Question 913545
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
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(1) {{{ 5n + 10d + 25q = 1490 }}} ( in cents )
(2) {{{ n = d + 2 }}}
(3) {{{ q = d - 8 }}}
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There are 3 equations and 3 unknowns
so it's solvable
Substitute (2) and (3) into (1)
(1) {{{ 5*( d+2 ) + 10d + 25*( d - 8 ) = 1490 }}}
(1) {{{ 5d + 10 + 10d + 25d - 200 = 1490 }}}
(1) {{{ 40d = 1490 + 190  }}}
(1) {{{ 40d = 1680 }}}
(1) {{{ d = 42 }}}
and
(3) {{{ q = d - 8 }}}
(3) {{{ q = 42 - 8 }}}
(3) {{{ q = 34 }}}
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She has 34 quarters
check:
(2) {{{ n = d + 2 }}}
(2) {{{ n = 42 + 2 }}}
(2) {{{ n = 44 }}}
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(1) {{{ 5*44 + 10*42 + 25*34 = 1490 }}} 
(1) {{{ 220 + 420 + 850 = 1490 }}}
(1) {{{ 1490 = 1490 }}}
OK