Maths question:

(2-t)/4-t=(2t-3)/5

(2-t)/4=(2t-3)/5+t

(2-t)/4=(2t-3)/5+5t/5

(2-t)/4=(2t-3+5t)/5

(2-t)/4=(7t-3)/5

4(7t-3)=5(2-t)

28t-12=10-5t

33t=22

t=22/33

t=2/3

your answer is correct...

(2-t)/4-t=(2t-3)/5
(2-t)/4=(2t-3)/5+t
2-t=((2t-3)/5+t)(4)
2-t=(8t-12)/5+4t
2-t-4t=(8t-12)/5
2-5t=(8t-12)/5
10-25t=8t-12
22=33t
t=2/3

(2-t)/(4-t)=(2t-3)/5
(2-t)/(4-t) X5(4-t)=(2t-3)/5 X5(4-t)
5(2-t)=(2t-3)(4-t)
8t^2-6t-2=0
4t^2-3t-1=0
(4t+1)(t-1)=0
t=-1/4, -1
肯定！

2006-11-03 22:40:40 補充：
t=-1/4, 1不是-1，SORRY！

(2-t)/(4-t) = (2t-3)/5
10-5t = (2t-3)(4-t)
10-5t = 8t - 2t^2 -12 + 3t
2t^2 -16t + 22 = 0

t^2 - 8t + 11 = 0

where, a = 1, b = -8, c = 11 (use quadratic formula)

t = [8 +/- √64 - 4 (11)]/2
t = (8 +/- √20)/2
t = (8 +/- 2√5)/2

t = 4 +/- √5

(2-t)/4-t=(2t-3)/5

5(2-t-4t)=4(2t-3)

10-5t-20t=8t-12

10-25t=8t-12

22=33t

t=3/2

2006-11-03 20:43:16 補充：
唔係呀,係2/3

(10-5t)/4=(2t-3)+5t
10-5t=28t-12
33t=22
t=2/3

(2-t)/4-t=(2t-3)/5
(2t-3)(4-t)=5(2-t)
-2t²+11t-12=10-5t
-2t²+16t-22=0
t²-8t+11=0

t=(8±√20)/2
=4±√5

1.)(2-t)/4-t=(2t-3)/5
(2-t)/4=(2t-3)/5+t
(2-t)/4=(2t-3)/5+5t/5
(2-t)/4=(2t-3+5t)/5
by timing 20(which is the L.C.M. of both 5 and 2)on the whole equation
5(2-t)=(28t-12)
10-5t-28t=-12
10+12=28t+5t
22=33t
t=2/3

2006-11-03 20:29:44 補充：
should be'which is the L.C.M. of both 5 and 4'