Question 647596
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
given:
(1) {{{ n + d + q = 22 }}}
(2) {{{ d = 4q }}}
(3) {{{ 5n + 10d + 25q = 230 }}} ( in cents )
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There are 3 unknowns and 3 equations, so it's solvable
substitute (2) into (1)
(1) {{{ n + 4q + q = 22 }}}
(1) {{{ n + 5q = 22 }}}
(1) {{{ n = 22 - 5q }}}
Now substitute this result and (2) into (3)
(3) {{{ 5*( 22 -5q ) + 10*4q + 25q = 230 }}}
(3) {{{ 110 - 25q + 40q + 25q = 230 }}}
(3) {{{ 40q = 230 - 110 }}}
(3) {{{ 40q = 120 }}}
(3) {{{ q = 3 }}}
and, since
(2) {{{ d = 4q }}}
(2) {{{ d = 4*3 }}}
(2) {{{ d = 12 }}}
and
(1) {{{ n + d + q = 22 }}}
(1) {{{ n + 12 + 3 = 22 }}}
(1) {{{ n + 15 = 22 }}}
(1) {{{ n = 22 - 15 }}}
(1) {{{ n = 7 }}}
There are 12 dimes in the bank
check:
(3) {{{ 5n + 10d + 25q = 230 }}}
(3) {{{ 5*7 + 10*12 + 25*3 = 230 }}}
(3) {{{ 35 + 120 + 75 = 230 }}}
(3) {{{ 155 + 75 = 230 }}}
(3) {{{ 230 = 230 }}}
OK