Question 693143
Let {{{ n }}} = the number of nickels he has
Let {{{ d }}} = the number of dimes he has
Let {{{ q }}} = the number of quarters he has
given:
(1) {{{ q = 2n }}}
(2) {{{ d = n + 3 }}}
(3) {{{ 5n + 10d + 25q = 420 }}} 
( the units for this equation is cents , for instance,
$4.20 is 420 cents )
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At this point it is important to know if you have 
enough information to arrive at a solution.
There are 3 unknowns and 3 equations, which
means a solution is possible.
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All you have to do is substitute (1) and (2) into (3)
so that the only variable is {{{ n }}}, then solve
(3) {{{ 5n + 10*( n + 3 ) + 25*(2n) = 420 }}}
(3) {{{ 5n + 10n + 30 + 50n = 420 }}}
(3) {{{ 65n + 30 = 420 }}}
Subtract {{{ 30 }}} from both sides
(3) {{{ 65n = 390 }}}
(3) {{{ n = 6 }}}
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And, since
(2) {{{ d = n + 3 }}}
(2) {{{ d = 6 + 3 }}}
(2) {{{ d = 9 }}}
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Also, since
(1) {{{ q = 2n }}}
(1) {{{ q = 2*6 }}}
(1) {{{ q = 12 }}}
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Mark has 6 nickels, 9 dimes, and 12 quarters
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check answer:
(3) {{{ 5n + 10d + 25q = 420 }}} 
(3) {{{ 5*6 + 10*9 + 25*12 = 420 }}} 
(3) {{{ 30 + 90 + 300 = 420 }}}
(3) {{{ 420 = 420 }}}
OK
(1) {{{ q = 2n }}}
(1) {{{ 12 = 2*6 }}}
(1) {{{ 12 = 12 }}}
OK
(2) {{{ d = n + 3 }}}
(2) {{{ 9 = 6 + 3 }}}
(2) {{{ 9 = 9 }}}
OK
Hope this helps