SOLUTION: Marge invests $7,000 , part at 6% and part at 9%. If she ears $493 interest how much is invested in each account.

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Question 1102971: Marge invests $7,000 , part at 6% and part at 9%. If she ears $493 interest how much is invested in each account.
Found 2 solutions by ikleyn, addingup:
Answer by ikleyn(52925)   (Show Source): You can put this solution on YOUR website!
.
0.06*x + 0.09*(7000-x) = 493.


Solve for x.


6x + 9*(7000-x) = 49300


6x + 63000 - 9x = 49300  ====>  -3x = 49300 - 63000 = -13700  ====>  x =  = 4566.67 dollars invested at 6%.


The rest, 7000 - 4566.67 = 2433.33 dollars invested at 9%.

On investment problems see the lesson
    - Using systems of equations to solve problems on investment
in this site.


Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
Of a total of 7000 she invests x money at 6%. And the rest, 7000 - x, she invests at 9%
------------------------
0.06x + 0.09(7000 - x) = 493
0.06x + 630 - 0.09x = 493
-0.03x = -137
x = 4566.67 this is the amount invested at 6x. The amount invested at 9% is 7000-4566.67 = 2433.33
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