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

Hence x+1=2x-2 or x+1=2-2x

So x=3 or 1/3...not sufficient

from 2
|x - 3| > 0 ie (x-3)>0 or (x-3)<0
so x<-3 or x>3 ..... sufficient

Ans option B

can you please confirm with OA?

Hey liferocks-

Thanks for the response.

Official answer is actually C

The second statement should be x>3 or x<3 so inefficient. Together, you know that x can't be 3 (based on statement 2) so x=1/3, which is less than 1.

I might be going crazy but what's the rule for absolute value inequality questions?

I usually just take the positive and the negative of the left side of the equation.

i.e., for |x + 1| = 2|x - 1|

--> -(x+1) = 2(x-1) -->-x-1 = 2x-2 -->-3x = -1, x = 1/3

AND

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

Again, I'm probably going crazy but this is right, right?

pkw209 wrote:Hey liferocks-

Thanks for the response.

Official answer is actually C

The second statement should be x>3 or x<3 so inefficient. Together, you know that x can't be 3 (based on statement 2) so x=1/3, which is less than 1.

I might be going crazy but what's the rule for absolute value inequality questions?

I usually just take the positive and the negative of the left side of the equation.

i.e., for |x + 1| = 2|x - 1|

--> -(x+1) = 2(x-1) -->-x-1 = 2x-2 -->-3x = -1, x = 1/3

AND

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

Again, I'm probably going crazy but this is right, right?

yup..you are absolutely right

This is the way I used to work for Inequakities :

st 2 : |x - 3| > 0

case 1 :

x-3 >0

>3

Case 2: -(x-3)>0

-x+3>0

-x>-3

Now Multiply using -1 on both sides to remove the negative sign of X.

x<3.

So from above 2 cases, we have x> 3 or x<3. To cross check try some random values plugged into the equation.

ST 1:

U have nicely followed the steps.

Combining st1 & st2, we know that X can't be 3. SO X has to be less than 3. So we have to look out for the value. From St 1 we have X= 1/3.

|1/3| <1. Solved!!

Thanks guys!