A sum of money is invested at 4% simple interest. $500 less than three times that amount is invested at 6% simple interest. If the total interest earned in one year from the two investments is $1686, how much was invested at 6%

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A sum of money is invested at 4% simple interest. $500 less than three times that amount is invested at 6% simple interest. If the total interest earned in one year from the two investments is $1686, how much was invested at 6%

Mathematics
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Remember the equation for exponential growth \[Pe ^{rt}\] where P is the principal (initial investment/amount) r is the rate of growth and t is the amount of time, usually in years for interest problems. With the given information we can write the equation, \[1686=Pe ^{.04t}+(3Pe ^{.06t}-500)\] The brackets are merely here to help us separate the two conditions, where 3 times the initial investment at 6% interest minus $500, plus the initial investment at 4% interest equals $1686. We can eliminate them and continue to solve for P. \[2186=P(e^{.04}+3e^{.06})\] We can assume that t=1 year and, in essence, remove it from the equation. Solve for P here, and then multiply by 3 to get the amount invested at 6% interest.

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