Question 560131
A certain sum of money is invested at 6.35%, and $4000 more than that amount is invested at 7.28%.
 If the annual interest from the two investments is $972.70, how much was invested at 6.35%? 
:
Let's do it this way
Let x = amt invested at 6.35%
then it says,"$4000 more than that amount, is invested at 7.28%." therefore
(x+4000) = amt invested at 7.28%
;
.0635x + .0728(x+4000) = 972.70; this is the equation you want
.0635x + .0728x + 291.20 = 972.70
.0635x + .0728x  = 972.70 - 291.20
.1363x = 681.50
x = {{{681.5/.1363}}}
x = $5000 invested at 6.35%
;
:
Check the solution by finding the amt invested on each one
.0635*5000 = 317.50
.0728*9000 = 655.20
===================
total int: = 972.70