solving three quadratic equations
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A=458.21
B=256.84
C=308.95
A=m*8/m*8+n*9+p*14
B=n*9/m*8+n*9+p*14
C=p*14/m*8+n*9+p*14
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Joseph Cheng
2014-7-9
substitution? not too hard of an series to solve by hand as the first A equation can simplified and substitute the 9n+14p = 394.21.
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