Question 884303
(a) He can't have an odd number of quarters, 
since that would make the total look like @@.@5
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(b) He can't have 23 or more quarters, since
that is $5.75 or greater
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Let {{{ d }}} = number of dimes he has
Let {{{ q }}} = number of quarters he has
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(1) {{{ d + q = 27 }}}
(2) {{{ 10d + 25q = 570 }}} ( in cents )
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Multiply both sides of (1) by {{{ 10 }}}
and subtract (1) from (2)
(2) {{{ 10d + 25q = 570 }}} 
(1) {{{ -10d - 10q = -270 }}}
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{{{ 15q = 300 }}}
{{{ q = 20 }}}
and, since
(1) {{{ d + q = 27 }}}
(1) {{{ d + 20 = 27 }}}
(1) {{{ d = 7 }}}
He has 7 dimes and 20 quarters
check:
(2) {{{ 10*7 + 25*20 = 570 }}} 
(2) {{{ 70 + 500 = 570 }}}
(2) {{{ 570 = 570 }}}
OK