Question 1122999
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xy\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2y\ =\ 11]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{1}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \frac{2}{x}\ =\ 11]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2\ =\ 11x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 11x\ +\ 2\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{11\ \pm\sqrt{121\ -\ (4)(1)(2)}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{11\ -\ \sqrt{113}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ \frac{11\ +\ \sqrt{113}}{2}]


You need to check both of these answers.  Since we introduced a squared term in the process of solving the problem, there is the potential that we introduced an extraneous root.  Hence, you either have 2, 1, or 0 answers to the problem.  I'll leave it to you to figure out which is correct.  Remember when you check these numbers in the second equation that *[tex \Large x] is presumed to be the larger of itself and its reciprocal.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

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