SOLUTION: Marie invested $12,000 in two different simple interest accounts. She invested part of the money at 8% and the rest at 6%. How much was invested at each amount, if the total inte

Algebra ->  Average -> SOLUTION: Marie invested $12,000 in two different simple interest accounts. She invested part of the money at 8% and the rest at 6%. How much was invested at each amount, if the total inte      Log On


   



Question 1027601: Marie invested $12,000 in two different simple interest accounts. She invested part of the money at 8% and the rest at 6%. How much was invested at each amount, if the total interest earned was $872? State what x represents, state the equation, and then state the answer.
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the amount invested at 8%. Thus 12000-x would be the amount invested at 6%. The setup looks like this:
.08x + .06(12000-x) = 872
Now solve for x.
.08x + 720 - .06x = 872
.02x + 720 = 872
.02x = 152
x = $7600 invested at 8%
12000-x = $4400 invested at 6%