Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0
I get easily tricked by absolute value type of problems. Can anyone explain easy way to get around such type of problems? Is the answer for the above problem C?
Aboslute values
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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.
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.
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thanks for the explaining... just a quick clarification: In the below step in 1.c, Actually -> x + 1 = 2x - 2 =>gives, x=3, which is consistent with the assumption. Hence statement 1 is not sufficient. Am I right to say that?
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.
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.