BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

PS Q

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Sun Jun 12, 2011 5:13 am
Hi,
Let A(x,y) be a point on the coordinate plane.
|x-1| is the normal distance of A from from the vertical line x=1
|x+2| is the normal distance of A from from the vertical line x=-2
So, for any -2<x<1, |x-1|+|x+2| will be the distance between lines x=1 and x+2=0. This distance is 3 units.
So, we need to find x which is 1 unit to the right of x-1 =0 or 1 unit to the left of x+2=0.
So, x=2 or x=-3.

(or)
|x-1|+|x+2|=5. Let a = x-1. So, |a|+|a+3| = 5 => |a+3| = 5 -|a| =>|a+3|^2 = (5-|a|)^2
i.e. a^2+6a+9 = a^2 -10|a| +25 =>6a+10|a| = 16.
case-1: If a>0, |a| = a. So, 16a=16 =>a=1
case-2: If a<0, |a| = -a. So, 6a-10a = 16 =>a = -4
So, a=1 or -4
x = a+1. So, x is 2 or -3.
Last edited by Frankenstein on Sun Jun 12, 2011 5:17 am, edited 1 time in total.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 29
Joined: Sat May 07, 2011 6:16 am

by vinayreguri » Sun Jun 12, 2011 5:17 am
I solved this other way around


x-1+x+2=5 and x-1+x+2=-5 because of mod and solving for x i got 2,-3.

Is this right?.

Will this work for others too.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Sun Jun 12, 2011 5:20 am
vinayreguri wrote:solve for x where |x-1|+|x+2|=5
For this problem the expression |x - 1| or |x + 2| changes their sign at x = -2 and x = 1. Hence, we have three ranges of values of x for which either of the expression changes sign.

1. x < -2 :
  • |x - 1| = -(x - 1) and |x + 2| = -(x + 2)
    --> |x - 1| + |x + 2| = 5
    --> -(x - 1) - (x + 2) = 5
    --> x = -3
1. -2 ≤ x < 1 :
  • |x - 1| = -(x - 1) and |x + 2| = (x + 2)
    --> |x - 1| + |x + 2| = 5
    --> -(x - 1) + (x + 2) = 5
    --> No Solution
1. 1 ≤ x:
  • |x - 1| = (x - 1) and |x + 2| = (x + 2)
    --> |x - 1| + |x + 2| = 5
    --> (x - 1) + (x + 2) = 5
    --> x = 2
Hence, the solution for the equation is x = -3 and x = 2.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Sun Jun 12, 2011 5:24 am
vinayreguri wrote:I solved this other way around


x-1+x+2=5 and x-1+x+2=-5 because of mod and solving for x i got 2,-3.

Is this right?.

Will this work for others too.
Hi,
This method is actually incorrect because you have effectively taken |(x-1)+(x+2)| = 5, which is incorrect.
|x-1|+|x+2| and |(x-1)+(x+2)| will be equal only if (x-1)&(x+2) are of the same sign.
Consider : |x-1|+|x+2| = 3.
Using you method: x-1+x+2=3 or x-1+x+2=-3.
Solving you get x = 1 or x = -2. But actually many more values of x(any value between -2 and 1) satisfy this equation.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Sun Jun 12, 2011 5:32 am
vinayreguri wrote:x-1+x+2=5 and x-1+x+2=-5 because of mod and solving for x i got 2,-3.

Is this right?.
Will this work for others too.
This is neither correct nor applicable to other problems.

Removing absolute values from the equation will result in four equations.
  • 1. (x - 1) + (x + 2) = 5
    2. -(x - 1) + (x + 2) = 5
    3. (x - 1) - (x + 2) = 5
    4. -(x - 1) - (x + 2) = 5 --> (x - 1) + (x + 2) = -5
Though 1 and 4 only will result in some feasible solution, 2 is also an interpretation of the given equation. But 2 has no solution. And from mechanical view, 3 is also an interpretation of the given equation but it is impractical.

For other problems, we have to consider all four of them and check which one is possible and which one is giving some feasible solution.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 934
Joined: Tue Nov 09, 2010 5:16 amFollowed by: 14 members
Location: AAMCHI MUMBAI LOCAL
Thanked: 63 times

by [email protected] » Sun Jun 12, 2011 6:50 am
Anurag could you please explain me the 1st explanation, especially the following line:

So, we need to find x which is 1 unit to the right of x-1 =0 or 1 unit to the left of x+2=0.
So, x=2 or x=-3.

I did not get this point...
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Sun Jun 12, 2011 7:25 am
[email protected] wrote:Anurag could you please explain me the 1st explanation, especially the following line:

So, we need to find x which is 1 unit to the right of x-1 =0 or 1 unit to the left of x+2=0.
So, x=2 or x=-3.

I did not get this point...
Hi,
I think you have understood up to the part that for any value of x between -2 and 1 , the sum of distances is zero.
Now, consider a point with x>1. Let the distance of this point from x=1 be a, The distance of this point from x=-2 is a + (distance between x=1 and x=-2) = a+3
So, total is a+(a+3) = 2a+3. We need that sum to be equal to 5. So, 2a+3=5 =>a=1.
So, x should be at a distance from x=1 to the right of it. So, x=2.
Similarly we will solve for x to the left of x=-2.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1