Question 1145873: Ms. Burke invested 29,000 in two accounts, one yielding 8% interest and the other yielding 10%. If she received a total of $2,600 in interest at the end of the year, how much did she invest in each account?
Answer by ikleyn(52788)   (Show Source):
You can put this solution on YOUR website! .
Let x be the amount invested at 10%;
then the amount invested at 8% is (29000-x) dollars
The total interest is the sum of partial interests, which gives you an equation
0.10*x + 0.08*(29000-x) = 2600 dollars.
From the equation, express x and calculate
x =  = 14000.
ANSWER. $14000 were invested at 10% and the rest, (29000-14000) = 15000 dolars were invested at 8%.
CHECK. 0.10*14000 + 0.08*15000 = 2600 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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