Question 633807
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It is nothing more than one rational expression divided by the other.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ \ \frac{b^2\ +\ 5b\ +\ 6}{3bc}\ \ \ }{\frac{b^2\ -\ 9}{6bc}}]


Which is the same as saying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{b^2\ +\ 5b\ +\ 6}{3bc}\ \div\ \frac{b^2\ -\ 9}{6bc}]


Just like dividing any other fractions, you invert and multiply. (That's right: You are NOT allowed to forget something you learned in the 4th grade just because you are now taking algebra)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{b^2\ +\ 5b\ +\ 6}{3bc}\ \times\ \frac{6bc}{b^2\ -\ 9}]


Hint: You will want to factor the numerator on the left and the denominator on the right because you will find a common factor that you can eliminate.  Right, reducing to lowest terms works just as well here as it does with plain numbers.


Let me know what you come up with.


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