Question 644755
Let {{{ a }}} = ounces of 1st alloy needed
Let {{{ b }}} = ounces of 2nd alloy needed
given:
{{{ .26a }}} = ounces of copper in 1st alloy
{{{ .54b }}} = ounces of copper in 2nd alloy
(1) {{{ ( .26a + .54b ) / 210 = .3 }}}
(2) {{{ a + b = 210 }}}
-----------------
(1) {{{ .26a + .54b = .3*210 }}}
(1) {{{ .26a + .54b = 63 }}}
(1) {{{ 26a + 54b = 6300 }}}
Multiply both sides of (2) by {{{ 26 }}}
and subtract (2) from (1)
(1) {{{ 26a + 54b = 6300 }}}
(2) {{{ -26a - 26b = -5460 }}}
{{{ 28b = 840 }}}
{{{ b = 30 }}}
and, since
(2) {{{ a + b = 210 }}}
(2) {{{ a = 210 - 30 }}}
(2) {{{ a = 180 }}}
180 ounces of 1st alloy are needed
30 ounces of 2nd alloy are needed
check:
(1) {{{ ( .26a + .54b ) / 210 = .3 }}}
(1) {{{ ( .26*180 + .54*30 ) / 210 = .3 }}}
(1) {{{ 46.8 + 16.2 = .3*210 }}}
(1) {{{ 63 = 63 }}}
OK