 ###### 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!

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# Solving "Less Than" Absolute Value Inequalities - Problem 1

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!

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Solving an absolute value inequality when the absolute value is less than another number. You first need to turn this into two statements; absolute value is less than 14. The 3x minus 1 less than 14, that will stay itself. And the other equation what we want to turn it into is the, absolute value stays the same. Now we want to flip the sign and take the opposite of this number.

Whenever we are dealing with the absolute value being less than, this turns into an end statement. This turns into an intersection. So what we are actually doing is taking the intersection between these two things.

We now have a intersection of two linear inequalities, solve them out as you normally would. So for this one, add one over 3x is less than or equal to 15, divide by 3, x is less than or equals to 5. Same idea over here, add one over 3x is greater than equal to -13. Divide by 3, x is greater than equal to -13 over 3.

To now sort of figure out where this intersection is we want to draw a number line. A number line x is less than or equal to 5. Solid circle because it's equal to and that is going to be number smaller. -13/3 is going to be on a number line and here we are shading up another closed circle. And we are dealing with the intersection, so we are dealing with where both of these exist.

On either extreme, there is just one that’s represented. In the middle, they are both there. The middle is where their intersection is going to be. Writing this in interval notation, hard brackets because we are equal two endpoints. -13/3 to 5, hard brackets.

So the main trick is to remember that when you are dealing with an absolute value less than, make your two equations and you are dealing with the intersection of those two inequalities.