Question 333680
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Missed it by <i><b>THAT</i></b> much (he said holding his thumb and forefinger apart with a scant millimeter between them).


You were absolutely on the right track, you simply stumbled over a little algebra toward the end.


You were absolutely correct up to and including the step where you said:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ 1\ =\ \frac{5}{3}\ +\ b] 


But then you treated *[tex \Large \frac{5}{3}] as if it were a coefficient of *[tex \Large b] rather than a separate term.  Rather than saying *[tex \Large b] is the reciprocal of *[tex \Large \frac{5}{3}], you needed to add *[tex \Large -\frac{5}{3}] to both sides of your equation, thus:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ 1\ -\ \frac{5}{3}\ =\ b]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ b\ =\ -\frac{2}{3}]


And finally you can write:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{1}{3}x\ -\ \frac{2}{3}]



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