I have found that the solutions that solve the first statement are: 3 and 2/3 and therefore statement 2 eliminates the first, hence C

however it took me 8 minutes...because I could not find many number to satisfy the first equation...any idea on how to make it quickly?

Carloblacksun wrote:I have found that the solutions that solve the first statement are: 3 and 2/3 and therefore statement 2 eliminates the first, hence C

however it took me 8 minutes...because I could not find many number to satisfy the first equation...any idea on how to make it quickly?

from stmt 1, I am getting x = 3 or 1/3

how did you get 2/3 ?

you are right, I meant 1/3, sorry... 3 AND 1/3...

but I still can't figure out how I got it...

stmt 1

|x+1| = 2|x-1|

4 possibilities

possibility 1:
x+1 = 2(x-1) => x+1 = 2x-2 => x = 3

possibility 2:
-(x+1) = 2(x-1) => -x-1 = 2x-2 => x = 1/3

possibility 3:
x+1 = -2(x-1) => x+1 = -2x+2 => x= 1/3

possibility 4:
-(x+1) = -2(x-1) => -x-1 = -2x+2 => x= 3
INSUFF

stmt 2

x != 3

INSUFF

combining both we get x = 1/3 SUFF

ans C.

Carloblacksun wrote:I have found that the solutions that solve the first statement are: 3 and 2/3 and therefore statement 2 eliminates the first, hence C

however it took me 8 minutes...because I could not find many number to satisfy the first equation...any idea on how to make it quickly?

Look at jimmiejaz's explanation on how to tacke abs questions.

https://www.beatthegmat.com/tricky-inequ ... 22180.html

HT helps

is this method correct? though m not getting the correct answer but really not able to understand where i m going wrong?
stmnt 1
|x+1| = 2|x-1|
|x+1| /|x-1| = 2
|(x-1)/(x-1)| =2
|((x-1)*(x+1)) / ((x-1)*(x-1))|=2
|(x^2-1)/(x-1)^2| =2 now we see numerator and denominator will be positive x^2 -1 will be positive & (x-1)^2 will be positive
(x^2-1)/(x-1)^2 =2
solving for x we get x=3 & x=1
m i not getting correct value of x ? it should be 3 & 1/3