Question 1136030
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            I will show you two methods of solving this problem.



<U>Solution 1</U>.  &nbsp;&nbsp;Using the system of 2 equations


<pre>
Let  x  be the amount invested at 5% and  y  be the amount invested at 7%.


Then you have these two equations


    x   +     y = 43000   dollars     (1)    (the sum of amounts is equal to the total investment)

  0.05x + 0.07y =  2430   dollars     (2)    (the sum of interests is equal to the total interest)


I will apply the Substitution method to solve the system.
For it, I express

    x = 43000 - y                     (3)


from the first equation and then substitute it into the second equation, replacing x. I will get

    0.05*(43000 - y) + 0.07y = 2430.  (4)


Simplify and solve this equation step by step:

    0.05*43000 - 0.05y + 0.07y = 2430

    0.07y - 0.05y = 2430 - 0.05*43000

    0.02y         = 280

        y         = {{{280/0.02}}} = 14000.


Thus we found that $14000 were invested at 7%.

To find x, use equation (3)


    x = 43000 - 14000 = 29000.


<U>ANSWER</U>.  $29000 invested at 5%  and  $14000 invested at 7%.


<U>CHECK</U>.   0.05*29000 + 0.07*14000 = 2430 dollars, the total interest.   ! Correct !
</pre>


<U>Solution 2</U>. &nbsp;&nbsp;Using single equation


<pre>
Let  y  be the amount invested at 7%.

Then the amount invested at 5% is the rest  (43000-y) dollars.


The 5% investment gives the interest of  0.05*(43000-y)  dollars.

The 7% investment gives the interest of 0.07*y dollars.


In total, these two investments give $2430.  Therefore, the "interest" equation is

    0.05*(43000-y) + 0.07*y = 2430.


It is the same equation (4) as you obtained in the <U>Solution 1</U>  after substitution.


So, the solution of this equation is the same, and it produces the same answer.
</pre>

My solutions and explanations are completed.


You learned two methods setuping the equations along with the methods of their solutions.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;H a p p y &nbsp;&nbsp;&nbsp;&nbsp;l e a r n i n g !



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It is a standard and typical problem on investments.


If you need more details,  &nbsp;or if you want to see other similar problems solved by different methods, &nbsp;look into 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.


You will find there different approaches &nbsp;(using one equation or a system of two equations in two unknowns), &nbsp;as well as 
different methods of solution to the equations &nbsp;(Substitution, &nbsp;Elimination).


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 lesson is the part of this online textbook under the topic &nbsp;"<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.