SOLUTION: If Javier has a handful of of coins consisting of dimes,nickles and pennies making a total of 2.63 how much of each coin does he have.. so far what I have is n= nickles p=pennies a
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Question 201467: If Javier has a handful of of coins consisting of dimes,nickles and pennies making a total of 2.63 how much of each coin does he have.. so far what I have is n= nickles p=pennies and d=dimes. The amount of nickles is 1 less than twice of pennies= 2p-1 and the amount of dimes is 3 more the amount of nickles= 2p-1+3d I don't know how to start its..I've tried writing it out like this: .05x(2 (2(.01)x)-1+.10x 2(.01x)-1+3(.10)=2.63
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If Javier has a handful of of coins consisting of dimes,nickles and pennies making a total of 2.63 how much of each coin does he have?
The amount of nickles is 1 less than twice of pennies --> n = 2p-1
The amount of dimes is 3 more the amount of nickels --> d = n + 3
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10d + 5n + p = 263
n = 2p - 1
d = n + 3
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d = n+3 = 2p+2
Sub for d and n in the 1st eqn
10(2p+2) + 5(2p-1) + p = 263
31p + 15 = 263
31p = 248
p = 8
-----
n = 15
------
d = 18
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