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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 = = 0.03.
ANSWER. $7000 invested at 3%; $14000 invested at 3.6%.
Solved.