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zmey
1 day ago
9

Thomas invested $8,500 for one year. Part of the money was invested at6% and the rest at 9%. The total interest earned was $667.

50. How much did Thomas invest at the 6% rate?
Mathematics
1 answer:
Zina [3.9K]1 day ago
4 0
Let x represent the amount invested at 6% and y the amount at 9%.
The equation x+y=8,500 leads to x=8500-y.
For the interest rates, we know 6%=0.06 and 9%=0.09.
The equation becomes 0.06x + 0.09y=667.5 (substituting for x to use only y).
Expanding yields: 0.06(8500-y)+0.09y=667.5.
Solving this gives us 510-0.06y+0.09y=667.5 (-510).
This simplifies to 0.03y=117.5 (/0.03), yielding y=$3916.67 for the 9% investment.
Thus, X=8500-Y results in x=$4583.33 for the 6% investment.
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