SOLUTION: You invest $7200 in two accounts paying 8% and 10% annual interest. At the end of the year, the accounts earn the same interest. How much was invested at each rate? I tried .08

Question 447273: You invest $7200 in two accounts paying 8% and 10% annual interest. At the end of the year, the accounts earn the same interest. How much was invested at each rate?
I tried .08x + .10(7200-x) = 7200 and that did not work. Having a tough time coming up with the equation that will allow me to solve this word problem. Any help would be greatly appreciated.
Found 2 solutions by mananth, jorel1380:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
8%----x
10%----7200-x
(you have equated the total interest to the total investment)
Interest same
0.08x=0.1(7200-x)
0.08x=720-0.1x
0.18x=820
x=720/0.18
$4000 @ 8%
Balance @10%
CHECK
0.08*4000=0.1*3200
320=320
m.ananth@hotmail.ca

Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
.08x=.1(7200-x)
.08x=720-.1x
.18x=720
x=720/.18=4000
You invested 4000 at 8% and 3200 at 10%..
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