How can I solve  [math]\frac{1}{x} +\frac{1}{y} =\frac{1}{3} [/math] and [math]\frac{x^2}{y} +\frac{y^2}{x} =12 [/math]?
If [math]\frac{x!}{(x-y)!*y!} ÷ \frac{x!}{(x-(y+1))!*(y+1)!} ÷ \frac{x!}{(x-(y+2))!*(y+2)!} = 1 ÷ 2 ÷ 3[/math], what does [math](x-y)[/math] equal to?