Can we simplify the following fraction as :-

$\displaystyle (\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0$

$\displaystyle (x^2 + y^2 +2) * (\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0 * (x^2 + y^2 +2)$

$\displaystyle

x^2 + 3xy + 2 = 0$

and then further simplification....

A friend of mine was contradicting this approach and was saying that if the denominator contains variables then you simply can't do it like this. His solution however was that :-

the solution would be that :-

Either, $\displaystyle x^2 + 3xy + 2 = 0$ or $\displaystyle x^2 + y^2 +2 = infinity$

What is the real solution ?