Question 1199905:  a person had $14,000 invested in two accounts, one paying 9% simple interest and one paying 10% simple interest. how much was invested in each account if the interest after 1 year is $1397? 
 Answer by ikleyn(52903)      (Show Source): 
You can  put this solution on YOUR website! . 
a person had $14,000 invested in two accounts, one paying 9% simple interest  
and one paying 10% simple interest. how much was invested in each account  
if the interest after 1 year is $1397? 
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Let x be amount invested at 10% annually.
Then the amount invested at 9% is the rest (14000-x) dollars.
10% investment generates the annual interest of 0.1x dollars.
9% investment generates the annual interest of 0.09*(14000-x).
The totall annual interest equation is
    0.1x + 0.09*(14000-x) = 1397  dollars.
Simplify this equation and find x
    0.1x + 0.09*14000 - 0.09x = 1397,
    0.1x - 0.09x = 1397 - 0.09*14000
        0.01x     =     137
             x     =     137/0.01 = 13700.
Thus $13700 invested at 10%  and the rest,  $14000 - $13700 = $300  invested at 0.09%.    ANSWER
CHECK.  0.1*13700 + 0.09*300 = 1397 dollars, the total annual interest.   ! correct !
 
Solved.
 
 
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