leonidas wrote:

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

When one solves this with 4 different conditions, one gets the value of x=3 and x=1/3. Is there a shorter way of identifying this equation and solving it only twice (instead of trying all 4 conditions) and get the two values?

I am trying to find some time saving strategies.

I tried the 4 conditions as:

(x+1) , (x-1)

-(x+1), -(x-1)

-(x+1), (x-1)

(x+1), -(x-1)

Hope I am making this clearer.

essentially these 4 conditions are in fact the same as 2 conditions:

1: if lx+1l and lx-1l both are +ve;

x+1 = 2x-2

x = 3

2: if lx+1l and lx-1l both are -ve;

-x-1 = -2x+2

x = 3

1 & 2 yeild the same value because if we change the side of 2, it becomes 1. lets do it again with if lx+1l and lx-1l both are -ve;

-x-1 = -2x+2

change the side

2x - 2 = x+1 ........ it is the same as 1.

x = 3

3: if lx+1l is +ve and lx-1l is -ve;

x+1 = -2x+2

x = 1/3

4: if lx+1l is -ve and lx-1l is +ve;

-x-1 = 2x-2

x = 1/3

the same applies to 3 and 4 too.

so do only twice i.e one for +ve and another for -ve.

_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html

Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

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