NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 22:48 It is currently 25 Apr 2024, 22:48
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

Which values of x are solutions |x + 1| + |x - 1| <= 2

SORT BY:
Date
Kudos
   1   2 
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
G
msk0657
Retired Moderator
Joined: 26 Nov 2012
Posts: 473
Own Kudos [?]: 493 [1]
Given Kudos: 46
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   17 May 2016, 08:51
1
Kudos
How many solutions does \(|x+3| - |4-x| = |8+x|\) have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), which is not a valid solution since we are considering \(x < -8\) range and -1 is out of it;

If \(-8\leq{x}\leq{-3}\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;

If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;

If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.

Hi Bunuel,

Can you please help me to understand when to change sign ( + / - )...as you did for the above four conditions.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619064 [2]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   17 May 2016, 08:58
2
Bookmarks
Expert Reply
msk0657 wrote:
How many solutions does \(|x+3| - |4-x| = |8+x|\) have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), which is not a valid solution since we are considering \(x < -8\) range and -1 is out of it;

If \(-8\leq{x}\leq{-3}\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;

If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;

If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.

Hi Bunuel,

Can you please help me to understand when to change sign ( + / - )...as you did for the above four conditions.


Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\)


Theory on Absolute Values: math-absolute-value-modulus-86462.html
The E-GMAT Question Series on ABSOLUTE VALUE: the-e-gmat-question-series-on-absolute-value-198503.html
Properties of Absolute Values on the GMAT: properties-of-absolute-values-on-the-gmat-191317.html
Absolute Value: Tips and hints: absolute-value-tips-and-hints-175002.html

DS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Absolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html
User avatar
S
OreoShake
Manager
Manager
Joined: 23 Jan 2016
Posts: 139
Own Kudos [?]: 81 [0]
Given Kudos: 509
Location: India
GPA: 3.2
Send PM
GMAT ToolKit User
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   11 Feb 2017, 07:03
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this



\(|x + 1|+|x - 1|\leq{2}\) --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If \(x<-1\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(-(x+1)-(x-1)\leq{2}\) --> \(x\geq{-1}\), not a valid range since we are considering \(x<-1\);

If \(-1\leq{x}\leq{1}\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(x+1-(x-1)\leq{2}\) --> \(0\leq{2}\) which is true, so for \(-1\leq{x}\leq{1}\) given inequality holds true;

If \(x>1\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(x+1+x-1\leq{2}\) --> \(x\leq{1}\), not a valid range since we are considering \(x>1\).

So, finally we have that \(|x + 1|+|x - 1|\leq{2}\) holds true for \(-1\leq{x}\leq{1}\).

For more check Absolute Value chapter of Math Book: https://gmatclub.com/forum/math-absolute ... 86462.html

Hope it helps.


Bunuel how did you get 0<=2 for the second range?? x gets crossed off so we get 2<=2 which becomes 0<=0 Could you please explain how you calculated that and how the second range is valid? Thank you.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619064 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   11 Feb 2017, 11:35
Expert Reply
OreoShake wrote:
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this



\(|x + 1|+|x - 1|\leq{2}\) --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If \(x<-1\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(-(x+1)-(x-1)\leq{2}\) --> \(x\geq{-1}\), not a valid range since we are considering \(x<-1\);

If \(-1\leq{x}\leq{1}\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(x+1-(x-1)\leq{2}\) --> \(2\leq{2}\) which is true, so for \(-1\leq{x}\leq{1}\) given inequality holds true;

If \(x>1\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(x+1+x-1\leq{2}\) --> \(x\leq{1}\), not a valid range since we are considering \(x>1\).

So, finally we have that \(|x + 1|+|x - 1|\leq{2}\) holds true for \(-1\leq{x}\leq{1}\).

For more check Absolute Value chapter of Math Book: https://gmatclub.com/forum/math-absolute ... 86462.html

Hope it helps.


Bunuel how did you get 0<=2 for the second range?? x gets crossed off so we get 2<=2 which becomes 0<=0 Could you please explain how you calculated that and how the second range is valid? Thank you.


You'll get \(2\leq{2}\). Since this inequality is true, then we can say that for the range we consider (\(-1\leq{x}\leq{1}\)) given inequality holds true.
avatar
B
umabharatigudipalli
Intern
Intern
Joined: 21 Jan 2017
Posts: 27
Own Kudos [?]: 5 [1]
Given Kudos: 45
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   23 Feb 2017, 23:56
1
Kudos
VeritasPrepKarishma wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this


You can solve this question using the number line approach too.

"Distance of x from -1" + "Distance of x from 1" <= 2


----------(-2)-----(-1)-----0-----(1)-----(2)--------

Note that distance between -1 and 1 is 2 so whenever x is between these two values, the sum of distance from -1 and 1 will be 2

----------(-2)-----(-1)---x--0-----(1)-----(2)--------

Hence all values between -1 and 1 (inclusive) will satisfy this condition.

When you go to the right of 1 or left of -1, the sum of distances from -1 and 1 will exceed 2. So the only range that satisfies the inequality is
-1 <= x <= 1





thanks a ton! that's pretty much clear. It would be great if you explain the range of values that satisfy |x+3| - |4-x| = |8+x| in the same method....


Thanks,
Uma
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14823
Own Kudos [?]: 64926 [1]
Given Kudos: 426
Location: Pune, India
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   24 Feb 2017, 03:26
1
Kudos
Expert Reply
umabharatigudipalli wrote:
VeritasPrepKarishma wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this


You can solve this question using the number line approach too.

"Distance of x from -1" + "Distance of x from 1" <= 2


----------(-2)-----(-1)-----0-----(1)-----(2)--------

Note that distance between -1 and 1 is 2 so whenever x is between these two values, the sum of distance from -1 and 1 will be 2

----------(-2)-----(-1)---x--0-----(1)-----(2)--------

Hence all values between -1 and 1 (inclusive) will satisfy this condition.

When you go to the right of 1 or left of -1, the sum of distances from -1 and 1 will exceed 2. So the only range that satisfies the inequality is
-1 <= x <= 1





thanks a ton! that's pretty much clear. It would be great if you explain the range of values that satisfy |x+3| - |4-x| = |8+x| in the same method....


Thanks,
Uma


This concept with 3 terms has been discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/1 ... es-part-v/

Review the post and then try this question. Ask for help if needed. Will provide the solution.
avatar
B
securegmat
Intern
Intern
Joined: 26 Apr 2017
Posts: 7
Own Kudos [?]: 5 [1]
Given Kudos: 1
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   27 Oct 2018, 11:49
Simple... :)

|x+1| + |x-1| <=2

|2x| <=2

2x <=2 or -2x <=2

x<=1 or x>=-1

Thus, -1<=x<=1
User avatar
P
twobagels
Manager
Manager
Joined: 12 Jul 2020
Posts: 137
Own Kudos [?]: 451 [0]
Given Kudos: 2892
Location: United States (CA)
WE:Science (Other)
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   23 Jan 2021, 16:36
You asked for a non-algebraic way to deduce the range of values of 'x' that satisfy the inequality. I initially graphed it on paper, but then used calc to graph it out.

since you have |x -1 | and | x+1| on the LHS as one gets smaller the other absolute value gets larger. Since the value of 'x' must must yield output less than equal to to, from looking at the graph, you can see the range is -1 to 1.
Attachments

absolute.JPG
absolute.JPG [ 71.19 KiB | Viewed 1500 times ]

avatar
B
Jaya6
Manager
Manager
Joined: 09 Jan 2021
Posts: 71
Own Kudos [?]: 12 [0]
Given Kudos: 142
Location: India
Schools: ISB '23 (S)
GPA: 3.2
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   19 Feb 2021, 09:47
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this



\(|x + 1|+|x - 1|\leq{2}\) --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If \(x<-1\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(-(x+1)-(x-1)\leq{2}\) --> \(x\geq{-1}\), not a valid range since we are considering \(x<-1\);

If \(-1\leq{x}\leq{1}\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(x+1-(x-1)\leq{2}\) --> \(2\leq{2}\) which is true, so for \(-1\leq{x}\leq{1}\) given inequality holds true;

If \(x>1\) then \(|x + 1|+|x - 1|\leq{2}\) expands as \(x+1+x-1\leq{2}\) --> \(x\leq{1}\), not a valid range since we are considering \(x>1\).

So, finally we have that \(|x + 1|+|x - 1|\leq{2}\) holds true for \(-1\leq{x}\leq{1}\).

For more check Absolute Value chapter of Math Book: https://gmatclub.com/forum/math-absolute ... 86462.html

Hope it helps.



what about option B? I'm getting the right answer for that as well.
User avatar
S
JohnGalt1
Intern
Intern
Joined: 10 Feb 2021
Posts: 43
Own Kudos [?]: 49 [0]
Given Kudos: 109
Location: Fiji
Schools: BU Stern '24 Kellogg '24 Wharton '24
GMAT 1: 580 Q44 V26
GMAT 2: 700 Q47 V40
GPA: 3.44
WE:Account Management (Consulting)
Send PM
GMAT ToolKit User
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   19 Feb 2021, 12:36
Super easy guessing method:
looking from equation we already know that 0 is one solution, so eliminate all variants which doesnt include 0, A, C, and E, are all wrong. Then test leftover B and D. We see that 2 is no solution so cross out B, and we are left with D.
User avatar
B
Ghostrider3147
Manager
Manager
Joined: 21 Jun 2021
Posts: 86
Own Kudos [?]: 20 [0]
Given Kudos: 210
GPA: 2.91
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   26 Mar 2022, 07:41
Put the end values of every option to see whether it satisfies the equation. correct answer D can be easily calculated.

Posted from my mobile device
avatar
B
Aravind04
Intern
Intern
Joined: 25 Nov 2021
Posts: 26
Own Kudos [?]: 1 [0]
Given Kudos: 893
Location: India
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   06 May 2022, 06:22
hi KarishmaB


Instead of Algebra can we solve this using Number line

How many solutions does |x+3|−|4−x|=|8+x||x+3|−|4−x|=|8+x| have?
User avatar
B
adityasuresh
Manager
Manager
Joined: 03 May 2020
Posts: 108
Own Kudos [?]: 33 [0]
Given Kudos: 512
Send PM
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   28 Mar 2023, 19:49
First we equate the expression within mod to zero, then we get x to be 1 and -1, we draw a number line which divides the values of x into 3 regions,<-1,between -1&1 &>1. While substituting values of x in the respective region, say in region 1(<-1) so the expression x +1 becomes negative so we multiply by -1, similarly the expression x -1 becomes negative so we multiply the expression by -1. We get -x -1 -x+1<=2, x>= -1.(dividing by -2 flips the sign),similarly we can calculate for other regions in region 2 the values of x are between -1 and 1 simplifying we get 2 <=2 so all values in this region satisfy the expression, finally in region 3 both expressions are positive so writing as is we get 2x <=2, x<=1
So combing the inequalities we get -1<=x<=1
D
Attachments

File comment: Pfa
IMG_6351.jpeg
IMG_6351.jpeg [ 881.73 KiB | Viewed 511 times ]

User avatar
S
HoudaSR
Manager
Manager
Joined: 08 Aug 2022
Posts: 75
Own Kudos [?]: 32 [0]
Given Kudos: 185
Location: Morocco
Schools: HBS '26 Oxford Said'25 Judge '25 Rotterdam'25 Yale '24
WE:Advertising (Non-Profit and Government)
Send PM
Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post   02 Jun 2023, 14:54
[quote="sandal85"]Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?


A. \(-2\leq{x}\leq{-1}\).

B. \(0\leq{x}\leq{2}\).

C. \(-7<{x}<{-6}\).

D. \(-1\leq{x}\leq{1}\).

E. \(2\leq{x}\leq{3}\).



I solved this question as following:

Ix+1I + Ix-1I <=2
Ix-1I <= 2- Ix+1I
And since absolute value is non negative, the last equation tells me that 2-Ix+1I>=0
Thus, - Ix+1I >=-2
Hence, Ix+1I <=2
-2<= x+1 <=2
Therefore, -1<= x <= 1

Correct answer is D.


Bunuel What do you think of this solving method?
GMAT Club Bot
gmatclubot
Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
02 Jun 2023, 14:54
   1   2 
Moderators:
Bunuel
Math Expert
92915 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions