SOLUTION: Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be $

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Question 193524: Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. How many dimes does he have?

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Joe has a collection of nickels and dimes that is worth $6.05.
If the number of dimes was doubled and the number of nickels was decreased by 10,
the value of the coins would be $9.85.
How many dimes does he have?
:
Write an equation for each statement:
:
"Joe has a collection of nickels and dimes that is worth $6.05."
.05n + .10d = 6.05
:
"If the number of dimes was doubled and the number of nickels was decreased by
10, the value of the coins would be $9.85."
.05(n-10) + .10(2d) = 9.85
.05n - .50 + .20d = 9.85
.05n + .20d = 9.85 + .50
.05n + .20d = 10.35
:
Subtract the 1st equation from the above:
.05n + .20d = 10.35
.05n + .10d = 6.05
---------------------Subtraction eliminates n find d
.10d = 4.30
d = 4.3%2F.1
d = 43 dimes is what he has
;
:
That's the answer they want, but to check the problem we have to find n
.05n + .10(43) = 6.05
.05n + 4.30 = 6.05
.05n = 6.05 - 4.30
.05n = 1.75
n = 1.75%2F.05
n = 35 dimes
:
Check solution in the 2nd equation
.05(n-10) + .10(2d) =
.05(25) + .10(86) =
1.25 + 8.60 = 9.85; confirms out solution of 43 dimes