有FORMULA的問題想請教高人

y = (3x + 1) / (x - 1)
y * (x - 1) = (3x + 1) / (x - 1) * (x - 1)
xy - y = 3x + 1
xy - 3x = 1 + y
x (y - 3) = 1 + y
x = (y + 1) / (y - 3)

2006-10-31 15:56:30 補充：
下面個位:Where did you get (y = 3x 1) from the question ??????You cannot get the values of two unknowns from one equation !!!!!!

若
y= 3x+1 over x-1
即
3x+1 over x-1= 3x+1 over x-1
左右同時 乘 (x-1)
3x+1 = 3x+1
故
y=3x+1
代入
3x+1 = 3x+1 over x-1
(3x+1)(x-1)= 3x+1
3x^2-3x+x-1=3x+1
3x^2-2x-1=3x+1
3x^2-5x-2=0
(3x+1)(x-2)=0
x=2
y=7

2006-10-31 19:57:01 補充：
阿上面個位 唔信代入公式計下

2006-10-31 20:06:23 補充：
阿上面個位用番你條公式放大眼睛lax = (y 1) / (y - 3) 2 = (7 1) / (7 - 3)2 = 8 / 42 = 2am i right ?

2006-10-31 20:11:04 補充：
唉兩個未知數的確不能求解但由頭到尾得一個 x 咋媽唔通x 1 = Ax 2 = Bx 3 = C又多三個 未知數呀