Question 990746
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
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(1) {{{ 5n + 10d + 25q = 600 }}} ( in cents )
(2) {{{ d = 2q }}}
(3) {{{ n = d + 87 }}}
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There are 3 equations and 3 unknowns, so it's solvable
(2) {{{ q = d/2 }}}
Substitute (2) and (3) into (1)
(1) {{{ 5*( d+87 ) + 10d + 25*(d/2) = 600 }}}
(1) {{{ 5d + 435 + 10d + (1/2)*25d = 600 }}}
Multiply both sids by {{{ 2 }}}
(1) {{{ 10d + 870 + 20d + 25d = 1200 }}}
(1) {{{ 55d = 330 }}}
(1) {{{ d = 6 }}}
and
(2) {{{ q = d/2 }}}
(2) {{{ q = 6/2 }}}
(2) {{{ q = 3 }}}
and
(3) {{{ n = d + 87 }}}
(3) {{{ n = 6 + 87 }}}
(3) {{{ n = 93 }}}
There are 93 nickels
There are 6 dimes
There are 3 quarters
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Check:
(1) {{{ 5n + 10d + 25q = 600 }}}
(1) {{{ 5*93 + 10*6 + 25*3 = 600 }}}
(1) {{{ 465 + 60 + 75 = 600 }}}
(1) {{{ 600 = 600 }}}
OK