Question 466174
Let {{{q}}} = number of quarters
Let {{{d}}} = number of dimes
Let {{{n}}} = number of nickels
given:
(1) {{{ 25q + 10d + 5n = 2415 }}} (in cents)
(2) {{{ n = 2q }}}
(3) {{{ d = n + 5 }}}
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From (2)
(2) {{{ q = n/2 }}}
Substitute (2) and (3) in (1)
(1) {{{ 25*(n/2) + 10*(n + 5) + 5n = 2415 }}}
(1) {{{ (25n)/2 + 10n + 50 + 5n = 2415 }}}
Multiply both sides by {{{2}}}
(1) {{{ 25n + 20n + 100 + 10n  = 4830 }}}
(1) {{{ 55n = 4730 }}}
(1) {{{ n = 86 }}}
and, since
(3) {{{ d = n + 5 }}}
(3) {{{ d = 86 + 5 }}}
(3) {{{ d = 91 }}}
and
(2) {{{ q = n/2 }}}
(2) {{{ q = 86/2 }}}
(2) {{{ q = 43 }}}
43 = number of quarters
91 = number of dimes
86 = number of nickels
check answer:
(1) {{{ 25*43 + 10*91 + 5*86 = 2415 }}}
(1) {{{ 1075 + 910 + 430 = 2415 }}}
(1) {{{ 2415 = 2415 }}}
OK