shirley has three types of coins:$1-coins ,$2-coins and$5-coins.
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-coin.
If shirley has 97 coins altogether, find the total value of the coin.
i want long steps and teach me how to do it plx very hurry!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
maths question
2008-01-02 9:05 pm
回答 (3)
2008-01-02 9:28 pm
✔ 最佳答案
Let the number of $1-coins = ythen, the number of $5-coins = y+3
the number of $2-coins = 2(the number of $5-coins)+ 8 = 2(y+3) + 8
so, the equation will be y + y +3 + 2(y+3) + 8 = 97
4y +17 = 97
4y = 80
y = 20
the number of $1-coins = y = 20
the number of $5-coins = y+3 =23
the number of $2-coins = 2(the number of $5-coins)+ 8 = 2(y+3) + 8 = 54
total value = 20 ($1) + 23 ($5) + 54 ($2) =20+115+108 = $ 243
參考: i hope u understand la~~~~
2008-01-02 9:15 pm
Let the number of $1-coins ,$2-coins and$5-coins be x,y and z respectively
z-x=3------>x=z-3--------1
y-2z=8------>y=8+2z------------2
x+y+z=97-----------3
Sub 1 and 2 into 3
z-3+8+2z+z=97
4z=92
z=23 and y=54 and x=20
Therefore , the total value of the coin
=20(1)+54(2)+23(5)
=$243
z-x=3------>x=z-3--------1
y-2z=8------>y=8+2z------------2
x+y+z=97-----------3
Sub 1 and 2 into 3
z-3+8+2z+z=97
4z=92
z=23 and y=54 and x=20
Therefore , the total value of the coin
=20(1)+54(2)+23(5)
=$243
2008-01-02 9:11 pm
lets no. of $1 = y
y + y+3 + 8(y+3) = 97
y + y+3 + 8(y+3) = 97
收錄日期: 2021-04-23 22:51:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080102000051KK01195