SOLUTION: 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

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(52775) (Show Source): You can put this solution on YOUR website!
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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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