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Question 1142577: One safe investment pays 10% per year, and a more risky investment pays 18% per year. A woman who has $138,600 to invest would like to have an income of $18,740 per year from her investments. How much should she invest at each rate?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! One safe investment pays 10% per year, and a more risky investment pays 18% per year.
A woman who has $138,600 to invest would like to have an income of $18,740 per year from her investments.
How much should she invest at each rate?
:
Let x = amt invested at 18%
then
(138600-x) = amt invested at 10%
:
.18x + .10(138600-x) = 18740
.18x + 13860 - .10x = 18740
.18x - .10x = 18740 - 13860
.08x = 4880
x = 4880/.08
x = $61,000 invested at 18%
then
138600 - 61000 = $77,600 invested at 10%
:
;
:
Check: find how much return on each investment
.18(61000) = 10980
.10(77600) = 7760
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total return: 18740
Answer by ikleyn(52855)  (Show Source):
You can put this solution on YOUR website! .
Let x = the amount invested at 10% and y = the amount invested at 18%.
Then from the condition you have these two equations in two unknowns
x + y = 138600 dollars (1) (total investment)
0.10x + 0.18y = 18740 dollars (2) (total interest)
Solve the system by the Substitution method.
For it, from equation (1) express x = 138000 - y and substitute it into equation (2). You will get
0.10*(138600-y) + 0.18y = 18740
Simplify and express y from the last equation; then calculate
y =  = 61000.
To find x, substitute the found value of y into equation (1). You will get then
x = 138600 - 61000 = 77600.
ANSWER. $77600 was invested at 10% and $61000 was invested at 18%.
CHECK. 0.10*77600 + 0.18*61000 = 18740 dollars. ! Correct !
Solved.
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It is a typical and standard problem on investment.
To see many other similar solved problems on investment, 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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