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Difficulty:
25%
(medium)
Question Stats:
78% (01:37) correct
22%
(01:48)
wrong
based on 734
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of x?
\(|x+2|+|x-3| = |2x+3|\)
A. -5
B. -2
C. 1
D. 4
E. 7
\(|x+2|+|x-3| = |2x+3|\)
A. -5
B. -2
C. 1
D. 4
E. 7
rajdavezz99
Joined: 28 May 2017
Last visit: 02 Feb 2018
Posts: 26
Location: India
GMAT 1: 750 Q50 V42
GPA: 3.03
WE:Management Consulting (Consulting)
div10
The ideal way to get the answer is by solving for different ranges of x. The above equation will take 4 different forms for the following ranges:
1. Greater than 3
2. From -1.5 to 3
3. From -2 to -1.5
4. Less than -2
Solving for each range separately, only one range gives a valid answer, which is 1.
A much easier way to solve this - substitute the options!
Hope this helps!
Sent from my iPhone using GMAT Club Forum mobile app
div10
Using the answer choices is the fastest.
If you do not know how to use critical values and intervals, see this post, below, with links to excellent material on complex absolute value questions.
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General Discussion
kanusha
Joined: 25 Mar 2013
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On substituting options
3+2 = 5
C
But on removing bar and considering two cases could not match with options
Case 1 and 2
( x + 2 ) + ( x-3) = 2x + 3
2x - 1 = 2x + 3
2x - 2x = 4
?
Can anyone explain. Thanks
3+2 = 5
C
But on removing bar and considering two cases could not match with options
Case 1 and 2
( x + 2 ) + ( x-3) = 2x + 3
2x - 1 = 2x + 3
2x - 2x = 4
?
Can anyone explain. Thanks
@genxer123 Can you explain how you go 1 in case 4?? isn't
Case 4
|x+2|+|x−3|=|2x+3|
−x−2−x+3=−2x−3
-2x+1 = -2x-3
1=-3
which is invalid???
Case 4
|x+2|+|x−3|=|2x+3|
−x−2−x+3=−2x−3
-2x+1 = -2x-3
1=-3
which is invalid???
RashmiT
Hi RashmiT
Yes you are correct in your observation. Case 4 by generis is incorrect.
This question is best solved through options. if you go for any other approach then you will have to work through multiple ranges.
Another option to remove mod is to square both sides but it will result in a complex polynomial equation.
RashmiT
RashmiT , you are correct.
My brain clearly was not working well.
Nice catch, +1, thank you, and my apologies.
With two absolute value terms in an equation, you can open mods.
With more than two mods?
If, as here, the answers are before you, use them.
If answers are not before you, use
"Critical Method" see chetan2u, Absolute Modulus, A Better Understanding
and Bunuel , "3-steps approach for complex problems," Absolute Value Modulus, 3-steps approach
And "Graphical Method," chetan2u , Absolute Modulus, A Better Understanding
(You probably know what to do when there are more than two absolute value expressions. I post these links just in case you might need them, but mostly to help people who have not seen the excellent posts to which the links refer.)
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gadde22
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broall
This is a very simple question should be solved in 1 min or else you can complete Quant session. You just write down the equation and start substituting values. Never ever think of solving by taking "x" greater than 3 , then x greater than -1.5, then x greater than -2.
So we start substituting ,
A . becomes 11 = 8 x is not equal to -5
B. Become 5 = 1. Not satisfies.
C. 5 = 5. So x =1 . Done.
broall
Fastest way is to utilize the answer choices
(C) is the only choice that works.
|1+2| + |1-3| = |2(1)+3|
3+2=|2+3|
5=5
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