SOLUTION: Things did not go quite as planned. You invested ​$21,700​, part of it in a stock that paid​ 12% annual interest.​ However, the rest of the money suffered a​ 5% loss. If

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Question 1152880: Things did not go quite as planned. You invested ​$21,700​, part of it in a stock that paid​ 12% annual interest.​ However, the rest of the money suffered a​ 5% loss. If the total annual income from both investments was ​$1924​, how much was invested at each​ rate?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Let x and y represent the amounts invested in the two places. Then we know x+y=21700.

I won't go any further with that setup for solving the problem, because I think it just adds extra work to solve the problem.

Instead, knowing that the sum of the two investments is $21,700, let the amounts be x and (21700-x).

Then the income/loss equation for amount x at 12% and amount (21700-x) at -5% is

.12%28x%29-.05%2821700-x%29+=+1924

Solve using basic algebra... although the calculations get a bit ugly.


Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let x = amount invested at 12%, in dollars.

Then the second amount, which suffered  the 5% loss,  is the rest (21700-x) dollars.



The interest from the 12% interest amount is 0.12*x  dollars.

The 5% loss from the second   amount is 0.05*(21700-x)   dollars.



Your equation is


    interest - loss           = final interest,   or


    0.12*x   - 0.05*(21700-x) = 1924   dollars.


From the equation, express x and calculate the answer


    x = %281924+%2B+0.05%2A21700%29%2F%280.12%2B0.05%29 = 17700.


Answer.  The amount at 12% is $18300;  the rest  $21700-$17700 = $4000 is the amount suffered 5% loss.


Check.   0.12*17700 - 0.05*4000 = 1924 dollars.   ! Precisely correct !

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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.