Question 572842
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You were correct up to this point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5(y\ +\ 2)\ -\ 10}{(y\ -\ 3)(y\ +\ 2)}\ =\ \frac{y^2\ -\ 3y}{(y\ -\ 3)(y\ +\ 2)}]


But *[tex \LARGE 5(y\ +\ 2)\ -\ 10\ =\ 5y\ +\ 10\ -\ 10\ =\ 5y\ \neq\ 5y\ +\ 20], so the next step looks like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5y}{(y\ -\ 3)(y\ +\ 2)}\ =\ \frac{y^2\ -\ 3y}{(y\ -\ 3)(y\ +\ 2)}]


Then multiply both sides by the denominator to eliminate the denominator leaving you with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5y\ =\ y^2\ -\ 3y]


Which becomes


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ -\ 8y\ =\ 0]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 8]


Check to be certain that neither root will drive any original denominator to zero before declaring either an element of the solution set.  You might want to check the work by substituting back into the original equation.


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