logan wrote:
hey Bunuel!! first i would like to thank you for posting such wonderful questions..
regarding a question that you posted above, i got a small doubt..
|x+2|=|y+2|
so lets say |x+2|=|y+2|=k (some 'k')
now |x+2|=k =====> x+2=+/- k
and x+2= +k, iff x>-2
x+2= -k, iff x<-2
also we have |y+2|=k =====> y+2=+/- k
and y+2= +k, iff y>-2
y+2= -k, iff y<-2
so x+2=y+2 ===> x=y , iff (x>-2 and y>-2) or (x<-2 and y<-2)--------eq1
and x+y=-4, iff (x<-2 and y>-2) or (x>-2 and y<-2)-------------------eq2
now coming to the options,
1) xy<0 i.e., (x=-ve and y=+ve) or (x=+ve and y=-ve)
(x=-ve and y=+ve): this also means that x and y can have values, x=-1 and y= some +ve value. so eq2 cannot be applied, x+y#-4. if x=-3 and y=some +ve value, x+y=-4. two cases. data insuff.
(x=+ve and y=-ve): this also means that x and y can have values, x=+ve value and y=-1.so eq2 cannot be applied, x+y#-4. if x=some +ve value and y=-3, x+y=-4. two cases. data insuff.
2)x>2,y>2 for this option too we cannot judge the value of x+y, with the limits of x and y being different in the question and the answer stem. so data insuff.
so i have a doubt that, why the answer cannot be E??
plz point out if i made any mistake..
Hi, Logan. Good way of thinking. Though I think that your solution is not correct.
Consider this:
We have |x+2|=|y+2|
(1) xy<0, hence x and y have opposite signs. Let's take x negative and y positive (obviously it doesn't matter which one we pick as equation is symmetric).
If y is positive RHS |y+2| will be positive as well and we can expend it as |y+2|=y+2.
Now for |x+2| we can have to cases:
A. -2<x<0 --> |x+2|=x+2=y+2 --> x=y. BUT: it's not valid solution as x and y have opposite signs and they can not be equal to each other.
B. x<=-2 --> |x+2|=-x-2=y+2 --> x+y=-4. Already clear and sufficient.
If we go one step further to see for which x and y is this solution is valid, we'll get:
As x+y=-4, x<=-2 and y>0 --> x must be <-4. If you substitute values of x<-4 you'll receive the values of y>0 and their sum will always be -4.
The same approach works for (2) as well.
Hope it's clear.
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