Homework Help: Algebra 1

Posted by Kris on Sunday, October 16, 2011 at 9:33pm.

In 2000, the internatonal price of gold was $270 per ounce, and the price of silver was $5 per ounce. How much silver could be mixed with 9 oz of gold to obtain a mixture that costs $164 per ounce?

Algebra 1 - Steve, Monday, October 17, 2011 at 1:04pm
total weight of silver + gold = s+g
total value of s+g = 5s + 270g

So, now for the mixture

5s + 270g = (s+g)*164
Bu, g=9

5s + 270*9 = 164(s+9)
5s + 2430 = 164s + 1476
159s = 954
s = 6

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