Question 807704
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
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(1) {{{ n + d = 90 }}}
(2) {{{ 5n + 10d = 640 }}} ( in cents )
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(2) {{{ n + 2d = 128 }}}
Subtract (1) from (2)
(2) {{{ n + 2d = 128 }}}
(1) {{{ -n - d = -90 }}}
{{{ d = 38 }}}
and, since
(1) {{{ n + d = 90 }}}
(1) {{{ n + 38 = 90 }}}
(1) {{{ n = 52 }}}
52 nickels are in the drawer
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check:
(2) {{{ 5n + 10d = 640 }}} 
(2) {{{ 5*52 + 10*38 = 640 }}} 
(2) {{{ 260 + 380 = 640 }}}
(2) {{{ 640 = 640 }}}
OK