mix (4) 700+

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!

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 ?

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