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Re: Interesting Absolute value Problem [#permalink]
13 Jun 2010, 06:01

9

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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. _________________

Re: Interesting Absolute value Problem [#permalink]
18 Jun 2010, 06:23

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.

Re: Interesting Absolute value Problem [#permalink]
18 Jun 2010, 06:28

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: \(x-y-y=6\) --> \(x-2y=6\). _________________

Re: Interesting Absolute value Problem [#permalink]
18 Jun 2010, 07:17

^ 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...

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
02 Jul 2013, 05:38

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

Re: Interesting Absolute value Problem [#permalink]
05 Jul 2013, 10:19

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?

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
03 Sep 2014, 10:18

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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
27 Sep 2015, 14:44

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
27 Sep 2015, 23:10

1

This post received KUDOS

Expert's post

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 _________________

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
24 Oct 2015, 06:05

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? _________________

If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
24 Oct 2015, 06:55

2

This post received KUDOS

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.

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
24 Oct 2015, 08:52

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! _________________

+1 Kudos if you liked my post! Thank you!

gmatclubot

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y =
[#permalink]
24 Oct 2015, 08:52

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