You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.
Number of nickels: n
Number of dimes: 10-n
For the main equation, multiply the number of coins by the value of each coin:
value of nickels + value of dimes = 80
5n + 10(10-n) = 80
Now solve the equation:
5n + 100 - 10n = 80
-5n = -20
5n = 20
n = 4
So, you have 4 nickels, and 6 dimes.
14 nickels = 70 cents 3 dimes = 30 cents 14 nickels + 3 dimes = 1.00
You will make 55 cents with 12 coins by using 5 pennies, 3 dimes, and 4 nickels.5 pennies, 3 dimes, and 4 nickels will make 55 cents with 12 coins.
You could have 6 dimes, 4 nickels and 2 pennies.
5 nickels, 2 dimes, and 3 pennies.
Three dimes, three nickels, five pennies
14 nickels = 70 cents 3 dimes = 30 cents 14 nickels + 3 dimes = 1.00
43 dimes and 30 nickels
You will make 55 cents with 12 coins by using 5 pennies, 3 dimes, and 4 nickels.5 pennies, 3 dimes, and 4 nickels will make 55 cents with 12 coins.
Two dimes, four nickels
4 dimes, 4 nickels
4 dimes and 7 nickels
3 dimes, 2 nickels, and 3 pennies
To make 75 cents out of 10 coins, you need 6 dimes and 3 nickels.
3 dimes, 13 nickels, and 14 cents
One half dollar, two nickels, two cents.
You could have 6 dimes, 4 nickels and 2 pennies.
2 Dimes 3 Nickels 2 Pennies