Is |x| < 1

1) |x+1| = 2|x-1|
2) |x-3| > 0

Can someone post a detailed explanation for this question

Thanks

Since (2) is the easiest to deal with, we will see what that implies first. On the face of it what it says is that |x-3| can't be zero since the magnitude (absolute value) of any number is greater than or equal to zero. The only way for |x-3| to be zero is for x to equal 3. So (2) just says x is not 3. If x were 96, the magnitude of x would not be less than 1.

For (1) we can use the fact that, given equations of type (1) with absolute values of both sides of the equations of the equal sign, there is only the two cases to consider rather than the 'normal' four cases. That is if
|a| = |b|
there are four cases to consider but those four cases reduce to the two cases a = b and a = -b since the other two cases -a = b and -a = -b reduce to the same thing two equalities. We note that 2|x-1| is the same as |2(x-1)|. So (1), with a = x + 1 and b = 2 (x-1), says
(1a) x + 1 = 2 (x - 1)
or
(1b) x +1 = -2 (x - 1)
(1a) says x = 3 and (2a) says x = 1/3. If x were 3 the magnitude of x would not be less than 1. But what if (2) were also true?

Very close, John. I should point out that in order for (2) to be insufficient, noting 96 as an example of |x| < 1 failing is not enough; you must also be able to identify a case where |x| < 1 holds, for instance when x = 1/2. Remember, if |x| < 1 is NEVER true, that means the statement is just as sufficient as when |x| < 1 is ALWAYS true. Note that the same is true of your analysis for (1)..

_________________
Tim Sanders
Manhattan GMAT Instructor