Let n, d, q represent the numbers of nickels, dimes, and quarters Sarah has. The problem statement gives rise to three equations:
... 5n +10d +25q = 1575 . . . . . . the collection's value in cents
... n + 10 = d . . . . . . . . . . . . . . . . the number of dimes is 10 more than the number of nickels
... q = 2d . . . . . . . . . . . . . . . . . . . she has twice as many quarters as dimes
Solving the second equation for n and substituting into the first equation, we have ...
... 5(d -10) + 10d + 25(2d) = 1575
... 65d -50 = 1575
... 65d = 1625
... d = 1625/65 = 25
... n = d -10 = 15
... q = 2d = 50
Sarah has 15 nickels, 25 dimes, and 50 quarters.
