I used substituting values as below and got D as answer:

S1)

|x+3| = 4x-3

when x=

-2 eq becomes 1 = -7 -->absurd

-1 eq becomes 2 = -7 -->absurd

0 eq becomes 3 = -3 -->absurd

1 eq becomes 5 = 5

2 eq becomes 6 = 9 -->absurd

so we can tell only x=2 satisifies the eq and hence S1 is sufficient.

similarly for S2)

|x+1| = 2x -1

when x =

-2 eq becomes 1 = -5 -->absurd

-1 eq becomes 0 = -3 -->absurd

0 eq becomes 1 = -1 -->absurd

1 eq becomes 2= 0 -->absurd

2 eq becomes 3 = 3

3 eq becomes 4 = 5 -->absurd

again we can tell x = 2 alone satisifes eq, hence S2 is sufficient.

Answer D.

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Thanks, Sri

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