SOLUTION: Suppose you invest money in two accounts. One of the accounts pay 9 % annual interest, whereas the other pays 11 % annual interest. If you have $ 4 , 000 more inve

Question 1146329: Suppose you invest money in two accounts. One of the accounts pay
9
%
annual interest, whereas the other pays
11
%
annual interest. If you have
$
4
,
000
more invested at
11
%
than you invested at
9
%
, how much do you have invested in each account if the total amount of interest you earn in a year is
$
840
?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Come on! You can make this a little more tutor friendly like I did here
:
Suppose you invest money in two accounts.
One of the accounts pay 9% annual interest,
whereas the other pays 11% annual interest.
If you have $4,000 more invested at 11% than you invested at 9%
how much do you have invested in each account if the total amount of interest
you earn in a year is $840?
:
let a = amt invested at 9%
"you have $4,000 more invested at 11% than you invested at 9%" therefore:
(a + 4000) = amt invested at 11%
:
9% interest + 11% interest = 840
.09a + .11(a+4000) = 840
.09a + .11a + 440 = 840
.20a = 840 - 440
a = 400/.2
a = $2000 invested at 9%
then
2000 + 4000 = $6000 invested at 11%
:
:
Check: find how much return from each
.09(2000) = 180
.11(6000) = 660
-----------------
total int: $840

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