Question 781991
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
----------------------------------
given:
(1) {{{ 5n + 10d + 25q = 610 }}} ( in cents )
(2) {{{ d = q - 5 }}}
(3) {{{ n = d - 7 }}}
---------------------
This is 3 equations with 3 unknowns, so it's solvable
Substitute (3) into (1)
(1) {{{ 5*( d-7 ) + 10d + 25q = 610 }}}
and
(2) {{{ d = q - 5 }}}
(2) {{{ q = d + 5 }}}
Substitute (2) into (1)
(1) {{{ 5*( d-7 ) + 10d + 25*( d+5 ) = 610 }}}
(1) {{{ 5d - 35 + 10d + 25d + 125 = 610 }}}
(1) {{{ 40d = 610 + 35 - 125 }}}
(1) {{{ 40d = 520 }}}
(1) {{{ d = 13 }}}
and
(3) {{{ n = d - 7 }}}
(3) {{{ n = 13 - 7 }}}
(3) {{{ n = 6 }}}
and
(2) {{{ q = d + 5 }}}
(2) {{{ q = 13 + 5 }}}
(2) {{{ q = 18 }}}
There are 6 nickels, 13 dimes, and 18 quarters
check:
(1) {{{ 5*6 + 10*13 + 25*18 = 610 }}}
(1) {{{ 30 + 130 + 450 = 610 }}}
(1) {{{ 610 = 610 }}}
OK