數學方程1

"1 / 10" 係 "十分之一"
係唔係 (y+2) / (y+3) = 2- (1-y) / 2 ?
(2y+4) / (y+3) = 4 - (1-y)
(2y+4) / (y+3) = y+3
(2y+4) = (y+3)^2
(2y+4) = (y^2 + 6y+9)
(y^2 + 4y + 5) = 0
沒有實數解. (只有複數解)

若然係 (y+3) / (y+2) = 2- 2 / (1-y) ...
(1-y) (y+3) / (y+2) = 2(1-y) - 2
(1-y) (y+3) / (y+2) = -2y
(1-y) (y+3) = -2y(y+2)
y^2 -2y +3 = -2y^2 - 4y
3y^2 + 2y + 3 = 0
沒有實數解. (只有複數解)

y+y+2=2-1-y
3 2
3y+y+6=4-1-2y
3 3 2 2 2
4y+6=3-2y
4y=3-2y-6
6y=-3
y=1/2

2006-12-07 14:49:01 補充：
y=-1/2

(y+3) / (y+2) = 0 / (1-y)
y +3 = 0
y = -3