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If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

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If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

A. 3
B. 4
C. 5
D. 6
E. 9
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Last edited by Bunuel on 14 Aug 2012, 00:31, edited 1 time in total.
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Re: Interesting Absolute value Problem [#permalink]

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I did it by brute force, considering all possibilities. But I'm sure someone can come up with a way to quickly identify the signs of x and y.

x + |x| + y = 7 ....1

x + |y| - y = 6 .....2

Consider x < 0
=> |x| = -x
Substitute in 1

x - x + y = 7
y = 7

Substitute in 2

x + 7 - 7 = 6
x = 6

But this violates x < 0 So our assumption was incorrect.

Consider x > 0 and y > 0
=> |x| = x and |y| = y

Substitute in 2

x + y - y = 6
x = 6

Substitute in 1

6 + 6 + y = 7
y = -5

This violates y > 0 so our assumption was incorrect.

Consider x > 0 and y < 0
=> |x| = x and |y| = -y

Substituting in 1 and 2:

2x + y = 7 .....3
x - 2y = 6 .....4

Solving we get x = 4 and y = -1
x + y = 3

Pick A.
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Re: Interesting Absolute value Problem [#permalink]

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Hussain15 wrote:
If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

A. 3
B. 4
C. 5
D. 6
E. 9


If \(x\leq{0}\), then \(x + |x| + y = 7\) becomes: \(x-x+y=7\) --> \(y=7>0\), but then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) --> hence \(x>0\).

Similarly if \(y\geq{0}\), then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), but then \(x + |x| + y = 7\) becomes: \(x+x+y=12+y=7\) --> \(y=-5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) --> hence \(y<0\).

So \(x>0\) and \(y<0\):
\(x+|x|+y=7\) becomes: \(x+x+y=7\) --> \(2x+y=7\);
\(x+|y|-y=6\) becomes: \(x-y-y=6\) --> \(x-2y=6\).

Solving: \(x=4\) and \(y=-1\) --> \(x+y=3\).

Answer: A.

I feel there is an easier way, but world cup makes it harder to concentrate.
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Re: Interesting Absolute value Problem [#permalink]

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New post 18 Jun 2010, 07:23
Bunuel wrote:
Hussain15 wrote:
If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

A. 3
B. 4
C. 5
D. 6
E. 9


If \(x\leq{0}\), then \(x + |x| + y = 7\) becomes: \(x-x+y=7\) --> \(y=7>0\), but then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) --> hence \(x>0\).

Similarly if \(y\geq{0}\), then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), but then \(x + |x| + y = 7\) becomes: \(x+x+y=12+y=7\) --> \(y=-5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) --> hence \(y<0\).

So \(x>0\) and \(y<0\):
\(x+|x|+y=7\) becomes: \(x-x+y=7\) --> \(2x+y=7\);
\(x+|y|-y=6\) becomes: \(x-y-y=6\) --> \(x-2y=6\).

Solving: \(x=4\) and \(y=-1\) --> \(x+y=3\).

Answer: A.

I feel there is an easier way, but world cup makes it harder to concentrate.



Why when y<0 do we get -2y?
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Re: Interesting Absolute value Problem [#permalink]

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New post 18 Jun 2010, 07:28
Expert's post
GMATBLACKBELT720 wrote:
Bunuel wrote:
Hussain15 wrote:
If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

A. 3
B. 4
C. 5
D. 6
E. 9


If \(x\leq{0}\), then \(x + |x| + y = 7\) becomes: \(x-x+y=7\) --> \(y=7>0\), but then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) --> hence \(x>0\).

Similarly if \(y\geq{0}\), then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), but then \(x + |x| + y = 7\) becomes: \(x+x+y=12+y=7\) --> \(y=-5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) --> hence \(y<0\).

So \(x>0\) and \(y<0\):
\(x+|x|+y=7\) becomes: \(x-x+y=7\) --> \(2x+y=7\);
\(x+|y|-y=6\) becomes: \(x-y-y=6\) --> \(x-2y=6\).

Solving: \(x=4\) and \(y=-1\) --> \(x+y=3\).

Answer: A.

I feel there is an easier way, but world cup makes it harder to concentrate.



Why when y<0 do we get -2y?


When \(y<0\), then \(|y|=-y\) and \(x+|y|-y=6\) becomes: \(x-y-y=6\) --> \(x-2y=6\).
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Re: Interesting Absolute value Problem [#permalink]

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New post 18 Jun 2010, 08:17
^
Wow... confusing as heck. Essentially saying that (hypothetical number here) |-3| = -(-3), thats fine. But I kept thinking you would apply this to -y and essentially make it +y since and make them both +y...
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New post 14 Jun 2011, 03:25
for x>0 and x < 0
2x+ y = 7 and y = 7

for y>0 and y<0
x=6 and x-2y=6

for x,y<0 solution exists.
solving x=4 and y = -1

3
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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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New post 02 Jul 2013, 06:38
If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

can be done in 2.3 mins :

there are 4 cases to be tested :
1) x is -ve and y is -ve
substituting in the equation , we get x-x+y=7 and x-y-y=6 solve for x and y we get x=20 and y=7 , so x+y=27 REJECT

2)x is +ve and y is +ve

substitute in the equation, we ger x+x+y=7 and x+y-y=6 solve for x and y we get x=6 and y=-5 ,therefore x+y=1 not on list so REJECT

3) x is -ve and y is +ve

substitute , we get x-x=y=7 and x+y-y=6 solve fo x and y we get x=6 and y=7, x+y=13 not on list so REJECT

4) x is +ve and y is -ve

substitute , we get x+x=y=7 and x-y-y=6 solve for x and y , we get x=4 and y= -1 ,x+y=3 , ANSWER CHOICE
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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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New post 02 Jul 2013, 11:22
I solved in 1.36 mins :)
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Re: Interesting Absolute value Problem [#permalink]

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New post 05 Jul 2013, 11:19
Bunuel wrote:
GMATBLACKBELT720 wrote:
Bunuel wrote:

If \(x\leq{0}\), then \(x + |x| + y = 7\) becomes: \(x-x+y=7\) --> \(y=7>0\), but then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), which contradicts the initial assumption \(x\leq{0}\). So \(x\) can not be \(\leq{0}\) --> hence \(x>0\).

Similarly if \(y\geq{0}\), then \(x + |y| - y =6\) becomes: \(x+y-y=6\) --> \(x=6>0\), but then \(x + |x| + y = 7\) becomes: \(x+x+y=12+y=7\) --> \(y=-5<0\), which contradicts the initial assumption \(y\geq{0}\). So \(y\) can not be \(\geq{0}\) --> hence \(y<0\).

So \(x>0\) and \(y<0\):
\(x+|x|+y=7\) becomes: \(x-x+y=7\) --> \(2x+y=7\);
\(x+|y|-y=6\) becomes: \(x-y-y=6\) --> \(x-2y=6\).

Solving: \(x=4\) and \(y=-1\) --> \(x+y=3\).

Answer: A.

I feel there is an easier way, but world cup makes it harder to concentrate.



Why when y<0 do we get -2y?


When \(y<0\), then \(|y|=-y\) and \(x+|y|-y=6\) becomes: \(x-y-y=6\) --> \(x-2y=6\).


Sorry, still confusing me.
I understand the first y i.e.
|y| = -y
But y < 0,
so wouldn't x + |y| - y = x - y - (-y), which would make it just x?
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New post 05 Jul 2013, 11:30
Expert's post
jjack0310 wrote:
Bunuel wrote:
GMATBLACKBELT720 wrote:

Why when y<0 do we get -2y?


When \(y<0\), then \(|y|=-y\) and \(x+|y|-y=6\) becomes: \(x-y-y=6\) --> \(x-2y=6\).


Sorry, still confusing me.
I understand the first y i.e.
|y| = -y
But y < 0,
so wouldn't x + |y| - y = x - y - (-y), which would make it just x?


When y<0, then |y|=-y: correct. But y must stay as it is.

Consider this, suppose we have only x-y=6, and I tell you that y is negative would you rewrite the equation as x-(-y)=6?

Hope it's clear.
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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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New post 09 Jul 2013, 16:24
If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

x>0, y>0

x + (x) + y = 7
2x+y=7

x + |y| - y = 6
x + y - y = 6
x=6

2x+y=7
2(6)+y=7
y=-5

INVALID as 6>0 but -5 is not > 0

x>0, y<0

x + (x) + y = 7
2x+y=7

x + |y| - y = 6
x + (-y) - y = 6
x - 2y = 6
x = 6+2y

2(6+2y) + y = 7
12+4y+y=7
12+5y=7
5y+-5
y=-1

2x+y=7
2x + (-1) = 7
2x - 1 = 7
2x=8
x=4

VALID as 4>0 and -1<0

therefore;

x+y = (4)+(-1) = 3

(A)
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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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Hussain15 wrote:
If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

A. 3
B. 4
C. 5
D. 6
E. 9



In such a question, you can use some brute force and get to the answer too. How long it takes depends on how quickly you observe the little things.

x + |x| + y = 7
x + |y| - y =6

Both equations yield about the same result though in one y is positive and in the other it is negative. |x| and |y| are positive and assuming x is positive, a negative y would pull down the first equation and pump up the second one to give almost equal values. The difference is very small also signifies that the negative variable might have a very small value.
Since the options give the value of x + y as 3/4/5... etc, it is likely that we are dealing with small number pairs such as (4, 2), (3, 2), (4, 1) etc.
Since the first equation has 7 as the result, both variables will not be even.
A couple of quick iterations brought me to (4, -1).
So x + y = 3
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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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New post 24 Oct 2015, 07:05
VeritasPrepKarishma wrote:
Hussain15 wrote:
If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

A. 3
B. 4
C. 5
D. 6
E. 9



In such a question, you can use some brute force and get to the answer too. How long it takes depends on how quickly you observe the little things.

x + |x| + y = 7
x + |y| - y =6

Both equations yield about the same result though in one y is positive and in the other it is negative. |x| and |y| are positive and assuming x is positive, a negative y would pull down the first equation and pump up the second one to give almost equal values. The difference is very small also signifies that the negative variable might have a very small value.
Since the options give the value of x + y as 3/4/5... etc, it is likely that we are dealing with small number pairs such as (4, 2), (3, 2), (4, 1) etc.
Since the first equation has 7 as the result, both variables will not be even.
A couple of quick iterations brought me to (4, -1).
So x + y = 3


Hi Karishma,

I tried another approach, but got stuck - can you hep me solve by this method?
x + |x| + y = 7 => x + y = 7 - |x| => Whatever be the value of x, |x| will be always non-negative => x + y has to be less than 7 => This eliminates option E
x + |y| - y = 6 => Now, if y is positive, then this equation becomes x = 6 => on substituting in above eqn, we get y = -5 => x+y = 1 => Not present in any of the options => y is negative => x - y - y = 6 => x - 2y = 6 => x = 6 + 2y
Now x + y = 7 - |6 + 2y|
If we take y = -1, we get x + y = 3 = Option A
If we take y = -2, we get x + y = 5 = Option C
I am getting both options here - where am I going wrong here? Or this approach incorrect?
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If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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sagar2911 wrote:


Hi Karishma,

I tried another approach, but got stuck - can you hep me solve by this method?
x + |x| + y = 7 => x + y = 7 - |x| => Whatever be the value of x, |x| will be always non-negative => x + y has to be less than 7 => This eliminates option E
x + |y| - y = 6 => Now, if y is positive, then this equation becomes x = 6 => on substituting in above eqn, we get y = -5 => x+y = 1 => Not present in any of the options => y is negative => x - y - y = 6 => x - 2y = 6 => x = 6 + 2y
Now x + y = 7 - |6 + 2y|
If we take y = -1, we get x + y = 3 = Option A
If we take y = -2, we get x + y = 5 = Option C
I am getting both options here - where am I going wrong here? Or this approach incorrect?


Let me try to answer.

You are making a mistake with using |6+2y|

After you get x=6+2y and substitute in equation (1), you get 6+2y+|6+2y|+y=7 ---->6+3y+|6+2y| = 7 ---> 3y+|6+2y| = 1. Now the point to note is that the 'nature' of |6+2y| changes at y=-3 and thus you need to evaluate |6+2y| for values smaller than -3 and for values greater than -3.

You have already assumed that y<0 in order to get x=6+2y, so your ranges to consider become -3 \(\le\)q y < 0 and y < -3

Case 1: -3 \(\le\) y < 0, giving you |6+2y| \(\geq\) 0 ----> 3y+|6+2y| = 1 ---> 3y+6+2y = 1 ---> 5y=-5 ---> y=-1. Acceptable value giving you x=6+2y = 4 --> x+y = 3

Case 2: y<-3 ---> |6+2y| = -(6+2y) ---> 3y+|6+2y| = 1 ---> 3y-6-2y = 1 ---> y=7 this contradicts the assumption that y<-3 , making this out of scope.

Thus the only value of x+y = 3.
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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

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New post 24 Oct 2015, 09:52
Engr2012 wrote:
sagar2911 wrote:


Hi Karishma,

I tried another approach, but got stuck - can you hep me solve by this method?
x + |x| + y = 7 => x + y = 7 - |x| => Whatever be the value of x, |x| will be always non-negative => x + y has to be less than 7 => This eliminates option E
x + |y| - y = 6 => Now, if y is positive, then this equation becomes x = 6 => on substituting in above eqn, we get y = -5 => x+y = 1 => Not present in any of the options => y is negative => x - y - y = 6 => x - 2y = 6 => x = 6 + 2y
Now x + y = 7 - |6 + 2y|
If we take y = -1, we get x + y = 3 = Option A
If we take y = -2, we get x + y = 5 = Option C
I am getting both options here - where am I going wrong here? Or this approach incorrect?


Let me try to answer.

You are making a mistake with using |6+2y|

After you get x=6+2y and substitute in equation (1), you get 6+2y+|6+2y|+y=7 ---->6+3y+|6+2y| = 7 ---> 3y+|6+2y| = 1. Now the point to note is that the 'nature' of |6+2y| changes at y=-3 and thus you need to evaluate |6+2y| for values smaller than -3 and for values greater than -3.

You have already assumed that y<0 in order to get x=6+2y, so your ranges to consider become -3 \(\le\)q y < 0 and y < -3

Case 1: -3 \(\le\) y < 0, giving you |6+2y| \(\geq\) 0 ----> 3y+|6+2y| = 1 ---> 3y+6+2y = 1 ---> 5y=-5 ---> y=-1. Acceptable value giving you x=6+2y = 4 --> x+y = 3

Case 2: y<-3 ---> |6+2y| = -(6+2y) ---> 3y+|6+2y| = 1 ---> 3y-6-2y = 1 ---> y=7 this contradicts the assumption that y<-3 , making this out of scope.

Thus the only value of x+y = 3.


Excellent. Thank you dude! :)
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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y =   [#permalink] 24 Oct 2015, 09:52
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