SOLUTION: A jar of coins contain Nickel, Dime and quarter totaling $6.75. There are three fewer quarter than nickels. They are three more dimes than the total number of nickels and quarter.

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Question 1169773: A jar of coins contain Nickel, Dime and quarter totaling $6.75. There are three fewer quarter than nickels. They are three more dimes than the total number of nickels and quarter. Find the number of each
Found 2 solutions by Boreal, ankor@dixie-net.com:
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
quarters=x
nickels=x+3
dimes=2x+6, the x+x+3+3
.25x+.05(x+3)+.10(2x+6)=6.75
.50x+.75=6.75
x=12 quarters 3.00
x+3=15 nickels 0.75
2x+6=30 dimes 3.00

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A jar of coins contains Nickels, Dimes and quarters totaling $6.75.
.05n + .10d + .25q = 6.75
There are three fewer quarters than nickels.
q = n - 3
They are three more dimes than the total number of nickels and quarters.
d = n + q + 3
replace q with (n-3)
d = n + (n-3) + 3
d = 2n
We know q = (n-3) and d = 2n, replace q and d in the first equation
.05n + .10(2n) + .25(n-3) = 6.75
.05n + .20n + .25n - .75 = 6.75
.50n = 6.75 + .75
.50n = 7.50
n = 7.50/.5
n = 15 nickels
then
d = 2(15) = 30 dimes
and
q = 15 - 3 = 12 quarters
:
:
See if that works
.05(15) + .10(30) + .25(12) = 6.75