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Hehe you'll find this funny. Decided I had enough of gmat shortcuts and dusted off my high school algebra book. Anyway there is a theorem regarding this type of problems

it gave the example |2x-3|=|7-3x| According to the book you can simplify it by simply:
2x-3=7-3x or 2x-3=-(7-3x)

And the OA is C but OE wasnt given downloaded one of the paper base gmat tests from this site only gave the answer key

in this problem 4 cases exist
a.RHS +ve ,LHS +ve
b.RHS -ve,LHS +ve
c.LHS +ve,RHS -ve
d.RHS -ve,LHS -ve

we can observe a and d are alike and similarly,b and c are alike
therefore,
we get (x+1)=2(x+1) or (x+1)= -2(x+1),
=>x+1=2x+2 or x+1= -2x-2
=>x= -1 or x= -1
hence,we can say A is sufficient

Re: Inequality with absolute value DS [#permalink]

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22 Aug 2006, 04:40

Guys sorry there is a typographical error. Question should read

Is |x|<1?

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

2) |x-3| not equal to zero
My fault the author of the test disabled the ability to cut and paste so I have to retype.
So from here C will stand.

And I really researched that particular theorem I was talking about. When face with two absolute values on either side of the equation. You dont need to break it down into its four scenarios. I did that as well and it took me too much time, which prompted me to research the topic. Thanks for the replies

for problem 2,
stmt 1 gives us 2 values of x -> 3, 1/3. thus we can't say if |x|<1
stmt 2 tells us that |x-3| <> 0, for which x can take any value other than 3.
combining these 2 stmts we can say that since x cannot be 3, it has to be 1/3 and |x| < 1 is true.
hence (C).
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