SOLUTION: John invests $6000, part at 12% interest and part at 8%. If he earned a total of $660 in interest for the year, how much did he invest at each rate?

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: John invests $6000, part at 12% interest and part at 8%. If he earned a total of $660 in interest for the year, how much did he invest at each rate?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1088591: John invests $6000, part at 12% interest and part at 8%. If he earned a total of $660 in interest for the year, how much did he invest at each rate?
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the amount of money invested at 12%
Let y be the amount of money invested at 8%
Since John invests $6000, then x%2By=6000
He earned a total of $660 in interest, hence 0.12x%2B0.08y=660
The system of equations is
system%28x%2By=6000%2C+0.12x%2B0.08y=660%29
system%28y=6000-x%2C+0.12x%2B0.08%286000-x%29=660%29
system%28y=6000-x%2C+0.12x%2B480-0.08x=660%29
system%28y=6000-x%2C+0.04x=180%29
system%28y=6000-x%2C+x=180%2F0.04%29
system%28y=6000-4500%2C+x=4500%29
system%28y=1500%2C+x=4500%29
John invested $4500 at 12% and $1500 at 8%
Answer: $4500 at 12% and $1500 at 8%