Question 201312
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
Let {{{q}}} = number of quarters
given:
{{{d = n + 6}}}
{{{q = 3d}}}
{{{5n + 10d + 25q = 1140}}} (in pennies)
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{{{d = n + 6}}}
{{{n = d - 6}}}
Then, by substituting,
{{{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}}}
Sam found 7 nickels, 13 dimes and 39 quarters
check answer:
{{{5n + 10d + 25q = 1140}}}
{{{5*7 + 10*13 + 25*39 = 1140}}}
{{{35 + 130 + 975 = 1140}}}
{{{1140 = 1140}}}
OK