SOLUTION: Lon Triah invested $5000 at a local bank. Part was invested at 6% simple interest and the remainder at 10% simple interest. At the end of one year, Lon had earned $428 interest. Ho

Question 1180729: Lon Triah invested $5000 at a local bank. Part was invested at 6% simple interest and the remainder at 10% simple interest. At the end of one year, Lon had earned $428 interest. How much was invested at each rate?
Answer by ikleyn(52802) (Show Source): You can put this solution on YOUR website!
.


x invested at 6%.


(5000-x) invested at 10%.


The total interest equation


    0.06x + 0.1(5000-x) = 428


From the equation


    x =  = 1800.


ANSWER.  $1800 invested at 6%  and the rest, (5000-1800) = 3200 dollars, invested at 10%.


CHECK.   0.06*1800 + 0.1*3200 = 428 dollars, total interest.   ! Correct !

Solved.

-------------------

It is a standard and typical problem on investments.

If you need more details, or if you want to see other similar problems solved by different methods, look into the lesson
- Using systems of equations to solve problems on investment
in this site.

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

Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".

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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