Question 770735
Let {{{ a }}} = gallons of 40% solution needed
Let {{{ b }}} = gallons of 80% solution needed
{{{ .4a }}} = gallons of alcohol in 40% solution
{{{ .8b }}} = gallons of alcohol in 80% solution
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(1) {{{ a + b = 40 }}}
(2) {{{ ( .4a + .8b ) / 40 = .5 }}}
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(2) {{{ .4a + .8b = .5*40 }}}
(2) {{{ .4a + .8b = 20 }}}
(2) {{{ 4a + 8b = 200 }}}
(2) {{{ a + 2b = 50 }}}
Subtract (1) from (2)
(2) {{{ a + 2b = 50 }}}
(1) {{{ -a - b = -40 }}}
{{{ b = 10 }}}
and, since
(1) {{{ a + b = 40 }}}
(1) {{{ a = 30 }}}
30 gallons of 40% solution are needed
10 gallons of 80% solution are needed
check:
(2) {{{ ( .4*30 + .8*10 ) / 40 = .5 }}}
(2) {{{ ( 12 + 8 ) / 40 = .5 }}}
(2) {{{ 20 = .5*40 }}}
(2) {{{ 20 = 20 }}}
OK