SOLUTION: Susan invests 3 times as much money at 11% as she does at 9%. If her total interest after 1 year is $1680, how much does she have invested at each rate?
Thank you !
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-> SOLUTION: Susan invests 3 times as much money at 11% as she does at 9%. If her total interest after 1 year is $1680, how much does she have invested at each rate?
Thank you !
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Question 910171: Susan invests 3 times as much money at 11% as she does at 9%. If her total interest after 1 year is $1680, how much does she have invested at each rate?
Thank you ! Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! We know that the interest for the two accounts is $1680
0.11*x+0.09*y=1680
We know that the account at 11% has 3 times the amount at 9%
We substitute for x
0.11*(3*y)+0.09*y=1680
We multiply out
(0.33+0.09)*y=1680
We combine like terms.
0.42*y=1680
Isolate y
y= 4000.00
y=$ 4000.00 at 9%
Calculate x
x=3*y
x=$12000.00 at 11%
Your total for x is $12000.00
Your total for y is $ 4000.00
Your total for x and y is $16000.00
Total invested $12000.00 + $ 4000.00=$16000.00
We check
0.11*12000.00+0.09* 4000.00=1680
1320.00+ 360.00=1680
1680=1680
Since this statement is TRUE and neither x nor y is negative then all is well.
