Question 635544
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I think the key to what you are doing wrong is in your statement "I distributed the 4 and -5 to (y + 2)..."  The 4 has nothing to do with the (y + 2) factor.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ -\ 5(y\ +\ 2)\ =\ 2(y\ -\ 1)\ -\ 7y\ -\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ -\ 5y\ -\ 10\ =\ 2y\ -\ 2\ -\ 7y\ -\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -5y\ -\ 6\ =\ -5y\ -\ 6]


Which is true for any real value of *[tex \LARGE y], hence the solution set is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left{y\ \in\ \mathbb{R}\right}]


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