Tough absolute value DS

[doclkk]

Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

OA: C

How to rephrase [spoiler]- explanation ? : We can rephrase the question by opening up the absolute value sign. In other words, we must solve all possible scenarios for the inequality, remembering that the absolute value is always a positive value. The two scenarios for the inequality are as follows:

If x > 0, the question becomes “Is x < 1?”
If x < 0, the question becomes: “Is x > -1?”
We can also combine the questions: “Is -1 < x < 1?”[/spoiler]

Can someone explain to me how it is -1 < x < 1. I don't understand how they came up with that.

[Stuart@KaplanGMAT]

Is |x| < 1 ?

Can someone explain to me how it is -1 < x < 1. I don't understand how they came up with that.

Let's ask, when will |x| < 1?

For absolute value, there are going to be two solutions: one in which the bracketed term is positive, one in which it's negative. In this case:

+(x) < 1

or

-(x) < 1

Solving each one:

+(x) < 1
x < 1

-(x) < 1
(divide both sides by -1, which reverses the inequality:)
x > -1

So, in order for "|x| < 1", -1 < x < 1.

We could also have just thought it through. For absolute value, the sign inside the bracket is irrelevant - after simplifying and then performing the absolute value operation, whatever was inside becomes positive.

Therefore, any negative fraction, 0 or any positive fraction will have an absolute value of less than 1. Another way we could rephrase the question:

"Is x a negative fraction, positive fraction or 0?"

[doclkk]

Is |x| < 1 ?

Can someone explain to me how it is -1 < x < 1. I don't understand how they came up with that.

Let's ask, when will |x| < 1?

For absolute value, there are going to be two solutions: one in which the bracketed term is positive, one in which it's negative. In this case:

+(x) < 1

or

-(x) < 1

Solving each one:

+(x) < 1
x < 1

-(x) < 1
(divide both sides by -1, which reverses the inequality:)
x > -1

So, in order for "|x| < 1", -1 < x < 1.

We could also have just thought it through. For absolute value, the sign inside the bracket is irrelevant - after simplifying and then performing the absolute value operation, whatever was inside becomes positive.

Therefore, any negative fraction, 0 or any positive fraction will have an absolute value of less than 1. Another way we could rephrase the question:

"Is x a negative fraction, positive fraction or 0?"

Thanks for the explanation!

[hk]

Statement (1), says |x + 1| = 2|x – 1|

Removing the absolute value sign, results in 2 scenarios:

1. (x+1) = 2 (x-1) if x is positive
2. -(x+1) = -2 (x-1) => am i wrong here??? in this case i'm always getting a value of 3 and hence i got the answer as A, since (2) is anyways insufficient!!

Can somebody explain this to me!!!

Thanks

[abhinav85]

2. -(x+1) = -2 (x-1) => am i wrong here???

In the second case over here it will be

either -(x+1) = 2 (x-1) or

(x+1) = -2 (x-1).

Note:The negative can only be at one side not both the sides.

[mike22629]

I always like to use the number line

So first rephrase the information so the two units in the absolute value sign are subtracted from one another

So:

A)
l x - (-1)l = 2*lx-1l

What does this actually say?
That the distance between x and -1 is twice the distance between x and 1

Use number the determine which numbers qualify under these constraints

1st possibility
---------(-1)--------0----(x = 1/3)-----1

Notice the distance between x and -1 = 4/3 and the distance between x and 1 = 2/3 (Hence twice the distance so this qualifies)

2nd possibility
---------(-1)-------0--------1-------2-------(x=3)

Notice the distance between x and -1 = 4 and the distance between x and 1 = 2 (Hence twice the distance so this also qualifies)

So A is insufficient

B) lx-3l > 0, which can be rephrased to say that x DOES NOT equal 3

Hence Answer is C, because this eliminates one of the two possibilities and x must equal 1/3