<PRE><B>Question 268140</B>
A person invested a sum of money at 9% annual simple interest and invested twice that amount at 11% annual simple interest.
 If the total yearly income from both investments was $1,395 how much was invested at each rate? 
:
Let x = amt invested at 9%
then
2x = amt invested at 11%
:
a total interest equation
.09x + .11(2x) = 1395
:
.09x + .22x = 1395
:
.31x = 1395
x = {{{1395/.31}}}
x = $4500 invested at 9%
and, obviously:
2(4500) = $9000 invested at 11%
:
:
Check this by finding the total interest with these values:
.09(4500) + .11(9000) = 
405 + 990 = 1395</PRE>