# SOLUTION: 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 in

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 Click here to see ALL problems on Money Word Problems 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 -TomThis question is from textbook introductory and intermediate algebra Answer by stanbon(57387)   (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? ------------------------------------------------------ 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%) ============================================= Cheers, Stan H.