Question 591482
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
given:
(1) {{{ 5n + 10d + 25q = 5725 }}} ( in change )
(2) {{{ d = 3n + 2 }}}
(3) {{{ q = 2n - 3 }}}
Substitute (2) and (3) into (1)
(1) {{{ 5n + 10*( 3n + 2 ) + 25*(  2n - 3 ) = 5725 }}}
(1) {{{ 5n + 30n + 20 + 50n - 75 = 5725 }}}
(1) {{{ 85n = 5725 + 75 - 20 }}}
(1) {{{ 85n = 5800 - 20 }}}
(1) {{{ 85n = 5780 }}}
(1) {{{ n = 68 }}}
and, since
(2) {{{ d = 3n + 2 }}}
(2) {{{ d = 3*68 + 2 }}}
(2) {{{ d = 204 + 2 }}}
(2) {{{ d = 206 }}}
and
(3) {{{ q = 2n - 3 }}}
(3) {{{ q = 2*68 - 3 }}}
(3) {{{ q = 136 - 3 }}}
(3) {{{ q = 133 }}}
There are 68 nickels, 206 dimes and 133 quarters
check:
(1) {{{ 5*68 + 10*206 + 25*133 = 5725 }}}
(1) {{{ 340 + 2060 + 3325 = 5725 }}}
(1) {{{ 5725 = 5725 }}}
OK