Question 1044420: You invested $29,000 in two accounts paying 6% and 9% annual interest, respectively.
If the total interest earned for the year was $1980, how much was invested at each rate?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620)   (Show Source):
Answer by MathTherapy(10555)  (Show Source):
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You invested $29,000 in two accounts paying 6% and 9% annual interest, respectively.
If the total interest earned for the year was $1980, how much was invested at each rate?
Let amount invested at 6% be S
Then amount invested at 9% = $29,000 - S
The following INTEREST equation is then formed: .06S + .09(29,000 - S) = 1,980
.06S + 2,610 - .09S = 1,980
.06S - .09S = 1,980 - 2,610
- .03S = - 630
S, or amount invested at 6% =  , or 
Amount invested at 9%: $29,000 - $21,000, or 
It is that simple!
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