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by Anurag@Gurome » Thu Jan 06, 2011 1:30 am
Night reader wrote:If |x+3|+|2y+6| ≤ 0, find x+y

A.-3
B. 0
C.-6
D. 6
E. 5
Note that both |x + 3| and |2y + 6| are non-negative number. Thus sum of these two can never be less than zero. Hence their sum is zero. Which is possible only if both of them are individually equal to zero.

Hence, |x + 3| = 0 and |2y + 6| = 0
=> x = -3 and y = -3

Hence (x + y) = -6

The correct answer is 6.
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by Night reader » Thu Jan 06, 2011 1:39 am
Thanks Anurag, from my solution I arrived to |x+3|+|2y+6| =< 0 is equivalent to |x+3+2y+6| =<0 which is x+3+2y+6 =<0 and -x-3-2y-6 =<0 OR x+3+2y+6 >= 0, Combining both we get solution area x+3+2y+6 = 0
Now x+2y+9 = 0 and I continued by testing all answer choices and plugging back the values of x and y into the original mode statement |x+3|+|2y+6| =< 0 , this way arrived to answer C (-6). I know this is tedious, and your method is as much useful as all math, thanks again!
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by gmat7202011 » Thu Jan 06, 2011 3:20 pm
Hi Anurag,

Your explanation :-

"
Hence (x + y) = -6

The correct answer is 6. " confused me a little, I got the answer as -6 so that is C

Is that correct ?
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by anshumishra » Thu Jan 06, 2011 5:16 pm
Night reader wrote:If |x+3|+|2y+6| =< 0, find x+y

A.-3
B. 0
C.-6
D. 6
E. 5
Here is an alternative way to look at this problem : |x-a| = d => x is at "d" distance from the point a.
Here we have : |x+3| +|2y+6| <=0
=> |x-(-3)| + 2 |y-(-3)| <=0
Now these two represent two distances; distance b/w two points can never be -ve. So, the only thing that is possible is, the distance between them {x and -3 AND y and -3} is 0. That means x and y are -3.

Hence; x+y = -6. C
Thanks
Anshu

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