Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
ghong14
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DS: Is |x| < 1 ?

by ghong14 Sun Sep 18, 2011 5:27 pm

Is |x| < 1 ?

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

(2) |x - 3| > 0

The explanation states that:

2. If -1 < x < 1, the value inside the absolute value symbols on the left side of the equation is positive, but the value on the right side of the equation is negative. Thus, only the value on the right side of the equation must be multiplied by -1:

|x + 1| = 2|x -1| x + 1 = 2(1 - x) x = 1/3


Why do we even have to consider the -1<x<1? I thought when we are presented with a absolute value equation we only had to worry about when X is negative or positive. I was able to solve everything but never thought to look for x=1/3. How do I determine in the future that I would need to consider when -1<X<1?

Thanks
StaceyKoprince
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Re: DS: Is |x| < 1 ?

by StaceyKoprince Fri Nov 11, 2011 10:02 pm

First, this is an extremely difficult question - so don't stress about it too much. There's a good chance you'd never have to do this on the real test. :)

Basically, there are three possibilities in general when you have two inequalities. The stuff inside both inequalities is positive, the stuff inside both is negative, or the stuff inside one is positive and the other is negative.

|x + 1| = 2|x - 1|

The key is what happens when you add 1 to x or subtract 1 from x, because the equation has x+1 and x-1.

What would need to be true of x for both quantities in the absolute values to be positive? Look at x-1: x would have to be bigger than 1 to make that positive, and then that would also make x+1 positive. So one scenario is when x > 1. (And that's scenario 3 in the explanation.)

What would need to be true of x for both quantities in the absolute values to be negative? Look at x+1: x would have to be smaller than -1, and that would also make x-1 negative. (That's scenario 1 in the explanation.)

But don't stop there - what would have to be true to make one of these pos and one neg? Look at what you've got so far: x > 1 and x < -1. What's left? x is between -1 and 1. Try 0. Ah, yes, that would make x+1 pos, but x-1 neg. Bingo, that's our third range (and scenario 2 in the explanation).

Again, you almost certainly will never have to do that on the real test!
Stacey Koprince
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