Question 230497
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The solution to this problem depends on whether you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\ -\ \frac{10}{y}\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{y\ -\ 10}{y}\right)]



Whichever, let *[tex \Large x\ =\ \left(y\ -\ \frac{10}{y}\right)] or *[tex \Large x\ =\ \left(\frac{y\ -\ 10}{y}\right)]


Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ - 27\ =\ 0]


which factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)(x\ +\ 9)\ =\ 0]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3\ \ ] or *[tex \LARGE \ \ x\ =\ -9]


Now replace the substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\ -\ \frac{10}{y}\right)\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ -\ 3y\ -\ 10\ = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y\ -\ 5)(y\ +\ 2)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 5\ \ ] or *[tex \LARGE \ \ y\ =\ -2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\ -\ \frac{10}{y}\right)\ =\ -9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ 9y\ -\ 10\ = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y\ -\ 1)(y\ +\ 10)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 1\ \ ] or *[tex \LARGE \ \ y\ =\ -10]


So the solution set is *[tex \LARGE \{-10,\,-2,\,1,\,5\}]


OR


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{y\ -\ 10}{y}\right)\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 10\ =\ 3y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -5]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{y\ -\ 10}{y}\right)\ =\ -9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 10\ =\ -9y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 1]


So the solution set is *[tex \LARGE \{-5,\,1\}]


Use parentheses next time so that I don't have to do twice the work.


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