This is a good question. Absolute values is a problem area for a lot of students.

Lets look at (1). It tells us that x/ |x| < x. Since |x| is positive we can multiply both sides by it and get x < |x|x. This does not tell us much because if x < 0 and we divide both sides by x we get |x| < 1 (we reverse the direction of the < when we divide by a negative number) However, if x >0 and we divide both sides by it we have |x| > 1. (1) is inconclusive. Now, lets look at (2). Suppose x >0, then x = |x|. However, if x < 0, then x < |x|. In other words, x must be a negative number. Again, the result is inconclusive.

Combining the two statements: Because x < 0 (according to (2)), then |x| < 1 (according to (1)).

Hey steve. I initally got (C) too. But then I overanalyzed and started substituting different values of x in the combines situation. what will happen when x = -0.5?

Yes, it still works. The question is a yes or no question. We want to know if |x| < 1 assuming that its not zero. If we use x= -0.5, and plug it in to our two statements we have the following:

(1) -.05 / |-.05| = -1 < -.05 (remember that in the negative world the less negative you are the bigger you are)

(2) |-.05| > -.05 (this is certainly true, positives are always bigger than negatives)

The value x= -.05 satisfies both (1) and (2) and it answers our question.....yes, when x = -.05 it satisfies the inequality |-.05| < 1.

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