Solving Absolute Value Inequalities
By S Taylor
You may recall that when solving an absolute value equation, you came up with two or more solutions. To review absolute value equations, click here
You may also recall that when solving a linear inequality, you came up with an interval rather than a single value for an answer. For more on solving linear inequalities, click here (linear inequalities.doc)
When solving absolute value inequalities, you are going to combine techniques used for solving absolute value equations as well as linear inequalities.
Think about the inequality |x| < 4. This means that whatever is in the absolute value symbols needs to be less than 4. So answers like 3, -3, 2, -2, 0, as well as many other possibilities will work. With so many possibilities, how do we go about finding them all?
With |x| < 4, any real number between -4 and 4 will make the inequality true. So we will set up the double inequality -4 < x < 4. In interval notation, this looks like
.
Suppose our inequality had been |x| > 4. In this case, we want the absolute value of x to be larger than 4, so obviously any number larger than 4 will work (5, 6, 7, etc.). But numbers such as -5, -6, -7 and so on will also work since the absolute value of all those numbers are positive and larger than 4. What we do in this situation is set up two separate inequalities and solve each one. For this problem, this will give us x < -4 and x > 4 which are already solved for x. In interval notation, we would have
In general, we will solve inequalities one of two ways depending on the type of problem.
- Given |expression|< k , set up a double inequality, -k<|expression|< k and solve.
- If the inequality has
instead of <, your procedure is still the same.
- Given |expression|> k , set up two separate inequalities, |expression|< -k or |expression|> k and solve each one.
- If the inequality has
instead of >, your procedure is still the same.
Let's Practice
- |x - 5| < 3
Set up the double inequality
and then solve.
In interval notation, the answer is
.
- |2x + 3|
8
This solution will involve setting up two separate inequalities and solving each.
orIn interval notation, the answer is
.
|5x| > 15
|x - 20|
4
|x - 7| < 6
|x + 12| + 3 > 17
|1 - 2x|
5
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