问题描述
说你有
3x + 2y = 11
2x - 3y = 16
您将如何在Java中计算 x和y
?
How would you work out x and y
in Java?
做完一些代数运算后,我发现 x = de-bf/ad-bc
和 y = af-ce/ad-bc
After doing some algebra I figured out that x = de-bf / ad-bc
and y = af-ce / ad-bc
这些显示字母是什么 a + b = e
和 c + d = f
These show what the letters area + b = e
and c + d = f
每当我编写代码时,它总是给我错误的答案,我不确定这是否是由于使用int而不是double或其他原因造成的.也可以解析方程式中的字母,例如
Whenever I write the code it always gives me the wrong answer, I am not sure if that is due to using int instead of doubles or what. Would it also be possible to parse the letters from the equation e.g
input: 5x - 3y = 5
parased as: a = 5, b = -3 and e = 5
这是没有解析的代码
public static void solveSimultaneousEquations(double a, double b, double c, double d, double e, double f) {
double det = 1/ ((a) * (d) - (b) * (c));
double x = ((d) * (e) - (b) * (f)) / det;
double y = ((a) * (f) - (c) * (e)) / det;
System.out.print("x=" + x + " y=" + y);
}
推荐答案
问题是您将行列式两次除法!
The problem is that you divide your determinant twice!
您的公式是
x = de-bf / ad-bc
y = af-ce / ad-bc
det = ad-bc
如此:
x = de-bf / det
y = af-ce / det
但是你计算:
double det = 1/ ((a) * (d) - (b) * (c));
因此您程序中的 det
不是来自公式的 det
,而是 1/det
!
so in your program det
is not det
from the formula, but 1/det
!
所以您可以纠正:
double det =((a)*(d)-(b)*(c));
double det = ((a) * (d) - (b) * (c));
或
double x =((d)*(e)-(b)*(f))* det;double y =((a)*(f)-(c)*(e))* det;
double x = ((d) * (e) - (b) * (f)) * det; double y = ((a) * (f) - (c) * (e)) * det;
我喜欢第一个:
public static void solveSimultaneousEquations(double a, double b, double c, double d, double e, double f) {
double det = ((a) * (d) - (b) * (c)); //instead of 1/
double x = ((d) * (e) - (b) * (f)) / det;
double y = ((a) * (f) - (c) * (e)) / det;
System.out.print("x=" + x + " y=" + y);
}
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