SOLUTION: A woman invests $39,000, part at 8% and the rest at 9.5% annual interest. If the 9.5% investment provides $467.50 more income than the 8% investment, how much is invested at each r

Question 1137065: A woman invests $39,000, part at 8% and the rest at 9.5% annual interest. If the 9.5% investment provides $467.50 more income than the 8% investment, how much is invested at each rate?
Answer by ikleyn(52748) (Show Source): You can put this solution on YOUR website!
.

Let x= amount invested at 9.5%.

Then invested at 8% is the rest, (39000-x) dollars.


The difference of interests is given


0.095x - 0.08*(39000-x) = 467.50.


It is your basic equation.  Its solution is


    x =  = 20500.


ANSWER.  $20500 invested at 9.5%.  The rest, (39000-20500) = 18500 dollars invested at 8%.


CHECK.   0.095*20500 - 0.08*18500 = 467.50 dollars.   ! Precisely correct !

Solved.

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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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