1. I like to break my absolute value equations into cases, depending on the sign. Please keep in mind that:
If x is negative, then |x| = -x. Example: |-1| = -(-1) = 1.
If x is positive or equal to zero, then |x| = x. Example: |4| = 4.
You've got three cases with this one.
a. x < - 1 means that x + 1 < 0, so |x + 1| = -(x + 1). Since x < -1, then x - 1 will also be smaller than zero, so |x - 1| = -(x - 1).
This means that:
-x - 1 = -2x + 2
x = 3.
However, you need to back check: x = 3 is not consistent with your initial assumption that x < -1, so this one is out.
b. x is between -1 and 1. This means that x + 1 has turned positive, so |x + 1| = x + 1. However, x - 1 is still negative, so |x - 1| = -(x - 1).
You get that:
x + 1 = -2x + 2
3x = 1
x = 1/3 - is consistent with x between -1 and 1, so this one's a winner.
c. x is greater than 1, when all of them are positive. This means that:
x + 1 = 2x - 2
x = -1 - not consistent with x greater than 1, so this one's out too.
You've got only one solution for this one, and that's x = 1/3. For this one, |x| < 1. So 1 is sufficient.
2. Absolute values are always positive, with only one exception: the absolute value of zero is 0. This means that for all values except 0, |x - 3| > 0. So 2 is insufficient.
Answer A.
