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Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).

(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.

(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.

Re: |x + 2| = |y + 2| what is the value of x + y ? [#permalink]

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24 Jun 2014, 05:25

Bunuel wrote:

|x + 2| = |y + 2| what is the value of x + y ?

Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).

(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.

(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.

Answer: D.

Hope it's clear.

hi bunuel,

following is my interpretation of the this problem . Could you please tell me where i am wrong

Stmt 1: XY <0

Case 1: XY <0 (POSITIVE) (NEGATIve)

Case 2: XY < 0 (Negative) (Positive)

Both cases satisy to give X+Y = -4

Stmt 1 Sufficient .

STMT2: X > 2 AND Y < 2

From above statement X is always postive and Y may be negative (or) Positive.

case 1: |x+2| = |y+2|

(x+2) = (-y+2)

case 2: |x+2| = |Y+2|

(X+2) = (Y+2)

How can we get X+Y = -4 from here on

and also from your above solution, i am unable to figure out if x and y are not equal, how come there will be only one solution .

Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).

(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.

(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.

Answer: D.

Hope it's clear.

hi bunuel,

following is my interpretation of the this problem . Could you please tell me where i am wrong

Stmt 1: XY <0

Case 1: XY <0 (POSITIVE) (NEGATIve)

Case 2: XY < 0 (Negative) (Positive)

Both cases satisy to give X+Y = -4

Stmt 1 Sufficient .

STMT2: X > 2 AND Y < 2

From above statement X is always postive and Y may be negative (or) Positive.

case 1: |x+2| = |y+2|

(x+2) = (-y+2)

case 2: |x+2| = |Y+2|

(X+2) = (Y+2)

How can we get X+Y = -4 from here on

and also from your above solution, i am unable to figure out if x and y are not equal, how come there will be only one solution .

Ex : if x= 5 (5>2) and y = 1 (1<2)

x+y = 6 and there are many values for x+y .

Help is appreciated. Thanks in advance.

Your logic for the first statement is not clear. How did you derive that since x and y have the opposite signs x + y must be -4?

Please re-read the solution above. |x + 2| = |y + 2| is true in two cases: (i) \(x=y\) (ii) \(x+y=-4\)

Now, let me ask you a question: is it possible to have x=y case for either of the statements? The answer is NO. Hence we must have the second case which says that x+y=-4.
_________________

Re: |x + 2| = |y + 2| what is the value of x + y ? [#permalink]

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04 Oct 2015, 22:42

Bunuel wrote:

|x + 2| = |y + 2| what is the value of x + y ?

Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).

(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.

(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.

Answer: D.

Hope it's clear.

Hi Bunuel, This was my way of understanding.

|x+2| = |y+2| So we have 2 possibilities either both sides the same sign or different sign. a) Both same sign : x+2 = y+2----->x=y b) Diff sign : x+2 = -y-2------> x+y=-4

Anything wrong in this?

Even from your explanation how can we derive that xy<0? For this to happen x and y need to be of different signs. But it is not necessary that they are of opp sign. It can be possible that x=y=-2 which implies xy>0. What is that I am missing her. Help!

Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).

(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.

(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.

Answer: D.

Hope it's clear.

Hi Bunuel, This was my way of understanding.

|x+2| = |y+2| So we have 2 possibilities either both sides the same sign or different sign. a) Both same sign : x+2 = y+2----->x=y b) Diff sign : x+2 = -y-2------> x+y=-4

Anything wrong in this?

Even from your explanation how can we derive that xy<0? For this to happen x and y need to be of different signs. But it is not necessary that they are of opp sign. It can be possible that x=y=-2 which implies xy>0. What is that I am missing her. Help!

|x + 2| = |y + 2| is true in two cases: (i) \(x=y\) (ii) \(x+y=-4\)

Next, (1) says that xy < 0. Is it possible to have the first case from above? If x = y, then xy = x^2 >= 0, not < 0, thus the first case is out and we get \(x+y=-4\).
_________________

|x + 2| = |y + 2| what is the value of x + y ? [#permalink]

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06 Oct 2015, 05:19

Bunuel wrote:

|x + 2| = |y + 2| what is the value of x + y ?

Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).

(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.

(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.

Answer: D.

Hope it's clear.

Bunuel, When is it right to square both sides of absolute question? and it is wrong? I appreciate if you can elaborate with examples the correct and wrong if there are.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If |x+2|=|y+2|, what is the value of x+y?

1) xy<0 2) x>2 and y<2

In the original condition, it becomes x+2=±(y+2). x=y or x+y=-4 are derived from x+2=y+2 or x+2=-(y+2). Then, there are 2 variables(x,y) and 1 equation(|x+2|=|y+2|), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), since x≠y, x+y=-4, which is unique and sufficient. For 2), since x≠y, x+y=-4, which is unique and sufficient. Therefore, the answer is D.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Given: |x+2|=|y+2| which is true only if case 1: when x = y case 2: When x+y = -4 e.g. x=2 and y=-6 or x=3 and y=-7 etc

Question : x+y=?

Statement 1: xy<0 i.e. x is not equal to y i.e. Case 2 is implied i.e. x+y=-4 SUFFICIENT

Statement 2: x>2 and y<2 i.e. x is not equal to y i.e. Case 2 is implied i.e. x+y=-4 SUFFICIENT

Answer: Option D

Engr2012 : This question is available on GMAT CLUB already.
_________________

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Re: |x + 2| = |y + 2| what is the value of x + y ? [#permalink]

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14 Feb 2017, 23:58

Another probable way to arrive at the solution!

Option D

|X + 2| = |Y + 2|; X + Y = ?

I: XY < 0 i.e., X & Y have opposite signs : X<0 & Y>0 or X>0 & Y<0 : For X<0 & Y>0 : -(X+2) = (Y+2) : X + Y =-4 : For X>0 & Y<0 : (X+2) = -(Y+2) : X + Y =-4 Sufficient!

II: X > 2 & Y < 2 : X >2 & 0<Y<2 OR X>2 & Y<0 : For X >2 & 0<Y<2: (X+2) = -(Y+2) : X + Y =-4 : For X>2 & Y<0: (X+2) = -(Y+2) : X + Y =-4 Sufficient!
_________________

Re: |x + 2| = |y + 2| what is the value of x + y ? [#permalink]

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08 Sep 2017, 02:32

guys i think we can solve this question in much easier way such questions are straight forward questions we can follow algebraic method according to that either sie of mod can have four conditions but in this case only two conditions are possible i.e. either both positive (i am talking about whole modulus functionnot just x and y signs ) or one of them will be negative

we get x=y in case one and in case two we get x+y=-4 this means either x will be equal to y which is not possible in both STATEMENTS so we left with only one case that is two in both statments we can get the value of x+y=-4 irrespective of x and y values we need to find the value of (x+y)

note: wrote case and statments dont undrstand them as same

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