Question 990770
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Your equation as rendered has two equally valid interpretations. One is a linear equation, and the other is not.  Repost using parentheses to make it clear whether the independent variable *[tex \Large x] is in the denominator or not.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \