Question 839801
Let {{{ n }}} = number of nickels she has
Let {{{ d }}} = number of dimes she has
Let {{{ q }}} = number of quarters she has
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(1) {{{ q = d + 4 }}}
(2) {{{ n = q + 5 }}}
(3) {{{ 5n + 10d + 25q = 465 }}} ( in cents )
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This is 3 equations and 3 unknowns, so it's solvable
(1) {{{ d = q - 4 }}}
Substitute (1) and (2) into (3)
(3) {{{ 5*( q+5 ) + 10*( q-4 ) + 25q = 465 }}} 
(3) {{{ 5q + 25 + 10q - 40 + 25q = 465 }}}
(3) {{{ 40q = 465 - 25 + 40 }}}
(3) {{{ 40q = 480 }}}
(3) {{{ q = 12 }}}
and
(1) {{{ d = q - 4 }}}
(1) {{{ d = 12 - 4 }}}
(1) {{{ d = 8 }}}
and
(2) {{{ n = q + 5 }}}
(2) {{{ n = 12 + 5 }}}
(2) {{{ n = 17 }}}
17= number of nickels she has
8 = number of dimes she has
12 = number of quarters she has
check:
(3) {{{ 5n + 10d + 25q = 465 }}} 
(3) {{{ 5*17 + 10*8 + 25*12 = 465 }}} 
(3) {{{ 85 + 80 + 300 = 465 }}}
(3) {{{ 465 = 465 }}}
OK