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
[tex]\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[/tex]
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
