Algebra 1

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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 -

    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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