Question

,,,Help Please.. Question in the file.

Answer 1

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`

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