Question 129650
Let {{{a}}}= the amount of the 1st alloy used in ounces
Let {{{b}}}= the amount of the 2nd alloy used in ounces
{{{((3/4)*a + (5/12)*b) / 6 = 2/3}}}
{{{a + b = 6}}}
{{{b = 6 - a}}}
{{{((3/4)*a + (5/12)(6 - a)) / 6 = 2/3}}}
multiply both sides by 6
{{{(3/4)*a + (5/12)(6 - a) = 12/3}}}
multiply both sides by 12
{{{9a + 30 - 5a = 48}}}
{{{4a = 18}}}
{{{a = 4.5}}}
{{{b = 6 - a}}}
{{{b = 1.5}}}
Use 4.5 oz of the 3/4 gold and 1.5 oz of the 5/12 gold
check
{{{((3/4)*a + (5/12)*b) / 6 = 2/3}}}
{{{(.75*4.5 + .4167*1.5) / 6 = .6667}}}
{{{(3.375 + .6251) / 6 = .6667}}}
{{{4.0001 / 6 = .6667}}}
{{{.66668 = .6667}}}
close enough