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
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
Subtracting 8, 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
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