solving three quadratic equations

benjamin

2014 7 9

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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.

benjamin 2014-7-10

thanks for the answers but the equation is a bit complicated not the usual substitution method is to be use but instead after running the equation in the matlab it has to return a total of 1024.

to clarify this example the first equation states A=m*8/m*8+n*9+p*14 == m*8/m*8+n*9+p*14=458.21 and the final equation should be A+ B+C=D (1024) thanks.

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