Solution to quadratic question of the form 0/0

Question

What are the possible values of $x$ for the following equation:

$$\frac{x - 1}{1 - x} = \frac1x$$

This equation is equivalent to $$x^2 - 1 = 0$$

which factors to $1, -1$.

However, is $1$ the correct answer to the original form of equation? Given that if one substitutes $1$ in the given form of equation, the LHS becomes $\frac{0}{0}$. In case, $1$ is actually the correct answer, please explain what is the catch here.

Answer 1

It is not equivalent to $x^2-1=0$. You can multiply both sides by $x(1-x)$ and get $x^2-1=0$, but that is valid only if the denominator $1-x$ is not $0$, and the denominator is $0$ when $x=1$, so you have to check separately whether $x=1$ is a solution to the original equation $\dfrac{x-1}{1-x}=\dfrac 1 x$. And it is not. So the original equation is actually equivalent to this: $$ x^2-1=0\text{ and } x\ne 1. $$

Answer 2

To write the first equation, there is an implicit condition in the moment you decide to write $\frac{x-1}{1-x}$ to avoid dividing by zero, which is $x\ne 1$. Thus, even though the equation you arrive at ($x^2-1=0$) has solutions $\pm 1$, the condition must be considered since you arrived at it from the first.

For that reason, the possible values of $x$ for the equation are all except $x=1$, and $x=0$ as well (because for the same reason regarding divisibility, on the right side $\frac{1}{x}$ can not be $x=0$).

Answer 3

Note that the equation $\frac{x-1}{1-x}=\frac{1}{x}$ only makes sense for $x \in \mathbb{R} \setminus \left\{0,1\right\}$, so $x=-1$ is the only root of the original equation.