二元一次聯立方程式 X+(5Y+4)=438 X+(2Y+5)=200 (正確答案 Y=40 ) 請問詳細步驟怎解呢? 我解都解不出來40阿... 快起笑了 = =?
回答 (2)
2017-03-26 2:09 pm
Sol
x+(5y+4)=438
x+(2y+5)=200
[x+(5y+4)]-[x+(2y+5)]=438-200
3y－1=238
3y=239
y=239/3………………………………..
x+(5*239/3+4)=438
x+1207/3=438
x=107/3
check
107/3+(478/3+5)
=107/3+493.3
=200
x+(5y+4)=438
x+(2y+5)=200
[x+(5y+4)]-[x+(2y+5)]=438-200
3y－1=238
3y=239
y=239/3………………………………..
x+(5*239/3+4)=438
x+1207/3=438
x=107/3
check
107/3+(478/3+5)
=107/3+493.3
=200
2017-03-24 10:41 pm
X+(5Y+4)=200 ...(1)
438 X+(2Y+5)=200 ...(2)
(1)*438 - (2)
[X+(5Y+4)]*438 - [438 X+(2Y+5)] = 200*438 - 200
2190Y+4*438 - 2Y - 5 = 200*437
2188Y=85653
Y=39.15
438 X+(2Y+5)=200 ...(2)
(1)*438 - (2)
[X+(5Y+4)]*438 - [438 X+(2Y+5)] = 200*438 - 200
2190Y+4*438 - 2Y - 5 = 200*437
2188Y=85653
Y=39.15
收錄日期: 2021-04-30 22:09:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170324141414AAvlNhh