利用加減消元法解聯立二元一次方程
{3x-5y=30
{4x+7y=-1
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一條好難的數學問題F.2 part2
2008-03-02 6:56 pm
回答 (3)
2008-03-02 10:54 pm
3x-5y=30.....(1)
4x+7y=-1.....(2)
(1)*4-(2)*3
12x-20y=120
-) 12x+21y=-3
----------------------
-41y=123
-y=3
y=-3..........(3)
把(3)代入(1)
4x+7y=-1
4x+7(-3)=-1
4x-21=-1
4x=-1+21
4x=20
x=5
4x+7y=-1.....(2)
(1)*4-(2)*3
12x-20y=120
-) 12x+21y=-3
----------------------
-41y=123
-y=3
y=-3..........(3)
把(3)代入(1)
4x+7y=-1
4x+7(-3)=-1
4x-21=-1
4x=-1+21
4x=20
x=5
參考: me
2008-03-02 7:28 pm
{3x-5y=30-------------1
{4x 7y=-1--------------2
1×4,
4(3x-5y)=4(30)
12x-20y=120----------3
2×3,
3(4x 7y)=3(-1)
12x 21y=-3------------4
4-3,
12x 21y-(12x-20y)=-3-120
41y=-123
y=-3
將y=-3代入2,
4x 7(-3)=-1
4x-21=-1
4x=20
x=5
∴{x=5
{y=-3
{4x 7y=-1--------------2
1×4,
4(3x-5y)=4(30)
12x-20y=120----------3
2×3,
3(4x 7y)=3(-1)
12x 21y=-3------------4
4-3,
12x 21y-(12x-20y)=-3-120
41y=-123
y=-3
將y=-3代入2,
4x 7(-3)=-1
4x-21=-1
4x=20
x=5
∴{x=5
{y=-3
參考: me
2008-03-02 7:09 pm
{3x-5y=30 [x4] [1]
{4x+7y=-1 [x3] [2]
12x-20y=120
(-) 12x+21y=-3
-41y=123
y=-3
sub y=-3 into [1]
3x-5y=30
3x-5(-3)=30
x= 5
{4x+7y=-1 [x3] [2]
12x-20y=120
(-) 12x+21y=-3
-41y=123
y=-3
sub y=-3 into [1]
3x-5y=30
3x-5(-3)=30
x= 5
參考: me
收錄日期: 2021-04-15 17:56:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080302000051KK00754