SOLUTION: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate? Can some

Algebra ->  Average -> SOLUTION: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate? Can some      Log On


   

Question 189886: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate? Can someone help me solve this problem I am confused.
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!


Let x = amount invested at 8%
And, $7,000 - x = amount inv. at 12%

Adding both to earn $764 interest:

0.08%28x%29%2B0.12%287000-x%29=764
0.08x%2B840-0.12x=764
Combine similar terms: (change signs, "+" or "-")
840-764=0.12x-0.08x
76=0.04x ----> cross%2876%291900%2Fcross%280.04%29=cross%280.04%29x%2Fcross%280.04%29
x= $1,900.00 amount inv. at 8%

$7,000.00 - $1,900.00 = $5,100.00, amount inv. at 12%


Let us check,
0.08%281900%29%2B0.12%285100%29=764
152%2B612=764
764=764

*Note: It's just make sense to invest more on higher interest



Thank you,
Jojo