tejal777 wrote:

I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,

a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1

b. when x<0,x=-5,|x| is still 5

how does this become x>-1??

Is

|x| < 1?

Is

|x| < 1, means is

x in the range (-1,1) or is

-1<x<1 true?

(1)

|x + 1| = 2|x - 1|Two key points:

x=-1 and

x=1 (key points are the values of x when absolute values equal to zero), thus three ranges to check:

---------{-1}--------{1}---------A.

x<-1 (blue range) -->

|x + 1| = 2|x - 1| becomes:

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

x=3, not OK, as this value is not in the range we are checking (

x<-1);

B.

-1\leq{x}\leq{1} (green range) -->

|x + 1| = 2|x - 1| becomes:

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

x=\frac{1}{3}. OK, as this value is in the range we are checking (

-1\leq{x}\leq{1});

C.

x>1 (red range) -->

|x + 1| = 2|x - 1| becomes:

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

x=3. OK, as this value is in the range we are checking (

x>1).

So we got TWO values of

x (two solutions):

\frac{1}{3} and

3, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2)

|x - 3|\neq{0}Just says that

x\neq{3}. But we don't know whether

x is in the range (-1,1) or not.

(1)+(2)

x=\frac{1}{3} or

x=3 AND

x\neq{3} --> means

x can have only value

\frac{1}{3}, which is in the range (-1,1). Sufficient.

Answer: C.

Hope it helps.