SOLUTION: bob invested $33,600, part at 5% interest and the remainder at 8% interest. If he earned twice as much from his 5% investment as his 8% investment, how much did he invest at each r

Algebra ->  Finance -> SOLUTION: bob invested $33,600, part at 5% interest and the remainder at 8% interest. If he earned twice as much from his 5% investment as his 8% investment, how much did he invest at each r      Log On


   

Question 163630: bob invested $33,600, part at 5% interest and the remainder at 8% interest. If he earned twice as much from his 5% investment as his 8% investment, how much did he invest at each rate?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
.05x=2[.08(33,600-x)]
.05x=2[2,688-.08x]
.05x=5,376-.16x
.05x+.16x=5,376
.21x=5,376
x=5,376/.21
x=25,600 is the amount invested @ 5%.
33,600-25,600=8,000 invested @ 8%.
Proof:
.05*25,600=2*.08*8,000
1,280=.16*8,000
1,280=1,280