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y>=0 (1) |x-3|>= y, x can be any number >=3, x<=3 Not suff 2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 Sufficinet

If y>=0 specified in the stem. Further mod of anything is positive. So, 0>=y according to 1. both y>=0 or y<=0, means y=0. So mod(x-3)>=0. Then x-3>=0 or x-3<=0 So, x>=3 or x<=3. Thus, x=3. Same thing for 2. So, D.

gmate2010 wrote:

y >= 0

1.) |x-3| >= y

=> -y >= (x-3) >= y

=> 3-y >= x >= 3+y

Now, if y = 0 , then, 3 >= x >= 3 => x can only be 3 but, if y = 1 , then, 2 >= x >= 4 => which is not possible for any value of x.

=> x can only be 3 when y = 0 ..sufficient..

2.) |x-3| <= -y

=> y <= (x-3) <= -y

=> 3+y <= x <= 3-y

when y = 0 ; then, 3 <= x <= 3 => x can only be 3.. when y = 1 ; then 4 <= x <= 2, which is not possible for any value of x.

O.K. as I see there has been kind of confusion about this question. I'm pretty sure answer is B.

I looked through the logic of people who think answer should be D and here is were they are making mistake:

(1) First of all: To check |x - 3|>=0 is sufficient or not just plug numbers A. x=5 y=1>0 and B. x=8 y=2>0; you'll see that both fits in |x - 3|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT.

|x - 3|>=y means that: x - 3>=y>=0 when x-3>0 --> x>3 OR (not and) -x+3>=y>=0 when x-3<0 --> x<3 Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT.

(2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT

In other words: (-y) is 0 or less, and the absolute value (|x-3|) must be at (0) or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0.

O.K. as I see there has been kind of confusion about this question. I'm pretty sure answer is B.

I looked through the logic of people who think answer should be D and here is were they are making mistake:

(1) First of all: To check |x - 3|>=0 is sufficient or not just plug numbers A. x=5 y=1>0 and B. x=8 y=2>0; you'll see that both fits in |x - 3|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT.

|x - 3|>=y means that: x - 3>=y>=0 when x-3>0 --> x>3 OR (not and) -x+3>=y>=0 when x-3<0 --> x<3 Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT.

(2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT

In other words: (-y) is 0 or less, and the absolute value (|x-3|) must be at (0) or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0.

So answer is B.

Got it finally! Thank you for clear explanation, Bunuel. B it is. Of course, A is not suff... _________________

Re: If y >= 0, what is the value of x? [#permalink]
05 Feb 2014, 15:24

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