Question 396395
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In a word, no.


Presuming you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{9}n\ -\ 5\ =\ 2\ +\ \frac{5}{9}n]


(and not *[tex \Large \frac{2}{9n}\ -\ 5\ =\ 2\ +\ \frac{5}{9n}], which is an equally valid interpretation of what you wrote)


Add *[tex \Large -\frac{5}{9}n] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{3}{9}n\ -\ 5\ =\ 2]


Add 5 to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{3}{9}n\ =\ 7]


Reduce the fraction:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{1}{3}n\ =\ 7]


Multiply both sides by -3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ =\ -21]


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