中一數學方程

(1)6x-4(x-5) = 7-x
6x-4x+20 = 7-x
6x-4x+x = 7-20
3x=-13
x=-13/3

(2)25/2-c=2(c+5)
25-2c=4(c+5)
25-2c=4c+20
6c=5
c=5/6

(3)d/3-(d+2)/4=8
4d-3(d+2)=96
4d-3d-6=96
d=102

其他自己努力吧

(1)6x-4(x-5) = 7-x
6x-4x+20 = 7-x
6x-4x+x = 7-20
3x=-13
x=-13/3

(2)25/2-c=2(c+5)
25-2c=4(c+5)
25-2c=4c+20
6c=5
c=5/6

(3)d/3-(d+2)/4=8
4d-3(d+2)=96
4d-3d-6=96
d=102
1)如果某2個連續偶數是30, 求這2個數字。
30/2=15 14(15-1) 16(15 1)=30
2)小明銀包內共有$54元硬幣,其中包括一些$2 和 一些$5 的硬幣,
如果他的銀包內共有15枚硬幣,問(她???)分別有多少枚$2 和 $5 ?
15個都是$5=$75
15個都是$2=$30
若1個$5換$2=75-3(5-2)=$72
再-3=$69…至第6次就=$54了。
即15-6=9($5硬幣)
$5硬幣有9枚,$2硬幣有6枚

1)
6x-4(x-5) = 7-x
6x-4x+20 = 7-x
6x-4x+x = 7-20
3x= -13
x= -13/3
2)
25/2-c=2(c+5)
25-2c=4(c+5)
25-2c=4c+20
6c=5
c=5/6
3)
d/3-(d+2)/4=8
4d-3(d+2)=96
4d-3d-6=96
d=102
long questions:
1)
let n be the first even num.
n+(n+2) = 30
2n = 28
n = 14
then the second even number is 14+2 = 16
so, the two nums. are 14 , 16
2)
let x be the amount of 2 in coin
let y be the amount of 5 in coin
2x+5y = 54 -------- (1)
x+y = 15 -------- (2)
(2) x 2
2x+2y = 30 -----------(3)
(3) - (1)
2x+2y-2x-5y = 30-54
-3y = -24
y = 8
put y=8 into (2)
x+8 = 15
x = 7
so, she?/he? has 7 of 2 in coins and 8 of 3in coins
Hope I can help you.

1)
6x-4(x-5) = 7-x
6x-4x+20 = 7-x
6x-4x+x = 7-20
3x= -13
x= -13/3

2)
25/2-c=2(c+5)
25-2c=4(c+5)
25-2c=4c+20
6c=5
c=5/6

3)
d/3-(d+2)/4=8
4d-3(d+2)=96
4d-3d-6=96
d=102

long questions:

1)
let n be the first even num.
n+(n+2) = 30
2n = 28
n = 14
then the second even number is 14+2 = 16
so, the two nums. are 14 , 16

2)
let x be the amount of 2 in coin
let y be the amount of 5 in coin
2x+5y = 54 -------- (1)
x+y = 15 -------- (2)
(2) x 2
2x+2y = 30 -----------(3)
(3) - (1)
2x+2y-2x-5y = 30-54
-3y = -24
y = 8
put y=8 into (2)
x+8 = 15
x = 7
so, she?/he? has 7 of 2 in coins and 8 of 3in coins

if you dont understand, you can ask for the detail

2008-11-05 22:18:12 補充：
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