Question 536826
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
Let {{{ q}}} = number of quarters
given:
(1) {{{ 5n + 10d + 25q = 6850 }}} ( in cents )
(2) {{{ n = q + 70 }}}
(3) {{{ d = n + 20 }}}
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(2) {{{ q = n - 70 }}}
Substitute (2) and (3) into (1)
(1) {{{ 5n + 10*( n + 20 ) + 25*( n - 70 ) = 6850 }}}
(1) {{{ 5n + 10n + 200 + 25n - 1750 = 6850 }}}
(1) {{{ 40n + 200 - 1750 = 6850 }}}
(1) {{{ 40n = 8400 }}}
(1) {{{ n = 210 }}}
and, since
(3) {{{ d = n + 20 }}}
(3) {{{ d = 230 }}}
and
(2) {{{ q = n - 70 }}}
(2) {{{ q = 210 - 70 }}}
(2) {{{ q = 140 }}}
He had 210 nickels,  230 dimes, and 140 quarters
check:
(1) {{{ 5*210 + 10*230 + 25*140 = 6850 }}} 
(1) {{{ 1050 + 2300 + 3500 = 6850 }}}
(1) {{{ 6850 = 6850 }}}
OK