IF there are four more nickels than dimes, and two more quarters, you can give each a variable; n, d and q. There are d+4=n, and q=2+n. Therefore, because we already know that n=d+4, then we can substitute this equation. Therefore, 2+d+4=q. We now know that there are 6 more quarters than nickels. This would mean that of 37 coins, d+n+q=37. But n=d+4, and q=2+n. Therefore d+d+4+2+d+4=37. So 3d=37-10. 3d=27. There are 9 dimes, 13 nickels, and 15 quarters.
To find the with divide 102 by 17, we get 6. Then add 6+6+17+17 because we know this is a rectangle and it has 2 with sides and 2 length sides. We get the answer of 36.
