Question 362058
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Yes, but not for the reason you state.  Your process is flawed and you just got lucky with this one.  Notice that the numerator of your fraction is the difference of two squares, so factor the numerator.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{64x^2\ -\ 100y^2}{8x\ +\ 10y}\ =\ \frac{(8x\ +\ 10y)(8x\ -\ 10y)}{8x\ +\ 10y}]


Then eliminate the common factor in the numerator and denominator.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ -\ 10y]



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