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1. 2[a-(1-a)]=3(a-1)+8
2(2a-1)=3a-3+8
4a-2=3a+5
a=7

4. k/30 +5=k/10
(k/30)x30+5x30=(k/10)x30
k+150=3k
150=2k
k=75

5. 9(2a-1)=3[4(1+2a)-5]-8a
18a-9=3[(4+8a)-5]-8a
18a-9=3(8a-1)-8a
18a-9=24a-3-8a
18a-9=16a-3
2a=6
a=3

6. 1/3(t+9)=1/3 -0.5(t+3)
[1/3(t+9)]x3=1/3x3-[-0.5(t+3)]x3
t+9=1-[-2(t+3)]
t+9=1-(-2t)-6
t+9=1+2t-6
t=14

7. (2n-5)/3=(n-10)/9
(2n-5)/3x9=(n-10)/9x9
(2n-5)x3=n-10
6n-15=n-10
5n=5
n=1

8. (x-1)/2 =(x+1)/3 +1
[(x-1)/2]x6=[(x+1)/3 +1]x6
(x-1)x3=(x+1)x2+6
3x-3=2x+2+6
3x-3=2x+8
x=11

9. 7u -(4u-1)/2=21
[7u -(4u-1)/2]x2=21x2
14u-(4u-1)=42
10u+1=42
u=4.1

10. (2-t)/4-t=(2t-3)/5
[(2-t)/4-t]x20=[(2t-3)/5]x20
(2-t)x5-20t=(2t-3)x4
10-5t-20t=8t-12
10-25t=8t-12
33t=22
t=2/3

11.Two trains are approaching each other from towns 600km apart.One train's speed is twice the other train's speed.They pass each other after 5 hours.Find their speeds.
In this problem,we note that :
distance=speed x time
first distance+second distance=total distance
Let the speed of the slower train be x km/h.
Then the speed of the other train=__2x km/h______.
In 5 hours,distance travelled by slower traim=-----5------- x ---------=-------600---km
Distance travelled by faster train=-----10------x---------------=------600----km
Given,total distance=----600---------km
Therefore,---5x--+----10x-=-----1800-------
x=130
2x=260
Therefore,the speeds of the trains are --130-----km/h and ---260-----km/h