Question 1196412
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The annual interest on a $14,000 investment exceeds 
the interest earned on a $7000 investment by $294. 
The $14,000 is invested at a 0.6% higher rate of interest than the $7000. 
What is the interest rate of each investment?
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Let x be the annual interest rate of the $7000 investment (in decimal form).

Then the annual interest rate of the $14000 investment is (x+0.006), 
according to the problem.


Having it, we write an equation expressing the difference of two annual
interest amounts


    14000*(x+0.006) - 7000x = 294  dollars.


Simplify it and find x

    14000x + 14000*0.006 - 7000x = 294

    7000x = 294 - 14000*0.006

    7000x = 210

        x = {{{210/7000}}} = 0.03.


<U>ANSWER</U>.  $7000 invested at 3%;  $14000 invested at 3.6%.
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Solved.