Question 680539
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How did you manage to get from *[tex \LARGE 6x\ -\ 18] to *[tex \LARGE 2(x\ -\ 3)]?


Factor out a 6 in the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6x\ -\ 18}{3\ -\ x}\ =\ \frac{6(x\ -\ 3)}{3\ -\ x}]


Ok, "so what?" he says


Here's a little hint, and perhaps the core point of this lesson.  For any real numbers *[tex \LARGE a] and *[tex \LARGE b],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ -(b\ -\ a)]


So

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 3\ =\ -(3\ -\ x)]


and then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6x\ -\ 18}{3\ -\ x}\ =\ \frac{6(x\ -\ 3)}{3\ -\ x}\ =\ \frac{-6(3\ -\ x)}{3\ -\ x}\ =\ -6] 


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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