Thomas invested $8,500 for one year. Part of the money was invested at 6% and the rest at 9%. The total interest earned was $667.50. How much did Thomas invest at the 6% rate?

Okay so say that x=amount invested with 6% and y=amount invested with 9% x+y=8,500 so > x=8500-y 6%=0.06 9%=0.09 0.06x +0.09y=667.5 (Substitute in x=8500-y so only numbers and y) 0.06(8500-y)+0.09y=667.5 (expand brackets) 510-0.06y+0.09y=667.5 (-510) 0.03y=117.5 (/0.03) $3916.67=Y ->9% X=8500-Y x=$4583.33 ->6%

Answered by Jerboa | 2024-06-10

Thomas invested $3300 at the 6% interest rate. The total investment amount was $8500, with the remainder $5200 invested at 9%. The interest earned from both investments totaled $667.50, confirming the correctness of the calculations.
;

Answered by Jerboa | 2025-01-17