Question 1136546
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The solution is in 3 steps.


<pre>
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  {{{(7200-200)/2}}} = 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 = {{{(306 - 0.06*3500)/(010-0.06)}}} = 2400.


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


<U>CHECK</U>.   0.13*3700 + 0.10*2400 + 0.06*1100 = 787 dollars as the total interest.    ! Correct !
</pre>

Solved.



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I'd like to make couple of comments to this solution.


The method of solution was to reduce the problem from 3 unknowns to only two unknowns.


Then the problem for two unknown investments was solved using single equation for only one unknown.


The method of reducing such problems from 3 unknowns to only two is very well known.


See the lesson 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/How-to-algebreze-and-to-solve-this-problem-on-2-eqns-in-2-unknwns.lesson>HOW TO algebreze and solve this problem on 2 equations in 2 unknowns</A> 

in this site.


On solution of investment problems for two accounts see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Using-systems-of-equations-to-solve-problems-on-investment.lesson>Using systems of equations to solve problems on investment</A>

in this site.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Systems of two linear equations in two unknowns</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.



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If you think that I hid from you the reducing step from 3 unknowns to 2 unknowns,

then I will present this step here EXPLICITLY.


<pre>
    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.
</pre>

Thus, now you have all the parts of the solution "in place".