imranmah wrote:

PareshGmat wrote:

\(\frac{p+q}{p-q}\)

Replace p with \(\frac{-1}{p}\) & q with \(\frac{-1}{q}\)

\(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)

\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}\)

\(\frac{\frac{p+q}{pq}}{\frac{q-p}{pq}}\)

\(\frac{p+q}{q-p}\)

Answer = A

Would you be able to explain how you went from: \(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)

to:\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}\)

I was just having difficulty following. Thank you for your help!

Take -1 common from both numerator and denominator.

\(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)

\(\frac{(-1)*(\frac{1}{p} + \frac{1}{q})}{(-1)*(-\frac{1}{q} + \frac{1}{p})}\)

Cancelling off the (-1) from the numerator with that from the denominator, you get

\(\frac{\frac{1}{p} + \frac{1}{q}}{-\frac{1}{q} + \frac{1}{p}}\)

\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p}-\frac{1}{q}}\)

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