x>y??

Hi,

You can consider doing that for statement 1.

x – y^2 > 0... for inequalities you can add a value to both sides.. so by adding y^2 to both sides:

x - y^2 + y^2 > 0 + y^2
x > y^2

Since y^2 is either 0 or positive.. and x > y^2 ..that tells us that:
1 - x must be a positive value
2 - x > y^2

Now.. since there is no restriction on the value of x and y.. there is a possibility for fractions here... so consider y = 1/3 -> y^2 = 1/9..
- x can be 1/4 and will satisfy the condition x > y^2...but x will be < y
- x can be 1/2 and will satisfy x > y^2 but x will be > y

Statement one alone is not sufficient but it sure tells you that x is a positive value.

Combining this with statement 2 will confirm that x > y

elegant really.

statement 1)

x-y^2>0
x>y^2
so abs(x)>y

can't verify if x>y. insufficient

statement 2)

xy<0

so x and y are opposite signs. but we don't know if x is negative or y is negative. so we don't know if x>y

insufficient.

combined

saying absolute value of x is greater than y whether x is negative or positive so x is greater than y.

c