Empty sym: 0-by-1

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ali kaptanoglu
ali kaptanoglu 2020 年 9 月 7 日
コメント済み: Walter Roberson 2020 年 9 月 9 日
clc;
clear;
syms x y z
a=-53.774+70+((30*(2*(37.839-x)-y))/((2*z-x)-y));
b=-54.827+70+((30*(2*(38.886-x)-y))/((2*z-x)-y));
c=-55.879+70+((30*(2*(39.932-x)-y))/((2*z-x)-y));
denklem=solve(a,b,c);
denklem.x

回答 (1 件)

Walter Roberson
Walter Roberson 2020 年 9 月 9 日
Your equations are inconsistent. There is no solution.
>> subs(c,solve([a,b],[x,y]))
ans =
-1/174500 == 0
Your third equation is not consistent with the first two.
  1 件のコメント
Walter Roberson
Walter Roberson 2020 年 9 月 9 日
The problem is in floating point round off.
syms x y z
T = sym(19501769)/sym(349000)
a=-53.774+70+((30*(2*(37.839-x)-y))/((2*z-x)-y));
b=-54.827+70+((30*(2*(38.886-x)-y))/((2*z-x)-y));
c=-T+70+((30*(2*(39.932-x)-y))/((2*z-x)-y));
subs(c,solve([a,b],[x,y]))
The result will be 0.
If you solve([a,b,c]) then you will get a numeric x and y, and z would be 0. This is not the full story. What the above tells you is that with that set of equations, you only have two independent variables, and the third equation will be satisfied if you know the values for any two of the variables.

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