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Bunuel
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What is the value of x? 25 Mar 2015, 03:59
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35% (medium)

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70% (01:41) correct 30% (01:54) wrong based on 490 sessions

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What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9

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A
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What is the value of x? 25 Mar 2015, 06:31
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Bunuel
What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9

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Statement 1:
6-3x=x-2
x=2

6-3x=-x+2
x=2
Sufficient

Statement 2:
5x+3=2x+9
x=2

5x+3=-2x-9
x= -12/7
Insufficient

Answer: A
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What is the value of x? 26 Mar 2015, 03:32
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(1) |6 - 3x| = x - 2

x-2 must be positive => x>2 => ||6 - 3x|=3x-6 => Just have one value of x => Suff

|5x + 3| = 2x + 9

x must be over -9/2 => 5x+3 can be positive or negative => Having 2 values of x => Insufficient

=> Answer: A
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What is the value of x? 28 Mar 2015, 08:59
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Bunuel
What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9

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Answer=A

1. 6-3X=X-2
X=2
OR
-6+3X=X-2
X=2. Statement 2 is sufficient
2. 5x+3=2x+9
x=2
OR
-5X-3=2X+9
7X=-12
X=-12/7. Statement 2 is Insufficient
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What is the value of x? 30 Mar 2015, 03:59
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I simply solved the 2 eqs. is there any better sol?
My sol :
1) |6x-3| = 6x-3 or -6x+3.
Thus putting in eq 1 : 6x-3=x-2 gives x = 2. Similarly, -6x+3=x-2 gives x=2. Thus we get a value for x. SUFF.

2) |5x+3| = 5x+3 or -5x-3.
Solving Similarly I got 2 diff values of x = 2 or -12/7. Thus NS.

Hence A.
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What is the value of x? 30 Mar 2015, 04:56
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What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9

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MAGOOSH OFFICIAL SOLUTION:
Attachment:
absolutes_text.png
absolutes_text.png [ 12.37 KiB | Viewed 5954 times ]
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What is the value of x? 15 Feb 2018, 11:55
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Hello chetan2u,

I Read your post about the absolute Modulus and accordingly I tried to solve this problem by critical value method. Hereunder my solution and please correct me if I did something wrong.

Statement1:-
the critical value is 2 so we are looking for values : x>=2 or x<2.

x>=2
If x>=2 ,let x=3 for example , the modulus would be negative. so by assigning -ive sign to the modulus we will get -(6-3x)=x-2 , so x=2, which is in the region x>=2 so we will accept this value.

x<2
If x<2, let x=-3 for example , the modulus would be positive. so by assigning +ve sign to the modulus we will get (6-3x)=x-2 , so x=2, which is not in the region x<2 , so this value is rejected.

The same as for Statement 2 , the critical value is -3/5 . so x>=-3/5 or x<-3/5 after solving it we will get 2 values , one accepted "x=2" and one rejected x=-12/17
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What is the value of x? 15 Feb 2018, 21:09
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ahmed.abumera
Hello chetan2u,

I Read your post about the absolute Modulus and accordingly I tried to solve this problem by critical value method. Hereunder my solution and please correct me if I did something wrong.

Statement1:-
the critical value is 2 so we are looking for values : x>=2 or x=2
If x>=2 ,let x=3 for example , the modulus would be negative. so by assigning -ive sign to the modulus we will get -(6-3x)=x-2 , so x=2, which is in the region x>=2 so we will accept this value.

x=-3/5 or x<-3/5 after solving it we will get 2 values , one accepted "x=2" and one rejected x=-12/17
Show more


hi,
you have found all other values correct but there is one problem..
|5x + 3| = 2x + 9...
take x as x<-3/5 and you have got an answer -12/17..
now -12/17 <-3/5..
so this is also accepted..
so two values for statement II hence insuff..

otherwise this is what you are required to do in a proper method
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What is the value of x? 24 Apr 2019, 00:13
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chetan2u
ahmed.abumera
Hello chetan2u,

I Read your post about the absolute Modulus and accordingly I tried to solve this problem by critical value method. Hereunder my solution and please correct me if I did something wrong.

Statement1:-
the critical value is 2 so we are looking for values : x>=2 or x=2
If x>=2 ,let x=3 for example , the modulus would be negative. so by assigning -ive sign to the modulus we will get -(6-3x)=x-2 , so x=2, which is in the region x>=2 so we will accept this value.

x=-3/5 or x<-3/5 after solving it we will get 2 values , one accepted "x=2" and one rejected x=-12/17


hi,
you have found all other values correct but there is one problem..
|5x + 3| = 2x + 9...
take x as x<-3/5 and you have got an answer -12/17..
now -12/17 <-3/5..
so this is also accepted..
so two values for statement II hence insuff..

otherwise this is what you are required to do in a proper method
Show more

chetan2u : What if -12/7 is not a valid solution or does not satisfy the equation |5x + 3| = 2x + 9. If only valid solution is x = 2, in that case, would the answer be D, as both st1 and st2 gives x = 2?
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What is the value of x? 24 Apr 2019, 06:53
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chetan2u
ahmed.abumera
Hello chetan2u,

I Read your post about the absolute Modulus and accordingly I tried to solve this problem by critical value method. Hereunder my solution and please correct me if I did something wrong.

Statement1:-
the critical value is 2 so we are looking for values : x>=2 or x=2
If x>=2 ,let x=3 for example , the modulus would be negative. so by assigning -ive sign to the modulus we will get -(6-3x)=x-2 , so x=2, which is in the region x>=2 so we will accept this value.

x=-3/5 or x<-3/5 after solving it we will get 2 values , one accepted "x=2" and one rejected x=-12/17


hi,
you have found all other values correct but there is one problem..
|5x + 3| = 2x + 9...
take x as x<-3/5 and you have got an answer -12/17..
now -12/17 <-3/5..
so this is also accepted..
so two values for statement II hence insuff..

otherwise this is what you are required to do in a proper method

chetan2u : What if -12/7 is not a valid solution or does not satisfy the equation |5x + 3| = 2x + 9. If only valid solution is x = 2, in that case, would the answer be D, as both st1 and st2 gives x = 2?
Show more

Yes, say instead of -12/17, you had something that did not fit in, then the answer would be D, as statement II gives one solution.
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What is the value of x? 15 Aug 2023, 09:51
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