SOLUTION: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?

Algebra ->  Expressions-with-variables -> SOLUTION: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?      Log On


   



Question 142584: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Total invested = 6000.
Total earned = 624.
Let x be the part invested at 0.09
Let y be the part invested at 0.11
6000+=+x+%2B+y
6000-x+=+y
0.09x+%2B+0.11y+=+624
0.09x+%2B+0.11%286000-x%29+=+624
0.09x+%2B+660+-+0.11x+=+624
-0.02x+=+-36
x+=+36%2F%280.02%29
x+=+1800
y+=+4200