Question 1169773
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
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See if that works
.05(15) + .10(30) + .25(12) = 6.75