SOLUTION: Yuan invested $1200, part at 4.5% per annum and the remainder at 6.5% per annum. After one year, Yuan had $70.60 in interest. Determine how much money she invested at each rate.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Yuan invested $1200, part at 4.5% per annum and the remainder at 6.5% per annum. After one year, Yuan had $70.60 in interest. Determine how much money she invested at each rate.      Log On


   

Question 1105298: Yuan invested $1200, part at 4.5% per annum and the remainder at 6.5% per annum. After one year, Yuan had $70.60 in interest. Determine how much money she invested at each rate.
Answer by ikleyn(52815) About Me  (Show Source):
You can put this solution on YOUR website!
.
interest + interest       = total interest


0.045*x  + 0.065*(1200-x) = 70.60

0.045x + 0.065*1200 - 0.065x = 70.60

-0.02x = 70.60 - 0.065*1200 = -7.4  ====>  x = %28-7.4%29%2F%28-0.02%29 = 370.


Answer.  $370 invested at 4.5%; the rest 1200-370 = 830 dollars invested at 6.5%.

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It is very standard problem on investments.

To see other solved problems of this kind (and other approaches and other solution methods used), look into the lesson
    - Using systems of equations to solve problems on investment
in this site.

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.