Question 177203This question is from textbook introductory and intermediate algebra
: This is my problem: Part of $34,000 is invested at an interest rate of 8% and the rest at an interest rate of 10%. if the total interest earned in 1 year was $3000, how much was invested at each rate?
I'd like to know what the equation is and how to solve it...
Thanks
-Tom
This question is from textbook introductory and intermediate algebra
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Part of $34,000 is invested at an interest rate of 8% and the rest at an interest rate of 10%. if the total interest earned in 1 year was $3000, how much was invested at each rate?
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Equation:
interext + interest = interest
0.08x + 0.10(34000-x) = 3000
Multiply thru by 100 to get:
8x + 10(34000-x) = 300000
8x + 340000 - 10x = 300000
-2x = -40,000
x = $20,000.00 (amt. invested at 8%)
34000-x = $14,000 (amt. invested at 10%)
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Cheers,
Stan H.
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