GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Oct 2018, 07:31

### 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

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

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

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 26 Nov 2012
Posts: 593
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

### Show Tags

17 May 2016, 08:51
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.
Math Expert
Joined: 02 Sep 2009
Posts: 50058
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

### Show Tags

17 May 2016, 08:58
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

_________________
Manager
Joined: 23 Jan 2016
Posts: 199
Location: India
GPA: 3.2
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

### Show Tags

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: http://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.
Math Expert
Joined: 02 Sep 2009
Posts: 50058
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

### Show Tags

11 Feb 2017, 11:35
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: http://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.
_________________
Intern
Joined: 21 Jan 2017
Posts: 33
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

### Show Tags

23 Feb 2017, 23:56
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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8418
Location: Pune, India
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

### Show Tags

24 Feb 2017, 03:26
1
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.veritasprep.com/blog/2016/1 ... es-part-v/

Review the post and then try this question. Ask for help if needed. Will provide the solution.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Non-Human User
Joined: 09 Sep 2013
Posts: 8541
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

### Show Tags

13 Mar 2018, 12:30
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.
_________________
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 &nbs [#permalink] 13 Mar 2018, 12:30

Go to page   Previous    1   2   [ 27 posts ]

Display posts from previous: Sort by