###### 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 - Problem 2

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

Solving an absolute value inequality where we have a lot more going on. First thing we always want to do, when we are solving any sort of equation, or any inequality like this, is to get whatever the main pieces by itself. So in this case we are dealing with an absolute value. First thing we want to do is get that by itself.

So first, add over 9, the 3 and the absolute values stay the same, and we get less than 24. Divide by 3, 2x plus 4 is less than 8. So we now have an absolute value less than a number. And here we need to make two different equations. First being 2x plus 4 is less than 8, that’s the easy one, and we also need to remember our other equation which is the inside the absolute value stays the same. Now we need to make sure we flip the sign, and take the opposite of the number itself.

And whenever we are dealing with the absolute value less than, this becomes an end statement, this becomes an intersection. Once we have this, it’s like solving any other intersection of two inequalities. Solve for x independently on each. So for this one subtract 4 over, 2x is less than 4, divide by 2. X is less than 2. Same thing over here, subtract 4 over, divide by 2, x is greater than 6. -6 I think I did something wrong didn’t I? Yes negatives just don’t disappear so this is a -6. We are still dealing with the intersection. Draw a number line to see where these two things over lap.

We have a 2, x is less than 2, not equal to, so we are going to have a open circle shading down. x is greater than -6. -6 not included so open circle shading up. And we are looking for the intersection, we are looking for both of these our present. And either end we just have 1 and the middle is where they overlap. So our answer then is from (-6 2) because we are not including those two endpoints.

So to solve this out; first thing we did was isolate our absolute value, made our two equations. We knew that it was intersection because it is a less than statement.