我想問一D數學既問題

(1) 9(2a-1)=3[4(1+2a)-5]-8a
18a-9=3[4+8a-5]-8a
18a-9=16a-3
2a=6
a=3
(2) (t+9)/3 = -0.5(t+3)/3
t+9 = -0.5t -1.5
1.5t= -10.5
t= -7
(3) (2n-5)/3 = (n-10)/9
3(2n-5) = n-10
6n-15 = n-10
n=1
(4) (x-1)/2 = (x+1)/3 +1
3(x-1) = 2(x+1) + 6
3x-3 = 2x+2+6
x = 11
(5) 7u - (4u-1)/2 = 21
14u-4u+1 = 42
10u =41
u=4.1 answer a bit strange, please check question again.
(6) (2-t)/4 - t = 2t/5 -3
5(2-t) -20t = 4(2t) - 3(20)
10 - 25t = 8t -60
33t = 70
t = 70/33 Again please check question again whether it is (2t-3)/5 instead of 2t/5 -3
if it were (2t-3)/5, then 5(2-t)-20t = 4(2t-3)
10 -25t = 8t-12
33t = 22
t=2/3

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

2. (1/3)(t+9)= (1/3) - 0.5(t+3)
2(t+9) = 2 - 3(t+3) both sides times 6
2t + 18 = 2 - 3t -9
5t = -25
t = -5

3. (2/3)n - 5 = (n/9) - 10
6n - 45 = n - 90
5n = -45
n= -9

4. (x/2) - 1 = (x+1)/3 + 1
3x - 6 = 2(x+1) + 6
3x - 6 = 2x + 2 +6
x = 14

5. 7u - (4u-1)/2 =21
14u - 4u+1 = 42
10u = 41
u= 41/10

6. (2-t)/4 - t = 2t/5 - 3
5(2-t) - 20t = 4(2t)- 60
10 - 5t - 20t = 8t - 60
33t = 70
t = 33/70