SOLUTION: a woman invested a sum of money at 6% and 3000 dollars more than 9%. If the total interest earned in one year is 4710 dollars, how much was invested at each rate?

Question 986580: a woman invested a sum of money at 6% and 3000 dollars more than 9%. If the total interest earned in one year is 4710 dollars, how much was invested at each rate?
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
x=amount invested at 6%
.
.06x+.09(x+$3000)=$4710
0.06x+0.09x+$270=$4710
0.15x=$4440
x=$29600
ANSWER 1:$29600 was invested at 6%.
Amount at 9%=x+$3000
Amount At 9%=$29600+$3000=$32600
ANSWER 2: $32600 was invested at 9%.
.
CHECK:
.06($29600)+0.09($32600)=$4710
$1776+$2934=$4710
$4710=$4710
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