SOLUTION: mary invests $24,000 in two seperate accounts. one earns 6% simple interest and the other 9%. if she earned a total of $1740, how much did she invest at each rte?

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: mary invests $24,000 in two seperate accounts. one earns 6% simple interest and the other 9%. if she earned a total of $1740, how much did she invest at each rte?      Log On


   

Question 160297: mary invests $24,000 in two seperate accounts. one earns 6% simple interest and the other 9%. if she earned a total of $1740, how much did she invest at each rte?
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!

Let "x", amount invested at 9%. So "24000-x" is the amount inv. at 6%.
Putting into eqn:
0.09x%2B0.06%2824000-x%29=1740 working eqn
0.09x%2B1440-0.06x=1740
0.03x=1740-1440
0.03x=300
cross%280.03%29x%2Fcross%280.03%29=cross%28300%2910000%2Fcross%280.03%29
x=10000 --------------> amt invested at 9%
24000-10000=14000-----> amt invested at 6%
check via working eqn:
0.09%2A10000%2B0.06%2A14000=1740
900%2B840=1740
1740=1740
Thank you,
Jojo