1.  One part of the condition says that the total amount is $7200.


    The other part of the condition says that the amount invested at 13% is 200 more than the combined amount invested at 6% and 10%.


    It implies immediately via simple logic, that the combined amount invested at 6% and 10% is   = 3500 dollars

    and the amount invested at 13% is  3500+200 = 3700 dollars.



2.  It implies that the interest produced by the amount at 13% is

        0.13*3700 = 481 dollars

    and the combined interest of two other accounts is  787 - 481 = 306 dollars.



3.  Let x be the amount invested at 10%.

    Then the amount invested at 6% is ((3500-x) dollars.


     The combined interest equation for these two accounts is


         0.10*x + 0.06*(3500-x) = 306.


     The solution to this equation is


         x =  = 2400.


Answer.  $3700 invested at 13%;  $2400 invested at 10%,  and  (3500-2400) = 1100 dollars invested at 6%.


CHECK.   0.13*3700 + 0.10*2400 + 0.06*1100 = 787 dollars as the total interest.    ! Correct !

    Let  A  be the amount invested at 13%  and  B  be  the combined amount invested ar 6% and 10%.


    We are given that 

        A + B = 7200
        A - B =  200
  ----------------------   Add these two equations. You will get


      2A      = 7400  ==============>  A = 7400/2 = 3700  dollars.


     Then  B = 7200 - 3700 = 3500  dollars, exactly as I stated in the core text.


     Using logical arguments,  take off $200 from $7200 - then you will get a total of $7000, consisting of two equal parts "reduced A" and "B",

      so "B"  is 7000/2 = 3500 dollars and "A" is 3500+200 = 3700.