Question 524851
Let {{{d}}} = number of dimes
Let {{{n}}} = number of nickels
Let {{{q}}} = number of quarters
given:
(1) {{{ d = 3n }}}
(2) {{{ q = d + 10 }}}
(3) {{{ 10d + 5n + 25q = 1130 }}} ( in cents )
This is 3 equations with 3 unknowns,
so it's solvable
-------------------
(1) {{{ n = d/3 }}}
Substitute (1) and (2) into (3)
(3) {{{ 10d + 5*(d/3) + 25*(d+10) = 1130 }}}
(3) {{{ 10d + (5d)/3 + 25d + 250 = 1130 }}}
Multiply both sides by {{{3}}}
(3) {{{ 30d + 5d + 75d + 750 = 3390 }}}
(3) {{{ 110d = 3390 - 750 }}}
(3) {{{ 110d = 2640 }}}
(3) {{{ d = 24 }}}
and, since
(1) {{{ n = d/3 }}}
(1) {{{ n = 24/3 }}}
(1) {{{ n = 8 }}}
also,
(2) {{{ q = d + 10 }}}
(2) {{{ q = 24 + 10 }}}
(2) {{{ q = 34 }}}
There are 24 dimes, 8 nickels, and 34 quarters
check:
(3) {{{ 10d + 5n + 25q = 1130 }}}
(3) {{{ 10*24 + 5*8 + 25*34 = 1130 }}}
(3) {{{ 240 + 40 + 850 = 1130 }}}
(3) {{{ 1130 = 1130 }}}
OK