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Question 1183674: Solve this application problem using a system of equations: $4400 is invested, part of it at 20% and part of it at 4%. For a certain year, the total yield is $600.00. How much was invested at 20%? How much was invested at 4%?
Answer by ikleyn(52873)   (Show Source):
You can put this solution on YOUR website! .
Solve this application problem using a system of equations: $4400 is invested, part of it at 20% and part of it at 4%.
For a certain year, the total yield is $600.00. How much was invested at 20%? How much was invested at 4%?
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Let x be the amount invested at 20%.
Then the amount invested at 4% is the rest, or (4400-x) dollars.
Next you write equation for the total interest
0.2x + 0.04*(4400-x) = 600 dollars.
From this equation, you find x
x =  = 2650 dollars.
ANSWER. $2650 was invested at 20%, and the rest (4400-2650) = 1750 dollars was invested at 4%.
CHECK. 0.2*2650 + 0.04*1750 = 600 dollars for the total interest. ! 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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