Question 1193974
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How can $69,000 be​ invested, part at 12% annual simple interest and the remainder at 11% annual simple​ interest, 
so that the interest earned by the two accounts will be​ equal?
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<pre>
x dollars invested at 12%;  the rest (69000-x) dollars invested at 11%.


Write equation as you read the problem

    0.12x = 0.11*(69000-x).


Simplify and find x

    0.12x = 0.11*69000 - 0.11x

    0.12x + 0.11x = 0.11*69000

         0.23x    = 7590

             x    = 7590/0.23 = 33000.


<U>ANSWER</U>.  $33000 invested at 12%;  the rest  69000-33000 = 36000 invested at 11%.


<U>CHECK</U>.  0.12*33000 = 3960 dollars, the annual interest from the 12% investment.

        0.11*36000 = 3960 dollars, the annual interest from the 11% investment.

        the interest is the same, so the answer is correct.
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Solved.