Answer:
He has 7 nickels, 14 dimes, and 21 quarters.
Explanation:
Let d = number of dimes.
Let n = number of nickels.
Let q = number of quarters.
The value of a dime is $0.10, so d dimes are worth 0.1d
The value of a nickel is $0.05, so n nickles are worth 0.05n
The value of a quarter is $0.25, so q quarters are worth 0.25q
The value of all coins combined is the sum of the values above, and it equals $7.00
0.1d + 0.05n + 0.25q = 7
The number of dimes = 2 times the number of nickels, so
d = 2n
The number of quarters = 3 times the number of nickels, so
q = 3n
Now we substitute 2n for d and 3n for q in our equation above.
0.1d + 0.05n + 0.25q = 7
0.1(2n) + 0.05n + 0.25(3n) = 7
0.2n + 0.05n + 0.75n = 7
n = 7
He has 7 nickels.
d = 2n = 2(7) = 14
He has 14 dimes.
q = 3n = 3(7) = 21
He has 21 quarters.
Answer: He has 7 nickels, 14 dimes, and 21 quarters.