Question 620375: $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.
Answer by lwsshak3(11628)   (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.
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x=number of pennies
y=number of nickels
z=number of dimes
..
z+4=number of nickels
z+(z+4)+x=53
2z+4+x=53
x=49-2z=number of pennies
..
pennies+nickels+dimes=$2.77
.01x+.05y+.1z=2.77
.01(49-2z)+.05(z+4)+(.1z)=2.77
.49-.02z+.05z+.2+.1z=2.77
.13z+.69=2.77
.13z=2.77-.69
.13z=2.08
z=2.08/.13
z=16
z+4=20
49-2z=17
..
number of pennies=17
number of nickels=20
number of dimes=16
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