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?
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A MGMAT Inequality
Source: Beat The GMAT — Data Sufficiency |
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?
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 rightpkw209 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?
"If you don't know where you are going, any road will get you there."
Lewis Carroll
Lewis Carroll
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!!
st 2 : |x - 3| > 0
case 1 :
x-3 >0
>3Case 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!!
