SOLUTION: Last year Carlos had $30000 to invest he invested some of it in an account that paid 10% simple interest per year and he invested the rest in an account that paid 8% simple interes

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Last year Carlos had $30000 to invest he invested some of it in an account that paid 10% simple interest per year and he invested the rest in an account that paid 8% simple interes      Log On


   

Question 1008497: Last year Carlos had $30000 to invest he invested some of it in an account that paid 10% simple interest per year and he invested the rest in an account that paid 8% simple interest per year. After one year he received a total of of $2800 in interest how much did he invest in each account
First account =
Second account =
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Last year Carlos had $30000 to invest he invested some of it in an account that paid 10% simple interest per year and he invested the rest in an account that paid 8% simple interest per year. After one year he received a total of of $2800 in interest how much did he invest in each account
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Quantity Eq:: t + e = 30000
Interest Eq::10t+8e = 280000
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Modify for elimination::
8t + 8e = 8*30000
10t+ 8e = 280000
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2t = 40000
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t = $20000 (amt invested at 10%)
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e = 30000- 20000
e = $10,000 (amt. invested at 8%)
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Cheers,
Stan H.
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