Quadratic equation : Two solutions or one solution? I have an equation to solve for y:
$$\frac{y^2}{y}=1$$
Normally, I would cancel out one $y$ and get $y=1$ as a single solution.
But If I think of it as quadratic equation
$$y^2=y$$
$$y^2-y=0$$
$$y(y-1)=0$$
$$y=0 \space \text{or} \space y=1$$
to have two solutions.
But when I put $y=0$ in original equation, I get $\frac{0}{0}$, so is $y=0$ a solution or not ?
If yes, then I get $\frac{0}{0}$.
If no, then how come this quadratic equation has $1$ solution ?
 A: This equation has two solutions, $y=0$ and $y=1$:
$$y^2=y \tag{1}$$
This equation has one solution, $y=1$:
$$\frac{y^2}{y} = 1 \tag{2}$$
The reason that it has one solution is that either of the operations "Cancel a $y$ from the top and bottom of the fraction" and "Multiply both sides of the equation by $y$" are only valid when $y\neq 0$. In particular, if you want to manipulate (2) to look like (1), you first have to assume that $y\neq 0$, which rules out one of the solutions of the quadratic equation.

A more sophisticated answer realises that when we are asked to "solve" an equation, what we are really doing is looking for a root of a particular function (i.e. a value of the argument at which evaluating the function gives zero). In the case of (1), the function is $f(y)=y^2-y$, and in the case of (2) the function is $f(y)=y^2/y-1$.
Now, functions have to have a domain. In the case of (1) the domain is $\mathbb{R}$ (the real numbers) which includes both 0 and 1. In the case of (2), the domain is $\mathbb{R}-\{0\}$ (the real numbers without 0) which doesn't include 0.
