Question 483799
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*[tex \LARGE\ \ \ \ \ \ \ \ \ \  \frac{1}{a}\ =\ \frac{1}{b}\ +\ \frac{1}{y}]


Add *[tex \Large -\frac{1}{b}] to both sides:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \  \frac{1}{y}\ =\ \frac{1}{a}\ -\ \frac{1}{b}]


LCD is *[tex \Large ab] so:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \  \frac{1}{y}\ =\ \frac{b\ -\ a}{ab}]


Take the reciprocal of both sides:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ y\ =\ \frac{ab}{b\ -\ a}]


provided


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ a\ \neq\ b]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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