Question 1199905
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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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<pre>
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%.    <U>ANSWER</U>


<U>CHECK</U>.  0.1*13700 + 0.09*300 = 1397 dollars, the total annual interest.   ! correct !
</pre>

Solved.


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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/equations/Typical-investment-problems.lesson>Typical investment problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Using-systems-of-equations-to-solve-problems-on-investment.lesson>Using systems of equations to solve problems on investment</A>

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