If x!=0, is (x^2 + 1)/x > y?

(1) x = y

(2) y > 0

What will be the correct answer to this question? I am confused.

Approach 1:

I tried to simplify this question as follows

x^2 + 1 - xy > 0

x(y-x) < 1

But we can't be sure of the sign of the inequality as I have cross multiplied by x (whose sign I do now know).

So in essence the question can be rephrased as:

x(y-x) [greater than, equal to or less than] 1 --> This is what is actually being asked.I want to know if i have made any fundamental mistake when doing this rephrasing. As I have already mentioned I know that the sign of the inequality should be reversed when multiplied by a negative number.

So going by this approach the answer is A

Approach 2:

I will consider both cases of x before cross multiplying.

Case 1: x is negative

x^2+1 < xy

x^2-xy + 1 < 0

x(x-y) + 1 <0Case 2: x is positive

x^2+1 > xy

x^2-xy + 1 > 0

x(x-y) + 1 > 0So based on these 2 equations we find that statement (1) alone is not sufficient because it gives different answers for Case 1 and 2 i.e. 1 < 0 and 1 > 0.

Statement 2 is also insufficient because it does not say anything about x.

Combining 1 and 2 -

y > 0

x = y

Therefore, x > 0

x(x-y) + 1 > 0Therefore C is the answer.

The second approach looks more methodical but I want to understand why the first approach is INVALID (if it is so).