Question 188424
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Let's get your terminology correct for starters.  You can't really <i><b>solve</b></i> anything here because you don't have an equation -- there is no equals sign.  You can, however, attempt to simplify this.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{rˆ2-5r+6}{r^2-4}]


Factor both numerator and denominator polynomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(r-2)(r-3)}{(r-2)(r+2)}]


Eliminate the common factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{r-3}{r+2}]


So you can now say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{rˆ2-5r+6}{r^2-4} = \frac{r-3}{r+2} \ \ \forall\ r\ \in\ \R\, |\, r\,\neq\,2]


So, why is there a restriction on the value of <i>r </i>?  You tell me.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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