Question 190977
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a - b + c = 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a - 2b = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3a + b - 3c = 2]


Look at the middle equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a - 2b = 0 \ \ \Rightarrow\ \ 2a = 2b \ \ \Rightarrow\ \ a = b]


Now use that fact in the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a - b + c = 1\text{ and }a = b  \ \ \Rightarrow\ \ 0 + c = 1  \ \ \Rightarrow\ \ c = 1]


Use this information in the last equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3a + b - 3c = 2 \ \ \Rightarrow\ \ 3a + a - 3(1) = 2  \ \ \Rightarrow\ \ 4a = 5  \ \ \Rightarrow\ \ a = \frac{5}{4}  \ \ \Rightarrow\ \ b = \frac{5}{4} ]


Check:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{4} - \frac{5}{4} + 1 = 1], yep


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\,\cdot\,\frac{5}{4} - 2\,\cdot\,\frac{5}{4} = 0], yep


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\frac{5}{4} + \frac{5}{4} - 3(1) = \frac{15}{4}+\frac{5}{4} - 3 = \frac{20}{4} - 3 = 5 - 3 = 2], yep



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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