1 nickel = 5 cents

1 dime = 10 cents

1 quarter = 25 cents

let

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

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

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

a box contains $7.00 = 700 cents of nickels, dimes, and quarters:

5n + 10d + 25q = 700 divide both sides by 5

n + 2d + 5q = 140

there are 41 coins in all:

n + d + q = 41

the sum of the nickels and dimes is 3 less then the number of quarters:

n + d = q - 3

by solving the system of equations:

n + 2d + 5q = 140

n + d + q = 41

n + d = q - 3

we find:

n = 8 nickles

d = 11 dimes

q = 22 quarters

click here to see the step by step solution of the system of equations:

there are 8 nickles, 11 dimes and 22 quarters.