From stmt1,

If X =+ve

X+1=2X-2; x=3

If X=-ve

The equation cannot be solved.[Its confusing here]

IMO A is insufficient

From stmt2 |x-3| ≠ 0

-->x cannot be equal to 3.

x can be any value like -1,0,2 except 3.

If x=0,|X|<1

If x=2,|x|>1

So B is insufficient


Answer is E

Statement 1:

1) -1< x < 1, then:
x+1 = -2x+2 or x=1/3

2) If x <- 1, then
-x-1 = -2x+2 or x=3

3) If x>1, then
x+1= 2x-2 or x=3

- insufficient


Statement 2: insufficient

Statement 1 & 2 - sufficient since it rules out x < -1 & x >1

Hence, C

---
edited to correct careless errors

my small thinking
is -1<x<1 ??
(1) |x+1|=2|x-1| here two way are possible one to square both parts and solving quadratic equation, and the second to open module step by step
_________ -1_________1 __________

|x+1|={(x+1) if x>-1. or -(x+1) if x<-1}

|x+1|={(x-1) if x>1, or -(x-1) if x<1}

the first interval
_________ -1, or x<-1

-(x+1)=-2(x-1)
x=3 but 3 does not fit in intrval x<-1
so we have no roots here

the second interval
(-1,1)

(x+1)=-2(x-1)
x=1/3 it can be root of given equation

the last interval x>1

x-3=2(x-1)
x=3 also can be root
so we have 2 answers x=3 not belong to (-1,1)
and 1/3 belong to (-1,1)
1st st insufficient

(2) x=/=3 insufficient

both 2st put off 3 as answer so we left with 1/3 within (-1,1)
the answer C

btw not sure if it would help but Manhattan GMAT's tutorial on absolute values really helped me in solving this one.

https://www.manhattangmat.com/tutorials/ ... -value.cfm

clock60 wrote:my small thinking
is -1<x<1 ??
(1) |x+1|=2|x-1| here two way are possible one to square both parts and solving quadratic equation, and the second to open module step by step
_________ -1_________1 __________

|x+1|={(x+1) if x>-1. or -(x+1) if x<-1}

|x+1|={(x-1) if x>1, or -(x-1) if x<1}

the first interval
_________ -1, or x<-1

-(x+1)=-2(x-1)
x=3 but 3 does not fit in intrval x<-1
so we have no roots here

the second interval
(-1,1)

(x+1)=-2(x-1)
x=1/3 it can be root of given equation

the last interval x>1

x-3=2(x-1)
x=3 also can be root
so we have 2 answers x=3 not belong to (-1,1)
and 1/3 belong to (-1,1)
1st st insufficient

(2) x=/=3 insufficient

both 2st put off 3 as answer so we left with 1/3 within (-1,1)
the answer C

ST 1 :1) |x + 1| = 2|x - 1|

case 1 :

(x+1) = 2( x-1)

x +1 = 2x-2

x = 3

case 2 :

-(x+1) = -2(x-1)

-x-1 = -2x+2

x=3

case 3 :

-(x+1) = 2(x-1)

-x-1 = 2x-2

-3x= -1

x=1/3

case 4:

(x+1)= -2(x-1)
x+1 = -2x +2

3x =1

x= 1/3


We have 2 values x = 3 or x = 1/3. Inconsistent as 3 > 1 & 1/3 <1

Insufficient

St 2:

|x - 3| Not equal to 0

So it means X is not equal to three

Still X can be any value. Say X can be 10 or 0 or 1. Inconsistent. SO insufficient

Now combining 2 statements:

we have X is NOT equal to 3 so we have x = 1/3 only

1/3 < 1. YES

Pick C..Move on !!

@gmatmachoman: in analyzing (1), you tested 2 too many cases! Because x=y and -x=-y are identical equations, your cases 1 and 2 are identical. Because x=-y and -x=y are identical equations, your cases 3 and 4 are identical.

Testluv wrote:@gmatmachoman: in analyzing (1), you tested 2 too many cases! Because x=y and -x=-y are identical equations, your cases 1 and 2 are identical. Because x=-y and -x=y are identical equations, your cases 3 and 4 are identical.

Yeah agreed!! Thx for that!! I am all worried about probabilities..somebody on this earth plz plz help me out in some way!!