Question 718478
Let {{{ q }}} = number of quarters
Let {{{ d }}} = number of dimes 
Let {{{ n }}} = number of nickels
given:
(1) {{{ n + d + q = 6 }}}
(2) {{{ 5n + 10d + 25q = 85 }}} ( in cents )
(3) {{{ 500n + 400d + 700q = 3100 }}} ( in cents )
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There are 3 equations and 3 unknowns, so 
it's solvable
Multiply both sides of (1) by {{{ 400 }}}
and subtract (1) from (3)
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(3) {{{ 500n + 400d + 700q = 3100 }}} 
(1) {{{ -400n - 400d - 400q = -2400 }}}
{{{ 100n + 300q = 700 }}}
{{{ n + 3q = 7 }}}
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Multiply both sides of (1) by {{{ 10 }}} and
subtract (1) from (2)
(2) {{{ 5n + 10d + 25q = 85 }}}
(1) {{{ -10n - 10d - 10q = -60 }}}
{{{ -5n + 15q = 25 }}}
{{{ -n + 3q = 5 }}}
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Add the results of these 2 operations:
{{{ n + 3q = 7 }}}
{{{ -n + 3q = 5 }}}
{{{ 6q = 12 }}}
{{{ q = 2 }}}
Two of the coins were quarters
check answer:
{{{ -n + 3q = 5 }}}
{{{ -n + 3*2 = 5 }}}
{{{ -n = -1 }}}
{{{ n = 1 }}}
and
(1) {{{ n + d + q = 6 }}}
(1) {{{ 1 + d + 2 = 6 }}}
(1) {{{ d = 3 }}}
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(3) {{{ 500n + 400d + 700q = 3100 }}}
(3) {{{ 5n + 4d + 7q = 31 }}}
(3) {{{ 5*1 + 4*3 + 7*2 = 31 }}}
(3) {{{ 5 + 12 + 14 = 31 }}}
(3) {{{ 31 = 31 }}}
OK