SOLUTION: Joe has a collection of nickels and dimes that is worth $6.20. If the number of dimes was doubled and the number of nickels was decreased by 7, the value of the coins would be $10

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Question 614212: Joe has a collection of nickels and dimes that is worth $6.20. If the number of dimes was doubled and the number of nickels was decreased by 7, the value of the coins would be $10.35. How many nickels and dimes does he have?
Found 2 solutions by richwmiller, lwsshak3:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
5n+10d=620;
10*(2d)+5*(n-7)=1035
d=45
n=34
check
5*(34)+10*(45)=620
170+450=620
ok
10*90+5*27=1035
900+135=1035
ok

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Joe has a collection of nickels and dimes that is worth $6.20. If the number of dimes was doubled and the number of nickels was decreased by 7, the value of the coins would be $10.35. How many nickels and dimes does he have?
**
let x= number of dimes
let y=number of nickels
===
.10x+.05y=6.20
.10(2x)+.05(y-7)=10.35
===
.10x+.05y=6.20
.20x+.05y-.35=10.35
===
10x+5y=620
20x+5y=1070
subtract
10x=450
x=45
5y=620-10x=620-450=170
y=170/5=34
ans:
number of dimes=45
number of nickels=34