SOLUTION: An investment club invest part of $8000 at 10% annual interest and the rest at 12%. If the annual income from these investment is $900, how much was invested at each rate.

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: An investment club invest part of $8000 at 10% annual interest and the rest at 12%. If the annual income from these investment is $900, how much was invested at each rate.      Log On


   

Question 839291: An investment club invest part of $8000 at 10% annual interest and the rest at 12%. If the annual income from these investment is $900, how much was invested at each rate.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Let x and y represent amount saved at 10% and 12% respectively

    x  + y = $8,000    0r  x+=+8000+-+y+

 .10x + .12y = $900

 .10(8,000 - y) + .12y = 900  |solving the system of EQs by substitution

             .02y = 900 - 800

              .02y = 100

                 y = $5000  and x = $3000

CHECKING our answer***

.10%2A3000+%2B+.12%2A5000+=+300+%2B+600 = $900

Wish You the Best in your Studies