<PRE><B>Question 180329</B>
Mike invested a total of $14,600, part at 9% simple interest and Part 6% interest for one year. If the interest in the 9% account was $300 greater than the interest in the 6% account, how much was invested in the 9% account? 
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Equation:
9% int = 6%int + 300

0.09x = 0.06(14000-x) + 300

Multiply both sides by 100 to get:

9x = 6(14000)-6x + 30000
15x = 30000 + 84000

x = $7600 (amt. invested at 9%)
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Cheers,
Stan H.</PRE>