SOLUTION: One of the accounts pay 9% annual interest, whereas the other pays 11% annual interest. If you have $700 more invested at 11% than you invested at 9%, how much do you have invested

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Question 923226: One of the accounts pay 9% annual interest, whereas the other pays 11% annual interest. If you have $700 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 $137?
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
We don't know the total invested.
We know the difference in the two accounts.
We know that the interest for the two accounts is $137
0.11*x+0.09*y=137
We know that the account at 11% has $700 more.
x=700+y
We substitute for x
0.11*(700+y)+0.09*y=137
We multiply out
77+0.11y+0.09*y=137
We combine like terms.
0.2*y=60
Isolate y
y=$300 at 9%
x=700+y
Calculate x
x=$1000 at 11%
Now,we know the total invested is: 1300
Total invested $1000+$300=$1300
We check
0.11*1000+0.09*300=137
110+27=137
137=137
Since this statement is TRUE and neither x nor y is negative all is well.
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