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

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.

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 \(x+y=-4\).

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

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.

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!

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

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!

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

Open Modulus (x+2) = (y+2) or (x+2) = -(y+2)


(1) xY < 0 means x and y have opposite signs so we must have the second case: x+2= -(y+2) and x+y = -4

suff

(2) x>2 and y<2. This can only be true if x and y have opposite signs, so we have case 2: x+y = -4

suff

OA is D

High quality question. It can be overly complicated, or incredibly simple depending whether you catch the trick or not.

Statement 1- Curve passes through 2nd and 4th Quadrant
We can see from the graph that only y=-x-4 Passes through those quadrant
That gives the sum of x+y=-4

Sufficient

Statement 2-
x>2 and y<2
we have to check for the curve that lies in 4 th quadrant
Again, only y=-x-4 passes through those quadrant


Sufficient.

Given: |x + 2| = |y + 2|
Key property: If |a| = |b|, then EITHER a = b OR a = -b
So, if |x + 2| = |y + 2|, then there are two possible cases:

case i: x + 2 = y + 2
Subtract 2 from both sides of the equation to get: x = y

case ii: x + 2 = -(y + 2)
Simplify: x + 2 = -y - 2
Subtract 2 from both sides of the equation to get: x = -y - 4
Add y to both sides to get: x + y = -4

Target question: What is the value of x + y?

Statement 1: xy < 0
If the product xy is negative, we can conclude that one of the values (x or y) is POSITIVE, and the other value is NEGATIVE.
This means that x cannot equal y, which means case i cannot be true, which means case ii (x + y = -4) MUST BE TRUE
The answer to the target question is x + y = -4
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x > 2 and y < 2
Combine inequalities to get: y < 2 < x
This means that x cannot equal y, which means case i cannot be true, which means case ii (x + y = -4) MUST BE TRUE
The answer to the target question is x + y = -4
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

@experts from the highlighted part would this also be an inference: x is equal to y (x=y) AND x+y is also equal to -4; which means 2y=-4 and y=-2

Hello Bunuel

I tried searching for this question before that I posted it as a new topic and it turned out, as you pointed it, that the question already existed in the questions bank. Is there any method to help me search efficiently for questions? Thank you.

You can use the search box to see if a problem has been previously discussed. However, locating certain questions, particularly algebra/inequality ones, can sometimes be challenging. Don't worry about reposting a question that has been posed before. If moderators discover previous discussions, they'll merge the topics, copy the solution from there, or provide a link to it.