Question 585807
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
given:
(1) {{{ 5n + 10d = 705 }}} ( in cents )
(2) {{{ 5*( n + 7 ) + 10*(2d) = 1330 }}}
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(2) {{{ 5n + 35 + 20d = 1330 }}}
(2) {{{ 5n + 20d = 1295 }}}
Subtract (1) from (2)
(2) {{{ 5n + 20d = 1295 }}}
(1) {{{ -5n - 10d = -705 }}}
{{{ 10d = 590 }}}
{{{ d = 59 }}}
and, since
(1) {{{ 5n + 10d = 705 }}} 
(1) {{{ 5n + 10*59 = 705 }}} 
(1) {{{ 5n = 705 - 590 }}}
(1) {{{ 5n = 115 }}}
(1) {{{ n = 23 }}}
He has 23 nickels and 59 dimes
check:
(2) {{{ 5*( 23 + 7 ) + 10*(2*59) = 1330 }}}
(2) {{{ 5*30 + 10*118 = 1330 }}}
(2) {{{ 150 + 1180 = 1330 }}}
(2) {{{ 1330 = 1330 }}}
OK