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# 10 coins consistinig of nickels and dimes equal 80 cents how many coins does she have?

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.

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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.