SOLUTION: Lane invested $16,000, part at 6% and part at 8%. If the total interest at the end of the year was $1,080, how much did she invest at 6%?

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Lane invested $16,000, part at 6% and part at 8%. If the total interest at the end of the year was $1,080, how much did she invest at 6%?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 456844: Lane invested $16,000, part at 6% and part at 8%. If the total interest at the end of the year was $1,080, how much did she invest at 6%?
Answer by benni1013(206) About Me  (Show Source):
You can put this solution on YOUR website!
Here is what we know:
1) The total interest is $1,080. We also know part was invested at 6% and the other at percent. This results in the first of two equations. .06x+.08y=$1,080.
2) We also know that together he invested $16,000. This is the second equation.
x+y=$16,000
This sets up the equations to solve in a system of equations computation problem. .06x+.08y=$1,080 and x+y=$16,000.
This should help quite a bit.