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A kind of rule to remember is to search the roots of the expression in each absolute value. It's similar to what I said in my response to u on another topic

As we have |x-3|, so we search the roots of x-3. Hence,
x-3 = 0 <=> x = 3

Thus, the domains to consider are x > 3 and x < 3.

Each domain implies a simplified equation that removes the absolute.

We do not consider x > 0 or x < 0 because nothing in the equation asks for it

Re: Absolute value DS [#permalink]
22 Jun 2010, 16:41

WE have [x-3]=3-x [] is absolute value if x>= 3 we have x-3 = 3-x. Solve x-3 = 3-x by itself we have x = 3. Put togethere with x>= 3 we have x= 3

if x=<3 we have 3-x = 3-x. it if right with all x =<3 ==> the solution without any additional condition is x =< 3

Let use the condition 1 only we can have x> 3 doesn't belong to the area x=<3==> insufficient Let use the condition 2 only we have x< 0 belong to area x=< 3 ==> sufficient

Re: DS: Absolute Value [#permalink]
28 Jun 2010, 00:15

i. Equality in question is TRUE if x is negative and FALSE if x is positive. Dual solution, hence NOT SUFFICIENT ii. |x| always positive. For -x to be greater than 0, x has to be negative. Single solution, hence SUFFICIENT.

gmatclubot

Re: DS: Absolute Value
[#permalink]
28 Jun 2010, 00:15