Re: If x does not equal y...
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20 Sep 2012, 00:32
I think the formatting of your solution (inside the spoiler tags) became illegible, but I may be able to guess at the mistake. When we replace 'x' with '1/x', and 'y' with '1/y', we have the following:
\(\frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x} - \frac{1}{y} }\)
I think you then rewrote this as \(\left( \frac{1}{x} + \frac{1}{y} \right) (x - y)\), and if so, that isn't right. If that's what you did, you were probably trying to apply the rule "when you divide by a fraction, multiply by its reciprocal". There are two problems with trying to do that right away: we aren't dividing by a simple fraction here, so we can't use that rule (in the denominator, we're subtracting one fraction from another), and regardless, x-y is not the reciprocal of (1/x) - (1/y). If those terms were reciprocals of each other, their product would be 1, and if you multiply them out, you'll see their product is not 1.
As a first step, since we have fractions in both our numerator and denominator, I'd just multiply the top and bottom by xy. We then need to multiply by -1 on top and bottom only because of how the question designer chose to write the correct answer choice:
\(\begin{align}\\
\frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x} - \frac{1}{y} } &= \left( \frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x} - \frac{1}{y} } \right) \left( \frac{xy}{xy} \right) \\\\
&= \frac{y + x}{y - x} \\\\
&= \left( \frac{x + y}{-x + y} \right) \left( \frac{-1}{-1} \right) \\\\
&= \frac{-(x+y)}{x-y}\\
\end{align}\)