解聯二元一次方程

1.
{ x = y ...[1]
{ 2x - y = 27 ...[2]
把 [1] 代入 [2],
2y - y = 27
y = 27
x = 27
{ x = 27
{ y = 27

2.
{ x = -2y ...[1]
{ 4x + 2y = 24 ...[2]
把 [1] 代入 [2],
4(-2y) + 2y = 24
-8y + 2y = 24
-6y = 24
y = -4
x = -2y = -2(-4) = 8
{ x = 8
{ y = -4

3.
{ x - y = 0 ...[1]
{ 4x - 3y = -8 ...[2]
從 [1] 可知 x = y ...[3]
把 [3] 代入 [2],
4y - 3y = -8
y = -8
x = -8
{ x = -8
{ y = -8

4.
{ x + y = 0 ...[1]
{ -2x + 5y = 63 ...[2]
從 [1] 可知 y = -x ...[3]
把 [3] 代入 [2],
-2x + 5(-x) = 63
-2x - 5x = 63
-7x = 63
x = -9
y = -(-9) = 9
{ x = -9
{ y = 9

5.
{ x - 5y = 0 ...[1]
{ 2x - 7y = 18 ...[2]
從 [1] 可知 x = 5y ...[3]
把 [3] 代入 [2],
2(5y) - 7y = 18
10y - 7y = 18
3y = 18
y = 6
x = 5(6) = 30
{ x = 30
{ y = 6


2014-09-14 19:22:47 補充：
噢～　不好意思～

再一次謝謝 Leo 那麼有風度！

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～～　Ｔｈａｎｋｓ！！！　～～
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1.
{ x = y ...[1]
{ 2x - y = 27 ...[2]
把 [1] 代入 [2],
2y - y = 27
y = 27
x = 27
{ x = 27
{ y = 27

2.
{ x = -2y ...[1]
{ 4x + 2y = 24 ...[2]
把 [1] 代入 [2],
4(-2y) + 2y = 24
-8y + 2y = 24
-6y = 24
y = -4
x = -2y = -2(-4) = 8
{ x = 8
{ y = -4

3.
{ x - y = 0 ...[1]
{ 4x - 3y = -8 ...[2]
從 [1] 可知 x = y ...[3]
把 [3] 代入 [2],
4y - 3y = -8
y = -8
x = -8
{ x = -8
{ y = -8

4.
{ x + y = 0 ...[1]
{ -2x + 5y = 63 ...[2]
從 [1] 可知 y = -x ...[3]
把 [3] 代入 [2],
-2x + 5(-x) = 63
-2x - 5x = 63
-7x = 63
x = -9
y = -(-9) = 9
{ x = -9
{ y = 9

5.
{ x - 5y = 0 ...[1]
{ 2x - 7y = 18 ...[2]
從 [1] 可知 x = 5y ...[3]
把 [3] 代入 [2],
2(5y) - 7y = 18
10y - 7y = 18
3y = 18
y = 6
x = 5(6) = 30
{ x = 30
{ y = 6

又迟咗.搬到意見柵

1.
x=y
2x - y =27
put x=y into 2x - y =27
2x - x =27
x =27
y=27

2.
x = -2y
4x + 2y = 24
put x= -2yinto 4x + 2y = 24
4(-2y) + 2y = 24
y=-4
x = -2(-4)=8

2014-09-14 19:19:09 補充：
3.
x - y = 0
4x - 3y = -8
From x - y = 0,
x = y
put x=y into 4x - 3y = -8
4x - 3x = -8
x = -8
y= -8

4.
x + y = 0
-2x + 5y = 63
From x + y = 0,
x = -y
put x=-y into -2x + 5y = 63
-2(-y) + 5y = 63
7y= 63
y=9
x=-9

2014-09-14 19:19:14 補充：
5.
x - 5y = 0
2x - 7y = 18
From x - 5y = 0,
x =5y
put x =5y into 2x - 7y = 18
2(5y) - 7y = 18
3y = 18
y = 6
x =5(6)=30