Question 755530
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
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given:
(1) {{{ n + d + q = 40 }}}
(2) {{{ n = 2q - 1 }}}
(3) {{{ 5n + 10d + 25q = 445 }}} ( in cents )
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You have 3 equations and 3 unknowns
so it's solvable
Substitute (2) into (1)
(1) {{{ 2q - 1 + d + q = 40 }}} 
(1) {{{ d + 3q = 41 }}}
(1) {{{ d = 41 - 3q }}}
Now substitute (1) and (2) into (3)
(3) {{{ 5*( 2q - 1 ) + 10*( 41 - 3q ) + 25q = 445 }}}
(3) {{{ 10q - 5 + 410 - 30q + 25q = 445 }}}
(3) {{{ -20q + 25q = 445 - 405 }}}
(3) {{{ 5q = 40 }}}
(3) {{{ q = 8 }}}
and, since
(1) {{{ d = 41 - 3q }}}
(1) {{{ d = 41 - 3*8 }}}
(1) {{{ d = 41 - 24 }}}
(1) {{{ d = 17 }}}
and
(2) {{{ n = 2q - 1 }}}
(2) {{{ n = 2*8 - 1 }}}
(2) {{{ n = 15 }}}
She has 15 nickels, 17 dimes, and 8 quarters
check:
(3) {{{ 5n + 10d + 25q = 445 }}} 
(3) {{{ 5*15 + 10*17 + 25*8 = 445 }}}  
(3) {{{ 75 + 170 + 200 = 445 }}}
(3) {{{ 445 = 445 }}}
OK