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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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    @AndréNicolas Ah, yes, I see. Thank you.2012-08-20

2 Answers 2

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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.

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If you take $x=1$, then you are dividing by $0$ in the fractions $\frac{x+1}{x-1}$ and $\frac{1}{x-1}$

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    This is the second time in a row I've been stuck on a problem simply because I forgot about the zero-divide no-no. :/ Thanks.2012-08-20