SOLUTION: $2.77 in a coin jar- all pennies(x), nickels(y), and dimes(z). If there are 53 coins in all, and 4 more nickels than dimes, how many of each coin do you have. Setup all 3 equatio

Question 419780: $2.77 in a coin jar- all pennies(x), nickels(y), and dimes(z). If there are 53 coins in all, and 4 more nickels than dimes, how many of each coin do you have. Setup all 3 equations in x,y,z and solve the equations to find x(pennies), y(nickels), and z(dimes). Note: 1 equation has only y and z.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
$2.77 in a coin jar- all pennies(x), nickels(y), and dimes(z). If there are 53 coins in all, and 4 more nickels than dimes, how many of each coin do you have. Setup all 3 equations in x,y,z and solve the equations to find x(pennies), y(nickels), and z(dimes). Note: 1 equation has only y and z.
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Equations:
10d + 5n + p = 277 cents
d + n + p = 53
n = d+4
----
Rearrange:
d + n + p = 53
d - n + 0 = -4
10d+ 5n + p = 277
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Solve by any method you know to get:
d = 16 (# of dimes
n = 20 (# of nickels
p = 17 (# of pennies
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Cheers,
Stan H.
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