young_gun wrote:

bkk145 wrote:

young_gun wrote:

Is |x| < 1?

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

(2) |x-3| ≠ 0

C.

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

x = 3 or x=1/3

INSUFFICIENT

(2) x ≠ 3

INSUFFICIENT

Together, x=1/3

SUFFICIENT

bkk, do you think you could show me how in (1) you solved x=3 or x=1/3? I'm not too great with abs values. Thanks!

Sure, you have

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

You know that

|x+1| is positive when x>-1 and negative when x<-1

|x-1| is positive when x>1 and negative when x<1

This means that in a number line, the intervals you have are:

x<-1, -1<x<1, x>1

Now, solve the equation according to those intervals:

For x<-1,

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

-x-1 = -2x+2

x = 3

For -1<x<1,

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

x+1 = -2x+2

x = 1/3

For x>1

x+1 = 2*(x-1)

x+1 = 2x-2

x = 3

x equals to 3 or 1/3

I know, all these signs are confusing! I attached a document for you, take a look. It helped me.