SOLUTION: A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelths pure gold. How many ounces of each should be mel

Question 111614: A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelths pure gold. How many ounces of each should be melted and mixed in order to obtain a 6-oz mixture that is two-thirds pure gold?
Answer by ptaylor(1894) (Show Source):
You can put this solution on YOUR website!

Let x=amount of 3/4 pure gold needed
Then 6-x=amount of 5/12 pure gold needed
Now we know that the amount of pure gold in the 3/4 alloy ((3/4)x) plus the amount of pure gold in the 5/12 alloy ((5/12)(6-x)) has to equal the amount of pure gold in the final alloy(6*(2/3)). So our equation to solve is:
(3/4)x+(5/12)(6-x)=6*(2/3) simplify
(3/4)x+(5/2)-(5/12)x=4 multiply each term by 12 (LCM)
9x+30-5x=48 subtract 30 from both sides
9x-5x+30-30=48-30 collect like terms
4x=18 divide both sides by 4
x=4.5 oz----------------------amount of 3/4 pure gold needed
6-x=6-4.5=1.5 oz-----------------------amount of 5/12 pure gold needed
CK
(4.5)(3/4)+(1.5)(5/12)=6(2/3)
3.375+0.625=4
4=4
Hope this helps---ptaylor