{{急急急!!!!!!}}數學天才請進***

1.
b/3 - (5-b) = 3
b-3(5-b) = 9
b - 15 + 3b = 9
b=6

2.
x/3 - x/2 = 1
2x - 3x = 6 ( 成條乘6)
x = -6

3.
(r+6)/3 = r/2 + 7/6
2(r+6) = 3r + 7 ( 成條乘6)
2r + 12 = 3r + 7
r = 5

4.
Let the 2 numbers be x and x+2
x + x + 2 = 20
x=9
therefore the smaller number is 9

5.
Let the number of $2 stamps be x and the number of $5 stamps be (15 - x)
2x + 5(15-x) = 54
2x + 75 - 5x = 54
x = 7
she bought 7 $2 stamps

1. b/3 -(5-b)=3
b/3=3+5-b
b=24-3b
4b=24
b=6

2. x/3 - x/2 =1
2x-3x=6
-x=6
x=-6


3. 3分之r+6 = r/2+7/6
(r+6)/3=r/2+7/6
2*(r+6)=3r+7
2r+12=3r+7
r=5


4.如果某兩個連續奇數的和是20，求較小的一個奇數
設較小的奇數為Ｘ，則另一個奇數為Ｘ＋２
　Ｘ＋Ｘ＋２＝２０
　２Ｘ＝１８
　　Ｘ＝９
較小的奇數為９



5.文娟買了一些$2及$5的郵票，如果她一共用去$54買了15枚郵票，她共買了多少枚$2郵票?
設文娟買了Ｘ枚＄２郵票，則＄５郵票有１５－Ｘ枚

據題意：２Ｘ＋５（１５－Ｘ）＝５４
　　　　２Ｘ＋７５－５Ｘ＝５４
　　　　３Ｘ＝２１
　　　　　Ｘ＝７

她共買了７枚$2郵票．

1. b/3 - (5-b) = 3

b - 3 * (5-b) = 9
b - 15 + 3b = 9
4b = 24
b = 6

2. x/3 - x/2 = 1

2x - 3x = 6
-x = 6
x = 16

3. (r+6)/3 = r/2 + 7/6

2 * (r+6) = 3r + 7
2r + 12 = 3r + 7
r = 5

4. 設較小的一個奇數 = a

a + (a+2) = 20
2a = 18
a = 9

5. 設a = $2的郵票數量, b = $5的郵票數量

算式1: 2 * a + 5 * b = 54
算式2: a + b = 15

算式2 ==> b = 15-a
算式1 ==> 2a + 5*(15-a) = 54
2a + 75 - 5a = 54
21 = 3a
a = 7

1.b/3-5+b=3
b/3=8-b
b=24-3b
4b=24
b=6
2.6(x/3-x/2)=6
2x-3x=6
-x=6
x=-6
3. r/3+6=r/2+7/6
r/3-r/2=7/6-6
6(r/3-r/2)=6(7/6-6)
2r-3r=7-36
-r=29
r=29
4. Let x be the smaller odd number
x+(x+2)=20
2x+2=20
2x=18
x=9
Therefore, the smaller odd number is 9.
5. Let x be the number of $2 stamps that she has bought.
2x+5(15-x)=54
2x+75-5x=54
-3x=-21
x=7
Therefore. she has bought 7 $2stamps.

2006-10-22 20:45:53 補充：
For question 3,if it is (r 6)/3=r/2 7/6 6(r 6)/3=6(r/2 7/6) 2r 12=3r 7 2r-3r=7-12 -r=-5 r=5