Question 1188035
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To start:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C\ +\ D\ +\ E\ =\ 410]


Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}C\ +\ D\ +\ \(E\ +\ \frac{1}{3}C)\ =\ 410]


Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}C\ +\ \frac{2}{5}D\ =\ \frac{3}{5}D\ +\ 20]


This simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}C\ -\ \frac{1}{5}D\ =\ 20]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ E\ +\ \frac{1}{3}C\ =\ \frac{3}{5}D\ +\ 30]


But from the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ E\ =\ 410\ -\ C\ -\ D]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 410\ -\ C\ -\ D\ +\ \frac{1}{3}C\ =\ \frac{3}{5}D\ +\ 30]


Which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{2}{3}C\ -\ \frac{8}{5}D\ =\ -380]


Solve the 2X2 System:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}C\ -\ \frac{1}{5}D\ =\ 20]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{2}{3}C\ -\ \frac{8}{5}D\ =\ -380]


for *[tex \Large C]

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

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