Math

$10,000 invested. part at 3% and part at 6% for a period of 1 year. how much was invested in each account if the interest earned in the 6% account was $357 greater than the 3% account?

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  1. If there was $x at 3%, then the rest (10000-x) was at 6%. Now just compare the two amounts, and find x:

    .03x + 357 = .06(10000-x)

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