3a+5b =4a+7b =4
2 3
請問有沒有人可以解答
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解答二元一次方程式
2011-02-11 2:46 am
回答 (2)
2011-02-11 2:56 am
✔ 最佳答案
(3a+5b)/2=(4a+7b)/3=4(3a+5b)/2=4 => 3a+5b=8...(1)
(4a+7b)/3=4 => 4a+7b=12...(2)
(1)*4-(2)*3:
(12a+20b)-(12a+21b)=8*4-12*3
-b=-4
b=4...(3)
Sub (3) into (1):
3a+5(4)=8
3a=8-20
3a=-12
a=-4
2011-02-11 2:57 am
(3a + 5b)/2 = (4a + 7b)/3 = 4
由 (3a + 5b)/2 = (4a + 7b)/3
3(3a + 5b) = 2(4a + 7b)
9a + 15b = 8a + 14b
a = -b
再代入 (4a + 7b)/3 = 4
3b/3 = 4
b = 4
因此a = -4
由 (3a + 5b)/2 = (4a + 7b)/3
3(3a + 5b) = 2(4a + 7b)
9a + 15b = 8a + 14b
a = -b
再代入 (4a + 7b)/3 = 4
3b/3 = 4
b = 4
因此a = -4
收錄日期: 2021-04-22 00:53:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110210000051KK01342