It is currently 17 Nov 2017, 20:22

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Retired Moderator
User avatar
Status: The last round
Joined: 18 Jun 2009
Posts: 1286

Kudos [?]: 1236 [0], given: 157

Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
GMAT ToolKit User
If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

Show Tags

New post 13 Jun 2010, 05:41
32
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

62% (02:22) correct 38% (02:32) wrong based on 558 sessions

HideShow timer Statistics

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

A. 3
B. 4
C. 5
D. 6
E. 9
[Reveal] Spoiler: OA

_________________

[ From 470 to 680-My Story ] [ My Last Month Before Test ]
[ GMAT Prep Analysis Tool ] [ US. Business School Dashboard ] [ Int. Business School Dashboard ]

I Can, I Will

GMAT Club Premium Membership - big benefits and savings


Last edited by Bunuel on 14 Aug 2012, 00:31, edited 1 time in total.
Edited the question.

Kudos [?]: 1236 [0], given: 157

3 KUDOS received
Manager
Manager
avatar
Joined: 04 May 2010
Posts: 86

Kudos [?]: 131 [3], given: 7

WE 1: 2 yrs - Oilfield Service
Reviews Badge
Re: Interesting Absolute value Problem [#permalink]

Show Tags

New post 13 Jun 2010, 06:42
3
This post received
KUDOS
1
This post was
BOOKMARKED
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.

Kudos [?]: 131 [3], given: 7

Expert Post
11 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42248

Kudos [?]: 132522 [11], given: 12324

Re: Interesting Absolute value Problem [#permalink]

Show Tags

New post 13 Jun 2010, 07:01
11
This post received
KUDOS
Expert's post
4
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 132522 [11], given: 12324

1 KUDOS received
Intern
Intern
User avatar
Affiliations: NYSSA
Joined: 07 Jun 2010
Posts: 33

Kudos [?]: 12 [1], given: 2

Location: New York City
Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program)
WE 1: Senior Associate - Thomson Reuters
WE 2: Analyst - TIAA CREF
Re: Interesting Absolute value Problem [#permalink]

Show Tags

New post 18 Jun 2010, 07:23
1
This post received
KUDOS
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?

Kudos [?]: 12 [1], given: 2

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42248

Kudos [?]: 132522 [0], given: 12324

Re: Interesting Absolute value Problem [#permalink]

Show Tags

New post 18 Jun 2010, 07:28
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: \(x-y-y=6\) --> \(x-2y=6\).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 132522 [0], given: 12324

Intern
Intern
User avatar
Affiliations: NYSSA
Joined: 07 Jun 2010
Posts: 33

Kudos [?]: 12 [0], given: 2

Location: New York City
Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program)
WE 1: Senior Associate - Thomson Reuters
WE 2: Analyst - TIAA CREF
Re: Interesting Absolute value Problem [#permalink]

Show Tags

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...

Kudos [?]: 12 [0], given: 2

VP
VP
avatar
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1283

Kudos [?]: 286 [0], given: 10

Re: Interesting Absolute value Problem [#permalink]

Show Tags

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
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Kudos [?]: 286 [0], given: 10

Intern
Intern
avatar
Joined: 09 Nov 2011
Posts: 1

Kudos [?]: [0], given: 1

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

Show Tags

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

Kudos [?]: [0], given: 1

Manager
Manager
avatar
Joined: 14 Jun 2011
Posts: 84

Kudos [?]: 50 [0], given: 15

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

Show Tags

New post 02 Jul 2013, 11:22
I solved in 1.36 mins :)
_________________

Kudos always encourages me

Kudos [?]: 50 [0], given: 15

Intern
Intern
avatar
Joined: 04 May 2013
Posts: 47

Kudos [?]: 9 [0], given: 7

Re: Interesting Absolute value Problem [#permalink]

Show Tags

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?

Kudos [?]: 9 [0], given: 7

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42248

Kudos [?]: 132522 [0], given: 12324

Re: Interesting Absolute value Problem [#permalink]

Show Tags

New post 05 Jul 2013, 11:30
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 132522 [0], given: 12324

Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 460

Kudos [?]: 201 [0], given: 134

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

Show Tags

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)

Kudos [?]: 201 [0], given: 134

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 15708

Kudos [?]: 281 [0], given: 0

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

Show Tags

New post 03 Sep 2014, 11:18
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 281 [0], given: 0

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 15708

Kudos [?]: 281 [0], given: 0

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

Show Tags

New post 27 Sep 2015, 15:44
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 281 [0], given: 0

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7736

Kudos [?]: 17766 [1], given: 235

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

Show Tags

New post 28 Sep 2015, 00:10
1
This post received
KUDOS
Expert's post
1
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



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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 17766 [1], given: 235

Manager
Manager
avatar
Joined: 08 Sep 2012
Posts: 66

Kudos [?]: 15 [0], given: 251

Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)
GMAT ToolKit User Reviews Badge
Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

Show Tags

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?
_________________

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

Kudos [?]: 15 [0], given: 251

2 KUDOS received
Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2676

Kudos [?]: 1767 [2], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

Show Tags

New post 24 Oct 2015, 07:55
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.

Thus the only value of x+y = 3.

Kudos [?]: 1767 [2], given: 794

Manager
Manager
avatar
Joined: 08 Sep 2012
Posts: 66

Kudos [?]: 15 [0], given: 251

Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)
GMAT ToolKit User Reviews Badge
Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]

Show Tags

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! :)
_________________

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

Kudos [?]: 15 [0], given: 251

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 15708

Kudos [?]: 281 [0], given: 0

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

Show Tags

New post 01 Jan 2017, 12:21
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 281 [0], given: 0

Intern
Intern
avatar
B
Joined: 12 May 2017
Posts: 14

Kudos [?]: 0 [0], given: 0

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

Show Tags

New post 12 Jun 2017, 13:55
Can someone help me please.

I understand that if you first that x= pos. and y= neg. you get x=6 and y=-5.

But then this wouldn't be right, because it would contradict the assumptions that x should be neg. and y should be pos.

I don't understand, where do i get these assumptions from?

Kudos [?]: 0 [0], given: 0

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y =   [#permalink] 12 Jun 2017, 13:55
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.