SOLUTION: Doris invested some money at 7% and some at 8%. She invested $6000 more at 8% then at 7%. Her total yearly interest from both investments was $1080, how much did she invest at each

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Doris invested some money at 7% and some at 8%. She invested $6000 more at 8% then at 7%. Her total yearly interest from both investments was $1080, how much did she invest at each      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 236200: Doris invested some money at 7% and some at 8%. She invested $6000 more at 8% then at 7%. Her total yearly interest from both investments was $1080, how much did she invest at each rate?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
.08(x+6,000)+.07x=1,080
.08x+480+.07x=1,080
.15x=1,080-480
.15x=600
x=600/.15
x=4,000 amount invested @ 7%.
4,000+6,000=10,000 invested @ 8%.
Proof:
.08*10,000+.07*4,000=1,080
800+280=1,080
1,080=1,080