x is greater than -4 again the opposite sign and the opposite number.
The difference with this from when the Absolute Value is greater than, is this now turns into a intersection this turns into an and statement and or intersection. Where I remember that is Absolute Value is less than, than is similar to than you could turn it to then so less than less then and and statement so that the way I remember it, whatever works for you. Okay so we now have this in a intersection form whenever I solve these out I do a little number line. We have x is less than 4 and 4 having going down, x is greater than -4 so open circle and -4 then going up and we're looking for the intersection we're looking for where both these exists. On either extreme we just have one set in the middle is where they both exist so the intersection is actually right in the middle so we're dealing from -4 to 4 intersection where they both exist at these end points we're always missing one so we're dealing from -4 to 4 soft brackets, so we turned our Absolute Value less than statement into a intersection statement plot in a number line to find your answer.