Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
25 Sep 2012, 09:21
Kudos
Bookmarks
Hi Experts,
I need help with inequalities.
I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.
I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.
Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.
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.
So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).
Thanks in anticipation.
I need help with inequalities.
I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.
I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.
Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.
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.
So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).
Thanks in anticipation.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
25 Sep 2012, 11:39
Kudos
Bookmarks
GetThisDone
Use the definition of the absolute value: \(|x| = x\) if \(x\geq0\) and \(|x|=-x\)if \(x<0.\)
Obviously, instead of \(x\) you can have any algebraic expression. So, for example \(|x+8|=x+8\) if \(x+8\geq0\) or \(x\geq-8\) and \(|x+8| = -(x+8)\) if \(x+8<0\) or \(x<-8\), etc.
Absolute value expresses a distance between two numbers and it is always non-negative.
\(|x|\) is the distance on the number line between \(x\) and \(0\). The distance between \(x\) and \(0\) is the same as the distance between \(-x\) and \(0.\)
\(|7|=7\) and \(|-7|=7\). How can we connect \(-7\) and \(7\)? There is no mathematical operation "just drop the minus sign..." But obviously \(7=(-1)*(-7)\).
souvik101990
Expert
Alum
Joined: 19 Mar 2012
Last visit: 11 Nov 2025
Posts: 4,314
Given Kudos: 2,326
Location: United States (WA)
Concentration: Leadership, General Management
Schools: Ross '20 (M)
GMAT 1: 760 Q50 V42
GMAT 2: 740 Q49 V42 (Online)
GMAT 3: 760 Q50 V42 (Online)
GPA: 3.8
WE:Marketing (Non-Profit and Government)
Products:
Expert reply
25 Sep 2012, 12:11
Kudos
Bookmarks
|x+3| - |4-x| = |8+x|
Hey
I would be happy to help though I did not get your question very clearly. Do you want to know how did the solution find out the ranges as given? Then I will try to help you out.
I assume you know the basic of the integer function which is:
|x|=x when x>=0
|x|=-x when x<0
So, you can see the checkpoints here are -8, -3 and 4
I hope till now you have followed.
Now lets look at each check points and see how the integer function behaves
When \(x<-8,\)
x+3 is negative, so \(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is negative, so \(|8+x|=-(8+x)\)
So the equation, in this case, becomes,
\(-(x+3)-(4-x)=-(8+x)\)
\(=> x=-1\) which violates the range we have considered
Similarly when\(-8<x<-3\)
x+3 is negative, so\(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so\(|8+x|=(8+x)\)
So equation becomes
\(-(x+3)-(4-x)=(8+x)\)
=>\(x=-15\) which violates the range we have considered
Again, when \(-3<x<4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
So the equation becomes,
\((x+3)-(4-x)=(8+x)\)
\(2x-1=8+x\)
\(X=9\) which violates the range we have considered
Last, when \(x>4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is negative, so \(|4-x|=-(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
\((x+3)+(4-x)=(8+x)\)
\(=>x=-1\) which again violates the range
SO no solution
Did this help?
Hey
I would be happy to help though I did not get your question very clearly. Do you want to know how did the solution find out the ranges as given? Then I will try to help you out.
I assume you know the basic of the integer function which is:
|x|=x when x>=0
|x|=-x when x<0
So, you can see the checkpoints here are -8, -3 and 4
I hope till now you have followed.
Now lets look at each check points and see how the integer function behaves
When \(x<-8,\)
x+3 is negative, so \(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is negative, so \(|8+x|=-(8+x)\)
So the equation, in this case, becomes,
\(-(x+3)-(4-x)=-(8+x)\)
\(=> x=-1\) which violates the range we have considered
Similarly when\(-8<x<-3\)
x+3 is negative, so\(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so\(|8+x|=(8+x)\)
So equation becomes
\(-(x+3)-(4-x)=(8+x)\)
=>\(x=-15\) which violates the range we have considered
Again, when \(-3<x<4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
So the equation becomes,
\((x+3)-(4-x)=(8+x)\)
\(2x-1=8+x\)
\(X=9\) which violates the range we have considered
Last, when \(x>4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is negative, so \(|4-x|=-(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
\((x+3)+(4-x)=(8+x)\)
\(=>x=-1\) which again violates the range
SO no solution
Did this help?
. If you could help me with the below, it will help me put this topic to rest. 
