Question 183783
In words:
(gold in the 3/4 alloy) + (gold in the 5/12 alloy) / (total amt of alloy)
= (gold in the final mixture)/(total amount of alloy)
Let {{{a}}} = the ounces of 3/4 alloy needed
Let {{{b}}}= the ounces of the 5/12 alloy needed
given:
Ounces of gold in 3/4 alloy = {{{(3/4)*a}}}
Ounces of gold in the 5/12 alloy = {{{(5/12)*b}}}
Ounces of alloy in the final mixture = {{{6}}}
Ounces of gold in the final mixture = {{{(2/3)*6 = 4}}}
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{{{((3/4)*a + (5/12)*b) / (a + b) = 4/6}}} (this is same as word description)
Multiply both sides by {{{a + b}}}
{{{(3/4)*a + (5/12)*b = (2/3)*(a + b)}}}
Multiply both sides by {{{12}}}
{{{9a + 5b = 8*(a + b)}}}
{{{9a + 5b = 8a + 8b}}}
(1) {{{a - 3b = 0}}}
(2) {{{a + b = 6}}}
Subtract (1) from (2)
(2) {{{a + b = 6}}}
(1) {{{-a + 3b = 0}}}
{{{4b = 6}}}
{{{b = 3/2}}}
And, since
{{{a + b = 6}}}
{{{a + 3/2 = 6}}}
{{{a = 12/2 - 3/2}}}
{{{a = 9/2}}}
9/2 ounces of 3/4 alloy should be melted and mixed
with 3/2 ounces of 5/12 alloy
check answer:
{{{((3/4)*a + (5/12)*b) / (a + b) = 4/6}}}
{{{((3/4)*(9/2) + (5/12)*(3/2)) / (3/2 + 9/2) = 4/6}}} 
{{{((3/4)*(9/2) + (5/12)*(3/2)) /6 = 4/6}}}
Multiply both sides by {{{6}}}
{{{(3/4)*(9/2) + (5/12)*(3/2) = 4}}}
Multiply both sides by {{{24}}} 
{{{3*3*9 + 3*5 = 96}}}
{{{81 + 15 = 96}}}
{{{96 = 96}}}
OK