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

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

(1) xy < 0

(2) x > 2 and y < 2

I don't have the official answer. Got this question off a video on youtube, i don't agree with the answer he gives so want your thoughts on this.
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Jun 2014, 13:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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

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New post 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 .

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.

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dheeraj24 wrote:
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 .

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.
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New post 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!

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

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New post 05 Oct 2015, 02:09
Johnjojop wrote:
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!


|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 \(x+y=-4\).
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|x + 2| = |y + 2| what is the value of x + y ? [#permalink]

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New post 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.

thanks for advance

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

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


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

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

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


* A solution will be posted in two days.



Given, \(|x+2| = |y+2|\)----> \(x+2 = \pm (y+2)\) ---> either x=y or x+y=-4

Per statement 1, xy<0 ---> x=y can not be true, leaving us with x+y=-4. Sufficient.

Per statement 2, x>2 and y<2 , again x=y can not be possible., leaving us with x+y=-4. Sufficient.

D is thus the correct answer.
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Re: If |x+2|=|y+2|, what is the value of x+y? [#permalink]

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New post 03 Feb 2016, 22:46
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.
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Re: If |x+2|=|y+2|, what is the value of x+y? [#permalink]

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New post 04 Feb 2016, 09:43
MathRevolution wrote:
If |x+2|=|y+2|, what is the value of x+y?

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


* A solution will be posted in two days.


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

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New post 06 Feb 2016, 12:11
Hi Engr2012

What am i missing in this?

|x+2| = |y+2| ==> Square both sides ==> x^2 + 4 + 4x = y^2 + 4 + 4y
==> x^2 - y^2 = -4x + 4y
==> (x-y)(x+y) = -4(x-y)
==> Cancelling x-y from both sides ==> x+y = -4

Please help me understand. Thanks in advance!


Engr2012 wrote:
MathRevolution wrote:
If |x+2|=|y+2|, what is the value of x+y?

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


* A solution will be posted in two days.



Given, \(|x+2| = |y+2|\)----> \(x+2 = \pm (y+2)\) ---> either x=y or x+y=-4

Per statement 1, xy<0 ---> x=y can not be true, leaving us with x+y=-4. Sufficient.

Per statement 2, x>2 and y<2 , again x=y can not be possible., leaving us with x+y=-4. Sufficient.

D is thus the correct answer.

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

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neeraj609 wrote:
Hi Engr2012

What am i missing in this?

|x+2| = |y+2| ==> Square both sides ==> x^2 + 4 + 4x = y^2 + 4 + 4y
==> x^2 - y^2 = -4x + 4y
==> (x-y)(x+y) = -4(x-y)
==> Cancelling x-y from both sides ==> x+y = -4

Please help me understand. Thanks in advance!


Engr2012 wrote:
MathRevolution wrote:
If |x+2|=|y+2|, what is the value of x+y?

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


* A solution will be posted in two days.



Given, \(|x+2| = |y+2|\)----> \(x+2 = \pm (y+2)\) ---> either x=y or x+y=-4

Per statement 1, xy<0 ---> x=y can not be true, leaving us with x+y=-4. Sufficient.

Per statement 2, x>2 and y<2 , again x=y can not be possible., leaving us with x+y=-4. Sufficient.

D is thus the correct answer.


You can not eliminate (x-y) unless categorically mentioned that x is not equal to y. When you cancel this factor, you are assuming that x= y.

This is one of the most common mistake people do in DS questions involving algebraic expressions.

Once you get (x-y)(x+y) = -4(x-y) ---> (x-y) (x+y+4)=0

So you get either x=y or x+y = -4

You need to use the given statements to eliminate one of the 2 above possible scenarios to evaluate sufficiency of the statements
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Re: |x + 2| = |y + 2| what is the value of x + y ? [#permalink]

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

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New post 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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Re: |x + 2| = |y + 2| what is the value of x + y ?   [#permalink] 08 Sep 2017, 02:32
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