Question 297266
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
Let {{{q}}} = number of quarters
given:
{{{d = 2n}}}
{{{q = d - 3}}}
{{{5n + 10d + 25q = 450}}} (in cents)
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This is 3 equations and 3 unknowns, so it's solvable
By substitutions:
{{{5n + 10d + 25*(d - 3) = 450}}}
And, substituting again:
{{{5*(d/2) + 10d + 25*(d - 3) = 450}}}
{{{5d/2 + 10d + 25d - 75 = 450}}}
{{{5d + 20d + 50d - 150 = 900}}}
{{{75d = 1050}}}
{{{d = 14}}}
and, since
{{{q = d - 3}}}
{{{q = 11}}}
and
{{{n = d/2}}}
{{{n = 7}}}
There are 14 dimes, 11 quarters, and 7 nickels
check:
{{{5n + 10d + 25q = 450}}} 
{{{5*7 + 10*14 + 25*11 = 450}}}
{{{35 + 140 + 275 = 450}}}
{{{450 = 450}}}
OK