係二元一次加減法計算ar!!
{6x-4y=11
{2x-4y=23
我想問下e條數學點做ar!!係二元一次
2009-05-11 4:03 am
回答 (3)
2009-05-11 6:04 am
6x - 4y = 11 (1)
2x - 4y = 23 (2)
==
(1) - (2),
4x = -12
x = -3
==
sub x = -3 into (1),
6(-3) - 4y = 11
-18 - 11 = 4y
-29 = 4y
y = -29/4
==
2x - 4y = 23 (2)
==
(1) - (2),
4x = -12
x = -3
==
sub x = -3 into (1),
6(-3) - 4y = 11
-18 - 11 = 4y
-29 = 4y
y = -29/4
==
2009-05-11 4:08 am
6x - 4y = 11 (1)
2x - 4y = 23 (2)
(1) - (2),
4x = -12
x = -3
sub x = -3 into (1),
6(-3) - 4y = 11
-18 - 11 = 4y
-29 = 4y
y = -29/4
2x - 4y = 23 (2)
(1) - (2),
4x = -12
x = -3
sub x = -3 into (1),
6(-3) - 4y = 11
-18 - 11 = 4y
-29 = 4y
y = -29/4
2009-05-11 4:06 am
{6x-4y=11- (1)
{2x-4y=23 - (2)
(1)-(2),
4x=-12
x=-3 - (3)
Sub. (3) into (1)
6(-3)-4y=11
-18-4y=11
-4y=29
y=-18/29
2009-05-10 21:00:35 補充：
傻左...打錯了...最尾應該係
y=-29/4
sorry~~
{2x-4y=23 - (2)
(1)-(2),
4x=-12
x=-3 - (3)
Sub. (3) into (1)
6(-3)-4y=11
-18-4y=11
-4y=29
y=-18/29
2009-05-10 21:00:35 補充：
傻左...打錯了...最尾應該係
y=-29/4
sorry~~
參考: myself
收錄日期: 2021-04-23 23:13:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090510000051KK01816