# SOLUTION: you invested \$29,000 in two accounts paying 8% and 9% annual interest respectively. if the total interest earned for the year was \$2350 how much was invested at each rate?

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 Question 522954: you invested \$29,000 in two accounts paying 8% and 9% annual interest respectively. if the total interest earned for the year was \$2350 how much was invested at each rate?Answer by Maths68(1474)   (Show Source): You can put this solution on YOUR website! Amount invested at 8% = x Amount invested at 9% = 29000-x Given Total interest on both investments = 2350 8% * (x) + 9% * (29000-x) = 2350 8/100 * (x) + 9/100 * (29000-x) = 2350 8/100 * (x) + 9/100 * (29000-x) = 2350 Multiply by 100 both sides 8 * (x) + 9 * (29000-x) = 2350 * 100 8x + 261000-9x = 235000 -x=235000-261000 -x=-26000 Multiply by -1 boths sides x=26000 Amount invested at 8% = x =26000 Amount invested at 9% =29000-x=29000-26000 = 3000 Check ===== 26000 * 8/100 +3000 * 9/100 = 2350 260*8+30*9=2350 2080+270=2350 2350=2350