Question 808802
Let {{{ a }}} = number of $1 coins
Let {{{ b }}} = number of $.50 coins
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(1) {{{ a + b = 80 }}}
(2) {{{ 1*a + .5*b = 50 }}}
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(1) {{{ b = 80 - a }}}
Substitute this into (2)
(2) {{{a + .5*( 80 - a ) = 50 }}}
(2) {{{ a + 40 - .5a = 50 }}}
(2) {{{ .5a = 50 - 40 }}}
(2) {{{ .5a = 10 }}}
(2) {{{ a = 20 }}}
and, since
(1) {{{ a + b = 80 }}}
(1) {{{ b = 60 }}}
There are 20 $1 coins and 60 $.50 coins
check:
(2) {{{ 1*20 + .5*60 = 50 }}}
(2) {{{ 20 + 30 = 50 }}}
(2) {{{ 50 = 50 }}}
OK