Question 997872
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ 2b\ +\ 2c\ =\ 64]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a\ +\ b\ +\ 2c\ =\ 62]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a\ +\ 2b\ +\ c\ =\ 59]
<pre>
    1    2    2      64
    2    1    2      62
    2    2    1      59

-2R1 + R2 -> R2
-2R1 + R3 -> R3

    1    2    2      64
    0   -3   -2     -66
    0   -2   -3     -69

-1/3R2 -> R2

    1    2    2      64
    0    1    2/3    22
    0   -2   -3     -69

2R2 + R3 -> R3

    1    2    2      64
    0    1    2/3    22
    0    0   -5/3   -25  

-3/5R3 -> R3

    1    2    2      64
    0    1    2/3    22
    0    0    1      15

-2/3R3 + R2 -> R2

    1    2    2      64
    0    1    0      12
    0    0    1      15

-2R2 - 2R3 + R1 -> R1

    1    0    0      10
    0    1    0      12
    0    0    1      15
</pre> 


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

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