SOLUTION: A woman invested an amount of money at 5% and twice as much money at 7%. If the total yearly income from the two investments is $760, How much was invested at 7%

Algebra ->  Human-and-algebraic-language -> SOLUTION: A woman invested an amount of money at 5% and twice as much money at 7%. If the total yearly income from the two investments is $760, How much was invested at 7%      Log On


   



Question 22101: A woman invested an amount of money at 5% and twice as much money at 7%. If the total yearly income from the two investments is $760, How much was invested at 7%
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = amount invested at 5%
2x = amount invested at 7%

Total interest=
.05x + .07(2x) = 760
.05x + .14x = 760
.19x = 760
%28.19x%29%2F.19+=+760%2F.19
x= $4000 at 5%
2x= $8000 at 7%

Check:
.05(4000)+.07(8000)
$200 + $560
$760
It checks!!

R^2 at SCC