## Absolute Value Equations & Inequalities: Exceptions (part 2 of 3)

This is the second part of a 3-part series on absolute value equations and inequalities. In this article I will talk about the exceptions to our strategies. I will address what to do when an absolute value that has been isolated is being compared to a negative number.

Case 1 – Absolute Value = Negative #

Suppose you are trying to solve the equation $|x-7|=-5$. No matter what number you plug in for x, $|x-7|$ will always be nonnegative. (If $x=7$, then the absolute value will equal 0, otherwise it will be a positive number.) So, can a nonnegative number ever equal a negative number? The answer is no, so this equation has no solution: $\emptyset$

Case 2 – Absolute Value < Negative #

Suppose you are trying to solve the inequality $|3x+2|<-9$. Again, $|3x+2|$ will always be nonnegative. So, can a nonnegative number ever be less than a negative number? The answer is no, so this inequality also has no solution: $\emptyset$

Case 3 – Absolute Value > Negative #

Here’s where many students get into trouble. We get to the point where we automatically think “No Solution” whenever we see an absolute value being compared to a negative number. But this is not true in this particular situation. Suppose you are trying to solve the inequality $|2x-5|>-13$. We now know that $|2x-5|$ will always be nonnegative. So, can a nonnegative number ever be greater than a negative number? Yes it can. As a matter of fact, it must  be greater than a negative number. So, no matter what number we substitute for x, the inequality will be true. The solution set to this inequality is the set of all real numbers, or $\mathbb{R}$.

Keeping It Straight

One way to keep these 3 cases straight is through memorization. You could make up a note card for each case and shuffle through them on a regular basis until you know which cases have no solution and which case has the set of all real numbers as its solution. You could also try to memorize which case has the set of all real numbers as its solution, knowing that the other two cases have no solution.

What I would recommend, however, is knowing why each case turns out the way it does. Then you will be able to reason your way through the problems you encounter.

In the next blog I will go over the typical stumbling blocks my students face with these problems..

– George

I am a math instructor at College of the Sequoias in Visalia, CA. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website – http://georgewoodbury.com.