Question 189726
Let {{{n}}}= number of nickels
Let {{{d}}}= number of dimes
Let {{{q}}}= number of quarters
given:
(1) {{{d = n + 6}}}
(2) {{{q = 3d}}}
(3) {{{5n + 10d + 25q = 1140}}} (in cents)
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From (1),
{{{n = d - 6}}}
Now I can rewrite (3)
(3) {{{5*(d - 6) + 10d + 25*3d = 1140}}}
{{{5*(d - 6) + 10d + 25*3d = 1140}}}
{{{5d - 30 + 10d + 75d = 1140}}}
{{{90d = 1140 + 30}}}
{{{90d = 1170}}}
{{{d = 13}}}
and from (1),
{{{n = d - 6}}}
{{{n = 13 - 6}}}
{{{n = 7}}}
From (2),
{{{q = 3d}}}
{{{q = 3*13}}}
{{{q = 39}}}
Sam found 7 nickels, 13 dimes, and 39 quarters 
check answer:
(3) {{{5n + 10d + 25q = 1140}}}
{{{5*7 + 10*13 + 25*39 = 1140}}}
{{{35 + 130 + 975 = 1140}}}
{{{1140 = 1140}}}
OK