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[m][fraction]x^{-1} - y^{-1} / (xy)^{-1} (x-y)[/fraction][/m] = [#permalink]
1
Expert Reply
Nothing fancy

first brush the exponents' rules https://gre.myprepclub.com/forum/gre-ma ... 24948.html

Now

\(\frac{x^{-1} - y^{-1} }{ (xy)^{-1} (x-y)}\) =

actually we have a double fraction one over the other

Solve the numerator

\(x^{-1} - y^{-1} \) \(= \frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\)

The denominator


\(\frac{1}{xy}*(x-y)\)

Combine in a fraction \(\frac{y-x}{xy}*\frac{xy}{x-y}\)

now simplify in cross but be careful you cannot cancel out \(y-x\) and \(x-y\) which have a different sign. so change sign in minus in the first one \(-(x-y)\)

In the end

\(-\frac{(x-y)}{xy} * \frac{xy}{(x-y)}= -\frac{1}{1}=-1\)

A is the answer
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Re: [m][fraction]x^{-1} - y^{-1} / (xy)^{-1} (x-y)[/fraction][/m] = [#permalink]
1
I see my mistake, thanks!

Carcass wrote:
Nothing fancy

first brush the exponents' rules https://gre.myprepclub.com/forum/gre-ma ... 24948.html

Now

\(\frac{x^{-1} - y^{-1} }{ (xy)^{-1} (x-y)}\) =

actually we have a double fraction one over the other

Solve the numerator

\(x^{-1} - y^{-1} \) \(= \frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\)

The denominator


\(\frac{1}{xy}*(x-y)\)

Combine in a fraction \(\frac{y-x}{xy}*\frac{xy}{x-y}\)

now simplify in cross but be careful you cannot cancel out y-x and x-y which have a different sign. so change sign in minus in the first one \(-(x-y)\)

In the end

\(-\frac{(x-y)}{xy} * \frac{xy}{(x-y)}= -\frac{1}{1}=-1\)

A is the answer
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