stmt 1 - there can be two solutions for x in this case.
x+1 =2(x-1)...... x= 3.
x+1 =2(1-x)....when modulus is used both conditions of difference need to be considered....In this case x = 1/3.....Since x has two values ...Insuff

stmt 2 - x not equal to 3 is the sol.....Insuff

Both - the only possible sol is x=1/3....Suff ...........So ans is C.

can you pls confirm the answer. Thanks

Hi

Can some one pls help me whit this one?

This is the way I solved statement 1)

lx+1l = 2lx-1l

so there are two options

Option 1)

(x+1) = 2 (x-1), in which case x = 3

Option 2)

-(x+1) = -2(x-1), in which case X = 3

Both the cases X > 1, so statement 1 is sufficient

However Jess has a diff ans. Can any one confirm the right approach and explain it?

Thanks

Hi ri2007,


Your option 1 is the same as the option 2.

When we have a modulus on a side that means it can have two values +/-. When there is a modulus on both sides that means that both sides will have a +/-

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

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

which leads to the following options:

x+1 = 2(x-1) or -(x+1) = - 2(x-1) (they are same)

AND

-(x+1) = 2(x-1) or x+1 = -2(x-1) (they are same as well)

Thanks so much us. Great explaination