二元一次方程

1.
y = x-2 -------- (1)
2x + y = 4 -------- (2)
以(1)代入(2), 得:
2x + (x-2) = 4
2x + x - 2 = 4
3x = 4+2
x = 6/3
x=2

以 x=2 代入(1), 得:
y = 2-2
y = 0

所以, x = 2, y = 0.

2.
x+2y = 1 -------- (1)
x-3y = 6 -------- (2)
由(1)得:
x = 1-2y -------- (3)
以(3)代入(2), 得:
(1-2y) - 3y = 6
1 - 5y = 6
-5y = 5
y = 5 / (-5)
y = -1

以 y=-1 代入(2), 得:
x - 3(-1) = 6
x + 3 = 6
x = 6 - 3
x = 3

所以, x = 3, y = -1.

3.
x + 2y = 5 -------- (1)
2x - 3y = -4 -------- (2)
由(1)得:
x = 5 - 2y -------- (3)
以(3)代入(2), 得:
2(5 - 2y) - 3y = -4
10 - 4y - 3y = -4
10 + 4 = 4y + 3y
14 = 7y
y = 2

以 y = 2 代入(3), 得:
x = 5 - 2(2)
x = 5 - 4
x = 1

所以, x = 1, y = 2.

4.
2x - y = 8 -------- (1)
3x - y = 6 -------- (2)
由(1)得:
y = 2x - 8 -------- (3)
以(3)代入(2), 得:
3x - (2x - 8) = 6
3x - 2x + 8 = 6
x = 6 - 8
x = -2

以 x = -2 代入(3), 得:
y = 2(-2) - 8
y = -4 -8
y = -12

所以, x = -2, y = -12.

5.
4x = 3y - 1 -------- (1)
3x = y - 7 -------- (2)
由(2)得:
y = 3x + 7 -------- (3)
以(3)代入(1), 得:
4x = 3(3x + 7) - 1
4x = 9x + 21 - 1
-5x = 20
x = -20/5
x = -4

以 x = -4 代入(3), 得:
y = 3(-4) + 7
y = -12 + 7
y = -5

所以, x = -4, y = -5.

6.
2x - 3y = 1 -------- (1)
3x + 4y = 10 -------- (2)
由(1)得:
2x = 1 + 3y
x = (1 + 3y)/2 -------- (3)
以(3)代入(2), 得:
3 [ (1 + 3y)/2 ] + 4y = 10
3 [1/2 + (3/2)y] + 4y = 10
3/2 + (9/2)y + 4y =10
(1/2)y = 10 - 3/2
(1/2)y = 7/2
y = (7/2)*2
y = 7

以 y = 7 代入(2), 得:
3x + 4(7) = 10
3x = 10 - 28
3x = -18
x = -18/3
x = -6

所以, x = -6, y = 7.

7.
x+5(y-2) = x+y-4 = 0
由上式得知:
x+5(y-2) = 0 -------- (1)
x+y-4 = 0 -------- (2)
由(2)得:
x = -y +4 -------- (3)
以(3)代入(1), 得:
(-y +4) + 5(y-2) = 0
- y + 4 + 5y - 10 =0
4y - 6 = 0
4y = 6
y = 6/4
y = 3/2
y = 1.5

以 y = 3/2 代入(3), 得:
x = -(3/2) + 4
x = 5/2
x = 2.5

所以, x = 5/2 = 2.5, y = 3/2 = 1.5.

2007-09-17 23:19:00 補充：
另一位解答者似乎"誤以為" 加減消去法 就是 代入消元法 的其中一種.加減消元法其實是指用加減運算來消去方程組內變元的方法.我所用的就全部都正是代入消元法了.

2007-09-17 23:21:38 補充：
另, 第六題出現了少許錯誤, 修正如下:由(1)得:2x = 1 + 3yx = (1 + 3y)/2 -------- (3)以(3)代入(2), 得:3 [ (1 + 3y)/2 ] + 4y = 103 [1/2 + (3/2)y] + 4y = 103/2 + (9/2)y + 4y =10(17/2)y = 10 - 3/2(17/2)y = 17/2y = 1以 y = 1 代入(2), 得:3x + 4(1) = 103x = 10 - 43x = 6x = 2所以, x = 2, y = 1.

以上回答者用的是代入法 Substitution, 是其中一個解決二元一次方程的方法之一

你想用的消元法 Elimination 是另一種

Answer to question 1
y = x - 2 ... (1)
2x + y = 4 ==> y = 4 - 2x ... (2)

(1) = (2)
x - 2 = 4 - 2x
3x = 6
x = 2 //

Sub x = 2 to (1)
y = 2 - 2
y = 0 //

Answer to question 2
x + 2y = 1 ==> x = 1 - 2y ... (1)
x - 3y = 6 ==> x = 6 + 3y... (2)

(1) = (2)
1 - 2y = 6 + 3y
-5 = 5y
y = -1 //

Sub y = -1 to (1)
x =1 - 2(-1)
x = 3 //

Answer to question 3
x + 2y = 5 ==> x = 5 - 2y ... (1)
2x - 3y = -4 ==> x = (3y - 4) / 2 ... (2)

(1) = (2)
5 - 2y = (3y - 4) / 2
10 - 4y = 3y - 4
14 = 7y
y = 2

Sub y = 2 to (1)
x = 5 - 2(2)
x = 1 //

Answer to question 4
2x - y = 8 ==> y = 2x - 8 ... (1)
3x - y = 6 ==> y = 3x - 6 ... (2)

(1) = (2)
2x - 8 = 3x - 6
x = -2 //

Sub x = -2 to (1)
y = 2 (-2) - 8
y = -12 //

Answer to question 5
4x = 3y - 1 ==> x = (3y - 1) / 4 ... (1)
3x = y - 7 ==> x = (y - 7) / 3 ... (2)

(1) = (2)
(3y - 1) / 4 = (y - 7) / 3
3 (3y - 1) = 4 (y - 7)
9y - 3 = 4y - 28
5y = -25
y = -5 //

Sub y = -5 to (1)
x = [3(-5) - 1] / 4
x = -4 //

Answer to question 6
2x - 3y = 1 ==> x = (1 + 3y) / 2... (1)
3x + 4y = 10 ==> x = (10 - 4y) / 3... (2)

(1) = (2)
(1 + 3y) / 2 = (10 - 4y) / 3
3 (1 + 3y) = 2 (10 - 4y)
3 + 9y = 20 - 8y
17y = 17
y = 1 //

Sub y = 1 to (1)
x = [1 + 3(1)] / 2
x = 2 //

Answer to question 7
x + 5(y - 2) = 0 ==> x = -5 (y - 2) ==> x = 10 - 5y ... (1)
x + y - 4 = 0 ==> x = 4 - y ... (2)

(1) = (2)
10 - 5y = 4 - y
-4y = -6
y = 1.5 //

Sub y = 1.5 to (2)
x = 4 - 1.5
x = 2.5 //

2007-09-17 00:05:35 補充：
消元法解決的關鍵在於將兩條題目公式改頭換面, 轉成以其中一個未知數 (例如是x)為主項的公程式然後因兩條式是相等, 就可以計到另一個未知數 (例如是y)最後將y的數值代入回原本題目內, 就可找到x

2007-09-17 00:16:37 補充：
代入法點做? 以上回答者已覆了你, 我不重覆了. 可是, 他計錯了第6題如果 x = -6, y = 7, 代入去佢第(3)條式就見到:x = (1 3y)/2 x = [1 3 (7)] / 2 x = 11明顯地, 同佢回覆 x = -6 是不同的...

2007-09-17 00:21:08 補充：
沒關係, 我幫他更正吧2x - 3y = 1 -------- (1)3x + 4y = 10 -------- (2)由(1)得:2x = 1 + 3yx = (1 + 3y)/2 -------- (3)以(3)代入(2), 得:3 [ (1 + 3y)/2 ] + 4y = 103 [0.5 + 1.5y] + 4y = 101.5 + 4.5y + 4y =108.5y = 8.5y = 1以 y = 1 代入(2), 得:3x + 4(1) = 103x = 10 - 43x = 6x = 2