What is the value of x?


(1) |6 - 3x| = x - 2......
Two cases..
a) 6-3x=x-2........4x=8.....X=2
b) -(6-3x)=x-2........3x-6=x-2.....2x=4.....X=2
Sufficient

(2) |5x + 3| = 2x + 9
Two cases
a) 5x+3=2x+9.......3x=6......X=2
b) -(5x+3)=2x+9.......5x+2x=-3-9......x=-12/7
So two values possible
Insufficient

A
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I have one silly question here/..
When look for the sign of |x-n| we always use these two conditions:
if x>n : |x-n| is positive so = x-n
If x<n then |x-n| is negative = -x+n
but what if x=n then ??
I am asking this because in statement one i got confused. when i took x>2, I got x=2 and again when x<2 solution was x=2 but thought there is no slution because in both the cases x=2 is not satifying the conditions i too for deciding the sign.
Please elaborate...


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Hi,
If this question has been asked like,
Is x > 0?
Then the solving part would have been very easy,
Because here statement are equations with one side modulus, so we can consider one side non-negative.
Say, for example, in the statement I here given is,
|6-3x| = x -2
We should have equated this to,
x-2 > = 0
Because modulus of anything will be non-negative.
But here the question asks the value of x,
So, we have to do it, either using the number line or by the definition of modulus,
Statement I is sufficient:
|6 - 3x| = x – 2
By the definition of modulus,
6 – 3x = x -2, whenever (6-3x) >= 0 i.e., whenever, x <= 2
Solving this we get,
x = 2
Which works according to the condition of x,
Now,
6-3x = -x +2, whenever (6-3x) <= 0 i.e., whenever, x > = 2
Solving this we get,
x = 2.
So only one x value works here.
So statement I is sufficient.

Statement II is insufficient:
|5x + 3| = 2x + 9
By the definition of modulus,
5x + 3 = 2x +9, whenever (5x+3) >= 0 i.e., whenever, x >= -3/5
Solving this we get,
x = 2
Which works according to the condition of x,
Now,
5x+3 = -2x-9, whenever (5x+3) <= 0 i.e., whenever, x < = -3/5
Solving this we get,
7x = 6.
x = -12/7
Which also works according to the condition of x,
Hence there are two values of x.
So statement II is insufficient.
So the answer is A(I alone).
Key to solving the modulus question, knowing the definition of modulus.
Hope this helps.

When opening modulus, you can take equal to with any of the inequality signs.
And the value you get then should fall in that range..
Say here you took X<=2..
So |6-3x|=x-2......so 3x-6=x-2.....X=2
So this value X=2 is within the range X<=2, so OK
Otherwise |6-3x|=x-2 it self means x-2 is 0 or >0 as left side is Modulus

Also if you had to solve this
|6-3x|=x-2......
3|2-x|=x-2....
Only possible of 2-x=0...X=2
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