Question 517053
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Break down each of the denominators into its prime factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9w^2\ =\ 3\,\cdot\,3\,\cdot\,w\,\cdot\,w]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6wt\ =\ 2\,\cdot\,3\,\cdot\,w\,\cdot\,t]


2 occurs most frequently in *[tex \LARGE 6wt] and it occurs 1 time


3 occurs most frequently in *[tex \LARGE 9w^2] and it occurs 2 times


*[tex \LARGE w] occurs most frequently in *[tex \LARGE 9w^2] and it occurs 2 times


*[tex \LARGE t] occurs most frequently in *[tex \LARGE 6wt] and it occurs 1 time.


The Lowest Common denominator then needs one 2, two 3s, 2 *[tex \LARGE w]s, and 1 *[tex \LARGE t], that is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\,\cdot\,3\,\cdot\,3\,\cdot\,w\,\cdot\,w\,\cdot\,t\ =\ 18w^2t]


Then since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{18w^2t}{9w^2}\ =\ 2t] we need to multiply the first fraction, *[tex \LARGE \frac{4t}{9w^2}] by 1 in the form of *[tex \LARGE \frac{2t}{2t}], to wit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{4t}{9w^2}\right)\left(\frac{2t}{2t}\right)\ =\ \frac{8t^2}{18w^2t}]


Do the other one the same way. 


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