What is the value of x?

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

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.

MAGOOSH OFFICIAL SOLUTION:

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

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?

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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