求一个二元一次方程组的详细解法

x+1.3*(y-x)=0.22x+0.9*(y-x)=0.16
2025-04-07 23:21:44
推荐回答(3个)
回答1:

x+1.3*(y-x)=0.22 (1)
x+0.9*(y-x)=0.16 (2)
(1)-(2)
0.4(y-x) = 0.06
y-x=0.15 (3)
from (1) and (3)
x+1.3*(y-x)=0.22
x+ 1.3*0.15 =0.22
x+0.195=0.22
x=0.025
from (3)
y-x=0.15
y-0.025=0.15
y=0.175

回答2:

二元一次方程组的解法!

回答3: