Question 1169899: Two investments totaling $33,000 produce an annual income of $2200. One investment yields 8% per year, while the other yields 6% per year. How much is invested at each rate?
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Let x be the amount invested at 8%;
then the amount invested at 6% is the rest 33000-x dollars.
To find x, you write the total interest equation
0.08x + 0.06*(33000-x) = 2200 dollars.
From the equation, you find
x =  = 11000 dollars.
ANSWER. $11000 invested at 8% and the rest, $33000-$1000 = $22000 invested at 6%.
CHECK. 0.08*11000 + 0.06*22000 = 2200 dollars is the total inderest. ! 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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