Question 517095
Let {{{q}}} = number of quarters
Let {{{ d }}} = number of dimes
Let {{{ n }}} = number of nickels
given:
(1) {{{ q + d + n = 223 }}}
(2) {{{ 25q + 10d + 5n = 2010 }}} ( in cents )
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Multiply both sides of (1) by {{{5}}}
and subtract (1) from (2)
(2) {{{ 25q + 10d + 5n = 2010 }}}
(1) {{{ -5q - 5d - 5n = -1115 }}}
{{{ 20q + 5d = 895 }}}
{{{ 5d = 895 - 20q }}}
{{{ d = 179 - 4q }}}
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Suppose there are 100 dimes
{{{ 100 = 179 - 4q }}}
{{{ 4q = 79 }}} ( doesn't work, {{{d}}} has to be odd)
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107 dimes:
{{{ 107 = 179 - 4q }}}
{{{ 4q = 72 }}}
{{{ q = 18 }}}
{{{ 107 + 18 + n = 223 }}}
{{{ n = 98 }}}
{{{ 25*18 + 10*107 + 5*98 = 450 + 1070 + 490 }}}
{{{ 2010 }}}
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Wow! That was just a guess.
18 quarters, 107 dimes and 98 nickels