Question 159285
    You invest $6000 in two accounts paying 9% and 6% anual interest. At the end of the year, the total interest from these investments was $480. How much was invested at each rate?
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Let x = amount invested at 9%
then because the total invested was $6000 then
6000-x = amount invested at 6%
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And, because the total interest amount at the end of the year was $480:
.09x + .06(6000-x) = 480
.09x + 360 - .06x = 480
.03x + 360 = 480
.03x = 120
x = $4000 (amount invested at 9%)
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Amount invested at 6%:
6000-x = 6000-4000 = $2000 (amount invested at 6%)