So hmmm there are 5pennies in a nickel, and 10pennies in a dime and 25pennies in a quarter
now, in $6.10, there are 610 pennies
n = amount of nickels coins
d = amount of dime coins
q = amount of quarter coins
so... we know, there are 5*n or 5n pennies in the nickels, and 10*d or 10d pennies in the dimes and 25*q or 25q in the quarter coins, and we know their sum is 610 pennies total, thus
5n + 10d + 25q = 610
now, there are 4 more "n" then "d", so whatever "d" is, "n" is 4 more than that
n = d + 4
and twice as many "q" than "n", so, whatever "n" is, then
q = 2n
![\bf 5n+10d+25q=610\implies n+2d+5q=122\qquad \begin{cases} n=d+4\\ q=2n\\ ------\\ q=2(d+4)\\ \qquad 2d+8 \end{cases} \\\\\\ \boxed{d+4}+2d+5\left( \boxed{2d+8} \right)=122](https://tex.z-dn.net/?f=%5Cbf%205n%2B10d%2B25q%3D610%5Cimplies%20n%2B2d%2B5q%3D122%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0An%3Dd%2B4%5C%5C%0Aq%3D2n%5C%5C%0A------%5C%5C%0Aq%3D2%28d%2B4%29%5C%5C%0A%5Cqquad%202d%2B8%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bd%2B4%7D%2B2d%2B5%5Cleft%28%20%5Cboxed%7B2d%2B8%7D%20%5Cright%29%3D122)
solve for "d", to see how many dimes are there
what about the nickels? well, n = d + 4
what about the quarters? well, q = 2d + 8