Number of solutions to an equation

Let $x=ky$ (obviously, $k$ is a real number). Then, $k=ky-y$ , $y=$ $\frac{k}{(k-1)}$ , $x$ $=$ $\frac{k^2}{(k-1)}$ , and the ordered pairs of the type ( $\frac{k^2}{(k-1)}$ , $\frac{k}{(k-1)}$ ) are the solutions of this problem. There are infinite number of $k$ , except 0 and 1. Therefore, there are infinite number of $\frac{k}{(k-1)}$ and ordered pairs of real numbers ( $x, y$ ) , which satisfy the above equation.

Rearrange the formula: $\frac{x}{y} = x-y$ $x = xy - y^2$ $y^2 - xy + x = 0$

We now have a quadratic, and the discriminant will reveal the range of real solutions. $\Delta = (-x)^2 - 4x$ $= (-1)^2 x^2 - 4x$ $= x^2 - 4x$ $\Delta = x(x - 4)$

For discriminant values $\geq 0$ real values exist. In this case, values $x > 4$ will have an infinite number of positive solutions, since both $x$ and $x - 4$ will be positive, and the product of two positive numbers is also positive. Values $x < 0$ will also have an infinite number of positive solutions, since both $x$ and $x - 4$ will be negative, and the product of two negatives is positive. Values for $x = 0$ and $x = 4$ will have $\Delta = 0$ . While $x = 0$ will yield a $\Delta = 0$ , it will lead to a contradiction, since $y \neq 0$ : $y^2 - xy + x = 0$ $y^2 - 0y + 0 = 0$ $y^2 = 0$ $y = 0$

The discriminant of the given function has an infinite amount of positive real solutions. Hence, the given function has an infinite amount of real solutions.