Algebra II For Dummies. Sterling Mary Jane

> 4. Dividing by –2, you have x < –2.

      Solving 7 – 2x < –11, you subtract 7 to get –2x < –18. Dividing by –2, you have x > 9. The solution of the absolute value inequality is x < –2 or x > 9. In interval notation, you write the solution as

      

or

.

      

Don’t ever write the solution x < –2 or x > 9 as 9 < x < –2. If you do, you indicate that some numbers can be bigger than 9 and smaller than –2 at the same time. It just isn’t so.

       Exposing an impossible inequality imposter

      The rules for solving absolute value inequalities are relatively straightforward. You change the format of the inequality and solve for the values of the variable that work in the problem. Sometimes, however, amid the flurry of following the rules, an impossible situation works its way in to try to catch you off guard.

      For example, say you have to solve the absolute value inequality

. It doesn’t look like such a big deal; you just subtract 8 from each side and then divide each side by 2. The dividing value is positive, so you don’t reverse the sense. After performing the initial steps, you use the rule where you change from an absolute value inequality to an inequality with the variable term sandwiched between inequalities. So, what’s wrong with that? Here are the steps:

      Subtracting 8, you get

, and dividing by 2, you get

.

      Under the format –c < ax + b < c, the inequality looks curious. Do you sandwich the variable term between –1 and 1 or between 1 and –1 (the first number on the left, and the second number on the right)? It turns out that neither works. First of all, you can throw out the option of writing 1 < 3x – 7 < –1. Nothing is bigger than 1 and smaller than –1 at the same time. The other version seems, at first, to have possibilities, so you try to solve –1 < 3x – 7 < 1 by adding 7 to each interval, giving you 6 < 3x < 8. Dividing each interval by 3, you have

.