代入消元法,求你幫幫忙!!

1.) x=2y......(1)
x-y=3......(2)

Sub (1) into (2)
x-y=3
2y-y=3
y=3 ... (3)

Sub (3) into (1)
x=2y
x=2x3
x=6

x=6, y=3

第2題
y-x=0......(1)
2x-y=6......(2)

From (1)
y-x=0
y=x ...(3)

Sub (3) into (2)
2x-y=6
2x-x=6
x=6 ...(4)

Sub (4) into (1)
y-x=0
y-6=0
y=6

x=6, y=6

10.)
x-y=5x-3y-4=0

x-y=0
x=y ...(1)

5x-3y-4=0 ...(2)

Sub (2) into (1)
5x-3y-4=0
5x-3x-4=0
2x=4
x=2 ...(3)

Sub (3) into (1)
x=y
y=2

x=2, y=2

1. x=2y......(1)
x-y=3......(2)
Sub (1) into (2),
2y-y=3
y=3
Sub y=3 into 1
x=2(3)
=6

2. y-x=0......(1)
2x-y=0......(2)
(1)x2,2y-2x=0.......(3)
(3)+(2),2y-2x+2x-y =0+0
y=0
Sub y=0 into (1),
0-x=0
-x=0
x=0

第10題計唔到......Sor







2011-02-21 21:57:19 補充：
第10題唔係y-y=5x-3y-4=0 咩??

2011-02-24 21:19:10 補充：
x-y=5x-3y-4=0
x-y=0 -------(1)
5x-3y-4=0--------(2)
(1),x=y
Sub x=y into (2)
5y-3y-4=0
2y=4
y=2
Sub y=2 into (1),
x-2=0
x=2

2011-02-24 21:21:24 補充：
2.y-x=0......(1)
2x-y=6......(2)
(1),y=x
Sub y into (2)
2x-x=6
x=6
Sub x=6 into (1)
y-6=0
y=6

請發問者自行檢查一次題目

2011-02-21 22:05:49 補充：
還有第三條
y-y=5x-3y-4=0
請問y - y = 0 要怎樣解...

2011-02-21 22:07:11 補充：
2. For y - x = 0 ---(1)
2x - y = 6 ---(2)
From (1), x = y ---(3)
Put (3) into (2),
2y - y = 6
y = 6
So put y = 6 into (3), x = 6
The solution is (6,6)