1 penny = 1 cent

1 nickle = 5 cents

1 dime = 10 cents

1 quarter = 25 cents

let

p = the number of pennies, p>0, p is integer

n = the number of nickels, n>0, n is integer

d = the number of dimes, d>0, d is integer

q = the number of quarters, q>0, q is integer

a = the number of $1 bills, a>0

b = the number of $5 bills, b>0

There are twice as many nickels as pennies,

n = 2p

p = (1/2)n

twice as many dimes as nickels,

d = 2n

n = (1/2)d

three fewer quarters than dimes

q = d - 3

q = 2n - 3

and the same number of one dollar bills and five dollar as there are nickels

a + b = n

The total value of the money I have is $93.82 = 9382 cents:

p + 5n + 10d + 25q + 100a + 500b = 9382

lets plug the values from above in the last equation:

(1/2)n + 5n + 10(2n) + 25(2n - 3) + 100n + 400b = 9283

by solving we find:

3/2(117n - 50) + 400b = 9283

plug n = (1/2)d into last equation and solve by d:

3/2(117(1/2)d - 50) + 400b = 9283

d = (8/351)(4679 - 200b) dimes