Question 1033040
(1) {{{ d = n + 5 }}}
(2) {{{ q = 2d - 3 }}}
(3) {{{ 5n + 10d + 25q = 1720 }}} ( in cents )
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This si 3 equations and 3 unknowns, so
it's solvable
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(1) {{{ n = d - 5 }}}
Substitute (1) and (2) into (3)
(3) {{{ 5*( d-5 ) + 10d + 25*( 2d-3 ) = 1720 }}}
(3) {{{ 5d - 25 + 10d + 50d - 75 = 1720 }}}
(3) {{{ 65d = 1820 }}}
(3) {{{ d = 28 }}}
and
(1) {{{ n = d - 5 }}}
(1) {{{ n = 28 - 5 }}}
(1) {{{ n = 23 }}}
and
(2) {{{ q = 2d - 3 }}}
(2) {{{ q = 2*28 - 3 }}}
(2) {{{ q = 56 - 3 }}}
(2) {{{ q = 53 }}}
there are 53 quarters
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check:
(3) {{{ 5n + 10d + 25q = 1720 }}}
(3) {{{ 5*23 + 10*28 + 25*53 = 1720 }}}
(3) {{{ 115 + 280 + 1325 = 1720 }}}
(3) {{{ 1720 = 1720 }}}
OK