Question 723641
given:
(1) {{{ n = q + d + 2 }}}
(2) {{{ d = q + 1 }}}
(3) {{{ 5n + 10d + 25q = 250 }}} ( in cents )
--------------
Substitute (1) into (3)
(3) {{{ 5*( d + q + 2 ) + 10d + 25q = 250 }}}
(3) {{{ 5d + 5q + 10 + 10d + 25q = 250 }}}
(3) {{{ 15d + 30q = 240 }}}
(3) {{{ 3d + 6q = 48 }}}
Substitute (2) into (3)
(3) {{{ 3*( q + 1 ) + 6q = 48 }}}
(3) {{{ 3q + 3 + 6q = 48 }}}
(3) {{{ 9q = 45 }}}
(3) {{{ q = 5 }}}
and, since
(2) {{{ d = q + 1 }}}
(2) {{{ d = 5 + 1 }}}
(2) {{{ d = 6 }}}
and
(1) {{{ n = q + d + 2 }}}
(1) {{{ n = 5 + 6 + 2 }}}
(1) {{{ n = 13 }}}
There are 13 nickels, 6 dimes and 5 quarters
check:
(3) {{{ 5n + 10d + 25q = 250 }}} 
(3) {{{ 5*13 + 10*6 + 25*5 = 250 }}} 
(3) {{{ 65 + 60 + 125 = 250 }}}
(3) {{{ 250 = 250 }}}
OK