Question 984043
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
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(1) {{{ 5n + 10d + 25q = 710 }}} ( in cents )
(2) {{{ q = d + 10 }}}
(3) {{{ n = 2q }}}
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You have 3 equations and 3 unknowns,
so it's solvable
(2) {{{ d = q - 10 }}}
Substitute (2) and (3) into (1)
(1) {{{ 5*2q + 10*( q-10 ) + 25q = 710 }}} 
(1) {{{ 10q + 10q- 100 + 25q = 710 }}}
(1) {{{ 45q = 810 }}}
(1) {{{ q = 18 }}}
and
(3) {{{ n = 2q }}}
(3) {{{ n = 2*18 }}}
(3) {{{ n = 36 }}}
and
(2) {{{ d = q - 10 }}}
(2) {{{ d = 18 - 10 }}}
(2) {{{ d = 8 }}}
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He has:
36 nickels
8 dimes
18 quarters
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check:
(1) {{{ 5n + 10d + 25q = 710 }}}
(1) {{{ 5*36 + 10*8 + 25*18 = 710 }}}
(1) {{{ 180 + 80 + 450 = 710 }}}
(1) {{{ 710 = 710 }}}
OK