有功課唔識

Shirley has three types of coins: The number of $1-coins, $2-coins and $5-coinss. The number of $5-coins is more than the number of $1-coins by 3, while the number of $2-coins is 8 more than twice the number of $5-coins. If Shirley has 97 coins altogether, find the total value of the coins.
設Shirley 有 x 個$5-coins，
則 $1-coins 有x - 3
及 $2-coins 有 2x + 8
依題意
x + (x - 3) + 2x + 8 = 97
4x +5 = 97
4x = 92
x = 23個
它共有
5*23 + (23-3) + (2*23 + 8)*2
= $243

let shirley has x $5- coins...y $1-coins and z $2 coins

x=y+3....(1)
z=2x+8....(2)
x+y+z=97...(3)

by...(1)...
y=x-3...(4)

put(4) into (3)

x+x-3+z=97
z=100-2x...(5)

put(5) into (2)

100-2x=2x+8
4x=92
x=23

x=23...y=20...z=54

total value of the coins = 23*5+20*1+54*2=$243

Suppose Shirley has x no. of $5 coins.

She has (x-3) $1 coins.
She has (2x + 8) $2 coins.

Totally, she has 97 coins.

97 = x + (x-3) + (2x + 8)
97 = 4x + 5
4x = 97 - 5 = 92
x = 23

The total value of coins
= $5 * x + $1 * (x-3) + $2 * (2x+8)
= $5 * 23 + $1 * (23-3) + $2 * (2*23 +8)
= $115 + $20 + $108
= $243

let: there are x $5-coins,

she has:
(x-3) $1 Coins
(2x+8) $2 Coins

when:
(x-3) + (2x+8) + x = 97
4x + 5 = 97
4x = 92, x = 23

Therefore, Shirley has:
23 $5 - coins
20 $1 - coins..........(x-3) = 23 - 3 = 20
34 $2 - coins..........(2x +8) = 23x2 + 8 = 34

2006-10-29 11:00:10 補充：
Total Value= 23($5) 20($1) 34 ($2) = $(115 20 68)= $203

Let x be the number of $5-coins

x+(x-3)+(2x+8)=97
x+x-3+2x+8=97
4x=97-8+3
4x=92
x=23

the number of $5-coins is 23,the number of $2-coins is 54 and the number of $1-coins is 20

Then(x=multiply)
23x5=$115
54x2=$108
20x1=$20

115+108+20=$243

the total value of the coins is $243