Question 497421
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
given:
(1) {{{ d = 2n }}}
(2) {{{ 5n + 10d = 3625 }}} (in cents)
---------------------
By substitution,
(2) {{{ 5n + 10*2n = 3625 }}}
(2) {{{ 25n = 3625 }}}
(2) {{{ n = 145 }}}
and, since,
(1) {{{ d = 2n }}}
(1) {{{ d = 2*145 }}}
(1) {{{ d = 290 }}}
The number of nickels is 145
The number of dimes is 290
----------
check:
(2) {{{ 5n + 10d = 3625 }}}
(2) {{{ 5*145 + 10*290 = 3625 }}}
(2) {{{ 725 + 2900 = 3625 }}}
(2) {{{ 3625 = 3625 }}}
OK