Question 190247
Let the number of nickels = {{{n}}}
Let the number of dimes = {{{d}}}
Let the number of quarters = {{{q}}}
given:
(1) {{{d = n + 6}}}
(2) {{{q = 3d}}}
(3) {{{5n + 10d + 25q = 1140}}} (in cents)
---------------------------
From (1) 
{{{n = d - 6}}}
therefore
(3) {{{5n + 10d + 25q = 1140}}}
{{{5*(d - 6) + 10d + 25*(3d) = 1140}}}
{{{5d - 30 + 10d + 75d = 1140}}}
{{{90d = 1170}}}
{{{d = 13}}}
and, since
{{{n = d - 6}}}
{{{n = 13 - 6}}}
{{{n = 7}}}
also
{{{q = 3d}}}
{{{q = 3*13}}}
{{{q = 39}}}
He found 7 nickels, 13 dimes and 39 quarters
check:
(3) {{{5n + 10d + 25q = 1140}}} 
You can check this