Question 798210
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
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(1) {{{ n + d + q = 13 }}}
(2) {{{ 5n + 10d + 25q = 150 }}} ( in cents )
(3) {{{ d = n }}}
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This is 3 equations and 3 unknowns,
so it's solvable
Substitute (3) into both (1) and (2)
(1) {{{ n + n + q = 13 }}}
(2) {{{ 5n + 10n + 25q = 150 }}}
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(1) {{{ 2n + q = 13 }}}
(2) {{{ 15n + 25q = 150 }}}
multiply both sides of (1) by {{{ 25 }}}
and subtract (2) from (1)
(1) {{{ 50n + 25q = 325 }}}
(2) {{{ -15n - 25q = -150 }}}
{{{ 35n = 175 }}}
{{{ n = 5 }}} 
and, since
(3) {{{ d = n }}}
{{{ d = 5 }}}
and
(1) {{{ 5 + 5 + q = 13 }}}
(1) {{{ 10 + q = 13 }}}
(1) {{{ q = 3 }}}
There are 5  nickels
There are 5  dimes
There are 3  quarters
check:
(2) {{{ 5*5 + 10*5 + 25*3 = 150 }}}
(2) {{{ 25 + 50 + 75 = 150 }}}
(2) {{{ 150 = 150 }}}
OK