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 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
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 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.
Updated on: 23 Jan 2021, 04:22
Originally posted by agarwalmanoj2000 on 10 Jan 2012, 21:33.
Last edited by Bunuel on 23 Jan 2021, 04:22, edited 2 times in total.
Last edited by Bunuel on 23 Jan 2021, 04:22, edited 2 times in total.
Kudos
Bookmarks
Why the value of x is different in the two scenarios? Is there any rule related to what side expression sign should be change?
Please advise.
|x+3| -7
so -7 < x < 1
2nd approach: changing sign on the RHS
positive case:
x+3 < 4
x < 1
negative case:
x+3 < -4
x < -7
so x < -7
Please advise.
|x+3| -7
so -7 < x < 1
2nd approach: changing sign on the RHS
positive case:
x+3 < 4
x < 1
negative case:
x+3 < -4
x < -7
so x < -7
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!
LalaB
Current Student
Joined: 23 Oct 2010
Last visit: 17 Jul 2016
Posts: 227
Own Kudos:
Given Kudos: 73
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
10 Jan 2012, 23:32
Kudos
Bookmarks
|x+3| < 4
ur 1st solution is good. the second one is not good
u stated that x<-7. please check it out. let x=-8 ,then u have got
|-8+3| < 4 |-5| < 4 or 5<4 it is obviously wrong
ur 1st solution is good. the second one is not good
u stated that x<-7. please check it out. let x=-8 ,then u have got
|-8+3| < 4 |-5| < 4 or 5<4 it is obviously wrong
11 Jan 2012, 01:34
Kudos
Bookmarks
In general, |a| = a when a is positive (or zero). When a is negative, then |a| = -a (because a is negative, -a will be positive). So if you see an expression like |x + 3|, there are two possibilities:
x + 3 > 0, so x > -3, in which case |x + 3| = x + 3
x + 3 < 0, so x < -3, in which case |x + 3| = -(x + 3)
When you break your problem into cases, you seem to be ignoring the assumptions you're making - that is, in the positive case, you're assuming in advance that x > -3, so you shouldn't be including any solutions less than -3 when you look only at that case.
Now, when you solve an equation like |x + 3| = 5, you can certainly break the problem into two cases. The 'positive case' is straightforward. In the negative case (where x < -3), we replace |x + 3| with -(x + 3). We then have
-(x + 3) = 5
Notice that since we can multiply an equation safely on both sides by -1, we can rewrite this as follows:
x + 3 = -5
We can *only* do this with an equation. If we instead had an inequality like |x + 3| < 5, then when we consider the 'negative case' where x < -3, we can still replace |x + 3| with -(x + 3) to get
-(x + 3) < 5
Notice however that we are *not* free to instead put the negative sign on the right side of the inequality. To do that, we'd need to multiply both sides by -1, and when you multiply on both sides of an inequality by a negative number, you need to *reverse* the inequality. That was the problem with your second solution.
All of that said, I don't like case-by-case approaches to absolute value. If you understand that |a - b| is the distance between a and b on the number line, then you can interpret the expression |x + 3| = |x - (-3)| as the distance between x and -3. So if |x - (-3)| is less than 4, then x is less than 4 away from -3, so must be between -7 and 1.
x + 3 > 0, so x > -3, in which case |x + 3| = x + 3
x + 3 < 0, so x < -3, in which case |x + 3| = -(x + 3)
When you break your problem into cases, you seem to be ignoring the assumptions you're making - that is, in the positive case, you're assuming in advance that x > -3, so you shouldn't be including any solutions less than -3 when you look only at that case.
Now, when you solve an equation like |x + 3| = 5, you can certainly break the problem into two cases. The 'positive case' is straightforward. In the negative case (where x < -3), we replace |x + 3| with -(x + 3). We then have
-(x + 3) = 5
Notice that since we can multiply an equation safely on both sides by -1, we can rewrite this as follows:
x + 3 = -5
We can *only* do this with an equation. If we instead had an inequality like |x + 3| < 5, then when we consider the 'negative case' where x < -3, we can still replace |x + 3| with -(x + 3) to get
-(x + 3) < 5
Notice however that we are *not* free to instead put the negative sign on the right side of the inequality. To do that, we'd need to multiply both sides by -1, and when you multiply on both sides of an inequality by a negative number, you need to *reverse* the inequality. That was the problem with your second solution.
All of that said, I don't like case-by-case approaches to absolute value. If you understand that |a - b| is the distance between a and b on the number line, then you can interpret the expression |x + 3| = |x - (-3)| as the distance between x and -3. So if |x - (-3)| is less than 4, then x is less than 4 away from -3, so must be between -7 and 1.
