gmattester wrote:Is lxl< 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l &#8800; 0

Besides plugging values, can someone tell me another method how to solve absolute values questions.

There is another way, which I use on pretty much every absolute value GMAT question, but you'd want to practice it a bit before beginning to use it on a test. Remember that |x| is the distance between x and zero on the number line, and |x-y| is the distance between x and y on the number line. If you understand these two facts, you can change many problems about absolute value into problems about distances on the number line, and you can bypass the algebra altogether. Doing that here:

Is |x| < 1? --> Is x less than 1 away from zero?

1) lx + 1l = 2lx - 1l ---> lx - (-1)l = 2lx - 1l

In words this says "the distance between x and -1 is twice the distance between x and 1", or phrased more simply, "x is twice as far from -1 as x is from 1". If you draw a number line, you'll see there are two places x could be: between -1 and 1 (but closer to 1), or to the right of 1. It's not difficult to see that x = 3 is one of the two solutions. Insufficient.

2) tells us x is not 3. Insufficient.

Together, there is only one possible solution, the point between -1 and 1, so the answer is yes, |x| < 1, and the two statements are sufficient together. C.