Bill's Two Investments

33)  With $35,000, Bill made two investments.  Since one had paid 8% interest,
and the other 12%, his total intererest-earning from both was $3,920, for the
year.  How much had he initially invested at each percent?

        x = 8% investment
        y = 12% investment

        x + y = 35,000  (The two investments and their total amount.)

        .08(x) + .12(y) = $3,920  (The interest-earning of each investment,
                                   and their combined annual yield.)

        8x + 12y = 392,000  (We eliminated the decimal places by multiplying
                            by 100.

         x + y  =   35,000
        8x + 12y = 392,000

      -8x  -8y = -280,000  (We multiplied the above equation by -8.)
       8x +12y =  392,000
       ------------------
            4y =  112,000

           y = 28,000  (We divided both sides by 4 to find the value of y.)

           Plugging the value of y (28,000) into the above formula, we can
       now solve for the value of x:

         x + 28,000 = 35,000

         x = 7,000

         So, now  we know that Bill had invested as follows:

                $28,000 (at 12%) = $3,360 (interest earned)
                 $7,000 (at 8%) =    $560 (interest earned)
                -------------------------
                $35,000            $3,920 (combined interest earned)

         We can also divide $3,920 by 35,000 to find his interest average
      for both investments: 11.2%