# Finding a value for $x$ when there should be none

I have an algebra problem in which I should find $x$. However, even though I manage to define $x = 1$, the text book says that there is no answer. Where am I going wrong in solving it?

$$\frac{\frac{x+1}{x-1}-1}{1+\frac{1}{x-1}} = 2$$ $$\iff \frac{x+1 - (x-1)}{(x-1)+1} = 2$$ $$\iff \frac{x+1 - x+1}{x-1+1} = 2$$ $$\iff \frac{2}{x} = 2$$ $$\iff 2 = 2x \iff x = 1$$

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Try substituting $x=1$ in the original equation and see what happens. –  Mark Bennet Aug 20 '12 at 15:05
The initial equation makes sense for $x\not =1$ and $x\not =0$. –  Raymond Manzoni Aug 20 '12 at 15:12
Your double arrows should be single arrows, unless you are careful about possibly multiplying by $0$. –  André Nicolas Aug 20 '12 at 15:12
@AndréNicolas You mean dividing by $0$? –  Quispiam Aug 20 '12 at 15:14
The first transformation involves multiplying by $\frac 0 0$ when $x=1$ ... –  Mark Bennet Aug 20 '12 at 15:14
Whenever you want to solve an equation like $$f(x)=0,$$ you must always write down any value of $x$ where $f$ cannot be computed. In other words, the unknown cannot live outside the largest set of numbers where $f$ is defined. In your case, there is a division by $x-1$, and you know that division by zero is not allowed. So you have the condition $x \neq 1$. All the double implications are true under this condition, and you must conclude that no value of $x$ is a solution.
If you take $x=1$, then you are dividing by $0$ in the fractions $\frac{x+1}{x-1}$ and $\frac{1}{x-1}$