Solution to quadratic question of the form 0/0

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.

• To get from $x^2-1=0$ to $(x-1)/(1-x)=1/x$, you have to assume that $x\neq 1$, or you are dividing by zero. – Alex S Jun 24 '15 at 3:21
• It would have been simpler to note that if $x\ne 1$ then $\frac{x-1}{1-x}=-1$. – André Nicolas Jun 24 '15 at 3:28
• Just because you can go from one equation to another doesn't mean they're equivalent. – anon Jun 24 '15 at 3:28

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: $$x^2-1=0\text{ and } x\ne 1.$$

• More precisely, although not directly influencing the results, the original equation is equivalent to this: $$x^2-1=0\, \mathrm{and}\,x\ne1\,\mathrm{and}\,x\ne0.$$ – Ruslan Jun 24 '15 at 10:41
• @Ruslan : But $x^2-1=0$ already implies that $x\ne0$. ${}\qquad{}$ – Michael Hardy Jul 29 '15 at 18:11

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.

• You should add an answer to the question "where is the catch", meaning, where in OP's argumentation is the error. – Alfe Jun 24 '15 at 8:25

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 $x=0$).

• @Quincunx then my English is wrong because I did not mean that, I agree with what you said. What I am stating is regarding the final equation, not the original expression, for the original expression are still $-1$ and $+1$. – iadvd Jun 24 '15 at 5:31
• @Quincunx just to clarify I have added some explanation, non English native speaker here... – iadvd Jun 24 '15 at 5:33
• Okay. I want to suggest that you mention that there is another value which $x$ cannot be: $0$. – Justin Jun 24 '15 at 6:02
• @Quincunx true! I was too focused on explaining the left side, thanks for that. – iadvd Jun 24 '15 at 6:15