Question 728696
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You are not dividing "1y" by "6y".  You have already taken care of the "y" once you get the exponent part straightend out.  You have a factor of 3 in the numerator and a factor of 6 in the denominator.  That reduces to 1 in the numerator and a 2 in the denominator just the same as when you worked with regular "numbers only" fractions in elementary school.


Now, as to the exponents on the y.  *[tex \LARGE y^4] in the numerator and *[tex \LARGE y^8] in the denominator.  You can look at this a couple of ways, but you always get to the same place.


You can take the *[tex \LARGE y^8] in the denominator, change the sign, and move it to the numerator, then you can add 4 and -8 to get *[tex \LARGE y^{-4}] in the numerator.  If you have a requirement to end up with all positive exponents, then you can simply take the *[tex \LARGE y^{-4}] in the numerator, change the sign, and move it to the denominator.


On the other hand, you can take the *[tex \LARGE y^4] in the numerator, change the sign, and move it to the denominator.  Then add 8 and -4 to get *[tex \LARGE y^4] in the denominator.  Same result as above.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x^2y^4}{6y^8}\ =\ \frac{x^2y^4y^{-8}}{2}\ =\ \frac{x^2y^{4\,-\,8}}{2}\ =\ \frac{x^2y^{-4}}{2}\ =\ \frac{x^2}{2y^4}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x^2y^4}{6y^8}\ =\ \frac{x^2}{2y^8y^{-4}}\ =\ \frac{x^2}{2y^{8\,-\,4}}\ =\ \frac{x^2}{2y^4}]


Whichever tickles your fancy, or more to the point, whichever pleases your instructor.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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