SOLUTION: 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 ea

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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 $1980, how much was invested at each​ rate?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39621)   (Show Source): You can put this solution on YOUR website!
You invested ​$p in two accounts paying x% and y% annual​ interest, respectively.
If the total interest earned for the year was $z, how much was invested at each​ rate?

s and t, the two quantities invested


Account for the amount of interest earned:



Solve this system for s and t:
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

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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