Question 803271
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
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(1) {{{ n + d + q = 58 }}}
(2) {{{ 5n + 10d + 25q = 440 }}} ( in cents )
(3) {{{ n = 3*( d + q ) - 2 }}}
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(3) {{{ n = 3d + 3q - 2 }}}
(3) {{{ -n + 3d + 3q = 2 }}}
Multiply both sides of (1) by {{{ -3 }}}
and add (1) and (3)
(1) {{{ -3n - 3d - 3q = -174 }}}
(3) {{{ -n + 3d + 3q = 2 }}}
{{{ -4n = -172 }}}
{{{ n = 43 }}}
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(2) {{{ 5*43 + 10d + 25q = 440 }}}
(2) {{{ 10d + 25q = 440 - 215 }}}
(2) {{{ 10d + 25q = 225 }}}
(2) {{{ 2d + 5q = 45 }}}
and
(3) {{{ -n + 3d + 3q = 2 }}}
(3) {{{ 3d + 3q = 43 + 2 }}}
(3) {{{ d + q = 15 }}}
(3) {{{ 2d + 2q = 30 }}}
Subtract (3) from (2)
(2) {{{ 2d + 5q = 45 }}}
(3) {{{ -2d - 2q = -30 }}}
{{{ 3q = 15 }}}
{{{ q = 5 }}}
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(1) {{{ n + d + q = 58 }}}
(1) {{{ 43 + d + 5 = 58 }}}
(1) {{{ d = 58 - 48 }}}
(1) {{{ d = 10 }}}
There are 10 dimes
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check:
(2) {{{ 5n + 10d + 25q = 440 }}} 
(2) {{{ 5*43 + 10*10 + 25*5 = 440 }}} 
(2) {{{ 215 + 100 + 125 = 440 }}}
(2) {{{ 440 = 440 }}}
OK